130A Quantum Physics

Part 2. Algebraic Methods

Introduction

Functions are Vectors

Wave Function

A function maps inputs to outputs: it takes in an input, does some processing, and returns an output. For example, a complex function

({{ψ:, ℝ, ->, ℂ}, { , x, ↦, ψ(x)}})

A vector is also a mapping: it maps indices to values, like looking up a table. For example, a complex vector

({{ψ:, ℤ, ->, ℂ}, { , i, ↦, (Subscript[ψ, i])}})

a vector (Subscript[ψ, i]), we call it a state vector.

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

Statistical interpretation: the value of the state vector describes the probability to observe the state (Ket[ψ]) behaving like the state (Ket[i]),

p(i|ψ)=((|(Subscript[ψ, i])|)^2).

a function ψ(x), we call it a wave function.

(Ket[ψ])=∫dx ψ(x)(Ket[x]).

Statistical interpretation: the value of the wave function describes the probability density to observe the state (Ket[ψ]) behaving like the state (Ket[x]) (continuously labeled),

p(x|ψ)=((|ψ(x)|)^2).

Dirac Delta

∀f:∫dx δ(x)f(x)=f(0).

By a shift of variable, Eq. () implies

Show Eq. () based on Eq. ().

∫dx δ(x-x')f(x)=∫dy δ(y)f(x'+y)=f(x'+0)=f(x').

This behavior is similar to

(∑_i)(Subscript[δ, i j])(Subscript[v, i])=(Subscript[v, j]).

Intuitively, δ(x) describes a unit impulse located at x=0. It can be considered as the limit of Gaussian functions at σ->0

δ(x)=(lim_(σ->0))(1/((Sqrt[2π])σ))(e^(-((x^2)/(2(σ^2))))).

Manipulate[Plot[Quiet[Exp[-x^2/(2σ^2)]/(Sqrt[2π]σ),General::munfl],{x,-3,3},PlotRange->{0,10},PlotRangePadding->Scaled[0.05],Filling->Axis,PlotStyle->Directive[AbsoluteThickness[1.5],Black],FrameLabel->{"\!\( x\)","\!\(f(x)\)"}],{{σ,0.1},0.01,1,ImageSize->Small},Paneled->False,LabelStyle->"InlineFormula"]

Loosely speaking, it looks like

δ(x)=({{{, ({{0, if x≠0,}, {∞, if x=0  (ill–defined).}})}})

while the area under the curve remains unity as the limit is taken

∫dx δ(x)=1.

Position and Momentum

Discrete v.s. Continuous

Examples: position x of a quantum particle.

where

(OverHat[x]) denotes the position operator (a Hermitian operator corresponding to the position observable of the particle)

x∈ℝ is the position eigenvalue.

(Ket[x]) the corresponding position eigenstate (the quantum state that describe the particle at the position x).

The Hilbert space dimension is infinite. It is always helpful to think about the continuous eigen spectrum as the limit of an infinitely dense discrete spectrum.

({{ , discrete, →, continuous}, {orthonormal basis, (BraKet[i, j])=(Subscript[δ, i j]), →, (BraKet[x, x'])=δ(x-x')}, {resolution of identity, (∑_i)(Ket[i])(Bra[i])=𝟙, →, ∫dx(Ket[x])(Bra[x])=𝟙}, {state decomposition, (Ket[v])=(∑_i)(Subscript[v, i])(Ket[i]), →, (Ket[ψ])=∫dx ψ(x)(Ket[x])}, { , (Subscript[v, i])=(BraKet[i, v]),  , ψ(x)=(BraKet[x, ψ])}, {state normalization, (∑_i)((|(Subscript[v, i])|)^2)=1, →, ∫dx((|ψ(x)|)^2)=1}, {scalar product, (BraKet[u, v])=(∑_i)(BraKet[u, i])(BraKet[i, v])
=(∑_i)(Subscript[u, i]^( * ))(Subscript[v, i]), →, (BraKet[ϕ, ψ])=∫dx(BraKet[ϕ, x])(BraKet[x, ψ])
=∫dx (SuperStar[ϕ(x)])ψ(x)}})

Position Operator

The position operator is diagonal in its own eigen basis.

Suppose the particle is in a state (Ket[ψ])

(Ket[ψ])=∫dx ψ(x)(Ket[x]),

described by the wave function ψ(x).

Applying the position operator,

(OverHat[x])(Ket[ψ])=∫dx (x ψ(x))(Ket[x]).

Translation

Suppose the particle was in the (Ket[x]) state (at position x).

After applying the translation operator, the particle is in a new state (Ket[x+a]) (at position x+a).

Therefore (OverHat[T])(a) translates the particle by displacement a.

Translation operator implements a basis transformation (from (Ket[x]) to (Ket[x+a])). Every basis transformation is unitary. So the translation operator is unitary.

Use the definition Eq. () to show that 
(SuperDagger[(OverHat[T])(a)])(OverHat[T])(a)=(OverHat[T])(a)(SuperDagger[(OverHat[T])(a)])=𝟙, 
thus the translation operator is unitary.

(SuperDagger[(OverHat[T])(a)])=∫dx (Ket[x])(Bra[x+a]).

(SuperDagger[(OverHat[T])(a)])(OverHat[T])(a)=∫dx' (Ket[x'])(Bra[x'+a])∫dx (Ket[x+a])(Bra[x])
=∫dx'∫dx (Ket[x'])(BraKet[x'+a, x+a])(Bra[x])
=∫dx'∫dx (Ket[x'])δ(x'+a-x-a)(Bra[x])
=∫dx'∫dx (Ket[x'])δ(x'-x)(Bra[x])
=∫dx (Ket[x])(Bra[x])
=𝟙.

