Quantum Mechanics

Part I. State and Operator

Quantum States

Ket and Bra

State as a Vector

Describe the state of the system: a set of variables encoding the relevant information of the system.

Predict (i) the observables (measurement outcomes) and (ii) their dynamics (time evolution).

State variables may not have “physical meaning”.

Choice of state variables may not be unique. (There can be more than one way to describe a system.)

Block[{dat},
dat=ColorNegate/@RandomChoice[#,3]&/@GroupBy[ResourceData["MNIST","TestData"],Last->First];Grid[Partition[RandomSample@Flatten@Values@dat,6]]]

Image file: brightness of each pixel. - describe a state by all possible observables.

Human: digits 0, 1, 2, …, 9. - describe a state by a name.

Machine learning: feature vectors in the latent space. - describe a state by a vector in a vector space. [This is the most close to what we do in quantum mechanics.]

Block[{dat,imgs,pts},
dat=ColorNegate/@RandomChoice[#,3]&/@GroupBy[ResourceData["MNIST","TestData"],Last->First];
imgs=Flatten@Values@dat;
SeedRandom[449];
pts=Flatten[Table[#+ϵ,{ϵ,0.2RandomVariate[NormalDistribution[],{3,2}]}]&/@RandomVariate[NormalDistribution[],{10,2}],1];
pts=pts/Max[Norm/@pts];
Graphics[{{Arrow[{#,0}&/@{-1,1.2}],Text["feature 1",{1.2,0},{1,1}],Arrow[{0,#}&/@{-1,1.2}],Text["feature 2",{0,1.2},{1,-1},{0,1}]},Inset[#1,#2,{0,0},0.15]&@@@Thread[{imgs,pts}]},PlotRange->All,ImageSize->180]]

The vector components are the state variables, and they may not need to have physical meanings. [They are also called probability amplitudes or wave amplitudes, but I don’t explain what is “waving” here.]

This particular (vector-based) approach of describing quantum states is not the only way. There are other ways to formulate quantum mechanics, just to name a few: density matrix formulation (matrix-based), classical shadow formulation (probability-based) [], quantum bootstrap (observable-based) [].

However, the vector description is a simple and efficient way to describe a (pure) state of a quantum system. So we will start from state vectors.

Information encoded in a quantum state is called quantum information. It provides the foundation for quantum computation/communication.

Hsin-Yuan Huang, Richard Kueng, John Preskill. arXiv:2002.08953.

Xizhi Han, Sean A. Hartnoll, Jorrit Kruthoff. arXiv:2004.10212.

Block[{state},
state[y0_]:=Graphics[{{Black,Rectangle[{0,0},{1,3},RoundingRadius->0.5]},Table[{Darker[<|1->Green,2->Yellow,3->Red|>[y],0.7-0.7Boole[y+y0==4]],Disk[{0.5,y-0.5},0.4]},{y,3}]},ImageSize->15];
Style[Column[Table[Row[{Ket[state[s]]," = ",MatrixForm[SparseArray[{s,1}->1,{3,1}]]}],{s,3}]],14,FontFamily->"CMU"]]

Distinct states. Two states are distinct, if they can be be distinguished by the different observation values of an observable.

Quantum superposition. What is peculiar about quantum states is that quantum mechanics allows us to write down linear combinations of states, which has no classical correspondence.

(1/(Sqrt[6]))(({{1}, {i}, {-2}}))=(1/(Sqrt[6]))((Ket)+i(Ket)-2(Ket)).

Mathematically this is ok, but physically what does it mean?

Well, the norm square of the superposition coefficients has the statistical interpretation of probability.

({{state, coeffient, probability}, {(Ket), (1/(Sqrt[6])), ((|(1/(Sqrt[6]))|)^2)=(1/6)}, {(Ket), (i/(Sqrt[6])), ((|(i/(Sqrt[6]))|)^2)=(1/6)}, {(Ket), ((-2)/(Sqrt[6])), ((|((-2)/(Sqrt[6]))|)^2)=(2/3)}})

The traffic light has 1/6 of the probability to be observed in red, and 1/6 in to be in yellow, and 2/3 to be in green.

But what about the relative sign or phase of the superposition coefficients? - The complex nature of the state vector component is a feature that enables quantum interference.

Ket Vector

(Ket[v])∈ℋ.

Vector space. The defining property of a vector space is that linear combinations of vectors are still in the vector space

∀(Ket[u]),(Ket[v])∈ℋ;α,β∈ℂ:
α(Ket[u])+β(Ket[v])∈ℋ.

The fact that ℋ is a complex vector space is reflected in the fact that α and  β are complex numbers.

This can be generalized to many vectors: any linear combinations of vectors is still a vector

∀{(Ket[(Subscript[u, n])])}⊂ℋ;{(Subscript[α, n])}⊂ℂ:
(∑_n)(Subscript[α, n])(Ket[(Subscript[u, n])])∈ℋ.

Vector representation: Each ket can be represented as a column vector

Note: “≏” implies the vector representation is basis dependent and the values of vector components may change if we view the same state in a different basis.

If (Ket[v])≏(({{(Subscript[v, 1])}, {(Subscript[v, 2])}})), (Ket[w])≏(({{(Subscript[w, 1])}, {(Subscript[w, 2])}})), what is the vector representation of the state (Ket[v])+(Ket[w]) and λ(Ket[v]) (λ∈ℂ is a complex scalar).

(Ket[v])+(Ket[w])≏(({{(Subscript[v, 1])}, {(Subscript[v, 2])}}))+(({{(Subscript[w, 1])}, {(Subscript[w, 2])}}))=(({{(Subscript[v, 1])+(Subscript[w, 1])}, {(Subscript[v, 2])+(Subscript[w, 2])}})).

λ(Ket[v])≏λ(({{(Subscript[v, 1])}, {(Subscript[v, 2])}}))=(({{λ (Subscript[v, 1])}, {λ (Subscript[v, 2])}}))

To write down the vector representation, we must specified a set of (orthonormal) basis vectors in the vector space ℋ, and represent them as one-hot unit vectors:

(Ket[1])≏(({{1}, {0}, {⋮}})),(Ket[2])≏(({{0}, {1}, {⋮}})), ….

Such that (Ket[v]) can be expressed as a linear combination of basis vectors

The ith vector component (Subscript[v, i]) is the linear combination coefficient in front of the ith basis vector (Ket[i]).

The two distinct states are (Ket[0]) and (Ket[1]).

We can choose (Ket[0]) and (Ket[1]) to be the basis vectors (like choosing a coordinate system) and write:

(Ket[0])≏(({{1}, {0}})), (Ket[1])≏(({{0}, {1}}))

The vector representation of a quantum state is also called a state vector.

By saying that a qubit is a two-state system, its state vector has two components.

A generic quantum state of a qubit is a complex linear superposition of the basis states

(Subscript[ψ, 0]),(Subscript[ψ, 1])∈ℂ are complex numbers. They parameterize the state (Ket[ψ]).

Conversely, every two-component complex vector describes a qubit state.

Statistical interpretation: ((|(Subscript[ψ, 0])|)^2) and ((|(Subscript[ψ, 1])|)^2) are respectively the probabilities to observe the qubit in the 0 and the 1 states.

Bra Vector (I)

If (Ket[u]) is a linear combination of (Ket[v]) and (Ket[w])

(Ket[u])=α(Ket[v])+β(Ket[w]),

then its dual is

Under duality: (i) ket is flipped to bra (and vice versa), (ii) every scalar coefficient gets complex conjugated.

Vector representation: Each bra can be represented as a row vector. If the ket vector (column vector) is

(Ket[v])≏(({{(Subscript[v, 1])}, {(Subscript[v, 2])}, {⋮}})),

then the bra vector (dual row vector) is simply the conjugate transpose of the ket vector

Every basis vector (Ket[i]) also has a dual basis vector (Bra[i]), the are represented as

(Bra[1])≏(({{1, 0, ⋯}})),
(Bra[2])≏(({{0, 1, ⋯}})),
⋯.

The dual basis vectors form a set of basis for the bra vector. In terms of basis vectors,

The ith vector component (Subscript[v, i]^( * )) is the linear combination coefficient in front of the ith dual basis vector (Bra[i]).

Scalar Product

({{(BraKet[·, ·]), :, ℋ×ℋ, →, ℂ}, { ,  , (Ket[u]),(Ket[v]), ↦, (BraKet[u, v])}})

Complex conjugation

∀(Ket[u]),(Ket[v])∈ℋ, (BraKet[u, v])=(SuperStar[(BraKet[v, u])])

Note: (BraKet[u, v])≠(BraKet[v, u]) in general, unless (BraKet[u, v])∈ℝ.

Linearity of the ket vector

∀(Ket[u]),(Ket[v]),(Ket[w])∈ℋ;α,β∈ℂ:
(Bra[u])(α(Ket[v])+β(Ket[w]))=α(BraKet[u, v])+β(BraKet[u, w]).

(Anti)linearity of the bra vector: using Eq. (), Eq. (), one can show

((SuperStar[α])(Bra[v])+(SuperStar[β])(Bra[w]))(Ket[u])=(SuperStar[α])(BraKet[v, u])+(SuperStar[β])(BraKet[w, u]).

Note: we follow the convention in Eq. () to denote the linear combination of bra vectors, which is different from the textbook notation (2.23).

Show that Eq. () is implied by Eq. () and Eq. ().

((SuperStar[α])(Bra[v])+(SuperStar[β])(Bra[w]))(Ket[u])
=(SuperStar[[(Bra[u])(α(Ket[v])+β(Ket[w]))]])
=(SuperStar[(α(BraKet[u, v])+β(BraKet[u, w]))])
=(SuperStar[α])(SuperStar[(BraKet[u, v])])+(SuperStar[β])(SuperStar[(BraKet[u, w])])
=(SuperStar[α])(BraKet[v, u])+(SuperStar[β])(BraKet[w, u]).

Scalar product of any vector (Ket[v]) with itself is real and positive definite,

(BraKet[v, v])>=0.

More specifically,

(BraKet[v, v])({{{, ({{=0, if (Ket[v])=0}, {>0, otherwise}})}}).

This implies the Cauchy-Schwarz inequality

((|(BraKet[u, v])|)^2)<=(BraKet[u, u])(BraKet[v, v]).

Prove Eq. ().

(BraKet[u, v])= (BraKet[v, v])=0,

(Ket[w])=(Ket[u])(BraKet[v, v])-(Ket[v])(BraKet[v, u]),

(BraKet[w, w])=((BraKet[v, v])(Bra[u])-(BraKet[u, v])(Bra[v]))((Ket[u])(BraKet[v, v])-(Ket[v])(BraKet[v, u]))
=(BraKet[v, v])(BraKet[u, u])(BraKet[v, v])-(BraKet[v, v])(BraKet[u, v])(BraKet[v, u])-(BraKet[u, v])(BraKet[v, u])(BraKet[v, v])+(BraKet[u, v])(BraKet[v, v])(BraKet[v, u])
=(BraKet[v, v])((BraKet[u, u])(BraKet[v, v])-(BraKet[u, v])(BraKet[v, u])).

(BraKet[u, u])(BraKet[v, v])-(BraKet[u, v])(BraKet[v, u])=((BraKet[w, w])/(BraKet[v, v]))>=0,

((|(BraKet[u, v])|)^2)=(BraKet[u, v])(BraKet[v, u])≤(BraKet[u, u])(BraKet[v, v]).

(Ket[v])≏(({{(Subscript[v, 1])}, {(Subscript[v, 2])}, {⋮}})), (Ket[w])≏(({{(Subscript[w, 1])}, {(Subscript[w, 2])}, {⋮}})),

Verify that Eq. () is consistent with the defining properties of the scalar product in Eq. (), Eq. (), Eq. ().

(BraKet[u, v])=(∑_i)(Subscript[u, i]^( * ))(Subscript[v, i])
=(∑_i)(SuperStar[((Subscript[u, i])(Subscript[v, i]^( * )))])
=(SuperStar[((∑_i)(Subscript[v, i]^( * ))(Subscript[u, i]))])
=(SuperStar[(BraKet[v, u])]).

(Bra[u])(α(Ket[v])+β(Ket[w]))
=(∑_i)(Subscript[u, i]^( * ))(α (Subscript[v, i])+β (Subscript[w, i]))
=(∑_i)(α (Subscript[u, i]^( * ))(Subscript[v, i])+β (Subscript[u, i]^( * ))(Subscript[w, i]))
=α(∑_i) (Subscript[u, i]^( * ))(Subscript[v, i])+β(∑_i) (Subscript[u, i]^( * ))(Subscript[w, i])
=α(BraKet[u, v])+β(BraKet[u, w]).

((SuperStar[α])(Bra[u])+(SuperStar[β])(Bra[v]))(Ket[w])
=(∑_i)(SuperStar[(α (Subscript[u, i])+β (Subscript[v, i]))])(Subscript[w, i])
=(∑_i)((SuperStar[α]) (Subscript[u, i]^( * ))(Subscript[w, i])+(SuperStar[β]) (Subscript[v, i]^( * ))(Subscript[w, i]))
=(SuperStar[α]) (∑_i)(Subscript[u, i]^( * ))(Subscript[w, i])+(SuperStar[β])(∑_i) (Subscript[v, i]^( * ))(Subscript[w, i])
=(SuperStar[α])(BraKet[u, w])+(SuperStar[β])(BraKet[v, w]).

Every quantum system is associated with a Hilbert space. Its different quantum states are different vectors in the same Hilbert space.

Bra Vector (II)

({{Φ, :, ℋ, →, ℂ}, { ,  , (Ket[v]), ↦, Φ((Ket[v]))=z}})

∀(Ket[v]),(Ket[w])∈ℋ;α,β∈ℂ:
Φ(α(Ket[v])+β(Ket[w]))=α Φ((Ket[v]))+β Φ((Ket[w]))

(Subscript[Φ, 1])((Ket[v]))=(Subscript[v, 1]).

