Quantum Mechanics

Part II. Quantum Entanglement

Information Theory

Classical Information

Probability Theory

The expectation value (mean, average) of a random variable X is the average of all possible values weighted by their probabilities

The variance of a random variable X is the expectation value of the squared deviation of X from its mean 𝔼[X].

p(x)=Pr(X=x)=({{{, ({{1, x=(SubStar[x])}, {0, x≠(SubStar[x])}})}}).

Observation Theory

Probability assignment is a means to describe our state of knowledge about the random variable X.

The probability assignment can be updated, if our state of knowledge is changed by observations (given new evidence from observations).

Before observing X=(SubStar[x]), the probability distribution p(x) is called the prior probability.

After observing X=(SubStar[x]), the probability distribution will collapse to the posterior probability p(x|X=(SubStar[x])) (reads: the probability of x given the observation X=(SubStar[x]))

If we continue to observe X again, we would obtain X=(SubStar[x]) for sure, based on the updated probability p(x|X=(SubStar[x])).

Prior probability:

({{x, 1, 2, 3, 4, 5, 6}, {p(x), (1/6), (1/6), (1/6), (1/6), (1/6), (1/6)}})

Posterior probability:

({{x, 1, 2, 3, 4, 5, 6}, {p(x|X=2), 0, 1, 0, 0, 0, 0}})

Information Theory

If p((SubStar[x]))=1, we already know for certain that X=(SubStar[x]), further observing X=(SubStar[x]) will not bring us new knowledge, thus we are not gaining more information from the observation, and I(X=(SubStar[x]))=0 in this case.

If p((SubStar[x]))→0, observing X=(SubStar[x]) is a rare event. But if the rare event happens, it would bring us a big surprise (a lot of knowledge), because base on the observation, a large amount of other possibilities (of X≠(SubStar[x])) can be ruled out, which is very informative. So I(X=(SubStar[x]))→+∞ in this case.

Tossing a coin (𝒳={head,tail}) and observe X=head.

({{x, head, tail}, {p(x), (1/2), (1/2)}})

Information gained from observation

I(X=head)=-log(1/2)=log 2=1 bit.

log 2 information is also called one bit of information (1 bit=log 2 is widely used as the information unit in information science), it is the amount of information obtained from knowing the answer of a yes-or-no question.

Observing an unknown bit string of length n=3 and obtain (say) X=010

({{x, 000, 001, 010, 011, 100, 101, 110, 111}, {p(x), (1/8), (1/8), (1/8), (1/8), (1/8), (1/8), (1/8), (1/8)}})

Information gained from observation

I(X=010)=-log(1/8)=3 log 2=3 bit.

The observation specifies one outcome from eight equally likely outcomes. To determine the bit string 010, equivalently, we need to ask three yes-or-no questions (is the 1st/2nd/3rd bit 0?). Answering each of the question will give us one bit of information about the unknown string. So the observation gives us totally three bits of information all at once.

In general, observing an unknown bit string of length n will provide n bits of information about the string, as it determines one outcome from (2^n) equally likely outcomes. The probability associated with the particular observation is

p=((((1/2)))^n)=>-log p=n log 2=n bit=I.

So the information I(X=(SubStar[x])) that we gain from observing X=(SubStar[x]) should be defined to be the negative logarithm of the probability p((SubStar[x])) for such observation event to happen.

I(X=(SubStar[x]))=-log p((SubStar[x]))

Why logarithm? Probability is multiplicative, while information is additive. The logarithm is only correct function to convert multiplication to addition.

Observing a random variable X (𝒳={a,b,c,d}) associated with the following (prior) probability

({{x, a, b, c, d}, {p(x), (1/2), (1/4), (1/8), (1/8)}}).

Information can be different for different observation outcomes

I(X=a)=-log(1/2)=log 2=1 bit,
I(X=b)=-log(1/4)=2log 2=2 bit,
I(X=c)=-log(1/8)=3 log 2=3 bit,
I(X=d)=-log(1/8)=3 log 2=3 bit.

However, different outcome happens with different probability. What is amount of information that we can obtained from observing X on average?

I(X)=I(X=a)p(a)+I(X=b)p(b)+I(X=c)p(c)+I(X=d)p(d)
=(1×(1/2)+2×(1/4)+3×(1/8)+3×(1/8))bit
=1.75 bit.

Entropy

Entropy is always non-negative (follows from 0<=p(x)<=1)

S(X)≥0.

S(X)=0 means the value of X is known for certain (no randomness).

Large S(X) indicates large uncertainty in X.

Entropy can change under observation, as observation can remove (or reduce) the uncertainty of a random variable.

A binary random variable X (with 𝒳={false,true})

p(false)=1-p, 
p(true)=p,

where 0<=p<=1. Entropy of X

S(X)=-p log p-(1-p)log (1-p).

Plot[-(p Log[p]+(1-p)Log[1-p])/Log[2],{p,0,1},PlotStyle->Black,FrameLabel->{"\!\( p\)","\!\(S(X)\) [bit]"},GridLines->{{0.5},{0,1}},ImageSize->180]

S(X)=0 when p=0 (X=false for sure) or p=1 (X=true for sure).

S(X) is maximized at p=1/2, where X is most uncertain. The maximum entropy of a binary random variable is 1 bit.

Rolling a dice (𝒳={1,2,3,4,5,6}) and observe X=2.

({{x, 1, 2, 3, 4, 5, 6,  }, {p(x), (1/6), (1/6), (1/6), (1/6), (1/6), (1/6), (prior)}, {p(x|X=2), 0, 1, 0, 0, 0, 0, (posterior)}})

Entropy of X before the observation (the prior entropy)

S((Subscript[X, prior]))=-((1/6)log((1/6)))×6=log 6.

Entropy of X after the observation (the posterior entropy)

S((Subscript[X, post]))=- ((0log 0)×5+1 log 1)=0.

Note: 0 log 0 =0 (as it should be understood as the limit (lim_(x->0))x log x=0), and 1log 1=0 (because log 1=0).

Expected information gain from the observation

I(X)=((1/6)log((1/6)))×6=log 6.

Observation changes the probability distribution (i.e. the state of knowledge)

({{ , (Subscript[X, prior]), (Overscript[→, observe]), (Subscript[X, post])}, {probability, p(x),  , p(x|X)}, {entropy, S((Subscript[X, prior])),  , S((Subscript[X, post]))}, { , more,  , less}})

The entropy change is the negative of the information gain

Δ S=S((Subscript[X, post]))-S((Subscript[X, prior]))=-I(X).

Quantum Information

Pure and Mixed States

a single state vector (ket state) (Ket[ψ]),

or the state projector (pure state density matrix)

The mixed state is modeled by a density matrix (density operator)  ρ, defined as the ensemble average of the state projectors

A density matrix should satisfy the following properties

Hermitian:  (SuperDagger[(OverHat[ρ])])=(OverHat[ρ]).

Normalization (trace one): Tr (OverHat[ρ])=1.

Positive (semi)definite: ∀(Ket[ψ]):(Bra[ψ])(OverHat[ρ])(Ket[ψ])≥0.

A pure state (Ket[ψ]) can be viewed a special limit of a mixed state  (OverHat[ρ]), when the density matrix factorizes as

(OverHat[ρ])= (Ket[ψ])(Bra[ψ]).

Meaning that the probability p((Ket[ϕ])) concentrated at on fixed state (Ket[ϕ])=(Ket[ψ]). Not every density matrix can be expressed in the form of (Ket[ψ])(Bra[ψ]) => A density matrix is more general than a state vector.