Momentum Operator

(OverHat[p])=i ℏ(∂_a)(OverHat[T])(a)(Subscript[|, a=0]),
=i ℏ (lim_(a→0))(((OverHat[T])(a)-(OverHat[T])(0))/a),

Suppose the particle is in a state (Ket[ψ])

(Ket[ψ])=∫dx ψ(x)(Ket[x]),

described by the wave function ψ(x).

Under translation,

(OverHat[T])(a)(Ket[ψ])=∫dx ψ(x)(Ket[x+a])
=∫dx ψ(x-a)(Ket[x]).

Applying the momentum operator,

(OverHat[p])(Ket[ψ])=i ℏ (lim_(a→0))(((OverHat[T])(a)(Ket[ψ])-(OverHat[T])(0)(Ket[ψ]))/a)
=i ℏ∫dx ( (lim_(a→0))((ψ(x-a)- ψ(x))/a))(Ket[x])
=∫dx ( -i ℏ(∂_x)ψ(x))(Ket[x]).

Show that Eq. () is consistent with Eq. () when acting on a state (Ket[ψ]).

(OverHat[p])(Ket[ψ])=-i ℏ∫dxdx'(Ket[x])(∂_x)δ(x-x')(Bra[x'])∫dx'' ψ(x'')(Ket[x''])
=-i ℏ∫dxdx'dx''(Ket[x])(∂_x)δ(x-x')ψ(x'')(BraKet[x', x''])
=-i ℏ∫dxdx'dx''(Ket[x])(∂_x)δ(x-x')ψ(x'')δ(x'-x'')
=-i ℏ∫dxdx'(Ket[x])(∂_x)δ(x-x')ψ(x')
=-i ℏ∫dx(Ket[x])(∂_x)ψ(x)
=∫dx (-i ℏ(∂_x)ψ(x))(Ket[x]).

Commutation Relation

(OverHat[x])≏x, (OverHat[p])≏-i ℏ(∂_x),

[(OverHat[x]),(OverHat[p])](Ket[ψ])≏[x,-i ℏ(∂_x)]ψ(x)
=-i ℏ(x(∂_x)-(∂_x)x)ψ(x)
=i ℏ ψ(x)
≏i ℏ(Ket[ψ]).

Harmonic Oscillator

Operator Algebra

Hamiltonian

(OverHat[H])=(1/(2m))((OverHat[p])^2)+(1/2)m (ω^2)((OverHat[x])^2).

[(OverHat[x]),(OverHat[p])]=i ℏ 𝟙.

(OverHat[p])->(OverHat[p])(Sqrt[ℏ m ω]), (OverHat[x])->(OverHat[x])(Sqrt[(ℏ/(m ω))]),

Show that the Hamiltonian in Eq. () gets rescaled to that in Eq. ().

(OverHat[H])=(1/(2m))((OverHat[p])^2)+(1/2)m (ω^2)((OverHat[x])^2)
->(1/(2m))((((OverHat[p])(Sqrt[ℏ m ω])))^2)+(1/2)m (((ω^2)((OverHat[x])(Sqrt[(ℏ/(m ω))])))^2)
=((ℏ m ω)/(2m))((OverHat[p])^2)+(1/2)m (ω^2) (ℏ/(m ω)) ((OverHat[x])^2)
=((ℏ ω)/2)((OverHat[p])^2)+((ℏ ω)/2) ((OverHat[x])^2)
=(1/2)ℏ ω (((OverHat[p])^2)+((OverHat[x])^2)).

Energy scale set by ℏ ω.

New operators (OverHat[x]) and (OverHat[p]) are dimensionless.

Concept of Bosons

(OverHat[H])(Ket[n])=(Subscript[E, n])(Ket[n]).

(Subscript[E, n])=ℏ ω (n+(1/2)),

({{ , oscillator, , boson}, { , state (Ket[n]), (Subscript[E, n])/ℏ ω, }, {ground state, (Ket[0]), 1/2, vacuum (0 bosons)}, {excited states, (Ket[1]), 3/2, 1 boson}, {⋮, (Ket[2]), 5/2, 2bosons}, {⋮, ⋮, ⋮, ⋮}})

Levels are equally spaced: oscillator can only absorb/emit energy in integer units of ℏ ω.

Each unit of energy (energy quantum) is carried by a boson --- an elementary excitation of the oscillator.

For mechanical oscillation (sound), the boson is called phonon.

For electromagnetic oscillation (light), the boson is called photon.

Bosons are indistinguishable particles, because each of them carries the same amount of energy (and nothing else to differentiate them).

created by the operator (SuperDagger[(OverHat[a])]): (Ket[0])->(Ket[1])->(Ket[2])->…,

annihilated by the operator (OverHat[a]): …->(Ket[2])->(Ket[1])->(Ket[0]).

Annihilation and Creation Operators

(OverHat[a]) and (SuperDagger[(OverHat[a])]) are Hermitian conjugate to each other.

Analogy: complex numbers z=x+i y, (SuperStar[z])=x-i y 
=> position (OverHat[x]) ∼ real part of (OverHat[a]), momentum (OverHat[p]) ∼ imaginary part of (OverHat[a]).

Prove Eq. () from Eq. ().

[(OverHat[a]),(SuperDagger[(OverHat[a])])]=[(1/(Sqrt[2]))((OverHat[x])+i (OverHat[p])),(1/(Sqrt[2]))((OverHat[x])-i (OverHat[p]))]
=(1/2)[(OverHat[x])+i (OverHat[p]),(OverHat[x])-i (OverHat[p])]
=(1/2)([(OverHat[x]),-i (OverHat[p])]+[i (OverHat[p]),(OverHat[x])])
=-i [(OverHat[x]), (OverHat[p])]
=(-i)i 𝟙
=𝟙.

(OverHat[a]) (SuperDagger[(OverHat[a])])=(SuperDagger[(OverHat[a])])(OverHat[a])+𝟙,
(SuperDagger[(OverHat[a])])(OverHat[a])=(OverHat[a]) (SuperDagger[(OverHat[a])])-𝟙.