This is a linear functional on the vector space.

(Subscript[Φ, 1])(α(Ket[v])+β(Ket[w]))=α (Subscript[v, 1])+β (Subscript[w, 1])=α (Subscript[Φ, 1])((Ket[v]))+β (Subscript[Φ, 1])((Ket[w])).

Denote the first basis vector as (Ket[1]), (Subscript[Φ, 1]) can be written as the scalar product:

(Subscript[Φ, 1])((Ket[v]))=(BraKet[1, v]).

Conversely, in mathematics, this corresponding linear functional is treated as the definition of the dual bra vector (Bra[u])=(Subscript[Φ, (Ket[u])]) of the original ket vector (Ket[u]).

Riesz theorem: any bounded linear functional acting on a Hilbert space ℋ can be represented as a scalar product of the form in Eq. ().

Dual vector space. The space of such functionals (bra vectors) is  called the dual vector space of ℋ, denoted as (SuperStar[ℋ]). The dual vector space (SuperStar[ℋ]) is the space of bra vectors.

Norm and Orthogonality

Norm

((‖v‖)^2)=(BraKet[v, v]).

Example: Consider a qubit state

(Ket[v])≏(({{(Subscript[v, 0])}, {(Subscript[v, 1])}})),

the normalization condition means

(BraKet[v, v])=(Subscript[v, 0]^( * ))(Subscript[v, 0])+(Subscript[v, 1]^( * ))(Subscript[v, 1])=1.

In general, the normalization condition means

(BraKet[v, v])=(∑_i)((|(BraKet[i, v])|)^2)=(∑_i)((|(Subscript[v, i])|)^2)=1.

According to the statistical interpretation of quantum state, ((|(Subscript[v, i])|)^2) is the probability to observe the system in the ith basis state. The normalization condition is simply a requirement that the probabilities must sum up to unity.

Normalization of a state: if a state (Ket[v]) was not normalized, it can be normalized by

(Ket[v])->((Ket[v])/(‖v‖))=(1/(Sqrt[(BraKet[v, v])]))(Ket[v]),

unless ‖v‖ is zero or infinity.

Normalize the vector (Ket[v])≏(({{1}, {2i}})).

(BraKet[v, v])=(({{1, -2i}}))(({{1}, {2i}}))=1+4=5.
(Sqrt[(BraKet[v, v])])=(Sqrt[5]).

(Ket[v])->(1/(Sqrt[(BraKet[v, v])]))(Ket[v])
≏(1/(Sqrt[5]))(({{1}, {2i}})).

Orthogonality

For example, the qubit states |0⟩ and |1⟩ (see Eq. ()) are orthogonal, as

(BraKet[0, 1])=(({{1, 0}}))(({{0}, {1}}))=0.

|0⟩ and |1⟩ are orthogonal for a good reason: they are distinct states of a qubit, i.e. if the qubit is in state 0, it is definitely not in state 1, vice versa.

∀(Ket[u])∈(Subscript[ℋ, 1]),(Ket[v])∈(Subscript[ℋ, 2]): (Bra[u])v⟩=0.

Complementary subspace: (Subscript[ℋ, 1,⊥]) is the complementary subspace of (Subscript[ℋ, 1]), if (Subscript[ℋ, 1]) and (Subscript[ℋ, 1,⊥]) are orthogonal and (Subscript[ℋ, 1])⊕(Subscript[ℋ, 1,⊥])=ℋ.

For each (Ket[v])∈ℋ, there is a unique decomposition:

(Ket[v])=(Ket[(Subscript[v, ∥])])+(Ket[(Subscript[v, ⊥])]),

such that (Ket[(Subscript[v, ∥])])∈(Subscript[ℋ, 1]) and (Ket[(Subscript[v, ⊥])])∈(Subscript[ℋ, 1,⊥]).

(Ket[(Subscript[v, ∥])]) is the projection of (Ket[v]) in (Subscript[ℋ, 1]).

Consider a Hilbert space ℋ spanned by three vectors 
(Ket[x])≏(({{1}, {0}, {0}})), (Ket[y])≏(({{0}, {1}, {0}})), (Ket[z])≏(({{0}, {0}, {1}})). 
(i) Let (Subscript[ℋ, ∥])=span{(Ket[x]),(Ket[y])} and (Subscript[ℋ, ⊥])=span{(Ket[z])}. Show that (Subscript[ℋ, ∥]) and (Subscript[ℋ, ⊥]) are orthogonal subspaces.
(ii) For (Ket[v])=(Subscript[v, x])(Ket[x])+(Subscript[v, y])(Ket[y])+(Subscript[v, z])(Ket[z]), find the unique decomposition (Ket[v])=(Ket[(Subscript[v, ∥])])+(Ket[(Subscript[v, ⊥])]), such that (Ket[(Subscript[v, ∥])])∈(Subscript[ℋ, ∥]) and (Ket[(Subscript[v, ⊥])])∈(Subscript[ℋ, ⊥]).

(Ket[u])=(Subscript[u, x])(Ket[x])+(Subscript[u, y])(Ket[y]).

(Ket[w])=(Subscript[w, z])(Ket[z]).

(BraKet[u, w])=((Subscript[u, x]^( * ))(Bra[x])+(Subscript[u, y]^( * ))(Bra[y]))(Subscript[w, z])(Ket[z])
=(Subscript[u, x]^( * ))(Subscript[w, z])(BraKet[x, z])+(Subscript[u, y]^( * ))(Subscript[w, z])(BraKet[y, z])
=0+0=0,

(Ket[(Subscript[v, ∥])])=(Subscript[v, 1])(Ket[x])+(Subscript[v, 2])(Ket[y]),
(Ket[(Subscript[v, ⊥])])=(Subscript[v, 3])(Ket[z]).

(Ket[v])=(Ket[(Subscript[v, ∥])])+(Ket[(Subscript[v, ⊥])])=(Subscript[v, 1])(Ket[x])+(Subscript[v, 2])(Ket[y])+(Subscript[v, 3])(Ket[z]).

(Ket[v])=(Subscript[v, x])(Ket[x])+(Subscript[v, y])(Ket[y])+(Subscript[v, z])(Ket[z]).

(Subscript[v, 1])=(Subscript[v, x]), (Subscript[v, 2])=(Subscript[v, y]), (Subscript[v, 3])=(Subscript[v, z]).

(Ket[(Subscript[v, ∥])])=(Subscript[v, x])(Ket[x])+(Subscript[v, y])(Ket[y]),
(Ket[(Subscript[v, ⊥])])=(Subscript[v, z])(Ket[z]).

Ray and Phase Ambiguity

(Ket[v'])=λ(Ket[v]),

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The square norm of the state vector does not affect the physical property of the quantum system, because it represent the total probability of observation outcomes, which can always be taken to 1.

To fix the norm ambiguity, we will always require a state vector to be normalized, i.e. (BraKet[v, v])=1.

The overall phase of the state vector also does not matter, as it turns out that all physical properties about the quantum system is always unchanged under (Ket[v])→(e^(i φ))(Ket[v]).

But there is no canonical way to fix the phase ambiguity. We must ensure that any prediction about the quantum system must not rely on the overall phase on the formulation level.

Fidelity

Fidelity is symmetric: F(u,v)=F(v,u).

Fidelity takes values in the range of

0<=F(u,v)≤1.

This follows from the Cauchy-Schwarz inequality of scalar product Eq. () that ((|(BraKet[u, v])|)^2)<=(BraKet[u, u])(BraKet[v, v]).

Statistical interpretation: If a quantum system is known to be in a state (Ket[v]), observing the system again may find the system in another state (Ket[u]) with probability

Detailed balance: the probability to observe one state on another is the same as the other way round

p(u|v)=p(v|u)=F(u,v)=((|(BraKet[u, v])|)^2).

Identical states. Two states (Ket[u]) and (Ket[v]) are identical iff the fidelity between them is one (fully overlap)

((|(BraKet[u, v])|)^2)=1.

This is only achievable when

(Ket[u])=(e^(i φ))(Ket[v]),

i.e. the two states are the same up to phase ambiguity.

Reality must be confirmable by repeated observations: if a quantum system is known to be in a state (Ket[v]), observing the system again will certainly confirm the state (Ket[v]) (with probability 1).

Distinct states. Two states (Ket[u]) and (Ket[v]) are distinct iff the fidelity between them is zero (no overlap)

((|(BraKet[u, v])|)^2)=0.

Orthogonal states ⇔ distinct realities.

Distinct realities are distinguishable by repeated observations: if a quantum system is known to be in a state (Ket[v]), observing the system again will certainly not find the system in another orthogonal state (Ket[u]).

Overlapping states. In general, two different states (Ket[u]) and (Ket[v]) may have partial overlap (they don’t need to be orthogonal), i.e. their fidelity falls between zero and one

0<((|(BraKet[u, v])|)^2)<1.

Realities can overlap: if two quantum states are more similar to (more overlapped with) each other, the probability to confuse them is higher.

(Ket[3])≏, (Ket[5])≏.

Therefore, there is generally some probability to observe one state given another state, if the two states has some similarity.

Basis

General Basis

(∑_i)(Subscript[c, i])(Ket[(Subscript[e, i])])=0⇔∀i:(Subscript[c, i])=0,

ℋ=span ℬ,

∀(Ket[v])∈ℋ,∃{(Subscript[v, i])}⊂ℂ:(Ket[v])=(∑_i)(Subscript[v, i])(Ket[(Subscript[e, i])]).

The coefficients (Subscript[v, i]) depend on what vector (Ket[v]) is being expanded.

The dimension of the vector space dim ℋ = the number of basis vectors = the maximal number of linearly independent vectors in the space.

The Hilbert space dimension of a quantum system can be finite or infinite. Example: a qubit - dim ℋ=2, ten qubits - dim ℋ=(2^10)=1024, a particle in a continuous space - dim ℋ=∞.

The summation in Eq. () becomes a infinite sum (countable) or an integration (uncountable) when the Hilbert space dimension is infinite.

Dimension of the Hilbert space is often a choice: we don’t really know how many independent states are there in a quantum system. We only care about the states that are relevant to us.

Orthonormal Basis

Each orthogonal basis state describes a distinct reality of the quantum system.

Example: (Ket[0]) and (Ket[1]) form an orthonormal basis of the qubit Hilbert space. They represent two distinct realities: if the qubit is in state (Ket[0]), it is definitely not in state (Ket[1]) (and vice versa).

Completeness: Any full set of distinct states in the Hilbert space ℋ forms a complete set of orthonormal basis ℬ, such that every state (Ket[v])∈ℋ can be expanded as a linear superposition of the basis states,

The superposition coefficient (Subscript[v, i]) are the components of the state vector, which can be extracted by the scalar product with the basis state,

The squared norm of a state (Ket[v]) is the sum of squared norm of its components on (any) orthonormal basis.

Using the orthonormal property (BraKet[i, j])=(Subscript[δ, i j]) to evaluate the norm of (Ket[v])=(∑_i)(Ket[i])(BraKet[i, v]).

((‖v‖)^2)=(BraKet[v, v])
=((∑_j)(BraKet[v, j])(Bra[j]))((∑_i)(Ket[i])(BraKet[i, v]))
=(∑_(i,j))(BraKet[v, j])(BraKet[i, j])(BraKet[i, v])
=(∑_(i,j))(BraKet[v, j])(Subscript[δ, i j])(BraKet[i, v])
=(∑_i)(BraKet[v, i])(BraKet[i, v])
=(∑_i)(Subscript[v, i]^( * ))(Subscript[v, i])
=(∑_i)((|(Subscript[v, i])|)^2).

Orthonormal basis states are represented by one-hot vectors, as they are normalized and orthogonal to each other

(Ket[1])≏(({{1}, {0}, {0}, {⋮}})), (Ket[2])≏(({{0}, {1}, {0}, {⋮}})), (Ket[3])≏(({{0}, {0}, {1}, {⋮}})), ….

Summary

States are vectors.

Ket and bra:

({{ , ket, bra (dual)}, {Hilbert space, ℋ, (SuperStar[ℋ])}, {basis, ℬ={(Ket[i])}, (SuperStar[ℬ])={(Bra[i])}}, {state, (Ket[v])=(∑_i)(Subscript[v, i])(Ket[i]), (Bra[v])=(∑_i)(Subscript[v, i]^( * ))(Bra[i])}, {vector, (({{(Subscript[v, 1])}, {(Subscript[v, 2])}, {⋮}})), (({{(Subscript[v, 1]^( * )), (Subscript[v, 2]^( * )), ⋯}}))}, {component, (Subscript[v, i])=(BraKet[i, v]), (Subscript[v, i]^( * ))=(BraKet[v, i])}})

Scalar product

(BraKet[u, v])=(∑_i)(BraKet[u, i])(BraKet[i, v])=(∑_i)(Subscript[u, i]^( * ))(Subscript[v, i]).

Normalized state (BraKet[v, v])=1. To normalize a state:

(Ket[v])→((Ket[v])/(Sqrt[(BraKet[v, v])])).

Orthogonal states (BraKet[u, v])=0.

Orthonormal basis ℬ={(Ket[i]):(BraKet[i, j])=(Subscript[δ, i j])}.

Fidelity (similarity between quantum states)

F(u,v)=((|(BraKet[u, v])|)^2)({{{, ({{=1, identical states}, {=0, distinct states}, {∈(0,1), overlapping states}})}}).

Statistical interpretation: p(u|v)=p(v|u)=F(u,v).

Quantum Operators

Operator and Matrix

Operator

({{(OverHat[O]), :, ℋ, →, ℋ}, { ,  , (Ket[v]), ↦, (Ket[w])=(OverHat[O])(Ket[v])}})

Distinction: operator is a vector-to-vector mapping, while functional is a vector-to-scalar mapping.

Linear operator: an operator (OverHat[O]) is say to be linear if it satisfies the linear property

∀(Ket[u]),(Ket[v])∈ℋ;α,β∈ℂ:
(OverHat[O])(α(Ket[u])+β(Ket[v]))=α(OverHat[O])(Ket[u])+β(OverHat[O])(Ket[v]).