Definition of mixed state: a mixed state is a quantum state described by a density matrix  (OverHat[ρ]) that can not be written as the pure state form (Ket[ψ])(Bra[ψ]).

Examples of pure states:

(σ^z) basis states: horizontal and vertical polarizations

(Ket)≏(({{1}, {0}})), (Ket)≏(({{0}, {1}})).

(σ^x) basis states: 45° polarizations

(Ket)≏(1/(Sqrt[2]))(({{1}, {1}})), (Ket)≏(1/(Sqrt[2]))(({{1}, {-1}})).

(σ^y) basis states: circular polarizations

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

Linear polarization along θ angle (with respect to x-axis)

(Ket[θ])≏(({{cos θ}, {sin θ}}))=>(Subscript[(OverHat[ρ]), θ])=(Ket[θ])(Bra[θ])≏(({{(cos^2)θ, cos θ sin θ}, {cos θ sin θ, (sin^2)θ}})).

Natural light: an ensemble of all possible polarizations with equal probability => maximally mixed state

(Subscript[(OverHat[ρ]), 𝒩])=∫((d θ)/(2π))(Subscript[(OverHat[ρ]), θ])≏(1/2)(({{1, 0}, {0, 1}})).

This is an example of a mixed state.

Show that it is impossible to express  (Subscript[(OverHat[ρ]), 𝒩]) as (Ket[ψ])(Bra[ψ]), hence  (Subscript[(OverHat[ρ]), 𝒩]) is indeed a mixed state.

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

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

(Subscript[(OverHat[ρ]), 𝒩])≏(1/2)(({{1, 0}, {0, 1}})).

((|(Subscript[ψ, 0])|)^2)=((|(Subscript[ψ, 1])|)^2)=(1/2),
(Subscript[ψ, 0])(Subscript[ψ, 1]^( * ))=(Subscript[ψ, 1])(Subscript[ψ, 0]^( * ))=0.

One one hand

((|(Subscript[ψ, 0])|)^2)((|(Subscript[ψ, 1])|)^2)=((|(Subscript[ψ, 0])|)^2)×((|(Subscript[ψ, 1])|)^2)=(1/2)×(1/2)=(1/4).

On the other hand

((|(Subscript[ψ, 0])|)^2)((|(Subscript[ψ, 1])|)^2)=((Subscript[ψ, 0])(Subscript[ψ, 1]^( * )))×((Subscript[ψ, 1])(Subscript[ψ, 0]^( * )))=0×0=0.

Observable Expectation Value

Let (OverHat[O]) be a Hermitian operator describing the observable O,

let  (OverHat[ρ]) be the density matrix describing the state of a quantum system,

observing O on the quantum system, the expectation value of O is given by

For pure state (OverHat[ρ])=(Ket[ψ])(Bra[ψ]),

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

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[{{EdgeForm[Directive[AbsoluteThickness[1],Dashed]],Lighter[Blue,0.9],Rectangle[{5.5,-1.2},{9.5,1.2},RoundingRadius->.3]},Line[{#,0}&/@{-2,2}],opH["O"],bra["ψ",{-2,0}],ket["ψ",{2,0}],Text["=",{4,0}],BSplineCurve[{{6,0},{6,0},{6,0},{5,0},{5,1.5},{6,1.5},{11,1.5},{12,1.5},{12,0},{11,0},{9,0},{9,0},{9,0}}],ket["ψ",{6.3,0}],bra["ψ",{8.7,0}],opH["O",{10.5,0}],Text["ρ",{7.5,-0.8}]},ImageSize->{Automatic,45}]]

For mixed state  (OverHat[ρ])=(∑_ϕ)p((Ket[ϕ]))(Ket[ϕ])(Bra[ϕ]),

⟨O⟩=(∑_ϕ)p((Ket[ϕ]))(Bra[ϕ])(OverHat[O])(Ket[ϕ])=(∑_ϕ)p((Ket[ϕ]))Tr(Ket[ϕ])(Bra[ϕ])(OverHat[O])=Tr (OverHat[ρ]) (OverHat[O]).

Measurement

Quantum state (as generally modeled by the density matrix  (OverHat[ρ])) describes our state of knowledge about the quantum system.

The density matrix can be updated (the quantum state can collapse), if our state of knowledge is changed by observations.

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

Before observing O=(Subscript[O, k]), the state of the quantum system is described by a prior density matrix  (Subscript[(OverHat[ρ]), prior]).

Based on this description, we can predict that the outcome (Subscript[O, k]) will appear with the probability given by

After observing O=(Subscript[O, k]), the quantum state will collapse to the posterior density matrix  (Subscript[(OverHat[ρ]), post]), which is given by

The denominator is a normalization factor to ensure Tr(Subscript[(OverHat[ρ]), post])=1.

Dynamics

(OverHat[U])(dt)=(e^(-i dt (OverHat[H]))),

(OverHat[ρ])(dt)=(e^(-i dt (OverHat[H])))(OverHat[ρ])(0)(e^(i dt (OverHat[H])))
=(OverHat[ρ])(0)-i dt [(OverHat[H]),(OverHat[ρ])(0)].

Here the density matrix is taken to be in the Schrödinger picture.

Even though the von Neumann equation looks like the Heisenberg equation i ℏ(∂_t)(OverHat[O])(t)=[(OverHat[O])(t),(OverHat[H])] (which governs the operator evolution in the Heisenberg picture), but there is a crucial sign difference.

However in the Heisenberg picture, the density matrix is time-independent, because the state does not evolve in the Heisenberg picture and the density matrix follows the state.

(OverHat[ρ])(0)≏(({{(Subscript[ρ, 00]), (Subscript[ρ, 01])}, {(Subscript[ρ, 10]), (Subscript[ρ, 11])}})).

(OverHat[ρ])(t)≏(({{(Subscript[ρ, 00]), (Subscript[ρ, 01])(e^(-i ω t))}, {(Subscript[ρ, 10])(e^(i ω t)), (Subscript[ρ, 11])}})).

(OverBar[ρ])=(1/T)(∫_0^∞)(OverHat[ρ])(t)(e^(-t/T))dt
≏(({{(Subscript[ρ, 00]), ((Subscript[ρ, 01])/(1+i ω T))}, {((Subscript[ρ, 10])/(1-i ω T)), (Subscript[ρ, 11])}})) (Overscript[⟶, ω T≫1])(({{(Subscript[ρ, 00]), 0}, {0, (Subscript[ρ, 11])}})).

Quantum Decoherence (brief idea): the loss of off-diagonal density matrix elements (quantum coherence) over time in the energy eigenbasis.

Decoherence generally takes a pure state to a mixed state (unless the pure state is an energy eigenstate) and is the fundamental source of entropy production (to be explained later).

Quantum Channel*

a unitary evolution (time evolution)

(OverHat[ρ])→(OverHat[U]) (OverHat[ρ]) (SuperDagger[(OverHat[U])]),

a projective measurement (measure and obtain a definite outcome)

(OverHat[ρ])→(OverHat[𝒫]) (OverHat[ρ]) (OverHat[𝒫]),   (without normalization)

(OverHat[ρ])→(((OverHat[𝒫]) (OverHat[ρ]) (OverHat[𝒫]))/(Tr((OverHat[𝒫]) (OverHat[ρ]) (OverHat[𝒫])))),  (with normalization)

the normalization factor Tr (OverHat[𝒫]) (OverHat[ρ]) (OverHat[𝒫])=Tr (OverHat[ρ]) (OverHat[𝒫]) is the probability to obtain the out come.