Number Operator

(OverHat[n])=(1/2)(((OverHat[p])^2)+((OverHat[x])^2))-(𝟙/2).

Verify Eq. () using Eq. ().

(OverHat[n])=(1/2)((OverHat[x])-i (OverHat[p]))((OverHat[x])+i (OverHat[p]))
=(1/2)(((OverHat[x])^2)+i (OverHat[x])(OverHat[p])-i (OverHat[p])(OverHat[x])+((OverHat[p])^2))
=(1/2)(((OverHat[x])^2)+i [(OverHat[x]),(OverHat[p])]+((OverHat[p])^2))
=(1/2)(((OverHat[x])^2)-𝟙+((OverHat[p])^2)).

(OverHat[H])(Ket[n])=ℏ ω (n+(1/2))(Ket[n]),

(Subscript[E, n])=ℏ ω (n+(1/2)).

Why must the eigenvalue n take quantized (discrete integer) values?

What are the eigenstates (Ket[n])? (What are the interrelations between the states (Ket[n]) under the action of creation/annihilation operators?)

Algebraic Relations

Commute (OverHat[a]) or (SuperDagger[(OverHat[a])])through (OverHat[n]):

Prove Eq. () using Eq. ().

(OverHat[n])(OverHat[a])(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])])(OverHat[a])(OverHat[a])
(Overscript[=, Eq. ()])((OverHat[a]) (SuperDagger[(OverHat[a])])-𝟙)(OverHat[a])
=(OverHat[a]) (SuperDagger[(OverHat[a])])(OverHat[a])-(OverHat[a])
(Overscript[=, Eq. ()])(OverHat[a]) (OverHat[n])-(OverHat[a])
=(OverHat[a]) ((OverHat[n])-𝟙).

(OverHat[n])(SuperDagger[(OverHat[a])])(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])])(OverHat[a])(SuperDagger[(OverHat[a])])
(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])])((SuperDagger[(OverHat[a])])(OverHat[a])+𝟙)
(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])]) ((OverHat[n])+𝟙).

Physical meaning: suppose (OverHat[n]) counts the number of bosons, counting it before or after the boson annihilation/creation process will result in a ±1 difference in the boson number results.

Consider any operator function of (OverHat[n])

f((OverHat[n])):=(Subscript[c, 0])𝟙+(Subscript[c, 1])(OverHat[n])+(Subscript[c, 2])((OverHat[n])^2)+…=(∑_(k=0)^∞)(Subscript[c, k])((OverHat[n])^k),

conjugating it as □->(SuperDagger[(OverHat[a])])□(OverHat[a]) shifts the operator (OverHat[n]) by 𝟙,

(SuperDagger[(OverHat[a])])f((OverHat[n]))(OverHat[a])=(OverHat[n])f((OverHat[n])-𝟙).

Show Eq. () based on Eq. ().

(SuperDagger[(OverHat[a])])((OverHat[n])^k)(OverHat[a])=(OverHat[n]) ((((OverHat[n])-𝟙))^k)

Base case (k=0): Eq. () reduces to

(SuperDagger[(OverHat[a])])(OverHat[a])=(OverHat[n]),

which is true by the definition of number operator.

Suppose Eq. () is true for some natural number k, we can show that it is also true for k+1:

(SuperDagger[(OverHat[a])])((OverHat[n])^(k+1))(OverHat[a])=(SuperDagger[(OverHat[a])])((OverHat[n])^k)(OverHat[n])(OverHat[a])
(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])])((OverHat[n])^k)(OverHat[a]) ((OverHat[n])-𝟙)
(Overscript[=, Eq. () for k])(OverHat[n]) ((((OverHat[n])-𝟙))^k)((OverHat[n])-𝟙)
=(OverHat[n]) ((((OverHat[n])-𝟙))^(k+1)).

(SuperDagger[(OverHat[a])])f((OverHat[n]))(OverHat[a])(Overscript[=, Eq. ()])(SuperDagger[(OverHat[a])])((∑_(k=0)^∞)(Subscript[c, k])((OverHat[n])^k))(OverHat[a])
=(∑_(k=0)^∞)(Subscript[c, k])(SuperDagger[(OverHat[a])])((OverHat[n])^k)(OverHat[a])
(Overscript[=, Eq. ()])(∑_(k=0)^∞)(Subscript[c, k])(OverHat[n]) ((((OverHat[n])-𝟙))^k)
=(OverHat[n]) (∑_(k=0)^∞)(Subscript[c, k]) ((((OverHat[n])-𝟙))^k)
(Overscript[=, Eq. ()])(OverHat[n]) f((OverHat[n])-𝟙).

Repeatedly applying Eq. (), we can show that

(SuperDagger[(OverHat[a])])(OverHat[a])=(OverHat[n]),
((((SuperDagger[(OverHat[a])])))^2)((OverHat[a])^2)=(OverHat[n]) ((OverHat[n])-𝟙),
((((SuperDagger[(OverHat[a])])))^3)((OverHat[a])^3)=(OverHat[n]) ((OverHat[n])-𝟙)((OverHat[n])-2𝟙),
…

from which we conclude (for m∈ℕ)

Quantum Bootstrap

General Principle

(OverHat[n])(Ket[n])=n(Ket[n]).

∀(OverHat[O]):(Bra[n])(SuperDagger[(OverHat[O])])(OverHat[O])(Ket[n])≥0.

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

Level Quantization

(Subscript[(OverHat[O]), m])=((OverHat[a])^m),

Substitute Eq. () into Eq. (), the positivity constraint requires

∀m:(((Bra[n])((SuperDagger[(OverHat[a])])))^m)((OverHat[a])^m)(Ket[n])≥0.