∀(Ket[v]): 𝟙(Ket[v])=(Ket[v]).

Operator as a Matrix

(Subscript[O, i,j])∈ℂ are complex coefficients, which can be extracted by

Prove Eq. () from Eq. ().

(Bra[i])(OverHat[O])(Ket[j])=(Bra[i])((∑_(k,l))(Ket[k])(Subscript[O, k,l])(Bra[l]))(Ket[j])
=(∑_(k,l))(BraKet[i, k])(Subscript[O, k,l])(BraKet[l, j])
=(∑_(k,l))(Subscript[δ, i,k])(Subscript[O, k,l])(Subscript[δ, l,j])
=(Subscript[O, i,j]).

(Ket[i])(Bra[j]) denotes the operator that take the system from state (Ket[j]) to state (Ket[i]), because

((Ket[i])(Bra[j]))(Ket[k])=(Ket[i])(BraKet[j, k])=(Ket[i])(Subscript[δ, j,k])
=({{{, ({{(Ket[i]), if k=j,}, {0, if k≠j.}})}})

Matrix representation. Every operator can be represented as a matrix

The ith row jth column matrix element (Subscript[O, i,j]) is the linear combination coefficient in front of the basis operator (Ket[i])(Bra[j]).

Use the vector representation of ket and bra basis to show Eq. () is the corresponding representation of  Eq. ().

(OverHat[O])=(∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j])
=(Ket[1])(Subscript[O, 1,1])(Bra[1])+(Ket[1])(Subscript[O, 1,2])(Bra[2])+(Ket[2])(Subscript[O, 2,1])(Bra[1])+(Ket[2])(Subscript[O, 22])(Bra[2])+…
=(({{1}, {0}, {⋮}}))(Subscript[O, 1,1])(({{1, 0, ⋯}}))+(({{1}, {0}, {⋮}}))(Subscript[O, 1,2])(({{0, 1, ⋯}}))+(({{0}, {1}, {⋮}}))(Subscript[O, 2,1])(({{1, 0, ⋯}}))+(({{0}, {1}, {⋮}}))(Subscript[O, 2,2])(({{0, 1, ⋯}}))+…
=(Subscript[O, 1,1])(({{1, 0, ⋯}, {0, 0, ⋯}, {⋮, ⋮, ⋱}}))+(Subscript[O, 1,2])(({{0, 1, ⋯}, {0, 0, ⋯}, {⋮, ⋮, ⋱}}))+(Subscript[O, 2,1])(({{0, 0, ⋯}, {1, 0, ⋯}, {⋮, ⋮, ⋱}}))+(Subscript[O, 2,2])(({{0, 0, ⋯}, {0, 1, ⋯}, {⋮, ⋮, ⋱}}))+…
=(({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}})).

(Subscript[𝟙, i j])=(Bra[i])𝟙(Ket[j])=(BraKet[i, j])=(Subscript[δ, i j])=({{{, ({{1, i=j}, {0, i≠j}})}}).

In matrix representation Eq. (),

𝟙=(({{1,  ,  }, { , 1,  }, { ,  , ⋱}})).

Using Dirac notation Eq. (),

𝟙=(∑_(i,j))(Ket[i])(Subscript[𝟙, i j])(Bra[j])=(∑_(i,))(Ket[i])(Bra[i]).

This is also call the resolution of identity.

((OverHat[σ])^x)=(Ket[1])(Bra[0])+(Ket[0])(Bra[1]),
((OverHat[σ])^y)=i (Ket[1])(Bra[0])-i(Ket[0])(Bra[1]),
((OverHat[σ])^z)=(Ket[0])(Bra[0])-(Ket[1])(Bra[1]),

𝟙=(Ket[0])(Bra[0])+(Ket[1])(Bra[1]),

Operator Acting on State

Consider an operator (OverHat[O]) acting on a state (Ket[v]) resulting in a state (Ket[w])

(Ket[w])=(OverHat[O])(Ket[v]),

where states and operators are expanded as

(Ket[v])=(∑_i)(Subscript[v, i])(Ket[i]), (Ket[w])=(∑_i)(Subscript[w, i])(Ket[i]),
 (OverHat[O])=(∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j]).

The expansion coefficients are related by the matrix-vector multiplication

(Subscript[w, i])=(∑_j)(Subscript[O, i,j])(Subscript[v, j]).

Prove the statement Eq. () using Eq. ().

(OverHat[O])(Ket[v])=((∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j]))((∑_k)(Subscript[v, k])(Ket[k]))
=(∑_(i,j,k))(Subscript[O, i,j])(Subscript[v, k])(Ket[i])(BraKet[j, k])
=(∑_(i,j,k))(Subscript[O, i,j])(Subscript[v, k])(Ket[i])(Subscript[δ, j,k])
=(∑_(i,j))(Subscript[O, i,j])(Subscript[v, j])(Ket[i])
=(∑_i)((∑_j)(Subscript[O, i,j])(Subscript[v, j]))(Ket[i]),

(Ket[w])=(∑_i)(Subscript[w, i])(Ket[i]).

(Subscript[w, i])=(∑_j)(Subscript[O, i,j])(Subscript[v, j]),

(({{(Subscript[w, 1])}, {(Subscript[w, 2])}, {⋮}}))=(({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}}))(({{(Subscript[v, 1])}, {(Subscript[v, 2])}, {⋮}})).

Tensor Network*

Each object (either state or operator) is a tensor (multi-dimensions array).

Vectors are rank-1 tensors, represented by an object with one leg

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{1,3}],ket["v",{3,0}]},ImageSize->{Automatic,25}]]

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{1,3}],ket["v",{3,0}],Text["\!\( i\)",{1,0},{-1,-0.5}],Text["\!\(=v\_i\)",{4,0},{-1,0}]},ImageSize->{Automatic,28}]]

Matrices are rank-2 tensors, represented by an object with two legs

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-2,2}],opH["O",{0,0}]},ImageSize->{Automatic,25}]]

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-2,2}],opH["O",{0,0}],Text["\!\( i\)",{-2,0},{-1,-0.5}],Text["\!\( j\)",{2,0},{1,-0.5}],Text["\!\(=O\_\(i j\)\)",{2,0},{-1,0}]},ImageSize->{Automatic,27}]]

Tensor contraction: indices on internal legs are automatically summed over. For example, matrix-vector multiplication can be expressed as a tensor contraction.

(Ket[w])=(OverHat[O])(Ket[v])

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-2,3}],opH["O"],ket["v",{3,0}],Line[{#,0}&/@{-7,-5}],ket["w",{-5,0}],Text["=",{-3,0}]},ImageSize->{Automatic,30}]]

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{{Line[{#,0}&/@{-7,-5}],ket["w",{-5,0}],Text["\!\( i\)",{-7,0},{-1,-0.5}]},Text["=",{-3,0}],{Line[{#,0}&/@{-2,3}],opH["O"],ket["v",{3,0}],Text["\!\( i\)",{-2,0},{-1,-0.5}],Text["\!\( j\)",{2,0},{1,-0.5}]}},ImageSize->{Automatic,30}]]

(Subscript[w, i])=(∑_j)(Subscript[O, i,j])(Subscript[v, j])

Operator Algebra

Operator Product

((OverHat[O])(OverHat[P]))(Ket[v])=((OverHat[O])((OverHat[P])(Ket[v]))).

Composing two operators ≏ multiplying two matrices.

Prove Eq. () using Eq. ().

(OverHat[O])(OverHat[P])=((∑_(i j))(Ket[i])(Subscript[O, i j])(Bra[j]))((∑_(k l))(Ket[k])(Subscript[P, k l])(Bra[l]))
=(∑_(i j,k l))(Ket[i])(Subscript[O, i j])(BraKet[j, k])(Subscript[P, k l])(Bra[l])
=(∑_(i j,k l))(Ket[i])(Subscript[O, i j])(Subscript[δ, j,k])(Subscript[P, k l])(Bra[l])
=(∑_(i j l))(Ket[i])(Subscript[O, i j])(Subscript[P, j l])(Bra[l])
=(∑_(i l))(Ket[i])((∑_j)(Subscript[O, i j])(Subscript[P, j l]))(Bra[l])
=(∑_(i j))(Ket[i])((∑_k)(Subscript[O, i k])(Subscript[P, k j]))(Bra[j]).

Tensor network representation

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-5,2}],opH["O",{-3,0}],opH["P",{0,0}]},ImageSize->{Automatic,25}]]

In component form:

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-5,2}],Text["\!\( i\)",{-5,0},{-1,-0.5}],opH["O",{-3,0}],Text["\!\( k\)",{-1.5,0},{0,-0.5}],opH["P",{0,0}],Text["\!\( j\)",{2,0},{1,-0.5}]},ImageSize->{Automatic,27}]]

(Subscript[(O P), i j])=(∑_k)(Subscript[O, i k])(Subscript[P, k j])

Operator product is non-commutative in general, i.e.

(OverHat[O])(OverHat[P])≠(OverHat[P])(OverHat[O]).

({{ , 𝟙, ((OverHat[σ])^x), ((OverHat[σ])^y), ((OverHat[σ])^z)}, {𝟙, 𝟙, ((OverHat[σ])^x), ((OverHat[σ])^y), ((OverHat[σ])^z)}, {((OverHat[σ])^x), ((OverHat[σ])^x), 𝟙, i ((OverHat[σ])^z), -i ((OverHat[σ])^y)}, {((OverHat[σ])^y), ((OverHat[σ])^y), -i ((OverHat[σ])^z), 𝟙, i ((OverHat[σ])^x)}, {((OverHat[σ])^z), ((OverHat[σ])^z), i ((OverHat[σ])^y), -i ((OverHat[σ])^x), 𝟙}})

Verify Eq. () by multiplying Pauli matrices defined in Eq. ().

TableForm[Table[MatrixForm[PauliMatrix[a].PauliMatrix[b]],{a,0,3},{b,0,3}],TableHeadings->({#,#}&@{"𝟙","\!\(σ\^x\)","\!\(σ\^y\)","\!\(σ\^z\)"})]

The table Eq. () can be summarized in a single formula: the product of Pauli matrices (as the defining property of Pauli matrices)

where a,b,c=x,y,z.

(δ^(a b)) denotes the Kronecker delta symbol, defined as

(δ^(a b))=({{{, ({{1, if a=b}, {0, if a≠b}})}})

(ϵ^(a b c)) denotes the Levi-Civita symbol, defined as

(ϵ^(a b c))=({{{, ({{1, if (a b c) is a cyclic of (x y z)}, {-1, if (a b c) is a cyclic of (z y x)}, {0, otherwise}})}})

Another version of Eq. () using vector notation

where m,n are three-component vectors (each component is a scalar).

The generalized vector (OverHat[σ]) should be understood as a vector of matrices, or as a three-dimensional tensor (shape: 3×2×2).

Here m·(OverHat[σ]) means

m·(OverHat[σ])=(Subscript[m, x])((OverHat[σ])^x)+(Subscript[m, y])((OverHat[σ])^y)+(Subscript[m, z])((OverHat[σ])^z)
≏(({{(Subscript[m, z]), (Subscript[m, x])-i (Subscript[m, y])}, {(Subscript[m, x])+i (Subscript[m, y]), -(Subscript[m, z])}})).

As we contract a 3-component vector m with a 3×2×2-component tensor (OverHat[σ]) along the first index (the dimension 3 index), the result is a 2×2 matrix.

Commutator

Commutator is antisymmetric, [(OverHat[O]),(OverHat[P])]=-[(OverHat[P]),(OverHat[O])].

As a result, commutator of an operator with itself always vanishes [(OverHat[O]),(OverHat[O])]=0.

If the commutator vanishes [(OverHat[O]),(OverHat[P])]=0, we say that the two operators (OverHat[O]) and (OverHat[P]) commute, i.e. (OverHat[O])(OverHat[P])=(OverHat[P])(OverHat[O]) (operators can pass though each other as if they were numbers) => it does not matter which operator is applied first, the consequence will be the same.

A: put on the socks,

B: put on the shoes,

C: put on the hat,

Bi-linearity

[(OverHat[O]),(OverHat[P])+(OverHat[Q])]=[(OverHat[O]),(OverHat[P])]+[(OverHat[O]),(OverHat[Q])],
[(OverHat[O])+(OverHat[P]),(OverHat[Q])]=[(OverHat[O]),(OverHat[Q])]+[(OverHat[P]),(OverHat[Q])].

Prove Eq. ().

[(OverHat[O]),(OverHat[P])+(OverHat[Q])]=(OverHat[O]) ((OverHat[P])+(OverHat[Q]))-((OverHat[P])+(OverHat[Q]))(OverHat[O])
=(OverHat[O]) (OverHat[P])+(OverHat[O]) (OverHat[Q])-(OverHat[P])(OverHat[O])-(OverHat[Q])(OverHat[O])
=(OverHat[O]) (OverHat[P])-(OverHat[P])(OverHat[O])+(OverHat[O]) (OverHat[Q])-(OverHat[Q])(OverHat[O])
=[(OverHat[O]),(OverHat[P])]+[(OverHat[O]),(OverHat[Q])].

Product rules

[(OverHat[O]),(OverHat[P]) (OverHat[Q])]=[(OverHat[O]),(OverHat[P])](OverHat[Q])+(OverHat[P]) [(OverHat[O]),(OverHat[Q])],
[(OverHat[O]) (OverHat[P]),(OverHat[Q])]=[(OverHat[O]),(OverHat[Q])](OverHat[P])+(OverHat[O]) [(OverHat[P]),(OverHat[Q])].

Prove Eq. ().