(OverHat[ρ])→(OverHat[K]) (OverHat[ρ]) (SuperDagger[(OverHat[K])]),   (without normalization)

(OverHat[ρ])→(((OverHat[K]) (OverHat[ρ]) (SuperDagger[(OverHat[K])]))/(Tr((OverHat[K]) (OverHat[ρ]) (SuperDagger[(OverHat[K])])))).  (with normalization)

Unitary evolution: phase retarders creates ϕ relative phase shift between horizontal and vertical polarizations

(Subscript[(OverHat[U]), ϕ])=(e^(i (ϕ/2) ((OverHat[σ])^z)))≏(({{(e^(i ϕ/2)), 0}, {0, (e^(-i ϕ/2))}})).

Projective measurement: polarizers oriented along θ angle axis

(Subscript[(OverHat[𝒫]), θ])=(Ket[θ])(Bra[θ])≏(({{(cos^2)θ, cos θ sin θ}, {cos θ sin θ, (sin^2)θ}})).

Graphics3D[{{Black,AbsoluteThickness[1],Line[{{-5,0,0},{5,0,0}}]},{EdgeForm[None],Opacity[0.3],Red,Cylinder[{{-5,0,0},{-2,0,0}},0.7]},
Translate[{{EdgeForm[None],LightGreen,Cylinder[{{-0.05,0,0},{0.05,0,0}},1]},
{Black,Arrowheads[{-0.05,0.05}],Arrow[Tube[{{0.1,-1,0},{0.1,1,0}}.RotationMatrix[-π/4,{1,0,0}],0.03]]}},{-2,0,0}],
{Red,Line[Table[{x,0.5Sin[4π x],0}.RotationMatrix[-π/4,{1,0,0}],{x,-1.9,1.9,0.01}]]},
Translate[{{EdgeForm[None],LightGreen,Cylinder[{{-0.05,0,0},{0.05,0,0}},1]},
{Black,Arrowheads[{-0.05,0.05}],Arrow[Tube[{{0.1,-1,0},{0.1,1,0}}.RotationMatrix[π/4,{1,0,0}],0.03]]}},{2,0,0}]},ViewPoint->{1.2,-2.4,0.5},Lighting->"Neutral",Boxed->False]

(OverHat[ρ])'=(Subscript[(OverHat[𝒫]), -π/4])(Subscript[(OverHat[𝒫]), π/4])(OverHat[ρ]) (Subscript[(OverHat[𝒫]), π/4])(Subscript[(OverHat[𝒫]), -π/4])
≏(({{(1/2), -(1/2)}, {-(1/2), (1/2)}}))(({{(1/2), (1/2)}, {(1/2), (1/2)}}))(({{(1/2), 0}, {0, (1/2)}}))(({{(1/2), (1/2)}, {(1/2), (1/2)}}))(({{(1/2), -(1/2)}, {-(1/2), (1/2)}}))
=(({{0, 0}, {0, 0}}))
=>Tr (OverHat[ρ])'=0.

Graphics3D[{{Black,AbsoluteThickness[1],Line[{{-5,0,0},{5,0,0}}]},{EdgeForm[None],Opacity[0.3],Red,Cylinder[{{-5,0,0},{-3,0,0}},0.7]},
Translate[{{EdgeForm[None],LightGreen,Cylinder[{{-0.05,0,0},{0.05,0,0}},1]},
{Black,Arrowheads[{-0.05,0.05}],Arrow[Tube[{{0.1,-1,0},{0.1,1,0}}.RotationMatrix[-π/4,{1,0,0}],0.03]]}},{-3,0,0}],
{Red,Line[Table[{x,0.5Sin[4π x],0}.RotationMatrix[-π/4,{1,0,0}],{x,-2.9,-0.1,0.01}]]},
Translate[{{EdgeForm[None],LightBlue,Cuboid[{-0.1,-0.7,-0.7},{0.1,0.7,0.7}]}},{0,0,0}],
{Red,Line[Table[{x,0.5Sin[4π x],0.5 Cos[4π x]},{x,0.1,2.9,0.01}]]},
Translate[{{EdgeForm[None],LightGreen,Cylinder[{{-0.05,0,0},{0.05,0,0}},1]},
{Black,Arrowheads[{-0.05,0.05}],Arrow[Tube[{{0.1,-1,0},{0.1,1,0}}.RotationMatrix[π/4,{1,0,0}],0.03]]}},{3,0,0}],
{Red,Line[Table[{x,0.35Sin[4π x],0}.RotationMatrix[π/4,{1,0,0}],{x,3.1,5,0.01}]]}},ViewPoint->{1.2,-2.4,0.5},Lighting->"Neutral",Boxed->False]

(OverHat[ρ])'=(Subscript[(OverHat[𝒫]), -π/4])(Subscript[(OverHat[U]), π/2])(Subscript[(OverHat[𝒫]), π/4])(OverHat[ρ]) (Subscript[(OverHat[𝒫]), π/4])(Subscript[(OverHat[U]), π/2]^†)(Subscript[(OverHat[𝒫]), -π/4])
≏(({{(1/2), -(1/2)}, {-(1/2), (1/2)}}))(({{(e^(i π/4)), 0}, {0, (e^(-i π/4))}}))(({{(1/2), (1/2)}, {(1/2), (1/2)}}))(({{(1/2), 0}, {0, (1/2)}}))(({{(1/2), (1/2)}, {(1/2), (1/2)}}))(({{(e^(-i π/4)), 0}, {0, (e^(i π/4))}}))(({{(1/2), -(1/2)}, {-(1/2), (1/2)}}))
=(({{(1/8), -(1/8)}, {-(1/8), (1/8)}}))
=>Tr (OverHat[ρ])'=(1/4).

von Neumann Entropy

When there is no degeneracy, eigen projector is simply (Subscript[(OverHat[𝒫]), ρ=(Subscript[p, k])])=(Ket[(Subscript[ϕ, k])])(Bra[(Subscript[ϕ, k])]).

The eigenvectors (Ket[(Subscript[ϕ, k])]) form an orthonormal basis.

The eigenvalues (Subscript[p, k]) has the physical meaning of probability, satisfying:

Hermitian:  (SuperDagger[(OverHat[ρ])])= (OverHat[ρ]) ⇔ (Subscript[p, k])∈ℝ.

Normalization (trace one): Tr  (OverHat[ρ])=1 ⇔ (∑_k)(Subscript[p, k])=1.

Positive (semi)definite: ∀(Ket[ψ]):(Bra[ψ]) (OverHat[ρ])(Ket[ψ])≥0 ⇔ (Subscript[p, k])≥0.

(OverHat[ρ]) is a pure state, iff there is only a single k for which (Subscript[p, k])=1 (and all the other (Subscript[p, j])=0 for j!=k). For example, if (Subscript[p, 1])=1 and (Subscript[p, 2])=(Subscript[p, 3])=…=0, then  (OverHat[ρ])=(Ket[(Subscript[ϕ, 1])])(Bra[(Subscript[ϕ, 1])]) is a pure state.

Otherwise, for generic distribution (Subscript[p, k]),  (OverHat[ρ]) is a mixed state.

S=-(∑_k)(Subscript[p, k])log (Subscript[p, k]).

Show that Eq. () is equivalent to Eq. () given Eq. ().

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

f((OverHat[ρ]))=(∑_k)f((Subscript[p, k]))(Subscript[(OverHat[𝒫]), ρ=(Subscript[p, k])]).

S((OverHat[ρ]))=-Tr (OverHat[ρ])log (OverHat[ρ])
=Tr(∑_k)(-(Subscript[p, k])log (Subscript[p, k]))(Subscript[(OverHat[𝒫]), ρ=(Subscript[p, k])])
=-(∑_k)(Subscript[p, k])log (Subscript[p, k])Tr(Subscript[(OverHat[𝒫]), ρ=(Subscript[p, k])])

S((OverHat[ρ]))=-(∑_k)(Subscript[p, k])log (Subscript[p, k]).