On the other hand,

(((Bra[n])((SuperDagger[(OverHat[a])])))^m)((OverHat[a])^m)(Ket[n])
(Overscript[=, Eq.()])(Bra[n])(∏_(l=0)^(m-1))((OverHat[n])-l 𝟙)(Ket[n])
(Overscript[=, Eq. ()])(∏_(l=0)^(m-1))(n-l)=((n!)/((n-m)!)).

m=0:1>=0,
m=1:n≥0,
m=2:n (n-1)≥0,
m=3:n (n-1)(n-2)≥0,
m=4:n (n-1)(n-2)(n-3)≥0,
…

Manipulate[Plot[n!/(n-m)!,{n,-1,5},Epilog->{Green,{AbsoluteThickness[3],HalfLine[{#,0}&/@{m-1,m}]},{AbsolutePointSize[8],Point[{#,0}]&/@Range[0,m-1]}},PlotStyle->Directive[AbsoluteThickness[1.5],Black],PlotRange->2,GridLines->{Range[-1,5],None},Filling->{1->{0,Lighter[#,0.8]&/@{Red,Green}}},FrameLabel->{"\!\( n\)","(Bra[n])((SuperDagger[(OverHat[a])]))^m((OverHat[a])^m)(Ket[n])"},PlotLabel->StringTemplate["\!\(m=`1`\)"][m],ImageSize->200],{{m,3},Range[0,6],ControlType->SetterBar},Paneled->False,LabelStyle->"InlineFormula"]

The eigenvalues n=0,1,2,… are discrete! For this reason, the operator (OverHat[n]) is called the number operator, which counts the number of bosons (elementary excitations).

The n=0 eigenstate, denoted as (Ket[0]), is also called the vacuum state, as it describes a state with no excitations. It is also the ground state of the Hamiltonian (OverHat[H]).

The eigenstates (Ket[n]) for n=1,2,… are excited states of the Hamiltonian (OverHat[H]), describing the states of n bosons occupying the oscillator mode (a mode = a slot that can host particles).

Fock Basis Representation

ℬ={(Ket[n])|n∈ℕ},

Applying the operator equality Eq. (), we have

(OverHat[n])(OverHat[a])(Ket[n])=(OverHat[a]) ((OverHat[n])-𝟙)(Ket[n])=(n-1)(OverHat[a])(Ket[n]),
(OverHat[n])(SuperDagger[(OverHat[a])])(Ket[n])=(SuperDagger[(OverHat[a])]) ((OverHat[n])+𝟙)(Ket[n])=(n+1)(SuperDagger[(OverHat[a])])(Ket[n]),

meaning that (OverHat[a])(Ket[n]) is an (unnormalized) eigenstate of (OverHat[n]) with eigenvalue (n-1), and (SuperDagger[(OverHat[a])])(Ket[n]) is an (unnormalized) eigenstate of (OverHat[n]) with eigenvalue (n+1).

Since we have already introduced (Ket[n±1]) to denote the normalized eigenstates corresponding to the n±1 eigenvalues, we must conclude that

(OverHat[a])(Ket[n])∝(Ket[n-1]),
(SuperDagger[(OverHat[a])])(Ket[n])∝(Ket[n+1]).

To determine the proportionality constant, we can compute the squared norms

(Bra[n])(SuperDagger[(OverHat[a])])(OverHat[a])(Ket[n])=(Bra[n])(OverHat[n])(Ket[n])=n,
(Bra[n])(OverHat[a])(SuperDagger[(OverHat[a])])(Ket[n])=(Bra[n])((SuperDagger[(OverHat[a])])(OverHat[a])+𝟙)(Ket[n])=(Bra[n])((OverHat[n])+𝟙)(Ket[n])=n+1.

Prove Eq. () using Eq. ().

Base case (n=0): Eq. () reduces to

(Ket[0])=(Ket[0]),

which holds trivially.

Suppose Eq. () holds for some non-negative integer n, we aim to show that it holds for n+1:

(Ket[n+1])(Overscript[=, Eq. ()])(((SuperDagger[(OverHat[a])])(Ket[n]))/(Sqrt[n+1]))
(Overscript[=, Eq. () for n])((SuperDagger[(OverHat[a])])/(Sqrt[n+1]))(1/(Sqrt[n!]))((((SuperDagger[a])))^n)(Ket[0])
=(1/(Sqrt[(n+1)!]))((((SuperDagger[a])))^(n+1))(Ket[0]).

Summary

Annihilation and creation operators

({{{, ({{(OverHat[a])=(1/(Sqrt[2]))((OverHat[x])+i (OverHat[p]))}, {(SuperDagger[(OverHat[a])])=(1/(Sqrt[2]))((OverHat[x])-i (OverHat[p]))}})}}), ({{{, ({{(OverHat[x])=(1/(Sqrt[2]))((OverHat[a])+(SuperDagger[(OverHat[a])]))}, {(OverHat[p])=(1/((Sqrt[2])i))((OverHat[a])-(SuperDagger[(OverHat[a])]))}})}}).

They satisfies the commutation relation

[(OverHat[x]),(OverHat[p])]=i 𝟙⇔[(OverHat[a]),(SuperDagger[(OverHat[a])])]=𝟙.

Number operator

(OverHat[n])=(SuperDagger[(OverHat[a])])(OverHat[a]).

It defines a discrete spectrum (OverHat[n])(Ket[n])=n(Ket[n]) for n∈ℕ. Such that

(OverHat[a])(Ket[n])=(Sqrt[n])(Ket[n-1]),
(SuperDagger[(OverHat[a])])(Ket[n])=(Sqrt[n+1])(Ket[n+1]).

Hamiltonian

(OverHat[H])=(1/2)ℏ ω (((OverHat[p])^2)+((OverHat[x])^2))=ℏ ω ((OverHat[n])+(𝟙/2)).

Eigen energies

(Subscript[E, n])=ℏ ω(n+(1/2)).

Every eigenstate (Ket[n]) can be raised from the ground state by

(Ket[n])=(1/(Sqrt[n!]))((((SuperDagger[(OverHat[a])])))^n)(Ket[0]).