[(OverHat[O]),(OverHat[P]) (OverHat[Q])]=(OverHat[O])(OverHat[P]) (OverHat[Q])-(OverHat[P]) (OverHat[Q])(OverHat[O])
=(OverHat[O])(OverHat[P]) (OverHat[Q])-(OverHat[P]) (OverHat[O])(OverHat[Q])+(OverHat[P]) (OverHat[O])(OverHat[Q])-(OverHat[P]) (OverHat[Q])(OverHat[O])
=((OverHat[O])(OverHat[P]) -(OverHat[P]) (OverHat[O]))(OverHat[Q])+(OverHat[P]) ((OverHat[O])(OverHat[Q])-(OverHat[Q])(OverHat[O]))
=[(OverHat[O]),(OverHat[P])](OverHat[Q])+(OverHat[P]) [(OverHat[O]),(OverHat[Q])]

Jacobi identity (as a replacement of associative law)

[(OverHat[O]),[(OverHat[P]),(OverHat[Q])]]+[(OverHat[P]),[(OverHat[Q]),(OverHat[O])]]+[(OverHat[Q]),[(OverHat[O]),(OverHat[P])]]=0,
[[(OverHat[O]),(OverHat[P])],(OverHat[Q])]+[[(OverHat[P]),(OverHat[Q])],(OverHat[O])]+[[(OverHat[Q]),(OverHat[O])],(OverHat[P])]=0.

[((OverHat[σ])^x),((OverHat[σ])^y)]=2i ((OverHat[σ])^z),
[((OverHat[σ])^y),((OverHat[σ])^z)]=2i ((OverHat[σ])^x),
[((OverHat[σ])^z),((OverHat[σ])^x)]=2i ((OverHat[σ])^y).

Operator Function

((OverHat[O])^n)=(OverHat[O])(OverHat[O])…(n times)…(OverHat[O]).

f(x)=(∑_n)(Subscript[c, n])(x^n),

f((OverHat[O])) is still an operator that can act on states in ℋ.

Operator exponential. Given the exponential function

(e^x)=1+x+((x^2)/(2!))+…=(∑_(n=0)^∞)(1/(n!))(x^n),

the exponential of an operator is defined as

(e^(OverHat[O]))=𝟙+(OverHat[O])+(((OverHat[O])^2)/(2!))+…=(∑_(n=0)^∞)(1/(n!))((OverHat[O])^n),

Note: exponentiating an matrix is NOT exponentiating each of the matrix element.

Given ((OverHat[σ])^y)≏(({{0, -i}, {i, 0}})), 
show that the matrix representation of (e^(i θ ((OverHat[σ])^y))) is 
(e^(i θ ((OverHat[σ])^y)))≏(({{cos θ, sin θ}, {-sin θ, cos θ}})).

(e^(i θ ((OverHat[σ])^y)))≏(e^((({{0, θ}, {-θ, 0}}))))=(({{cos θ, sin θ}, {-sin θ, cos θ}})).

MatrixExp[{{0,θ},{-θ,0}}]

(e^(i θ ((OverHat[σ])^y)))=(∑_n)(1/(n!))(((i θ ((OverHat[σ])^y)))^n)
=(∑_n)((((i θ))^n)/(n!))(((((OverHat[σ])^y)))^n)

(((((OverHat[σ])^y)))^n)=({{{, ({{𝟙, if n is even,}, {((OverHat[σ])^y), if n is odd.}})}})

(e^(i θ ((OverHat[σ])^y)))=(∑_m)((((i θ))^(2m))/((2m)!))(((((OverHat[σ])^y)))^(2m))+(∑_m)((((i θ))^(2m+1))/((2m+1)!))(((((OverHat[σ])^y)))^(2m+1))
=(∑_m)((((i θ))^(2m))/((2m)!))𝟙+(∑_m)((((i θ))^(2m+1))/((2m+1)!))((OverHat[σ])^y)
=cos θ 𝟙+i sin θ ((OverHat[σ])^y)
≏(({{cos θ, sin θ}, {-sin θ, cos θ}})).

Operator Trace

On the matrix level, taking the trace is simply summing over diagonal matrix elements

Tr(({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}}))=(Subscript[O, 11])+(Subscript[O, 22])+…=(∑_i)(Subscript[O, i,i]).

Tensor network representation (tracing = closing the legs)

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-1,1}],Line[{#,1.6}&/@{-1,1}],Circle[{-1,0.8},0.8,{π/2,3π/2}],Circle[{1,0.8},0.8,{-π/2,π/2}],opH["O",{0,0}],(*Text["\!\( i\)",{1,0},{-1,-0.9}]*)},ImageSize->{Automatic,36}]]

Linear property: trace is a linear functional of operators.

Tr (α(OverHat[O])+β (OverHat[P]))=a Tr (OverHat[O])+β Tr (OverHat[P]).

Cyclic property: the trace of a product of operators is invariant under the cyclic permutation of the operators.

Tr ((OverHat[O]) (OverHat[P]))=Tr ((OverHat[P])(OverHat[O])),
Tr ((OverHat[O]) (OverHat[P])(OverHat[Q]))=Tr ((OverHat[P])(OverHat[Q]) (OverHat[O]))=Tr ((OverHat[Q])(OverHat[O]) (OverHat[P])),
…

Prove Eq. ().

Tr ((OverHat[O]) (OverHat[P]))=(∑_i)(Bra[i])(OverHat[O])(OverHat[P])(Ket[i])
=(∑_(i,j))(Bra[i])(OverHat[O])(Ket[j])(Bra[j])(OverHat[P])(Ket[i])
=(∑_(i,j))(Bra[j])(OverHat[P])(Ket[i])(Bra[i])(OverHat[O])(Ket[j])
=(∑_j)(Bra[j])(OverHat[P])(OverHat[O])(Ket[j])
=Tr ((OverHat[P])(OverHat[O])).

Tr ((OverHat[O]) (OverHat[P])(OverHat[Q]))=Tr (((OverHat[O]) (OverHat[P]))(OverHat[Q]))=Tr ((OverHat[Q ])((OverHat[O]) (OverHat[P])))=Tr ((OverHat[Q])(OverHat[O]) (OverHat[P])).

Tr ((OverHat[σ])^x)=Tr ((OverHat[σ])^y)=Tr ((OverHat[σ])^z)=0.

Projection Operators

Projectors

(Subscript[(OverHat[𝒫]), v])(Ket[u])=(Ket[v])(BraKet[v, u]).

Block[{θ=π/3},
Graphics[{{AbsoluteThickness[1],Gray,{Dashed,Line[{{Cos[θ],0},{Cos[θ],Sin[θ]}}]},Translate[Line[0.07{{1,0},{1,1},{0,1}}],{Cos[θ],0}],Circle[{0,0},0.1,{0,θ}]},Text["θ",{0.1,0},{-1,-0.8}],{Arrowheads[0.08],Arrow[{{0,0},#}],Text[#2,#1,#3]}&@@@{{{1,0},"\!\( v\)",{-1,0}},{{Cos[θ],Sin[θ]},"\!\( u\)",{0,-0.5}},{{Cos[θ],0},"\!\(v cos θ\)",{0,1}}}
},ImageSize->120]]

v cos θ=v (v·u).

Basis projectors satisfies the following properties

(Subscript[(OverHat[𝒫]), i]^2)=(Subscript[(OverHat[𝒫]), i]),
(Subscript[(OverHat[𝒫]), i])(Subscript[(OverHat[𝒫]), j])=0    if i≠j.

Verify Eq. (), following the definition Eq. ().

(Subscript[(OverHat[𝒫]), i]^2)=(Ket[i])(BraKet[i, i])(Bra[i])
=(Ket[i])(Bra[i])
=(Subscript[(OverHat[𝒫]), i]).

(Subscript[(OverHat[𝒫]), i])(Subscript[(OverHat[𝒫]), j])=(Ket[i])(BraKet[i, j])(Bra[j])
(Overscript[=, i≠j])(Ket[i])0(Bra[j])
=0.

Given the one-hot representation of orthonormal basis

(Ket[1])≏(({{1}, {0}, {0}, {⋮}})), (Ket[2])≏(({{0}, {1}, {0}, {⋮}})), ….

Matrix representations of the corresponding basis projectors are

(Subscript[(OverHat[𝒫]), 1])≏(({{1,  ,  ,  }, { , 0,  ,  }, { ,  , 0,  }, { ,  ,  , ⋱}})), (Subscript[(OverHat[𝒫]), 2])≏(({{0,  ,  ,  }, { , 1,  ,  }, { ,  , 0,  }, { ,  ,  , ⋱}})), ….

Dimension of the subspace (Subscript[ℋ, 1]) is given by the trace of its projector (as it counts the number of basis states in (Subscript[ℬ, 1]))

Tr (Subscript[(OverHat[𝒫]), (Subscript[ℋ, 1])])=dim (Subscript[ℋ, 1]).

Completeness relation: full Hilbert space projector ≡ identity operator

dim ℋ=Tr (Subscript[(OverHat[𝒫]), ℋ])=Tr 𝟙.

Quantum State Collapse*

Given the prior knowledge that a quantum system is described by a state (Ket[v]).

An observer comes with an objective to test the hypothesis that "the quantum system is in a target state (Ket[u])".

For this purpose, two projection operators are defined

(Subscript[(OverHat[𝒫]), u])=(Ket[u])(Bra[u]), (projector to the state (Ket[u]))
(Subscript[(OverHat[𝒫]), (OverBar[u])])=𝟙-(Subscript[(OverHat[𝒫]), u]).  (projector to the complement subspace)

Quantum mechanics predicts that the observation will accept the hypothesis with the probability

and reject the hypothesis with the (complement) probability

p((OverBar[u])|v)=(Bra[v])(Subscript[(OverHat[𝒫]), (OverBar[u])])(Ket[v])=1-((|(BraKet[u, v])|)^2).

After observation, if the hypothesis is accepted (the state (Ket[u]) is observed), we must update our knowledge about the quantum system to the posterior knowledge described by the state (Ket[u]). In this sense, the original quantum state (Ket[v]) collapses to

However, if the hypothesis is rejected (the state (Ket[u]) is not observed), we must also update our posterior knowledge by collapsing to

(Ket[v])→(((Subscript[(OverHat[𝒫]), (OverBar[u])])(Ket[v]))/(Sqrt[p((OverBar[u])|v)]))=(((Subscript[(OverHat[𝒫]), (OverBar[u])])(Ket[v]))/(Sqrt[(Bra[v])(Subscript[(OverHat[𝒫]), (OverBar[u])])(Ket[v])])).

Verify Eq. (), Eq. () and Eq. ().

(Bra[v])(Subscript[(OverHat[𝒫]), u])(Ket[v])=(BraKet[v, u])(BraKet[u, v])=((|(BraKet[u, v])|)^2).

(Bra[v])(Subscript[(OverHat[𝒫]), (OverBar[u])])(Ket[v])=(Bra[v])𝟙-(Subscript[(OverHat[𝒫]), u])(Ket[v])=(BraKet[v, v])-(Bra[v])(Subscript[(OverHat[𝒫]), u])(Ket[v])=1-((|(BraKet[u, v])|)^2).

(((Subscript[(OverHat[𝒫]), u])(Ket[v]))/(Sqrt[(Bra[v])(Subscript[(OverHat[𝒫]), u])(Ket[v])]))=(((Ket[u])(BraKet[u, v]))/(Sqrt[((|(BraKet[u, v])|)^2)]))=((BraKet[u, v])/(|(BraKet[u, v])|))(Ket[u]),

As an observable, it gives the probability p(u|v) for the state collapse to happen.

As an operator, it implements the state collapse (Ket[v])→(Ket[u]) (Note: this is not a linear operation, as (Ket[v]) appears on both the numerator and the denominator).

(Ket[ψ])≏(({{(Subscript[ψ, 0])}, {(Subscript[ψ, 1])}})).

(Ket[0])≏(({{1}, {0}}))=>(Subscript[(OverHat[𝒫]), 0])=(Ket[0])(Bra[0])=(({{1}, {0}}))(({{1, 0}}))=(({{1, 0}, {0, 0}}))

p(0|ψ)=(Bra[ψ])(Subscript[(OverHat[𝒫]), 0])(Ket[ψ])
=(({{(Subscript[ψ, 0]^( * )), (Subscript[ψ, 1]^( * ))}}))(({{1, 0}, {0, 0}}))(({{(Subscript[ψ, 0])}, {(Subscript[ψ, 1])}}))
=(Subscript[ψ, 0]^( * ))(Subscript[ψ, 0])=((|(Subscript[ψ, 0])|)^2).

(((Subscript[(OverHat[𝒫]), 0])(Ket[ψ]))/(Sqrt[p(0|ψ)]))≏(1/(Sqrt[((|(Subscript[ψ, 0])|)^2)]))(({{1, 0}, {0, 0}}))(({{(Subscript[ψ, 0])}, {(Subscript[ψ, 1])}}))
=((Subscript[ψ, 0])/(|(Subscript[ψ, 0])|))(({{1}, {0}}))∼(({{1}, {0}}))≏(Ket[0]).