(OverHat[ρ])≏(1/2)(({{1, m}, {m, 1}})),

Step 1: diagonalize the density matrix (find the eigenvalues)

(OverHat[ρ])≏(1/2)(({{1, m}, {m, 1}}))→(({{((1+m)/2), 0}, {0, ((1-m)/2)}}))

Step 2: extract eigenvalues and arrange them into a probability distribution

(Subscript[p, 1])=((1+m)/2), (Subscript[p, 2])=((1-m)/2).

Make sure that (Subscript[p, 1])+(Subscript[p, 2])=1.

Step 3: evaluate the entropy using Eq. (),

S((OverHat[ρ]))=-((Subscript[p, 1])log (Subscript[p, 1])+(Subscript[p, 2])log (Subscript[p, 2]))
=-(((1+m)/2)log ((1+m)/2)+((1-m)/2)log ((1-m)/2))
=log 2-(1/2)log(1-(m^2))-m arctanh m.

Plot[-((1+m)/2Log2[(1+m)/2]+(1-m)/2Log2[(1-m)/2]),{m,-1,1},PlotStyle->Black,FrameLabel->{"\!\( m\)","\!\(S(ρ\&^)\) [bit]"}]

Purity and Rényi Entropy

(OverHat[ρ]) is ({{{, ({{pure, if Tr  ((OverHat[ρ])^2)=1,}, {mixed, if Tr  ((OverHat[ρ])^2)<1.}})}})

(S^((n)))=(((1-n))^(-1))log (∑_i)(Subscript[p, i]^n).

n is the Rényi index.

n=0: max-entropy, simply counts the log of the Hilbert space dimension (S^((0)))=log dim ℋ.

n->1 limit: equivalent to the von Neumann entropy, i.e. S((OverHat[ρ]))=(lim_(n->1))(S^((n)))((OverHat[ρ])).

Show that in the n->1 limit, the Rényi entropy reduces to the von Neumann entropy.

(lim_(n->1))(S^((n)))((OverHat[ρ]))=(lim_(n->1))(1/(1-n))ln Tr ((OverHat[ρ])^n)
=-(lim_(n->1))(∂_n)ln Tr ((OverHat[ρ])^n)
=-(lim_(n->1))((Tr ((OverHat[ρ])^n)ln (OverHat[ρ]))/(Tr ((OverHat[ρ])^n)))
=-((Tr (OverHat[ρ]) ln (OverHat[ρ]))/(Tr (OverHat[ρ])))
=-Tr (OverHat[ρ]) ln (OverHat[ρ])=S((OverHat[ρ])).

n=2: the 2nd Rényi entropy is directly related to purity by (S^((2)))=-log Tr ((OverHat[ρ])^2).

n=∞: min-entropy, lower bound of all Rényi entropies, (S^((∞)))=-log (max_i) (Subscript[p, i]).

The spectrum of the density matrix, i.e. all eigenvalues (Subscript[p, i]), can be reconstructed from the family of Rényi entropies (by solving the following equations, in principle).

(∑_i)(Subscript[p, i]^n)=(e^((1-n)(S^((n)))))   (for n=1,2,…,dim ℋ).

Entropy and Knowledge

a pure state, we know that the system is in a definite quantum state => we have the full knowledge about the system => the system has no entropy, i.e. (S^((n)))((OverHat[ρ]))=0 (for all n).

a mixed state, there are several possible states that the system can take (we are not sure) => we only have partial knowledge about the system => our ignorance gives rise to the entropy of the system.

a maximally mixed state  (OverHat[ρ])=𝟙/Tr(OverHat[ρ]), we have no knowledge about the system, every state in the Hilbert space are equally likely =>  the entropy of the system is maximized.

(OverHat[ρ]) is ({{{, ({{pure, if (S^((n)))((OverHat[ρ]))=0,}, {mixed, if (S^((n)))((OverHat[ρ]))>0,}})}}) for n=1,2,….

Decoherence generally takes a pure state to a mixed state (unless the pure state is an energy eigenstate) and is the fundamental source of entropy production (to be explained later).

ln dim ℋ=(S^((0)))≥(S^((1)))≥(S^((2)))≥…≥(S^((∞)))≥0.

∀i:(Subscript[p, i])=(1/(dim ℋ))=>∀n≥0:(S^((n)))=log dim ℋ.

(OverHat[ρ])=(1/(dim ℋ))𝟙.

(OverHat[ρ])=(𝟙/2)≏(1/2)(({{1, 0}, {0, 1}})).

Application: if the qubit basis corresponds to the photon polarization, then the density matrix in Eq. () describes the natural light ensemble of photons.

All Rényi entropies are identically log 2 for a maximally mixed qubit,

(S^((n)))=(1/(1-n))log ((1/(2^n))+(1/(2^n)))=log 2=1bit.

This is the maximal entropy that a qubit could have: our ignorance about a qubit is at most 1 bit. This is why a qubit is called a quantum bit.

({{state (OverHat[ρ]), pure, ↔, maximally mixed}, {purity Tr ((OverHat[ρ])^2), 1, ↔, 1/dim ℋ}, {entropy (S^((n)))((OverHat[ρ])), 0, ↔, log dim ℋ}, {knowledge, max, ↔, none}})

Quantum Entanglement

Two-Qubit Systems

Tensor Product Hilbert Space

(Subscript[ℋ, A])=span{(Subscript[(Ket[i]), A])}, (Subscript[ℋ, B])=span{(Subscript[(Ket[j]), B])},

(Subscript[ℋ, A⋃B])=(Subscript[ℋ, A])⊗(Subscript[ℋ, B])=span{(Subscript[(Ket[i]), A])⊗(Subscript[(Ket[j]), B])}=span{(Ket[i j])}.

Hilbert space tensor product => Hilbert space dimension multiplies

dim (Subscript[ℋ, A⋃B])=dim (Subscript[ℋ, A])dim (Subscript[ℋ, B]).

Rule of scalar product: scalar product of two tensor product states equals the product of scalar products in each tensor Hilbert space

(BraKet[i j, k l])=(Subscript[(Bra[j]), B])⊗(Subscript[(Bra[i]), A])(Subscript[(Ket[k]), A])⊗(Subscript[(Ket[l]), B])=(Subscript[(BraKet[i, k]), A])(Subscript[(BraKet[j, l]), B])=(Subscript[δ, i k])(Subscript[δ, j l]).

The tensor product basis is still orthonormal.

Generic states in (Subscript[ℋ, A⋃B])

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

(Subscript[v, i j])=(BraKet[i j, v]).

Generic operators in (Subscript[ℋ, A⋃B])

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

where the matrix (tensor) element

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

Tensor product of states. Suppose (Ket[u])=(∑_i)(Subscript[u, i])(Subscript[(Ket[i]), A]), (Ket[v])=(∑_j)(Subscript[v, j])(Subscript[(Ket[j]), B])

(Ket[u])⊗(Ket[v])=(∑_(i,j))(Subscript[u, i])(Subscript[v, j])(Subscript[(Ket[i]), A])⊗(Subscript[(Ket[j]), B])=(∑_(i,j))(Subscript[u, i])(Subscript[v, j])(Ket[i j]).

Note: the double index i j labels a single state (Ket[i j]).