Representation Theory

Fock State Representations

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

Correspondingly, the creation and annihilation operators are represented as matrices

(Subscript[a, m n])=(Bra[m])(OverHat[a])(Ket[n])=(Sqrt[n])(BraKet[m, n-1])=(Sqrt[n])(Subscript[δ, m,n-1]),(Subscript[((SuperDagger[a])), m n])=(Bra[m])(SuperDagger[(OverHat[a])])(Ket[n])=(Sqrt[n+1])(BraKet[m, n+1])=(Sqrt[n+1])(Subscript[δ, m,n+1]),

or more explicitly as

(OverHat[a])≏(({{0, (Sqrt[1]),  ,  }, { , 0, (Sqrt[2]),  }, { ,  , 0, ⋱}, { ,  ,  , ⋱}})),(SuperDagger[(OverHat[a])])≏(({{0,  ,  ,  }, {(Sqrt[1]), 0,  ,  }, { , (Sqrt[2]), 0,  }, { ,  , ⋱, ⋱}})).

By matrix multiplication, one can verify that (SuperDagger[(OverHat[a])])(OverHat[a]) indeed acts as the number operator (OverHat[n])

(OverHat[n])=(SuperDagger[(OverHat[a])])(OverHat[a])≏(({{0,  ,  ,  }, { , 1,  ,  }, { ,  , 2,  }, { ,  ,  , ⋱}})),

which is diagonal in its own eigen basis (i.e. the Fock basis).

The coordinate and momentum operators are represented as

(OverHat[x])=(((OverHat[a])+(SuperDagger[(OverHat[a])]))/(Sqrt[2]))≏(1/(Sqrt[2]))(({{0, (Sqrt[1]),  ,  }, {(Sqrt[1]), 0, (Sqrt[2]),  }, { , (Sqrt[2]), 0, ⋱}, { ,  , ⋱, ⋱}})),
(OverHat[p])=(((OverHat[a])-(SuperDagger[(OverHat[a])]))/((Sqrt[2])i))≏(1/((Sqrt[2])i))(({{0, (Sqrt[1]),  ,  }, {-(Sqrt[1]), 0, (Sqrt[2]),  }, { , -(Sqrt[2]), 0, ⋱}, { ,  , ⋱, ⋱}})).

To have some fun, let us check

(OverHat[x]) (OverHat[p])≏(i/2)(({{1, 0, -(Sqrt[2]),  ,  }, {0, 1, 0, -(Sqrt[6]),  }, {(Sqrt[2]), 0, 1, 0, ⋱}, { , (Sqrt[6]), 0, 1, ⋱}, { ,  , ⋱, ⋱, ⋱}})),
(OverHat[p]) (OverHat[x])≏(i/2)(({{-1, 0, -(Sqrt[2]),  ,  }, {0, -1, 0, -(Sqrt[6]),  }, {(Sqrt[2]), 0, -1, 0, ⋱}, { , (Sqrt[6]), 0, -1, ⋱}, { ,  , ⋱, ⋱, ⋱}})),

Check the matrix multiplication in Eq. ().

Manipulate[Block[{a,ad,x,p},
a=Normal@SparseArray[{Band[{1,2}]->Sqrt[Range[n]]},{n+1,n+1}];
ad=Transpose[a];
x=(a+ad)/Sqrt[2];
p=(a-ad)/(Sqrt[2]I);Style[Column[Prepend[Row[{StringTemplate["\!\(`1` = I\/2\)"][#1],#2}]&@@@Thread[{{"x\&^ p\&^","p\&^ x\&^"},MatrixForm/@{x.p/(I/2),p.x/(I/2)}}],StringTemplate["\!\(n\_cut=`1`\)"][n]]],"InlineFormula"]],{{n,3},Range[1,6],ControlType->SetterBar},Paneled->False,LabelStyle->"InlineFormula"]

so we indeed verify the commutation relation [(OverHat[x]),(OverHat[p])]=i 𝟙, as

[(OverHat[x]),(OverHat[p])]≏(({{i,  ,  ,  }, { , i,  ,  }, { ,  , i,  }, { ,  ,  , ⋱}}))=i 𝟙.

Uncertainty Relation

(std x) (std p)≥(1/2)|[(OverHat[x]),(OverHat[p])]|=(1/2),

(OverHat[x])=(((OverHat[a])+(SuperDagger[(OverHat[a])]))/(Sqrt[2])), (OverHat[p])=(((OverHat[a])-(SuperDagger[(OverHat[a])]))/((Sqrt[2])i)).

(Bra[n])(OverHat[x])(Ket[n])=(1/(Sqrt[2]))(Bra[n])(OverHat[a])+(SuperDagger[(OverHat[a])])(Ket[n])
=(1/(Sqrt[2]))((Sqrt[n])(BraKet[n, n-1])+(Sqrt[n+1])(BraKet[n, n+1]))
=0,

(Bra[n])((OverHat[x])^2)(Ket[n])=(1/2)(((Bra[n])((OverHat[a])+(SuperDagger[(OverHat[a])])))^2)(Ket[n])
=(1/2)((Bra[n])(OverHat[a]) (OverHat[a])(Ket[n])+(Bra[n])(OverHat[a]) (SuperDagger[(OverHat[a])])(Ket[n])+(Bra[n])(SuperDagger[(OverHat[a])]) (OverHat[a])(Ket[n])+(Bra[n])(SuperDagger[(OverHat[a])]) (SuperDagger[(OverHat[a])])(Ket[n]))
=(1/2)(0+(n+1)+n+0)
=n+1/2.

(((std x))^2)=(Bra[n])((OverHat[x])^2)(Ket[n])-(Bra[n])(OverHat[x])((Ket[n])^2)=n+1/2.

(((std p))^2)=(Bra[n])((OverHat[p])^2)(Ket[n])-(Bra[n])(OverHat[p])((Ket[n])^2)=n+1/2.