Measurement

Hermitian Operators

Hermitian Conjugate

({{ , ket, bra (dual)}, {Hilbert space, ℋ, (SuperStar[ℋ])}, {basis, ℬ={(Ket[i])}, (SuperStar[ℬ])={(Bra[i])}}, {state, (Ket[v])=(∑_i)(Subscript[v, i])(Ket[i]), (Bra[v])=(∑_i)(Subscript[v, i]^( * ))(Bra[i])}, {vector, (({{(Subscript[v, 1])}, {(Subscript[v, 2])}, {⋮}})), (({{(Subscript[v, 1]^( * )), (Subscript[v, 2]^( * )), ⋯}}))}, {component, (Subscript[v, i])=(BraKet[i, v]), (Subscript[v, i]^( * ))=(BraKet[v, i])}, {operator, (OverHat[O])=(∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j]), (SuperDagger[(OverHat[O])])=(∑_(i,j))(Ket[i])(Subscript[O, j,i]^( * ))(Bra[j])}, {matrix, (({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}})), (({{(Subscript[O, 11]^( * )), (Subscript[O, 21]^( * )), ⋯}, {(Subscript[O, 12]^( * )), (Subscript[O, 22]^( * )), ⋯}, {⋮, ⋮, ⋱}}))}, {component, (Subscript[O, i,j])=(Bra[i])(OverHat[O])(Ket[j]), (Subscript[O, i,j]^( * ))=(Bra[j])(SuperDagger[(OverHat[O])])(Ket[i])}, {action, (Ket[w])=(OverHat[O])(Ket[v]), (Bra[w])=(Bra[v])(SuperDagger[(OverHat[O])])}})

Just like the bra (Bra[v]) is the dual of the ket (Ket[u]), the Hermitian conjugate operator (SuperDagger[(OverHat[O])]) is the dual of the original operator (OverHat[O]), such that

if the operator (OverHat[O]) takes (Ket[v]) to (Ket[w]):

({{(OverHat[O]), :, ℋ, →, ℋ}, { ,  , (Ket[v]), ↦, (Ket[w])=(OverHat[O])(Ket[v])}})

then the operator (SuperDagger[(OverHat[O])]) takes (Bra[v]) to (Bra[w]):

({{(SuperDagger[(OverHat[O])]), :, (SuperStar[ℋ]), →, (SuperStar[ℋ])}, { ,  , (Bra[v]), ↦, (Bra[w])=(Bra[v])(SuperDagger[(OverHat[O])])}})

Tensor network representation

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-2,3}],opH["O"],ket["v",{3,0}],Line[{#,0}&/@{6,8}],ket["w",{8,0}],Text["=",{5,0}]},ImageSize->{Automatic,30}]]

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{2,-3}],opH["O\^†"],bra["v",{-3,0}],Line[{#,0}&/@{5,7}],bra["w",{5,0}],Text["=",{3,0}]},ImageSize->{Automatic,30}]]

Given an orthonormal basis  ℬ={(Ket[i]):i=1,2,…} of the Hilbert space ℋ, if (OverHat[O]) is given by

(OverHat[O])=(∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j]),

then (SuperDagger[(OverHat[O])]) should be given by

Verify that Eq. () is consistent with the definition Eq. ().

(Bra[v])=(∑_i)(Subscript[v, i]^( * ))(Bra[i]), (Bra[w])=(∑_i)(Subscript[w, i]^( * ))(Bra[i]),
(SuperDagger[(OverHat[O])])=(∑_(i,j))(Ket[i])(Subscript[O, j,i]^( * ))(Bra[j]).

(Bra[v])(SuperDagger[(OverHat[O])])=((∑_i)(Subscript[v, i]^( * ))(Bra[i]))((∑_(j,k))(Ket[j])(Subscript[O, k,j]^( * ))(Bra[k]))
=(∑_(i ,j,k))(Subscript[v, i]^( * ))(Subscript[O, k,j]^( * ))(BraKet[i, j])(Bra[k])
=(∑_(i ,j,k))(Subscript[v, i]^( * ))(Subscript[O, k,j]^( * ))(Subscript[δ, i,j])(Bra[k])
=(∑_(,j,k))(Subscript[v, j]^( * ))(Subscript[O, k,j]^( * ))(Bra[k])
=(∑_(i,j))(Subscript[v, j]^( * ))(Subscript[O, i,j]^( * ))(Bra[i]),

(Subscript[w, i]^( * ))=(∑_(i,j))(Subscript[v, j]^( * ))(Subscript[O, i,j]^( * )).

(Subscript[w, i])=(∑_j)(Subscript[O, i,j])(Subscript[v, j]),

In terms of matrix representation, the Hermitian conjugate acts as

matrix transpose (interchanges the rows and columns),

followed by complex conjugation of each matrix element.

(SuperDagger[(({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}}))])=(({{(Subscript[O, 11]^( * )), (Subscript[O, 21]^( * )), ⋯}, {(Subscript[O, 12]^( * )), (Subscript[O, 22]^( * )), ⋯}, {⋮, ⋮, ⋱}})).

Duality: suppose (OverHat[O]) is an operator

(SuperDagger[((SuperDagger[(OverHat[O])]))])=(OverHat[O]).

Linearity: suppose (OverHat[O]) and (OverHat[P]) are operators, α and  β are complex numbers,

(SuperDagger[(α (OverHat[O])+β (OverHat[P]))])=(SuperStar[α])(SuperDagger[(OverHat[O])])+(SuperStar[β])(SuperDagger[(OverHat[P])]).

Transpose Property: suppose (OverHat[O]) and (OverHat[P]) are operators

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-5,2}],opH["O",{-3,0}],opH["P",{0,0}]},ImageSize->{Automatic,25}]]

Block[{ket,bra,opH,opU},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-5,2}],opH["P\^†",{-3,0}],opH["O\^†",{0,0}]},ImageSize->{Automatic,25}]]

Prove the property Eq. ().

(OverHat[O])=(∑_(i,j))(Ket[i])(Subscript[O, i,j])(Bra[j]), (OverHat[P])=(∑_(i,j))(Ket[i])(Subscript[P, i,j])(Bra[j]),

(SuperDagger[((OverHat[O]) (OverHat[P]))])=(SuperDagger[((∑_(i j))(Ket[i])((∑_k)(Subscript[O, i k])(Subscript[P, k j]))(Bra[j]))])
=(∑_(i j))(Ket[j])((∑_k)(Subscript[O, i k]^( * ))(Subscript[P, k j]^( * )))(Bra[i])
=(∑_(i j,k))(Subscript[O, i k]^( * ))(Subscript[P, k j]^( * ))(Ket[j])(Bra[i])
=(∑_(i j,k,l))(Subscript[O, i k]^( * ))(Subscript[P, l j]^( * ))(Ket[j])(BraKet[l, k])(Bra[i])
=(∑_(i j,k,l))(Ket[j])(Subscript[P, l j]^( * ))(BraKet[l, k])(Subscript[O, i k]^( * ))(Bra[i])
=((∑_(j,l))(Ket[j])(Subscript[P, l j]^( * ))(Bra[l]))((∑_(i k))(Ket[k])(Subscript[O, i k]^( * ))(Bra[i]))
=(SuperDagger[(OverHat[P])])(SuperDagger[(OverHat[O])]).

Hermitian Operator

A real number is a number whose complex conjugation is itself.

A real operator Hermitian operator is an linear operator whose Hermitian conjugate is itself.

(Subscript[O, i j])=(Subscript[O, j i]^( * )).

Given a complex number z, real part: Re z=(z+(SuperStar[z]))/2, imaginary part: Im z=(z-(SuperStar[z]))/(2i). Similarity, given a generally non-Hermitian operator (OverHat[P])

Re (OverHat[P])=(1/2)((OverHat[P])+(SuperDagger[(OverHat[P])])), Im (OverHat[P])=(1/(2i))((OverHat[P])-(SuperDagger[(OverHat[P])])).

Both Re (OverHat[P]) and Im (OverHat[P]) are Hermitian operators.

Eigenvalues and Eigenvectors (General)

Eq. () is called the eigen equation of an operator (OverHat[O]).

The eigenvalues can be found by solving the algebraic (polynomial) equation for O∈ℝ

det((OverHat[O])-O 𝟙)=0.

For each solution of eigenvalue  O=(Subscript[O, k]), the corresponding eigenvector (Ket[(Subscript[O, k])]) is found by solving the linear equation

((OverHat[O])-(Subscript[O, k]) 𝟙)(Ket[(Subscript[O, k])])=0.

Use Mathematica to solve the eigen problem (recommended)

Eigensystem[{{1,-1},{1,1}}]

Since any linear combination of the degenerated eigenvectors

(∑_(m=1)^(Subscript[g, k]))(Subscript[α, m])(Ket[(Subscript[O, k]),m]), (with (Subscript[α, k])∈ℂ)

is still an eigenvector of the same eigenvalue (Subscript[O, k]), thus the degenerated eigenvectors forms a degenerated subspace (Subscript[ℋ, O=(Subscript[O, k])]) (the subspace of all eigenvectors of the same eigenvalue (Subscript[O, k]))

(Subscript[ℋ, O=(Subscript[O, k])])=span{(Ket[(Subscript[O, k]),m]):m=1,2,…,(Subscript[g, k])}.

Eigen projector: projection operator to the degenerated subspace associated with an eigenvalue (Subscript[O, k])

Eigenvalues and Eigenvectors (Hermitian Operator)

(OverHat[O])(Ket[(Subscript[O, k])])=(Subscript[O, k])(Ket[(Subscript[O, k])]), (k=1,2,…).

Eigenvalues are real.

(OverHat[O])=(SuperDagger[(OverHat[O])])=>(Subscript[O, k])∈ℝ.

Eigenvectors form a complete set of basis. (Any vector  can be expanded as a sum of these eigenvectors.)

Eigenvectors of different eigenvalues are orthogonal (automatically)

(Subscript[O, k])≠(Subscript[O, l])=>(BraKet[(Subscript[O, k]), (Subscript[O, l])])=0.

Eigenvectors of the same eigenvalue can be made orthogonal (by orthogonalization, e.g. Gram-Schmidt procedure).

Orthogonalize[{{1,2},{3,4}}]

For bounded Hermitian operators (e.g. finite matrices in finite dimensional Hilbert space), eigenvectors can be normalized.

Prove Eq. () and Eq. ().

(OverHat[O])(Ket[(Subscript[O, k])])=(Subscript[O, k])(Ket[(Subscript[O, k])]),

(OverHat[O])(Ket[(Subscript[O, l])])=(Subscript[O, l])(Ket[(Subscript[O, l])]).

(Bra[(Subscript[O, k])])(SuperDagger[(OverHat[O])])=(Bra[(Subscript[O, k])])(OverHat[O])=(Bra[k])(Subscript[O, k]^( * )),

(Bra[(Subscript[O, k])])(OverHat[O])(Ket[(Subscript[O, l])])=(Subscript[O, k]^( * ))(BraKet[(Subscript[O, k]), (Subscript[O, l])]),
(Bra[(Subscript[O, k])])(OverHat[O])(Ket[(Subscript[O, l])])=(Subscript[O, l])(BraKet[(Subscript[O, k]), (Subscript[O, l])]).

If k=l, Eq. () implies

(Bra[(Subscript[O, k])])(OverHat[O])(Ket[(Subscript[O, k])])=(Subscript[O, k]^( * ))(BraKet[(Subscript[O, k]), (Subscript[O, k])])=(Subscript[O, k])(BraKet[(Subscript[O, k]), k]),

i.e. (Subscript[O, k]^( * ))=(Subscript[O, k]), so (Subscript[O, k])∈ℝ. This proves Eq. ().

If k≠l, Eq. () implies

0=((Subscript[O, k])-(Subscript[O, l]))(BraKet[(Subscript[O, k]), (Subscript[O, l])]).

If (Subscript[O, k])!=(Subscript[O, l]), Eq. () implies (BraKet[(Subscript[O, k]), (Subscript[O, l])])=0. This proves Eq. ().

If (Subscript[O, k])=(Subscript[O, l]), then (Ket[(Subscript[O, k])]) and (Ket[(Subscript[O, l])]) are degenerated. Degenerated states can be arbitrarily recombined within the degenerate subspace to form a set of orthogonal basis.

Hermitian operator (OverHat[O]) can always be represented in its own eigenbasis, leading to the spectral decomposition

Note: unlike a generic matrix representation (OverHat[O])=(∑_(i j))(Ket[i])(Subscript[O, i j])(Bra[j]), in the spectral decomposition Eq. (), the summation only run through the eigenbasis once.

In the eigenbasis, the Hermitian operator is represented as a diagonal matrix.

(OverHat[O])≏(({{(Subscript[O, 1]),  ,  }, { , (Subscript[O, 2]),  }, { ,  , ⋱}})).

So the procedure of bring the matrix representation to its diagonal form by transforming to its eigenbasis is called diagonalization. (We will discuss more about it later.)

More generally, when there are degeneracies, Eq. () should be written to

where the eigen projector (Subscript[(OverHat[𝒫]), O=(Subscript[O, k])]) was defined in Eq. ().

The fact that all eigenvectors of (OverHat[O]) form a complete set of orthonormal basis can be rephrases as the following properties of eigen projectors

Orthonormality

Meaning that (Subscript[(OverHat[𝒫]), O=(Subscript[O, k])]^2)=(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])]) and (Subscript[(OverHat[𝒫]), O=(Subscript[O, k])])(Subscript[(OverHat[𝒫]), O=(Subscript[O, l])])=0 if k≠l (i.e. (Subscript[O, k])≠(Subscript[O, l])).

Completeness

Pauli-x

((OverHat[σ])^x)=(Subscript[(OverHat[𝒫]), (σ^x)=+1])-(Subscript[(OverHat[𝒫]), (σ^x)=-1]),

with projection operators

(Subscript[(OverHat[𝒫]), (σ^x)=+1])=(Ket[+])(Bra[+])=((𝟙+((OverHat[σ])^x))/2),
(Subscript[(OverHat[𝒫]), (σ^x)=-1])=(Ket[-])(Bra[-])=((𝟙-((OverHat[σ])^x))/2).

Pauli-y

((OverHat[σ])^y)=(Subscript[(OverHat[𝒫]), (σ^y)=+1])-(Subscript[(OverHat[𝒫]), (σ^y)=-1]),

with projection operators

(Subscript[(OverHat[𝒫]), (σ^y)=+1])=(Ket[i])(Bra[i])=((𝟙+((OverHat[σ])^y))/2),
(Subscript[(OverHat[𝒫]), (σ^y)=-1])=(Ket[(OverBar[i])])(Bra[(OverBar[i])])=((𝟙-((OverHat[σ])^y))/2).

Pauli-z

((OverHat[σ])^z)=(Subscript[(OverHat[𝒫]), (σ^z)=+1])-(Subscript[(OverHat[𝒫]), (σ^z)=-1]),

with projection operators

(Subscript[(OverHat[𝒫]), (σ^z)=+1])=(Ket[0])(Bra[0])=((𝟙+((OverHat[σ])^z))/2),
(Subscript[(OverHat[𝒫]), (σ^z)=-1])=(Ket[1])(Bra[1])=((𝟙-((OverHat[σ])^z))/2).