Take two qubits for example

(({{(Subscript[u, 0])}, {(Subscript[u, 1])}}))⊗(({{(Subscript[v, 0])}, {(Subscript[v, 1])}}))=(({{(Subscript[u, 0])(({{(Subscript[v, 0])}, {(Subscript[v, 1])}}))}, {(Subscript[u, 1])(({{(Subscript[v, 0])}, {(Subscript[v, 1])}}))}}))=(({{(Subscript[u, 0])(Subscript[v, 0])}, {(Subscript[u, 0])(Subscript[v, 1])}, {(Subscript[u, 1])(Subscript[v, 0])}, {(Subscript[u, 1])(Subscript[v, 1])}})).

Tensor product of operators. Suppose A=(∑_(i,j))(Subscript[(Ket[i]), A])(Subscript[A, i j])(Subscript[(Bra[j]), A]), B=(∑_(k,l))(Subscript[(Ket[k]), B])(Subscript[B, k l])(Subscript[(Bra[l]), B]),

(OverHat[A])⊗(OverHat[B])=(∑_(i,j,k,l))(Subscript[A, i j])(Subscript[B, k l])(Subscript[(Ket[i]), A])(Subscript[(Ket[k]), B])(Subscript[(Bra[j]), A])(Subscript[(Bra[l]), B])
=(∑_(i,j,k,l))(Subscript[A, i j])(Subscript[B, k l])(Ket[i k])(Bra[j l]).

Take two qubits for example

(({{(Subscript[A, 00]), (Subscript[A, 01])}, {(Subscript[A, 10]), (Subscript[A, 11])}}))⊗(({{(Subscript[B, 00]), (Subscript[B, 01])}, {(Subscript[B, 10]), (Subscript[B, 11])}}))
=(({{(Subscript[A, 00])(({{(Subscript[B, 00]), (Subscript[B, 01])}, {(Subscript[B, 10]), (Subscript[B, 11])}})), (Subscript[A, 01])(({{(Subscript[B, 00]), (Subscript[B, 01])}, {(Subscript[B, 10]), (Subscript[B, 11])}}))}, {(Subscript[A, 10])(({{(Subscript[B, 00]), (Subscript[B, 01])}, {(Subscript[B, 10]), (Subscript[B, 11])}})), (Subscript[A, 11])(({{(Subscript[B, 00]), (Subscript[B, 01])}, {(Subscript[B, 10]), (Subscript[B, 11])}}))}}))
=(({{(Subscript[A, 00])(Subscript[B, 00]), (Subscript[A, 00])(Subscript[B, 01]), (Subscript[A, 01])(Subscript[B, 00]), (Subscript[A, 01])(Subscript[B, 01])}, {(Subscript[A, 00])(Subscript[B, 10]), (Subscript[A, 00])(Subscript[B, 11]), (Subscript[A, 01])(Subscript[B, 10]), (Subscript[A, 01])(Subscript[B, 11])}, {(Subscript[A, 10])(Subscript[B, 00]), (Subscript[A, 10])(Subscript[B, 01]), (Subscript[A, 11])(Subscript[B, 00]), (Subscript[A, 11])(Subscript[B, 01])}, {(Subscript[A, 10])(Subscript[B, 10]), (Subscript[A, 10])(Subscript[B, 11]), (Subscript[A, 11])(Subscript[B, 10]), (Subscript[A, 11])(Subscript[B, 11])}})).

Two-Qubit States

({{ , , qubit B, }, {, , (Ket[0]), (Ket[1])}, {qubit A, (Ket[0]), (Ket[00]), (Ket[01])}, {, (Ket[1]), (Ket[10]), (Ket[11])}})

(Ket[00])=(Subscript[(Ket[0]), A])⊗(Subscript[(Ket[0]), B])≏(({{1}, {0}}))⊗(({{1}, {0}}))=(({{1}, {0}, {0}, {0}})).

(Ket[01])=(Subscript[(Ket[0]), A])⊗(Subscript[(Ket[1]), B])≏(({{1}, {0}}))⊗(({{0}, {1}}))=(({{0}, {1}, {0}, {0}})),
(Ket[10])=(Subscript[(Ket[1]), A])⊗(Subscript[(Ket[0]), B])≏(({{0}, {1}}))⊗(({{1}, {0}}))=(({{0}, {0}, {1}, {0}})),
(Ket[11])=(Subscript[(Ket[1]), A])⊗(Subscript[(Ket[0]), B])≏(({{0}, {1}}))⊗(({{0}, {1}}))=(({{0}, {0}, {0}, {1}})).

(Ket[ψ])=(Subscript[ψ, 0])(Ket[00])+(Subscript[ψ, 1])(Ket[01])+(Subscript[ψ, 2])(Ket[10])+(Subscript[ψ, 3])(Ket[11])≏(({{(Subscript[ψ, 0])}, {(Subscript[ψ, 1])}, {(Subscript[ψ, 2])}, {(Subscript[ψ, 3])}})).

Product state: a state that can be factorized as a tensor product of single-qubit states.

(Ket[u])⊗(Ket[v])=((Subscript[u, 0])(Ket[0])+(Subscript[u, 1])(Ket[1]))⊗((Subscript[v, 0])(Ket[0])+(Subscript[v, 1])(Ket[1]))
=(Subscript[u, 0])(Subscript[v, 0])(Ket[00])+(Subscript[u, 0])(Subscript[v, 1])(Ket[01])+(Subscript[u, 1])(Subscript[v, 0])(Ket[10])+(Subscript[u, 1])(Subscript[v, 1])(Ket[11]).

A generic state as (Ket[ψ]) in Eq. () has six real parameters. 4×2-1-1=6.

A generic product state as (Ket[u])⊗(Ket[v]) in Eq. () has only four real parameters. (2×2-1-1)×2=4.

Entangled state: any state that can not be factorized to product states are entangled.

Prove that (1/(Sqrt[2]))((Ket[01])-(Ket[10])) can not be written as a product state.

(Subscript[u, 0])(Subscript[v, 0])=0,(Subscript[u, 0])(Subscript[v, 1])=(1/(Sqrt[2])),(Subscript[u, 1])(Subscript[v, 0])=-(1/(Sqrt[2])),(Subscript[u, 1])(Subscript[v, 1])=0.

Two-Qubit Operators

Single-qubit observables (each as a 3×4×4 tensor, extended from the 3×2×2 Pauli tensor):

(Subscript[(OverHat[σ]), A])=((Subscript[(OverHat[σ]), A]^x),(Subscript[(OverHat[σ]), A]^y),(Subscript[(OverHat[σ]), A]^z)),
(Subscript[(OverHat[σ]), B])=((Subscript[(OverHat[σ]), B]^x),(Subscript[(OverHat[σ]), B]^y),(Subscript[(OverHat[σ]), B]^z)).

Two-qubit observables (joint measurements, 3×3×4×4 tensor):

(Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B])=(({{(Subscript[(OverHat[σ]), A]^x)(Subscript[(OverHat[σ]), B]^x), (Subscript[(OverHat[σ]), A]^y)(Subscript[(OverHat[σ]), B]^x), (Subscript[(OverHat[σ]), A]^z)(Subscript[(OverHat[σ]), B]^x)}, {(Subscript[(OverHat[σ]), A]^x)(Subscript[(OverHat[σ]), B]^y), (Subscript[(OverHat[σ]), A]^y)(Subscript[(OverHat[σ]), B]^y), (Subscript[(OverHat[σ]), A]^z)(Subscript[(OverHat[σ]), B]^y)}, {(Subscript[(OverHat[σ]), A]^x)(Subscript[(OverHat[σ]), B]^z), (Subscript[(OverHat[σ]), A]^y)(Subscript[(OverHat[σ]), B]^z), (Subscript[(OverHat[σ]), A]^z)(Subscript[(OverHat[σ]), B]^z)}})).