(std x) (std p)=n+1/2≥1/2.

⟨(x^2)⟩⟨(p^2)⟩≥(1/4)((|[(OverHat[x]),(OverHat[p])]|)^2)=(1/4),

E=⟨H⟩=((ℏ ω)/2)(⟨(p^2)⟩+⟨(x^2)⟩)≥((ℏ ω)/2)(⟨(p^2)⟩+(1/(4⟨(p^2)⟩)))≥((ℏ ω)/2).

Wave Function Representation

(OverHat[a])(Ket[0])=0

((OverHat[x])+i (OverHat[p]))(Ket[0])=0.

(Ket[0])≏(Subscript[ψ, 0])(x) the (coordinate space) wave function corresponds to the state (Ket[0]),

(OverHat[x])≏x when acting on the wave function in the coordinate space,

(OverHat[p])≏-i(∂_x) when acting on the wave function in the coordinate space.

(x+(∂_x))(Subscript[ψ, 0])(x)=0,

(Subscript[ψ, 0])(x)=(1/(π^(1/4)))(e^(-(1/2)(x^2))).

Show that Eq. () is a normalized solution of the differential equation Eq. ().

DSolve[x ψ0[x]+ψ0'[x]==0,ψ0[x],x]

Integrate[(Exp[-x^2/2]/π^(1/4))^2,{x,-∞,∞}]

(Subscript[ψ, 1])(x)=(1/(Sqrt[2]))(x-(∂_x))(Subscript[ψ, 0])(x)=(1/(π^(1/4)))(Sqrt[2])x (e^(-(1/2)(x^2))).

Check Eq. () by applying the differential operator.

creation=(x#-D[#,x])/Sqrt[2]&;

creation[Exp[-x^2/2]/π^(1/4)]

Nest[creation,Exp[-x^2/2]/π^(1/4),3]//Simplify

(Subscript[ψ, n])(x)=(1/((π^(1/4))(Sqrt[(2^n)n!])))(Subscript[H, n])(x)(e^(-(1/2)(x^2))),

Style[TableForm[Table[Factor@{n,Nest[(x#-D[#,x])/Sqrt[2]&,Exp[-x^2/2]/π^(1/4),n]/Sqrt[n!],HermiteH[n,x],Plot[HermiteH[n,x]Exp[-x^2/2]/π^(1/4)/Sqrt[2^n n!],{x,-6,6},Ticks->None,Frame->None,PlotStyle->Black,PlotRange->{-1,1},AspectRatio->1/3,ImageSize->80]},{n,0,5}],TableHeadings->{None,{"\!\( n\)","\!\(ψ\_n(x)\)","\!\(H\_n(x)\)","plot of \!\(ψ\_n(x)\)"}}],"InlineFormula"]

Angular Momentum

Operator Algebra

Definition

(Subscript[ϵ, a b c]) is the Levi-Civita symbol: the sign of the a b c permutation (e.g. (Subscript[ϵ, 123])=+1, (Subscript[ϵ, 321])=-1).

Repeated indices are automatically contracted in Eq. ().

Equivalently, in vector form, (OverHat[J])×(OverHat[J])=i (OverHat[J]).

Orbital angular momentum of a particle.

(OverHat[x])=((Subscript[(OverHat[x]), 1]),(Subscript[(OverHat[x]), 2]),(Subscript[(OverHat[x]), 3])) and (OverHat[p])=((Subscript[(OverHat[p]), 1]),(Subscript[(OverHat[p]), 2]),(Subscript[(OverHat[p]), 3])) are position and momentum operators in 3D space.

In component form, (Subscript[(OverHat[L]), a])=(Subscript[ϵ, a b c])(Subscript[(OverHat[x]), b])(Subscript[(OverHat[p]), c]).

From [(Subscript[(OverHat[x]), a]),(Subscript[(OverHat[p]), b])]=i (Subscript[δ, a b])𝟙 (set ℏ=1 for simplicity), one can verify that

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]).

Verify Eq. ().

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=[(Subscript[ϵ, a c d])(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), d]),(Subscript[ϵ, b e f])(Subscript[(OverHat[x]), e])(Subscript[(OverHat[p]), f])]
=(Subscript[ϵ, a c d])(Subscript[ϵ, b e f])[(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), d]),(Subscript[(OverHat[x]), e])(Subscript[(OverHat[p]), f])]
=(Subscript[ϵ, a c d])(Subscript[ϵ, b e f])((Subscript[(OverHat[x]), e])[(Subscript[(OverHat[x]), c]),(Subscript[(OverHat[p]), f])](Subscript[(OverHat[p]), d])+(Subscript[(OverHat[x]), c])[(Subscript[(OverHat[p]), d]),(Subscript[(OverHat[x]), e])](Subscript[(OverHat[p]), f]))
=(Subscript[ϵ, a c d])(Subscript[ϵ, b e f])(i (Subscript[δ, c f])(Subscript[(OverHat[x]), e])(Subscript[(OverHat[p]), d])-i (Subscript[δ, d e])(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), f]))
=i ((Subscript[ϵ, a c d])(Subscript[ϵ, b e c])(Subscript[(OverHat[x]), e])(Subscript[(OverHat[p]), d])-(Subscript[ϵ, a c d])(Subscript[ϵ, b d f])(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), f]))
=i (((Subscript[δ, a e])(Subscript[δ, b d])-(Subscript[δ, a b])(Subscript[δ, d e]))(Subscript[(OverHat[x]), e])(Subscript[(OverHat[p]), d])-((Subscript[δ, a f])(Subscript[δ, b c])-(Subscript[δ, a b])(Subscript[δ, c f]))(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), f]))
=i ((Subscript[(OverHat[x]), a])(Subscript[(OverHat[p]), b])-(Subscript[δ, a b])(Subscript[(OverHat[x]), d])(Subscript[(OverHat[p]), d])-(Subscript[(OverHat[x]), b])(Subscript[(OverHat[p]), a])+(Subscript[δ, a b])(Subscript[(OverHat[x]), c])(Subscript[(OverHat[p]), c]))
=i ((Subscript[(OverHat[x]), a])(Subscript[(OverHat[p]), b])-(Subscript[(OverHat[x]), b])(Subscript[(OverHat[p]), a]))
=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]).