Prove Eq. () and Eq. ().

n·(OverHat[σ])≏(({{(Subscript[λ, 1]), 0}, {0, (Subscript[λ, 2])}}))

(Subscript[λ, 1])+(Subscript[λ, 2])=Tr (n·(OverHat[σ]))=0,
(({{(Subscript[λ, 1]^2), 0}, {0, (Subscript[λ, 2]^2)}}))≏(((n·(OverHat[σ])))^2)=𝟙,

(Subscript[λ, 1]^2)=(Subscript[λ, 2]^2)=1=>(Subscript[λ, 1]),(Subscript[λ, 2])=±1,
(Subscript[λ, 1])+(Subscript[λ, 2])=0=>(Subscript[λ, 1])=-(Subscript[λ, 2]).

n·(OverHat[σ])=(Subscript[(OverHat[𝒫]), n·σ=+1])-(Subscript[(OverHat[𝒫]), n·σ=-1]).

n·(OverHat[σ])≏(({{1, 0}, {0, -1}})).

(Subscript[(OverHat[𝒫]), n·σ=+1])≏(({{1, 0}, {0, 0}})), (Subscript[(OverHat[𝒫]), n·σ=-1])≏(({{0, 0}, {0, 1}})).

(Subscript[(OverHat[𝒫]), n·σ=±1])=((𝟙±n·(OverHat[σ]))/2).

Observables

Physical Observable

(OverHat[O])=(∑_k)(Subscript[O, k])(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])]),

(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])])=(∑_(m=1)^(Subscript[g, k]))(Ket[(Subscript[O, k]),m])(Bra[(Subscript[O, k]),m]).

All possible measurement outcomes (or observation values) of the observable O are given by (and only by) the eigenvalues (Subscript[O, k]).

The measurement projects (collapses) the quantum state to the eigenspace (Subscript[ℋ, k]) of the corresponding measurement outcome (Subscript[O, k]). The state collapse is implemented by the eigen projector (Subscript[(OverHat[𝒫]), (Subscript[ℋ, k])]).

Measurement Postulate

Non-degenerated case: Suppose (Subscript[O, k]) has no degeneracy,

(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])])=(Ket[(Subscript[O, k])])(Bra[(Subscript[O, k])]).

The probability to observe (Subscript[O, k]) is given by the squared overlap between the prior state (Ket[ψ]) and the eigenstate (Ket[(Subscript[O, k])])

p((Subscript[O, k])|ψ)=((|(BraKet[(Subscript[O, k]), ψ])|)^2).

After the measurement, if (Subscript[O, k]) is observed, the system collapses to the posterior state (Ket[(Subscript[O, k])])

(Ket[ψ])(Overscript[⟶, measure O, observe (Subscript[O, k])])(Ket[(Subscript[O, k])]).

Degenerated case (generic): With degeneracy, the eigen projector is

(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])])=(∑_(m=1)^(Subscript[g, k]))(Ket[(Subscript[O, k]),m])(Bra[(Subscript[O, k]),m]).

The probability to observe (Subscript[O, k]) is given by the squared overlap between the prior state (Ket[ψ]) and all eigenstates (Ket[(Subscript[O, k]),m]) in the degenerate subspace

After the measurement, if (Subscript[O, k]) is observed, the system collapses to the posterior state, given by a linear superposition of (Ket[(Subscript[O, k]),m])

where the coefficients (Subscript[α, m]) are given by

(Subscript[α, m])=((BraKet[(Subscript[O, k]),m, ψ])/(Sqrt[p((Subscript[O, k])|ψ)])).

Derive Eq. () and Eq. () from Eq. ().

p((Subscript[O, k])|ψ)=(Bra[ψ])(Subscript[(OverHat[𝒫]), (Subscript[ℋ, k])])(Ket[ψ])
=(Bra[ψ])((∑_(m=1)^(Subscript[g, k]))(Ket[(Subscript[O, k]),m])(Bra[(Subscript[O, k]),m]))(Ket[ψ])
=(∑_(m=1)^(Subscript[g, k]))(BraKet[ψ, (Subscript[O, k]),m])(BraKet[(Subscript[O, k]),m, ψ])
=(∑_(m=1)^(Subscript[g, k]))((|(BraKet[(Subscript[O, k]),m, ψ])|)^2).

(((Subscript[(OverHat[𝒫]), (Subscript[ℋ, k])])(Ket[ψ]))/(Sqrt[p((Subscript[O, k])|ψ)]))=(1/(Sqrt[p((Subscript[O, k])|ψ)]))((∑_(m=1)^(Subscript[g, k]))(Ket[(Subscript[O, k]),m])(Bra[(Subscript[O, k]),m]))(Ket[ψ])
=(1/(Sqrt[p((Subscript[O, k])|ψ)]))(∑_(m=1)^(Subscript[g, k]))(Ket[(Subscript[O, k]),m])(BraKet[(Subscript[O, k]),m, ψ])
=(∑_(m=1)^(Subscript[g, k]))((BraKet[(Subscript[O, k]),m, ψ])/(Sqrt[p((Subscript[O, k])|ψ)]))(Ket[(Subscript[O, k]),m]).

(Subscript[α, m])=((BraKet[(Subscript[O, k]),m, ψ])/(Sqrt[p((Subscript[O, k])|ψ)])).

Expectation Value

⟨O⟩:=(∑_k)(Subscript[O, k])p((Subscript[O, k])|ψ)
=(∑_k)(Subscript[O, k])(Bra[ψ])(Subscript[(OverHat[𝒫]), O=(Subscript[O, k])])(Ket[ψ]).

The answer is a real scalar (as (OverHat[O]) is Hermitian).

Represented as vectors and matrices,

⟨O⟩=(({{(Subscript[ψ, 1]^( * )), (Subscript[ψ, 2]^( * )), ⋯}}))(({{(Subscript[O, 11]), (Subscript[O, 12]), ⋯}, {(Subscript[O, 21]), (Subscript[O, 22]), ⋯}, {⋮, ⋮, ⋱}}))(({{(Subscript[ψ, 1])}, {(Subscript[ψ, 2])}, {⋮}})).

Tensor network representation

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-3,3}],bra["ψ",{-3,0}],Text["\!\( i\)",{-1.5,0},{0,-0.5}],opH["O"],Text["\!\( j\)",{1.5,0},{0,-0.5}],ket["ψ",{3,0}]},ImageSize->{Automatic,27}]]

⟨O⟩=(∑_(i,j))(Subscript[ψ, i]^( * ))(Subscript[O, i,j])(Subscript[ψ, j]).

The equivalence between Eq. () and Eq. () is self-evident from the tensor network diagrams

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Translate[{Line[{#,0}&/@{-2,2}],bra["ψ",{-2,0}],opH["O"],ket["ψ",{2,0}]},{-4,0}],Text["=",{0,0}],Translate[{Circle[{-3.2,0.8},0.8,{π/2,3π/2}],Line[{#,0}&/@{-3.2,0}],opH["O",{-2,0}],ket["ψ",{0,0}],bra["ψ",{2.5,0}],Line[{#,0}&/@{3,3.2}],Circle[{3.2,0.8},0.8,{-π/2,π/2}],Line[{#,1.6}&/@{3.2,-3.2}]},{5,0}]},ImageSize->{Automatic,38}]]

The advantage of this approach is to circumvent solving for (Ket[ψ]) explicitly (sometimes the state projector is easier to construct than the state vector).

Sequential Measurements

Commuting Operators

Commuting operators: Two operators (OverHat[A]) and (OverHat[B]) commute, if (OverHat[A])(OverHat[B])=(OverHat[B])(OverHat[A]) (operators can commute through each other), i.e.

[(OverHat[A]),(OverHat[B])]=0.

(Here 0 denotes the zero operator, a matrix whose elements are all zeros.)

Common eigenbasis: Two operators (OverHat[A]) and (OverHat[B]) share a common eigenbasis, if there exist a complete set of basis ℬ={(Ket[(Subscript[A, k]),(Subscript[B, k])]):k=1,2,…,dim ℋ} such that both operators are diagonal in the basis ℬ, i.e.

(OverHat[A])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[A, k])(Ket[(Subscript[A, k]),(Subscript[B, k])]),
(OverHat[B])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[B, k])(Ket[(Subscript[A, k]),(Subscript[B, k])]).

(OverHat[A])(OverHat[B])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[A, k])(Subscript[B, k])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[B, k])(Subscript[A, k])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(OverHat[B])(OverHat[A])(Ket[(Subscript[A, k]),(Subscript[B, k])]).

(OverHat[A])≏(({{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}})), (OverHat[B])≏(({{0, -i, 0, 0}, {i, 0, 0, 0}, {0, 0, 0, i}, {0, 0, -i, 0}})).

A={{0,0,0,1},{0,0,1,0},{0,1,0,0},{1,0,0,0}};
B={{0,-I,0,0},{I,0,0,0},{0,0,0,I},{0,0,-I,0}};
A//MatrixForm
B//MatrixForm

A.B-B.A//MatrixForm

Construct a random operator (OverHat[H])=a (OverHat[A])+b (OverHat[B]) by combining (OverHat[A]) and (OverHat[B]) with random numbers a and b.

(H=0.2A+0.5B)//MatrixForm

Find the eigenbasis of (OverHat[H])

{vals,vecs}=Chop@Eigensystem[H]

(Ket[1])=(1/2)(({{-1}, {-i}, {-i}, {-1}})), (Ket[2])=(1/2)(({{-1}, {+i}, {-i}, {+1}})), (Ket[3])=(1/2)(({{-1}, {-i}, {+i}, {+1}})), (Ket[4])=(1/2)(({{-1}, {+i}, {+i}, {-1}}))

With probability 1, the eigenbasis of (OverHat[H]) is the common eigenbasis of (OverHat[A]) and (OverHat[B]), such that both operators are diagonal in this basis.

Chop[vecs.A.ConjugateTranspose[vecs]]//MatrixForm
Chop[vecs.B.ConjugateTranspose[vecs]]//MatrixForm

(Ket[1])→(Ket[+1,-1]), (Ket[2])→(Ket[-1,+1]),(Ket[3])→(Ket[-1,-1]),(Ket[4])→(Ket[+1,+1]);

(OverHat[A])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[A, k])(Ket[(Subscript[A, k]),(Subscript[B, k])]),
(OverHat[B])(Ket[(Subscript[A, k]),(Subscript[B, k])])=(Subscript[B, k])(Ket[(Subscript[A, k]),(Subscript[B, k])]);

((Subscript[A, 1]),(Subscript[A, 2]),(Subscript[A, 3]),(Subscript[A, 4]))=(+1,-1,-1,+1),
((Subscript[B, 1]),(Subscript[B, 2]),(Subscript[B, 3]),(Subscript[B, 4]))=(-1,+1,-1,+1).

Compatible Observables

measure A,

measure B,

measure A.

In step 1, we measure A and suppose that we obtain the outcome (Subscript[A, 1]) (as one eigenvalue of (OverHat[A])), the system has collapse to the state (Ket[(Subscript[A, 1])]).

In step 2, we measure B and suppose that we obtain the outcome (Subscript[B, 1]) (as one eigenvalue of (OverHat[B])), the system will collapse to the state (Ket[(Subscript[B, 1])]). (This will happen with probability ((|(BraKet[(Subscript[B, 1]), (Subscript[A, 1])])|)^2)).

In step 3, we measure A again. There is no guarantee that the previous outcome (Subscript[A, 1]) will appear again. Instead we may obtain a different outcome (Subscript[A, 2]) with probability ((|(BraKet[(Subscript[A, 2]), (Subscript[B, 1])])|)^2) in general.

In order to obtain (Subscript[A, 1]) again with probability 1, we must require

((|(BraKet[(Subscript[A, 1]), (Subscript[B, 1])])|)^2)=1,

i.e. (Ket[(Subscript[A, 1])]) and (Ket[(Subscript[B, 1])]) labels the identical state. Since (Ket[(Subscript[A, 1])]) is an eigenstate of (OverHat[A]) with eigenvalue (Subscript[A, 1]) and (Ket[(Subscript[B, 1])]) is an eigenstate of (OverHat[B]) with eigenvalue (Subscript[B, 1]), the state must be a common eigenstate (Ket[(Subscript[A, 1]),(Subscript[B, 1])]) of both (OverHat[A]) and (OverHat[B]), s.t.

(OverHat[A])(Ket[(Subscript[A, 1]),(Subscript[B, 1])])=(Subscript[A, 1])(Ket[(Subscript[A, 1]),(Subscript[B, 1])]),
(OverHat[B])(Ket[(Subscript[A, 1]),(Subscript[B, 1])])=(Subscript[B, 1])(Ket[(Subscript[A, 1]),(Subscript[B, 1])]).

For this scenario to always happen regardless the choice of ((Subscript[A, 1]),(Subscript[B, 1])), (OverHat[A]) and (OverHat[B]) share a set of common eigenbasis (Ket[(Subscript[A, k]),(Subscript[B, k])]) (k=1,2,…,dim ℋ).

A and B are compatible observables;

(OverHat[A]) and (OverHat[B]) share common eigenbasis;

(OverHat[A]) and (OverHat[B]) are commuting operators: [(OverHat[A]),(OverHat[B])]=0.

Repeated Measurements

Statistics of Measurements

prepare a quantum system in the state (Ket[ψ]) (e.g. by measuring (Subscript[(OverHat[𝒫]), ψ])=(Ket[ψ])(Bra[ψ]) and post-select);

measure O on the state (Ket[ψ]) (the outcome (Subscript[O, k]) will be obtained with probability p((Subscript[O, k])|ψ)=(Bra[ψ])(Subscript[O, k])(Ket[ψ]))

discard the quantum system (or reset the state).