(Subscript[(OverHat[σ]), A]^x)⊗(Subscript[𝟙, B])≏(({{0, 1}, {1, 0}}))⊗(({{1, 0}, {0, 1}}))=(({{0, 0, 1, 0}, {0, 0, 0, 1}, {1, 0, 0, 0}, {0, 1, 0, 0}})).

(Subscript[(OverHat[σ]), A]^z)⊗(Subscript[(OverHat[σ]), B]^y)≏(({{1, 0}, {0, -1}}))⊗(({{0, -i}, {i, 0}}))=(({{0, -i, 0, 0}, {i, 0, 0, 0}, {0, 0, 0, i}, {0, 0, -i, 0}})).

A Two-Qubit Model

H=(J/4) (Subscript[(OverHat[σ]), A])·(Subscript[(OverHat[σ]), B])=(J/4)((Subscript[σ, A]^x)(Subscript[σ, B]^x)+(Subscript[σ, A]^y)(Subscript[σ, B]^y)+(Subscript[σ, A]^z)(Subscript[σ, B]^z)).

H≏(J/4)(({{1, 0, 0, 0}, {0, -1, 2, 0}, {0, 2, -1, 0}, {0, 0, 0, 1}})).

Eigenvalue (Subscript[E, s])=-3J/4: a unique eigenstate => spin-singlet state

(Ket[s])=(1/(Sqrt[2]))((Ket[01])-(Ket[10])).

Eigenvalue (Subscript[E, t])=J/4: three degenerated eigenstates => spin-triplet states (there is a basis freedom here, we make the following choice)

(Ket[(Subscript[t, 1])])=(1/(Sqrt[2]))((Ket[01])+(Ket[10])),
(Ket[(Subscript[t, 2])])=(1/(Sqrt[2]))((Ket[00])+(Ket[11])),
(Ket[(Subscript[t, 3])])=(1/(Sqrt[2]))((Ket[00])-(Ket[11])).

The Spin-Singlet State

(Ket[s])=(1/(Sqrt[2]))((Ket[01])-(Ket[10]))≏(1/(Sqrt[2]))(({{0, 1, -1, 0}}))ᵀ.

Expectation value of single-qubit observables

(Bra[s])(Subscript[(OverHat[σ]), A])(Ket[s])=(0,0,0),
(Bra[s])(Subscript[(OverHat[σ]), B])(Ket[s])=(0,0,0).

Expectation value of two-qubit observables

(Bra[s])(Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B])(Ket[s])=(({{-1, 0, 0}, {0, -1, 0}, {0, 0, -1}})).

(Ket[s]) is a pure state of the two-qubit system => the system is in a definite quantum state, entropy of the entire system = 0 => we have the full knowledge about the system.

However (Bra[s])(Subscript[(OverHat[σ]), A])(Ket[s])=0 implies nothing is know about qubit A, because qubit A is in a maximally mixed state with maximal entropy of the subsystem (1bit) => we are completely ignorant about the subsystems. (Same argument applies for qubit B)

Classical information is stored locally (bit-by-bit) in every single classical bit. Knowing the entire system = knowing the state of every classical bit.

Quantum information can be stored jointly in the interrelations among qubits, but not locally in single qubits. Knowing the entire system does not imply the knowledge of its subsystem.

Entanglement Entropy

(Ket[s])=(1/(Sqrt[2]))((Ket[↑↓])-(Ket[↓↑])).

Full density matrix

(OverHat[ρ])=(Ket[s])(Bra[s])≏(1/2)(({{0}, {1}, {-1}, {0}}))(({{0, 1, -1, 0}}))=(1/2)(({{0, 0, 0, 0}, {0, 1, -1, 0}, {0, -1, 1, 0}, {0, 0, 0, 0}})).

Partial trace over qubit B => reduced density matrix of qubit A

(Subscript[(OverHat[ρ]), A])=(Subscript[Tr, B])(Ket[s])(Bra[s])
≏(1/2)(({{tr(({{0, 0}, {0, 1}})), tr(({{0, 0}, {-1, 0}}))}, {tr(({{0, -1}, {0, 0}})), tr(({{1, 0}, {0, 0}}))}}))=(1/2)(({{1, 0}, {0, 1}})).

Note that (Subscript[ρ, A]) indeed describes a maximally mixed qubit.

Compute the entropy of the reduced density matrix,

S( (Subscript[(OverHat[ρ]), A]))=-Tr (Subscript[(OverHat[ρ]), A])log (Subscript[(OverHat[ρ]), A])=log 2=1 bit.

(Ket[ψ])=(1/2)((Ket[00])+(Ket[01])+(Ket[10])+(Ket[11])).

Full density matrix

(OverHat[ρ])=(Ket[ψ])(Bra[ψ])≏(1/4)(({{1}, {1}, {1}, {1}}))(({{1, 1, 1, 1}}))=(1/4)(({{1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}})).

Partial trace over qubit B => reduced density matrix of qubit A

(Subscript[(OverHat[ρ]), A])=(Subscript[Tr, B])(OverHat[ρ])
≏(1/4)(({{tr(({{1, 1}, {1, 1}})), tr(({{1, 1}, {1, 1}}))}, {tr(({{1, 1}, {1, 1}})), tr(({{1, 1}, {1, 1}}))}}))=(1/2)(({{1, 1}, {1, 1}})).

Compute the entropy of the reduced density matrix,

S( (Subscript[(OverHat[ρ]), A]))=-Tr  (Subscript[(OverHat[ρ]), A])log  (Subscript[(OverHat[ρ]), A])=-(0log 0+1 log 1)=0 bit.

({{pure state (Ket[ψ]), product, ↔, maximally entangled}, {(Subscript[(OverHat[ρ]), A]), pure, ↔, maximally mixed}, {EE (S^((n)))( (Subscript[(OverHat[ρ]), A])), 0, ↔, log dim (Subscript[ℋ, A])}, {entanglement, none, ↔, max}})

EPR Pair and Bell Inequality

(Ket[EPR])=(1/(Sqrt[2]))((Ket[01])-(Ket[10])).

Graphics[{{LightGray,Disk[{0,5},{6,1}]},Text["EPR state",{0,5}],{Arrowheads[{{0.1,0.8}}],Arrow[{{0,0},{5,5}}],Arrow[{{0,0},{-5,5}}]},{Disk[#,0.5]&/@{{-5,5},{5,5}}},{EdgeForm[Black],LightGray,Rectangle[{-1,-1},{1,1},RoundingRadius->.5]},
{Text["Alice",{-5,5},{0,-1.2}],Text["Bob",{5,5},{0,-1.2}],Text["source",{0,-1},{0,1}]}},ImageSize->120]

Alice defines her set of observables:

(Subscript[(OverHat[σ]), A])=((Subscript[(OverHat[σ]), A]^x),(Subscript[(OverHat[σ]), A]^y),(Subscript[(OverHat[σ]), A]^z)).

Bob defines his set of observables:

(Subscript[(OverHat[σ]), B])=((Subscript[(OverHat[σ]), B]^x),(Subscript[(OverHat[σ]), B]^y),(Subscript[(OverHat[σ]), B]^z)).

The observables are perfectly anti-correlated between Alice and Bob

(Bra[EPR])(Subscript[(OverHat[σ]), A])(Ket[EPR])=(Bra[EPR])(Subscript[(OverHat[σ]), B])(Ket[EPR])=(0,0,0),
(Bra[EPR])(Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B])(Ket[EPR])=(({{-1, 0, 0}, {0, -1, 0}, {0, 0, -1}})).