Spin angular momentum of a qubit.

(OverHat[σ])=(((OverHat[σ])^x),((OverHat[σ])^y),((OverHat[σ])^z)) are the Pauli matrices.

The commutation relation of Pauli matrices implies

[(Subscript[(OverHat[S]), a]),(Subscript[(OverHat[S]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[S]), c]).

Casimir Operator

((OverHat[J])^2) is Hermitian.

By Eq. (), one can verify that (for a=1,2,3)

[((OverHat[J])^2),(Subscript[(OverHat[J]), a])]=0.

Prove Eq. ().

[((OverHat[J])^2),(Subscript[(OverHat[J]), 3])]=[(Subscript[(OverHat[J]), 1]^2),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 2]^2),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 3]^2),(Subscript[(OverHat[J]), 3])]

[(Subscript[(OverHat[J]), 1]^2),(Subscript[(OverHat[J]), 3])]=(Subscript[(OverHat[J]), 1])[(Subscript[(OverHat[J]), 1]),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 1]),(Subscript[(OverHat[J]), 3])](Subscript[(OverHat[J]), 1])
=-i (Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])-i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]),

[(Subscript[(OverHat[J]), 2]^2),(Subscript[(OverHat[J]), 3])]=(Subscript[(OverHat[J]), 2])[(Subscript[(OverHat[J]), 2]),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 2]),(Subscript[(OverHat[J]), 3])](Subscript[(OverHat[J]), 2])
=i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1])+i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]),

[(Subscript[(OverHat[J]), 3]^2),(Subscript[(OverHat[J]), 3])]=0.

Raising and Lowering Operators

In analogy to (e^(±i θ))=cos θ±i sin θ.

(Subscript[(OverHat[J]), ±]) are not Hermitian. Under Hermitian conjugate: (Subscript[(OverHat[J]), ±]^†)=(Subscript[(OverHat[J]), ∓]).

Prove Eq. ().

(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])=((Subscript[(OverHat[J]), 1])∓i (Subscript[(OverHat[J]), 2]))((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))
=(Subscript[(OverHat[J]), 1]^2)∓i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])+(Subscript[(OverHat[J]), 2]^2)
=((Subscript[(OverHat[J]), 1]^2)+(Subscript[(OverHat[J]), 2]^2))±i ((Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])-(Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]))
=(((OverHat[J])^2)-(Subscript[(OverHat[J]), 3]^2))∓(Subscript[(OverHat[J]), 3])
=((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±𝟙).

Algebraic Relations

Commute (Subscript[(OverHat[J]), ±]) through ((OverHat[J])^2) and (Subscript[(OverHat[J]), 3])

Prove Eq. ().

[((OverHat[J])^2),(Subscript[(OverHat[J]), ±])]=0,

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±])=(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))
=(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), 2])
=((Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 3])+i (Subscript[(OverHat[J]), 2]))±i ((Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 3])-i (Subscript[(OverHat[J]), 1]))
=(Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 3])+i (Subscript[(OverHat[J]), 2])±i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 3])± (Subscript[(OverHat[J]), 1])
=(Subscript[(OverHat[J]), 1])((Subscript[(OverHat[J]), 3])±𝟙)±i (Subscript[(OverHat[J]), 2])((Subscript[(OverHat[J]), 3])±𝟙)
=((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))((Subscript[(OverHat[J]), 3])±𝟙)
=(Subscript[(OverHat[J]), ±])((Subscript[(OverHat[J]), 3])±𝟙).

Physical meaning: (Subscript[(OverHat[J]), 3]) corresponds to the angular momentum in the z direction, associated to rotations in the x y-plane, while (Subscript[(OverHat[J]), ±]) spins up/down the x y-plane rotation. Observing (Subscript[(OverHat[J]), 3]) before or after (Subscript[(OverHat[J]), ±]) will result in a ±1 difference. However, ((OverHat[J])^2) commutes with (Subscript[(OverHat[J]), ±]), so it does not care whether (Subscript[(OverHat[J]), ±]) is applied before or after, a consequence of the total angular momentum conservation.

Consider any operator function f of ((OverHat[J])^2) and (Subscript[(OverHat[J]), 3]), conjugating by (Subscript[(OverHat[J]), ±]) shifts only the (Subscript[(OverHat[J]), 3]) operator by ±𝟙,

(Subscript[(OverHat[J]), ∓])f(((OverHat[J])^2),(Subscript[(OverHat[J]), 3]))(Subscript[(OverHat[J]), ±])=(((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±𝟙))f(((OverHat[J])^2),(Subscript[(OverHat[J]), 3])±𝟙),

which can be shown by combining Eq. () and Eq. ().

Repeatedly applying Eq. (), we can show that

Prove Eq. ().

Trivial case (l=0): Eq. () reduces to 𝟙=𝟙, obviously true.

Base case (l=1): Eq. () reduces to

(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])=((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±𝟙),

which holds by Eq. ().

Suppose Eq. () holds for some positive integer l, we aim to show that it holds for l+1

(Subscript[(OverHat[J]), ∓]^(l+1))(Subscript[(OverHat[J]), ±]^(l+1))=(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Subscript[(OverHat[J]), ±])
(Overscript[=, Eq. () for l])(Subscript[(OverHat[J]), ∓])(∏_(k=0)^(l-1))(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±k 𝟙)((Subscript[(OverHat[J]), 3])±(k+1)𝟙))(Subscript[(OverHat[J]), ±])
(Overscript[=, Eq. ()])(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(∏_(k=0)^(l-1))(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±(k+1) 𝟙)((Subscript[(OverHat[J]), 3])±(k+2)𝟙))
=(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(∏_(k=1)^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±k 𝟙)((Subscript[(OverHat[J]), 3])±(k+1)𝟙))
(Overscript[=, Eq. ()])(((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±𝟙))(∏_(k=1)^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±k 𝟙)((Subscript[(OverHat[J]), 3])±(k+1)𝟙))
=(∏_(k=0)^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±k 𝟙)((Subscript[(OverHat[J]), 3])±(k+1)𝟙)).