{(Subscript[O, k]):p((Subscript[O, k])|ψ)=((|(BraKet[(Subscript[O, k]), ψ])|)^2)}.

Mean (expectation value):

⟨O⟩=(∑_k)(Subscript[O, k]) p((Subscript[O, k])|ψ)=(Bra[ψ])(OverHat[O])(Ket[ψ]).

Variance (2nd moment):

var O=(∑_k)((((Subscript[O, k])-⟨O⟩))^2) p((Subscript[O, k])|ψ)=(((Bra[ψ])((OverHat[O])-⟨O⟩𝟙))^2)(Ket[ψ]).

Introduce the observable Δ O (the deviation of O from its expectation value)

Δ O=O-⟨O⟩𝟙,

The variance can be written as var O=⟨(((Δ O))^2)⟩.

Standard deviation: characterizes the uncertainty of the measurement of O

std O=(((var O))^(1/2))=((⟨(((Δ O))^2)⟩)^(1/2)).

Uncertainty Relation

In words, the product of the uncertainties cannot be smaller than half of the magnitude of the expectation value of the commutator.

For commuting observables ([A,B]=0), (std A)(std B)≥0, it is possible to have std A=std B=0 simultaneously, i.e. A and B can be jointly measured with perfect certainty.

For non-commuting observables, there exists a state on which |⟨[A,B]⟩|≠0. Then on such state, it is impossible to have std A=std B=0 simultaneously, i.e. A and B can not be jointly measured with certainty.

(Bra[ψ])((OverHat[A])-i x (OverHat[B]))((OverHat[A])+i x (OverHat[B]))(Ket[ψ])=(BraKet[ϕ, ϕ])≥0.

(Bra[ψ])((OverHat[A])-i x (OverHat[B]))((OverHat[A])+i x (OverHat[B]))(Ket[ψ])
=(Bra[ψ])((OverHat[A])^2)+i x [(OverHat[A]),(OverHat[B])]+(x^2)((OverHat[B])^2)(Ket[ψ])
=⟨(B^2)⟩(x^2)+i⟨[A,B]⟩x+⟨(A^2)⟩≥0.

Δ=(((i⟨[A,B]⟩))^2)-4⟨(B^2)⟩⟨(A^2)⟩≤0.

((⟨(A^2)⟩)^(1/2))((⟨(B^2)⟩)^(1/2))≥(1/2)|⟨[A,B]⟩|.

Dynamics

Unitary Operators

Basis Transformation

ℬ={(Ket[i]):i=1,2,…,dim ℋ},
ℬ'={(Ket[i])':i=1,2,…,dim ℋ}.

The same state (Ket[v]) can have different vector representations in different bases

(Subscript[v, i])=(Bra[i])(Ket[v]), (Subscript[v, i]^′)=(Bra[i])'(Ket[v]).

The same operator (OverHat[O]) can have different matrix representations in different bases

(Subscript[O, i j])=(Bra[i])(OverHat[O])(Ket[j]), (Subscript[O, i j]^′)=(Bra[i])'(OverHat[O])(Ket[j])'.

(Subscript[v, i]^′)=(∑_j)(Subscript[U, i j])(Subscript[v, j]),
(Subscript[O, i j]^′)=(∑_(k,l))(Subscript[U, i k])(Subscript[O, k l])(Subscript[U, j l]^( * )).

Using Eq. () to prove that Eq. () is compatible with Eq. () and Eq. ().

(Subscript[v, i]^′)=(∑_j)(Subscript[U, i j])(Subscript[v, j])
=(∑_j)(Bra[i])'(Ket[j])(Bra[j])(Ket[v])
=(Bra[i])'𝟙(Ket[v])
=(Bra[i])'(Ket[v]).

(Subscript[O, i j]^′)=(∑_(k,l))(Subscript[U, i k])(Subscript[O, k l])(Subscript[U, j l]^( * ))
=(∑_(k,l))(Bra[i])'(Ket[k])(Bra[k])(OverHat[O])(Ket[l])(Bra[l])(Ket[j])'
=(Bra[i])'𝟙(OverHat[O])𝟙(Ket[j])'
=(Bra[i])'(OverHat[O])(Ket[j])'.

Check that the matrix element of (OverHat[U]) in Eq. () is indeed given by Eq. (), when represented in either the basis ℬ or ℬ'.

(Subscript[U, i j])=(Bra[i])(OverHat[U])(Ket[j])
=(∑_k)(BraKet[i, k])(Bra[k])'(Ket[j])
=(∑_k)(Subscript[δ, i k])(Bra[k])'(Ket[j])
=(Bra[i])'(Ket[j]).

(Subscript[U, i j]^′)=(Bra[i])'(OverHat[U])(Ket[j])'
=(∑_k)(Bra[i])'(Ket[k])(Bra[k])'(Ket[j])'
=(∑_k)(Bra[i])'(Ket[k])(Subscript[δ, k j])
=(Bra[i])'(Ket[j]).

Check that Eq. () satisfies the defining property Eq. () for unitary operator.

(SuperDagger[(OverHat[U])])(OverHat[U])=(∑_i)(Ket[i])'(Bra[i])(∑_j)(Ket[j])(Bra[j])'
=(∑_(i,j))(Ket[i])'(BraKet[i, j])(Bra[j])'
=(∑_(i,j))(Ket[i])'(Subscript[δ, i j])(Bra[j])'
=(∑_(i,j))(Ket[i])'(Bra[i])'=𝟙.

The inverse of a unitary operator is its Hermitian conjugate

((OverHat[U])^(-1))=(SuperDagger[(OverHat[U])]).

The operator (basis transformation) implemented by (OverHat[U]) is reversed by that of (SuperDagger[(OverHat[U])]), and vice versa.

When the two sets of basis (Ket[i]) and (Ket[i])' are identical, U=𝟙 becomes the identity operator (which is also unitary).

({{for ket state:, (Ket[v])->(OverHat[U])(Ket[v]),}, {for bra state:, (Bra[v])→(Bra[v])(SuperDagger[(OverHat[U])]),}, {for operator:, (OverHat[O])→(OverHat[U]) (OverHat[O]) (SuperDagger[(OverHat[U])]).}})

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-4,0}],ket["v"],opU["U",{-2,0}],Line[{#,0}&/@{3,7}],bra["v",{3,0}],opU["U\^†",{5,0}]},ImageSize->{Automatic,30}]]

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-4,4}],opH["O"],opU["U",{-2,0}],opU["U\^†",{2,0}]},ImageSize->{Automatic,30}]]

The operator (OverHat[O]) is also made of ket and bra states, so the unitary operator must be applied from both sides, when transforming an operator.

The expectation value of an observable is invariant under basis transformation. (Physical reality should be basis-independent.)

⟨O⟩=(Bra[ψ])(OverHat[O])(Ket[ψ])→(Bra[ψ])(SuperDagger[(OverHat[U])]) (OverHat[U]) (OverHat[O]) (SuperDagger[(OverHat[U])])(OverHat[U])(Ket[ψ])=(Bra[ψ])𝟙 (OverHat[O]) 𝟙(Ket[ψ])=⟨O⟩.

Matrix Diagonalization

(OverHat[O])=(∑_k)(Ket[(Subscript[O, k])])(Subscript[O, k])(Bra[(Subscript[O, k])]),
(OverHat[U])=(∑_k)(Ket[k])(Bra[(Subscript[O, k])]),

(OverHat[Λ])=(OverHat[U]) (OverHat[O]) (SuperDagger[(OverHat[U])])=(∑_k)(Ket[k])(Subscript[O, k])(Bra[k])≏(({{(Subscript[O, 1]),  ,  }, { , (Subscript[O, 2]),  }, { ,  , ⋱}}))

Every Hermitian matrix can be written as

(OverHat[O])=(SuperDagger[(OverHat[U])])(OverHat[Λ]) (OverHat[U]),

with (OverHat[Λ]) being diagonal and (OverHat[U]) being unitary.

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-2,2}],opH["O"],Line[{#,0}&/@{4,12}],opU["U\^†",{6,0}],opD["Λ",{8,0}],opU["U",{10,0}],Text["=",{3,0}]},ImageSize->{Automatic,30}]]

Or equivalently, the unitary transformation (OverHat[U]) brings the Hermitian matrix to its diagonal form,

(OverHat[U]) (OverHat[O]) (SuperDagger[(OverHat[U])])=(OverHat[Λ]).

Block[{ket,bra,opH,opU,opD},
ket[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,0},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
bra[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Blue,0.8],Polygon[CirclePoints[{1,π},3]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opH[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Red,0.8],Disk[{0,0},0.8]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opU[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Green,0.8],Polygon[CirclePoints[4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
opD[txt_,pos_:{0,0}]:=Translate[{{EdgeForm[Black],Lighter[Yellow,0.8],Polygon[CirclePoints[{0.8,0},4]]},Text["\!\( "<>txt<>"\)",{0,0},{0,0.2}]},pos];
Graphics[{Line[{#,0}&/@{-4,4}],opH["O"],opU["U",{-2,0}],opU["U\^†",{2,0}],Line[{#,0}&/@{6,10}],opD["Λ",{8,0}],Text["=",{5,0}]},ImageSize->{Automatic,30}]]

(Subscript[(OverHat[U]), H])=(({{(Bra[+])}, {(Bra[-])}}))≏(1/(Sqrt[2]))(({{1, 1}, {1, -1}})).

This unitary operation (Subscript[(OverHat[U]), H]) is also known as the Hadamard gate in quantum information, an example of single-qubit gate.

Under the unitary transformation, ((OverHat[σ])^x) is brought to its diagonal form, which is ((OverHat[σ])^z)

(Subscript[(OverHat[U]), H]) ((OverHat[σ])^x) (Subscript[(OverHat[U]), H]^†)≏(1/(Sqrt[2]))(({{1, 1}, {1, -1}}))(({{0, 1}, {1, 0}}))(1/(Sqrt[2]))(({{1, 1}, {1, -1}}))
=(({{1, 0}, {0, -1}}))≏((OverHat[σ])^z).

Hermitian Generators

(SuperStar[z])z=z (SuperStar[z])=((|z|)^2)=1.

For complex numbers, a phase factor can be written as z=(e^(i θ)), where θ∈ℝ is a real phase angle.

Similar ideas apply to unitary operators: every unitary operator can be generated by a Hermitian operator (OverHat[Θ]) in the form of

(OverHat[Θ])=(∑_k)(Subscript[Θ, k])(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])

either by operator Taylor expansion Eq. ()

(e^( i (OverHat[Θ])))=𝟙+i (OverHat[Θ])+((((i (OverHat[Θ])))^2)/(2!))+((((i (OverHat[Θ])))^3)/(3!))+….

or by spectral decomposition (HW )

(e^( i (OverHat[Θ])))=(∑_k)(e^(i (Subscript[Θ, k])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])

Use Eq. () to show that (OverHat[U])=(e^( i (OverHat[Θ]))) is unitary as long as (OverHat[Θ]) is Hermitian.

(SuperDagger[(OverHat[U])])(OverHat[U])=(SuperDagger[((e^( i (OverHat[Θ]))))])(e^( i (OverHat[Θ])))
=(SuperDagger[((∑_k)(e^(i (Subscript[Θ, k])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])]))])((∑_l)(e^(i (Subscript[Θ, l])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, l])]))
=(∑_(k,l))(e^(-i (Subscript[Θ, k])))(e^(i (Subscript[Θ, l])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, l])])
=(∑_(k,l))(e^(-i (Subscript[Θ, k])))(e^(i (Subscript[Θ, l])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])(Subscript[δ, k l])
=(∑_k)(e^(-i (Subscript[Θ, k])))(e^(i (Subscript[Θ, k])))(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])
=(∑_k)(Subscript[(OverHat[𝒫]), Θ=(Subscript[Θ, k])])
=𝟙.

(OverHat[U])(θ)=(e^(i θ ((OverHat[σ])^y)))≏(({{cos θ, sin θ}, {-sin θ, cos θ}})).

(OverHat[U])(θ)(Ket[0])≏(({{cos θ, sin θ}, {-sin θ, cos θ}}))(({{1}, {0}}))=(({{cos θ}, {-sin θ}})).

(OverHat[U])(Δ θ)=(OverHat[U])(0)+(OverHat[U])'(0)Δ θ+…=𝟙+(OverHat[U])'(0)Δ θ+…,

(OverHat[U])'(0) is also an operator (matrix), usually denoted as (OverHat[U])'(0)=i (OverHat[G]). We call (OverHat[G]) the generator of the rotation / unitary operator, because it generates an infinitesimal rotation

(OverHat[U])(Δ θ)=𝟙+i Δ θ (OverHat[G])+....

(OverHat[U])(Δ θ) is unitary => (OverHat[G]) is Hermitian.

(SuperDagger[U(Δ θ)])U(Δ θ)
=(𝟙-i Δ θ (SuperDagger[(OverHat[G])])+...)(𝟙+i Δ θ (OverHat[G])+...)
=𝟙+i Δ θ((OverHat[G])-(SuperDagger[(OverHat[G])]))+…=𝟙.

Large rotations can be accumulated from small rotations.

(OverHat[U])(N Δ θ)=(((OverHat[U])(Δ θ))^N)=(((𝟙+i Δ θ (OverHat[G])))^N).

As Δ θ is small (but N can be large, s.t. θ=N Δ θ is finite),

ln (OverHat[U])(N Δ θ)=N ln(𝟙+i Δ θ (OverHat[G]))=i N Δ θ (OverHat[G]),

So (OverHat[U])(N Δ θ)=(e^(i N Δ θ (OverHat[G]))), we obtain the exponential form

Time Evolution

Time-Evolution is Unitary

start out in different states => remains in different states (towards both future and past).

start out in the same state => follow identical evolution (towards both future and past).