(Subscript[σ, A]^z)=-(Subscript[σ, B]^z)=({{{, ({{+1, p=1/2}, {-1, p=1/2}})}}).

Quantum explanation: can be inferred from ⟨(Subscript[σ, A]^z)⟩=⟨(Subscript[σ, B]^z)⟩=0 and ⟨(Subscript[σ, A]^z)(Subscript[σ, B]^z)⟩=-1.

This is not too surprising: just a perfect anti-correlation between two random variables. Classically, one may model the perfect anti-correlation by a hidden variable:

Block[{bg,str},
bg={{Arrowheads[{{0.05,0.85}}],Arrow[{{0,0},{5,5}}],Arrow[{{0,0},{-5,5}}]},{Disk[#,0.5]&/@{{-5,5},{5,5}}},{EdgeForm[Black],LightGray,Rectangle[{-1,-1},{1,1},RoundingRadius->.5]},
{Text["Alice",{-5,5},{0,-1.2}],Text["Bob",{5,5},{0,-1.2}]},
Text["\!\(p=1/2\)",{0,-1},{0,1}]};
str[bs_]:=Inset[Grid[{#1},Alignment->{Center,Center},Frame->All,FrameStyle->Thin,Spacings->{0.05,0}],{#2,3},{-Sign[#2],1}]&@@@Thread[{bs,{-3,3}}];
Graphics[{Translate[{bg,str[{{"0"},{"1"}}]},{-8,0}],Translate[{bg,str[{{"1"},{"0"}}]},{8,0}]},ImageSize->240]]

(Subscript[σ, A]^x)=(Subscript[σ, B]^x)=({{{, ({{+1, p=1/2}, {-1, p=1/2}})}}).

Quantum explanation: can be inferred from ⟨(Subscript[σ, A]^x)⟩=⟨(Subscript[σ, B]^x)⟩=0 and ⟨(Subscript[σ, A]^x)(Subscript[σ, B]^x)⟩=1.

To model this classically: we will need to introduce another hidden variable to encode the perfect correlation in (σ^x) channel.

Block[{bg,str},
bg={{Arrowheads[{{0.05,0.85}}],Arrow[{{0,0},{5,5}}],Arrow[{{0,0},{-5,5}}]},{Disk[#,0.5]&/@{{-5,5},{5,5}}},{EdgeForm[Black],LightGray,Rectangle[{-1,-1},{1,1},RoundingRadius->.5]},
{Text["Alice",{-5,5},{0,-1.2}],Text["Bob",{5,5},{0,-1.2}]},
Text["\!\(p=1/2\)",{0,-1},{0,1}]};
str[bs_]:=Inset[Grid[{#1},Alignment->{Center,Center},Frame->All,FrameStyle->Thin,Spacings->{0.05,0}],{#2,3},{-Sign[#2],1}]&@@@Thread[{bs,{-3,3}}];
Graphics[{Translate[{bg,str[{{"?","0"},{"?","1"}}]},{-8,0}],Translate[{bg,str[{{"?","1"},{"?","0"}}]},{8,0}]},ImageSize->240]]

(Subscript[σ, A]^z)=({{{, ({{+1, p=1/2}, {-1, p=1/2}})}}), (Subscript[σ, B]^x)=({{{, ({{+1, p=1/2}, {-1, p=1/2}})}}).

Quantum explanation: can be inferred from ⟨(Subscript[σ, A]^z)⟩=⟨(Subscript[σ, B]^x)⟩=0 and ⟨(Subscript[σ, A]^z)(Subscript[σ, B]^x)⟩=0.

The classical hidden variables can reproduce this behavior only if they follow the joint distribution

({{Alice, Bob, p}, {00, 11, 1/4}, {01, 10, 1/4}, {10, 01, 1/4}, {11, 00, 1/4}})

({{(Subscript[n, A])·(Subscript[σ, A]), (Subscript[n, B])·(Subscript[σ, B]), p}, {+1, -1, (1+(Subscript[n, A])·(Subscript[n, B]))/4}, {+1, +1, (1-(Subscript[n, A])·(Subscript[n, B]))/4}, {-1, -1, (1-(Subscript[n, A])·(Subscript[n, B]))/4}, {-1, +1, (1+(Subscript[n, A])·(Subscript[n, B]))/4}})

p((Subscript[n, A])·(Subscript[σ, A])=(Subscript[n, B])·(Subscript[σ, B]))=((1+(Subscript[n, A])·(Subscript[n, B]))/2).

Quantum explanation: can be inferred from ⟨(Subscript[n, A])·(Subscript[σ, A])⟩=⟨(Subscript[n, B])·(Subscript[σ, B])⟩=0 and ⟨(Subscript[n, A])·(Subscript[σ, A])(Subscript[n, B])·(Subscript[σ, B])⟩=(Subscript[n, A])·(Subscript[n, B]).

Classically, to reproduce all these, we will need many (could be infinitely many) hidden variables. (This is ugly but not fatal yet.)

Block[{bg,str},
bg={{Arrowheads[{{0.12,0.85}}],Arrow[{{0,0},{5,5}}],Arrow[{{0,0},{-5,5}}]},{Disk[#,0.5]&/@{{-5,5},{5,5}}},{EdgeForm[Black],LightGray,Rectangle[{-1,-1},{1,1},RoundingRadius->.5]},
{Text["Alice",{-5,5},{0,-1.2}],Text["Bob",{5,5},{0,-1.2}]}};
str[bs_]:=Inset[Grid[{#1},Alignment->{Center,Center},Frame->All,FrameStyle->Thin,Spacings->{0.05,0}],{#2,3},{-Sign[#2],1}]&@@@Thread[{bs,{-3,3}}];
Graphics[{bg,str[{{"1","0","0","1","0","…"},{"0","1","1","0","1","…"}}],Text["\!\(p[10010…]\)",{0,-1},{0,1}]},ImageSize->160]]

({{Alice, Bob, p}, {…00…, …11…, (1+(Subscript[n, 1])·(Subscript[n, 2]))/4}, {…01…, …10…, (1-(Subscript[n, 1])·(Subscript[n, 2]))/4}, {…10…, …01…, (1-(Subscript[n, 1])·(Subscript[n, 2]))/4}, {…11…, …00…, (1+(Subscript[n, 1])·(Subscript[n, 2]))/4}}).

Classically, there must be three hidden variables associated with the three observables, following some marginal distribution

({{Alice, Bob, p}, {…000…, …111…, (Subscript[p, 1])}, {…001…, …110…, (Subscript[p, 2])}, {…010…, …101…, (Subscript[p, 3])}, {…011…, …100…, (Subscript[p, 4])}, {…100…, …011…, (Subscript[p, 5])}, {…101…, …010…, (Subscript[p, 6])}, {…110…, …001…, (Subscript[p, 7])}, {…111…, …000…, (Subscript[p, 8])}}).

The probability must sum up to 1, i.e.

(Subscript[p, 1])+(Subscript[p, 2])+…+(Subscript[p, 8])=1.

If Alice measures (Subscript[n, 1])·(Subscript[σ, A]) and Bob measures (Subscript[n, 2])·(Subscript[σ, B]), the probability that they obtain the opposite outcome is

p((Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B]))=(Subscript[p, 1])+(Subscript[p, 2])+(Subscript[p, 7])+(Subscript[p, 8]).

If Alice measures (Subscript[n, 2])·(Subscript[σ, A]) and Bob measures (Subscript[n, 3])·(Subscript[σ, B]), the probability that they obtain the opposite outcome is

p((Subscript[n, 2])·(Subscript[σ, A])=-(Subscript[n, 3])·(Subscript[σ, B]))=(Subscript[p, 1])+(Subscript[p, 4])+(Subscript[p, 5])+(Subscript[p, 8]).