Distinct states remain distinct:

(BraKet[ϕ(0), ψ(0)])=0=>(BraKet[ϕ(t), ψ(t)])=(Bra[ϕ(0)])(SuperDagger[(OverHat[U])(t)])(OverHat[U])(t)(Ket[ψ(0)])=0.

Identical states remain the identical:

(BraKet[ψ(0), ψ(0)])=1=>(BraKet[ψ(t), ψ(t)])=(Bra[ψ(0)])(SuperDagger[(OverHat[U])(t)])(OverHat[U])(t)(Ket[ψ(0)])=1.

Or, the fact that the probability adds up to 1 must be preserved.

(Bra[i])(SuperDagger[(OverHat[U])(t)])(OverHat[U])(t)(Ket[j])=(Subscript[δ, i j])=>(SuperDagger[(OverHat[U])(t)])(OverHat[U])(t)=𝟙.

Hamiltonian

(OverHat[H])=i (OverHat[U])'(0)=i (∂_t)(OverHat[U])(t)(Subscript[|, t=0]).

(OverHat[U])(Δ t)=𝟙-i (OverHat[H]) Δ t+...,

(Ket[ψ(Δ t)])=(OverHat[U])(Δ t)(Ket[ψ(0)])=(Ket[ψ(0)])-i Δ t (OverHat[H])(Ket[ψ(0)]),

i (∂_t)(Ket[ψ(0)])=i (((Ket[ψ(Δ t)])-(Ket[ψ(0)]))/(Δ t))=(OverHat[H])(Ket[ψ(0)]).

i(∂_t)(Ket[ψ(t)])=(OverHat[H])(Ket[ψ(t)]).

The Hamiltonian (OverHat[H])(t)=i (OverHat[U])'(t) can be time-dependent in general.

But in many cases, we consider (OverHat[H]) to be time-independent, by assuming the time-translation symmetry.

ℏ=(h/(2π))=1.0545718(13)×(10^(-34))J s.

i ℏ(∂_t)(Ket[ψ(t)])=(OverHat[H])(Ket[ψ(t)]).

Schrödinger Equation: State Dynamics

(Ket[(Subscript[E, k])(t)])=(e^(-(i/ℏ)(Subscript[E, k]) t))(Ket[(Subscript[E, k])]).

(Ket[(Subscript[E, k])]) form a complete set of orthonormal basis, called energy eigenbasis.

Verify that Eq. () is a solution of Eq. ():

i ℏ(∂_t)(Ket[(Subscript[E, k])(t)])=i ℏ(∂_t)((e^(-(i/ℏ)(Subscript[E, k]) t))(Ket[(Subscript[E, k])]))
=i ℏ(∂_t)((e^(-(i/ℏ)(Subscript[E, k]) t)))(Ket[(Subscript[E, k])])
=(Subscript[E, k])(e^(-(i/ℏ)(Subscript[E, k]) t))(Ket[(Subscript[E, k])])
=(Subscript[E, k])(Ket[(Subscript[E, k])(t)]),

(OverHat[H])(Ket[(Subscript[E, k])(t)])=(e^(-(i/ℏ)(Subscript[E, k]) t))(OverHat[H])(Ket[(Subscript[E, k])])
=(e^(-(i/ℏ)(Subscript[E, k]) t))(Subscript[E, k])(Ket[(Subscript[E, k])])
=(Subscript[E, k])(Ket[(Subscript[E, k])(t)]).

(Ket[ψ(t)])=(∑_i)(e^(-(i/ℏ)(Subscript[E, i]) t))(Ket[(Subscript[E, i])])(BraKet[(Subscript[E, i]), ψ(0)])
=(e^(-(i/ℏ) (OverHat[H]) t))(Ket[ψ(0)]).

(OverHat[U])(t)=𝒯 exp(-(i/ℏ)(∫_0^t)(OverHat[H])(t')dt').

Larmor Precession and Rabi Oscillation

derive it from experiment,

borrow it from some theory we like,

pick one and see what happens.☜

(OverHat[H])=(Subscript[h, 0])𝟙+(Subscript[h, x])((OverHat[σ])^x)+(Subscript[h, y])((OverHat[σ])^y)+(Subscript[h, z])((OverHat[σ])^z)
=(Subscript[h, 0])𝟙+h·(OverHat[σ]),

The time-evolution operator (set ℏ=1 in the following)

(OverHat[U])(t)=(e^(-i (OverHat[H]) t))
=(e^(-i (Subscript[h, 0])t))(cos(|h|t)𝟙-i sin(|h|t)(OverTilde[h])·(OverHat[σ])),

where |h|=(Sqrt[h·h]) and (OverTilde[h])=h/|h|.

Derive Eq. () from Eq. ().

(OverHat[U])(t)=(e^(-i (OverHat[H]) t))
=(e^(-i ((Subscript[h, 0])𝟙+h·(OverHat[σ])) t))
=(e^(-i (Subscript[h, 0])t))exp(-i |h|(OverTilde[h])·(OverHat[σ]) t)
=(e^(-i (Subscript[h, 0])t))(cos(|h|t)𝟙-i sin(|h|t)(OverTilde[h])·(OverHat[σ])).

A state (Ket[ψ(0)]) will evolve with time following

(Ket[ψ(t)])=(OverHat[U])(t)(Ket[ψ(0)])
=(e^(-i (Subscript[h, 0])t))(cos(|h|t)𝟙-i sin(|h|t)(OverTilde[h])·(OverHat[σ]))(Ket[ψ(0)]).

If we measure σ on the state (Ket[ψ(t)]), the expectation value will be given by

(Subscript[⟨σ⟩, t])=(Bra[ψ(t)])(OverHat[σ])(Ket[ψ(t)])
=cos(2|h|t)(Subscript[⟨σ⟩, 0])+sin(2|h|t)(OverTilde[h])×(Subscript[⟨σ⟩, 0])+(1-cos(2|h|t))(OverTilde[h])((OverTilde[h])·(Subscript[⟨σ⟩, 0])).

which also evolves with time.

Derive Eq. () from Eq. ().
Hint: Eq. () can make life much more easier.

m·(Subscript[⟨σ⟩, t])=(Bra[ψ(t)])m·(OverHat[σ])(Ket[ψ(t)])=(Bra[ψ(0)])(cos(|h|t)𝟙+i sin(|h|t)(OverTilde[h])·(OverHat[σ]))m·σ(cos(|h|t)𝟙-i sin(|h|t)(OverTilde[h])·(OverHat[σ]))(Ket[ψ(0)]).

(cos(|h|t)𝟙+i sin(|h|t)(OverTilde[h])·(OverHat[σ]))m·(OverHat[σ])(cos(|h|t)𝟙-i sin(|h|t)(OverTilde[h])·(OverHat[σ]))
=(cos^2)(|h|t)m·(OverHat[σ])+i sin(|h|t)cos(|h|t)((OverTilde[h])·(OverHat[σ]) m·(OverHat[σ])-m·(OverHat[σ])(OverTilde[h])·(OverHat[σ]))+(sin^2)(|h|t)(OverTilde[h])·(OverHat[σ]) m·(OverHat[σ])(OverTilde[h])·(OverHat[σ]).

(OverTilde[h])·(OverHat[σ]) m·(OverHat[σ])=(OverTilde[h])·m 𝟙+i ((OverTilde[h])×m)·(OverHat[σ]),
m·(OverHat[σ])(OverTilde[h])·(OverHat[σ])=(OverHat[h])·m 𝟙-i ((OverTilde[h])×m)·(OverHat[σ]),
(OverTilde[h])·(OverHat[σ]) m·(OverHat[σ])(OverTilde[h])·(OverHat[σ])=(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])+i ((OverTilde[h])×m)·(OverHat[σ])(OverTilde[h])·(OverHat[σ])
=(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])+i(((OverTilde[h])×m)·(OverTilde[h])𝟙+i(((OverTilde[h])×m)×(OverTilde[h]))·(OverHat[σ]))
=(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])-(((OverTilde[h])×m)×(OverTilde[h]))·(OverHat[σ])
=(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])-((OverTilde[h])·(OverTilde[h])m·(OverHat[σ])-(OverTilde[h])·m(OverTilde[h])·(OverHat[σ]))
=2(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])-m·(OverHat[σ]).

(cos^2)(|h|t)m·(OverHat[σ])-2 sin(|h|t)cos(|h|t) ((OverTilde[h])×m)·(OverHat[σ])+(sin^2)(|h|t)(2(OverTilde[h])·m (OverTilde[h])·(OverHat[σ])-m·(OverHat[σ]))
=cos(2|h|t)m·(OverHat[σ])+sin(2|h|t)m· ((OverTilde[h])×(OverHat[σ]))+(1-cos(2|h|t))(OverTilde[h])·m (OverTilde[h])·(OverHat[σ]).

m·(Subscript[⟨σ⟩, t])=cos(2|h|t)m·(Subscript[⟨σ⟩, 0])+sin(2|h|t)m· ((OverTilde[h])×(Subscript[⟨σ⟩, 0]))+(1-cos(2|h|t))(OverTilde[h])·m (OverTilde[h])·(Subscript[⟨σ⟩, 0]).

(Subscript[⟨σ⟩, t])=(sin (Subscript[θ, 0])cos ((Subscript[φ, 0])+2(Subscript[h, z])t), sin (Subscript[θ, 0]) sin((Subscript[φ, 0])+2(Subscript[h, z])t),cos (Subscript[θ, 0])),

The solution describes the spin ⟨σ⟩ precessing around the axis of the Zeeman field h.

Block[{θ=π/4},Graphics3D[{EdgeForm[None],{Red,Sphere[{0,0,0},0.4],Tube[#{-Sin[θ],0,Cos[θ]}&/@{-1,0.6},0.1],Cone[#{-Sin[θ],0,Cos[θ]}&/@{0.6,1.5},0.2],Text["\!\(⟨σ⟩\)",1.5{-Sin[θ],0,Cos[θ]},{1,-0.5}]},
{Blue,Tube[{0,0,#}&/@{-1.5,1.5},0.05],Cone[{0,0,#}&/@{1.5,2},0.1],Text[Style["\!\( h\)",Bold],{0,0,2},{0,-0.5}]},
{Arrowheads[{0.08,#}&/@{0.25,0.5,0.75}],Dashed,Arrow[Table[1.5{-Sin[θ]Cos[ϕ],-Sin[θ]Sin[ϕ],Cos[θ]},{ϕ,0,2π,π/32}]]}},Lighting->"Neutral",Boxed->False,ImageSize->80]]

The precession frequency ω=2|h| is called the Larmor frequency. It can be used to probe the local Zeeman field strength, which has applications in nuclear magnetic resonance (NMR) and nitrogen-vacancy (NV) center.

Energy of a spin in the Zeeman field is ⟨H⟩=-h·⟨σ⟩ (up to some constant energy shift (Subscript[h, 0])).

(OverHat[H])=Ω ((OverHat[σ])^x)+Δ ((OverHat[σ])^z)≏(({{Δ, Ω}, {Ω, -Δ}})),

(Subscript[p(1|0), t])=(Subscript[⟨(Subscript[𝒫, 1])⟩, t])=((1-(Subscript[⟨(σ^z)⟩, t]))/2)=(((sin^2)(ω t/2))/(1+(((Δ/Ω))^2))),

Row@{Plot[Evaluate@Table[Sin[Sqrt[1+Δ^2] Ωt/2]^2/(1+Δ^2),{Δ,{0,1,2}}],{Ωt,0,4π},FrameLabel->{"\!\(Ω t\)","\!\(p(1|0)\_t\)"},FrameTicks->{{Automatic,Automatic},{π Range[0,4],None}},GridLines->{π Range[0,4],{0,1}},BaselinePosition->Scaled[0.64],ImageSize->180],
Grid[{Graphics[{#1,Line[{{0,0},{1,0}}]},AspectRatio->0.1,ImageSize->20],Style[StringTemplate["\!\(Δ/Ω=`1`\)"][#2],FontFamily->"CMU"]}&@@@{{Blue,0},{Red,1},{Green,2}},Spacings->{Automatic, 0.1}]}

Rabi π-Pulse: flipping (Ket[0]) to (Ket[1]) (and vice versa) by a π-pulse (turn on the driving field Ω for time t=π/Ω and turn off) at resonance Δ=0. This implements a NOT gate (or X gate) on a single qubit.

Heisenberg Equation: Operator Dynamics

Schrödinger picture: state evolves in time, operator remains fixed,

⟨O(t)⟩=(Bra[ψ(t)])(OverHat[O])(Ket[ψ(t)]).

Heisenberg picture: operator evolves in time, state remains fixed,

⟨O(t)⟩=(Bra[ψ])(OverHat[O])(t)(Ket[ψ]).

(Ket[ψ(t)])=(OverHat[U])(t)(Ket[ψ])=>(OverHat[O])(t)=(SuperDagger[(OverHat[U])(t)])(OverHat[O]) (OverHat[U])(t),

⟨O(t)⟩=(Bra[ψ])(SuperDagger[(OverHat[U])(t)])(OverHat[O]) (OverHat[U])(t)(Ket[ψ]).

Derive Eq. () from Eq. ().

(OverHat[O])(Δ t)=(SuperDagger[(OverHat[U])(Δ t)])(OverHat[O]) (OverHat[U])(Δ t)
=(e^(i (OverHat[H]) Δ t))(OverHat[O]) (e^(-i (OverHat[H]) Δ t))
=(𝟙+i (OverHat[H]) Δ t+…)(OverHat[O]) (𝟙-i (OverHat[H]) Δ t+…)
=(OverHat[O])+i ((OverHat[H]) (OverHat[O])-(OverHat[O]) (OverHat[H])) Δ t+…
=(OverHat[O])-i [(OverHat[O]),(OverHat[H])] Δ t+…

i (∂_t)(OverHat[O])=i(((OverHat[O])(Δ t)-(OverHat[O]))/(Δ t))=[(OverHat[O]),(OverHat[H])].

i ℏ(∂_t)⟨O(t)⟩=⟨[(OverHat[O])(t),(OverHat[H])]⟩.