If Alice measures (Subscript[n, 3])·(Subscript[σ, A]) and Bob measures (Subscript[n, 1])·(Subscript[σ, B]), the probability that they obtain the opposite outcome is

p((Subscript[n, 3])·(Subscript[σ, A])=-(Subscript[n, 1])·(Subscript[σ, B]))=(Subscript[p, 1])+(Subscript[p, 3])+(Subscript[p, 6])+(Subscript[p, 8]).

Put together,

p((Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B]))+p((Subscript[n, 2])·(Subscript[σ, A])=-(Subscript[n, 3])·(Subscript[σ, B]))+p((Subscript[n, 3])·(Subscript[σ, A])=-(Subscript[n, 1])·(Subscript[σ, B]))
=3(Subscript[p, 1])+(Subscript[p, 2])+(Subscript[p, 3])+(Subscript[p, 4])+(Subscript[p, 5])+(Subscript[p, 6])+(Subscript[p, 7])+3(Subscript[p, 8])
=1+2(Subscript[p, 1])+2(Subscript[p, 8])

This leads to a (version of) Bell inequality.

Now what is the quantum mechanical prediction? Recall the quantum result in Eq. (), the Bell inequality would require

((1+(Subscript[n, 1])·(Subscript[n, 2]))/2)+((1+(Subscript[n, 2])·(Subscript[n, 3]))/2)+((1+(Subscript[n, 3])·(Subscript[n, 1]))/2)≥1,

for three unit vectors (Subscript[n, 1]), (Subscript[n, 2]) and (Subscript[n, 3]).

Graphics[{Arrowheads[0.2],{Arrow[{{0,0},#1}],Text["\!\(n\_"<>#2<>"\)",#1,-#1]}&@@@Thread[{CirclePoints[3],ToString/@Range[3]}]},ImageSize->80]

(Subscript[n, 1])·(Subscript[n, 2])=(Subscript[n, 2])·(Subscript[n, 3])=(Subscript[n, 3])·(Subscript[n, 1])=-1/2.

(1/4)+(1/4)+(1/4)=(3/4)≥1,

(Ket[ψ])=cos α(Ket[01])-sin α(Ket[10])≏(({{0}, {cos α}, {-sin α}, {0}})).

ψ={0,Cos[α],-Sin[α],0};

There are two limits:

α=π/4 => cos α=sin α=(1/(Sqrt[2])) => (Ket[ψ])=(1/(Sqrt[2]))((Ket[01])-(Ket[10])) is the EPR state (maximally entangled).

α=0 => cos α=1 and sin α=0 => (Ket[ψ])=(Ket[01]) reduces to the product state (no entanglement).

Entanglement entropy

S((Subscript[(OverHat[ρ]), A]))=-(cos^2)α log (cos^2)α-(sin^2)α log (sin^2)α

ρ=ψ⊗ψ;
ρA=TensorContract[ArrayReshape[ρ,{2,2,2,2}],{{2,4}}];
SA=Total[-# Log[#]&/@Eigenvalues@ρA]

Plot[SA,{α,0,π/4},FrameLabel->{"\!\(α\)","\!\(S(ρ\&^\_A)\)"},PlotStyle->Black]

Observable expectation values

(Bra[ψ])(Subscript[(OverHat[σ]), A])(Ket[ψ])=(0,0,cos 2α),
(Bra[ψ])(Subscript[(OverHat[σ]), B])(Ket[ψ])=(0,0,-cos 2α),
(Bra[ψ])(Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B])(Ket[ψ])=(({{-sin 2α, 0, 0}, {0, -sin 2α, 0}, {0, 0, -1}})).

Table[ψ.KroneckerProduct[PauliMatrix[a],PauliMatrix[b]].ψ,{a,0,3},{b,0,3}]//FullSimplify//TableForm

If Alice measures (Subscript[n, 1])·(Subscript[σ, A]) and Bob measures (Subscript[n, 2])·(Subscript[σ, B]), the probability for their measurement outcomes to be opposite is given by

p((Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B]))=(Bra[ψ])(Subscript[(OverHat[𝒫]), (Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B])])(Ket[ψ]),

where the projection operator (Subscript[(OverHat[𝒫]), (Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B])]) is given by

(Subscript[(OverHat[𝒫]), (Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B])])=(∑_(s=±1))(Subscript[(OverHat[𝒫]), (Subscript[n, 1])·(Subscript[σ, A])=s])(Subscript[(OverHat[𝒫]), (Subscript[n, 2])·(Subscript[σ, B])=-s])
=(∑_(s=±1))((𝟙+s (Subscript[n, 1])·(Subscript[(OverHat[σ]), A]))/2)((𝟙-s (Subscript[n, 2])·(Subscript[(OverHat[σ]), B]))/2)
=((𝟙- (Subscript[n, 1])·((Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B]))·(Subscript[n, 2]))/2).

Therefore

p((Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B]))=((1- (Subscript[n, 1])·(Bra[ψ])(Subscript[(OverHat[σ]), A])⊗(Subscript[(OverHat[σ]), B])(Ket[ψ])·(Subscript[n, 2]))/2).

Suppose the unit vectors (Subscript[n, 1]), (Subscript[n, 2]), (Subscript[n, 3]) are still placed in the x z-plane with 120° to each other

(Subscript[n, 1])=(cos θ,0,sin θ),
(Subscript[n, 2])=(cos(θ+2π/3),0,sin(θ+2π/3)),
(Subscript[n, 2])=(cos(θ-2π/3),0,sin(θ-2π/3)).

Then the left-hand-side (l.h.s.) of the Bell inequality reads

p((Subscript[n, 1])·(Subscript[σ, A])=-(Subscript[n, 2])·(Subscript[σ, B]))+p((Subscript[n, 2])·(Subscript[σ, A])=-(Subscript[n, 3])·(Subscript[σ, B]))+p((Subscript[n, 3])·(Subscript[σ, A])=-(Subscript[n, 1])·(Subscript[σ, B]))
=(3/8)(3-sin 2α).

Simplify@Total[(1-#1.DiagonalMatrix[{-Sin[2α],-1}].#2)/2&@@@Thread@{#,RotateLeft[#]}&@CirclePoints[{1,θ},3]]

It turns out that this result is independent of the choice of θ.

Plot[3/8(3-Sin[2α]),{α,0,π/4},FrameLabel->{"\!\(α\)","Bell's l.h.s."},PlotStyle->Black,GridLines->{None,{1}}]

ParametricPlot[{SA/Log[2],3/8(3-Sin[2α])},{α,0,π/4},AspectRatio->1,PlotStyle->Black,PlotRangePadding->0.01,Prolog->{Opacity[0.1],{Red,Polygon[{{0,1},{1,1},{1,0},{0,0}}]},{Green,Polygon[{{0,1},{1,1},{1,2},{0,2}}]}},GridLines->{None,{1}},
FrameLabel->{"\!\(S(ρ\&^\_A)\) [bit]","Bell's l.h.s."},
Epilog->{{Red,Text["violate",{0.95,1},{1,1}]},{Green,Text["obey",{0.95,1},{1,-1}]}}]

For pure state, such as (Ket[ψ]) in the above example, entanglement entropy S((Subscript[(OverHat[ρ]), A]))>0 as long as α>0 ⇔ the state is entangled. But the Bell inequality is not always violated. => It is an entanglement witness.

For mixed state, entropy no longer provides a good measure of quantum entanglement. We had to rely on Bell inequalities and other entanglement witness.