Quantum Bootstrap

Problem Setup

((OverHat[J])^2)(Ket[j,m])=(Subscript[λ, j])(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=(Subscript[λ, m])(Ket[j,m]),

(Subscript[λ, j]) is the the eigenvalue of ((OverHat[J])^2) of the (Ket[j,m]) state,

(Subscript[λ, m]) is the the eigenvalue of (Subscript[(OverHat[J]), 3]) of the (Ket[j,m]) state.

∀O:(Bra[j,m])(SuperDagger[(OverHat[O])]) (OverHat[O])(Ket[j,m])≥0.

Angular Momentum Quantization

(Subscript[O, l])=(Subscript[(OverHat[J]), ±]^l),

Substitute Eq. () into Eq. (), the positivity constraint requires

∀l:(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Ket[j,m])≥0.

On the other hand,

(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Ket[j,m])
(Overscript[=, Eq. ()])(Bra[j,m])(∏_(k=0)^(l-1))(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±k 𝟙)((Subscript[(OverHat[J]), 3])±(k+1)𝟙))(Ket[j,m])
(Overscript[=, Eq. ()])(∏_(k=0)^(l-1))((Subscript[λ, j])-((Subscript[λ, m])±k)((Subscript[λ, m])±(k+1))).

l=0:1>=0,
l=1:(Subscript[λ, j])-(Subscript[λ, m])((Subscript[λ, m])±1)≥0,
l=2:((Subscript[λ, j])-(Subscript[λ, m])((Subscript[λ, m])±1))((Subscript[λ, j])-((Subscript[λ, m])±1)((Subscript[λ, m])±2))≥0,
…

Manipulate[Block[{pltrng={{-3,3},{0,14}},region},
SetAttributes[region,Listable];
region[λj==y_,λm==x_]:=Point[{x,y}];
region[Inequality[y1_,LessEqual,λj,Less,y2_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,y2,0.1}]],Point[{x/.λj->y1,y1}]};
region[Inequality[y1_,Less,λj,Less,y2_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,y2,0.1}]]};
region[Inequality[λj,Greater,y1_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,Max[Last@pltrng],0.1}]]};
region[Inequality[λj,Greater,y1_],Inequality[x1_,LessEqual,λm,LessEqual,x2_]]:=Block[{pts1,pts2},
{pts1,pts2}=Table[{#/.λj->y,y},{y,y1,Max[Last@pltrng],0.1}]&/@{x1,x2};
{{Opacity[0.2],Polygon[Join[pts1,Reverse[pts2]]]},
Line[pts1],Line[pts2]}];
Graphics[{AbsolutePointSize[5],ReplaceRepeated[Reduce[And@@Flatten@Table[{Product[λj-(λm-k)(λm-(k+1)),{k,0,l-1}]>=0,Product[λj-(λm+k)(λm+(k+1)),{k,0,l-1}]>=0},{l,1,lmax}],{λj,λm}],{And->region,Or->List}]},
GridLines->{Range[-3,3,1/2],#(#+1)&/@Range[0,3,1/2]},Frame->True,FrameLabel->{"\!\(λ\_m\)","\!\(λ\_j\)"},PlotRange->pltrng,PlotRangeClipping->True,PlotRangePadding->{{0.2,0.2},{0.5,0}},PlotLabel->Style[StringTemplate["\!\(l\_max=`1`\)"][lmax],14,Black],AspectRatio->1,ImageSize->180]],{{lmax,3},Range[1,8,1],ImageSize->Small},Paneled->False]

For orbital angular momentum j takes integer values. 
For spin angular momentum j can also be half-integers.

The eigen equations in Eq. () become

((OverHat[J])^2)(Ket[j,m])=j(j+1)(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=m(Ket[j,m]).

Operator Representation

Applying the operator equality Eq. (), we have

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Subscript[(OverHat[J]), ±])((Subscript[(OverHat[J]), 3])±𝟙)(Ket[j,m])
=(m±1)(Subscript[(OverHat[J]), ±])(Ket[j,m])

=> the state (Subscript[(OverHat[J]), ±])(Ket[j,m]) (as long as it is not zero) is also an eigenstate of (Subscript[(OverHat[J]), 3]) but with the eigenvalue (m±1) => (Subscript[(OverHat[J]), ±])(Ket[j,m]) is just the (Ket[j,m±1]) state (up to overall coefficient)

(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Subscript[c, j,m]^±)(Ket[j,m±1]).

To determine the coefficient (Subscript[c, j,m]^±), use Eq. (),

(Bra[j,m])(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(Ket[j,m])=j(j+1)-m(m±1).

On the other hand

(Bra[j,m])(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(Ket[j,m])=((((Subscript[c, j,m]^±)))^2)(BraKet[j,m±1, j,m±1])=((((Subscript[c, j,m]^±)))^2).

Combining Eq. () and Eq. (), (Subscript[c, j,m]^±) can be solved

(Subscript[c, j,m]^±)=(Sqrt[j(j+1)-m(m±1)]).

either raised from the lowest weight state,

(Ket[j,m])=((((((j-m)!)/((2j)!(j+m)!))))^(1/2))(Subscript[(OverHat[J]), +]^(j+m))(Ket[j,-j]),

or lowered from the highest weight state,

(Ket[j,m])=((((((j+m)!)/((2j)!(j-m)!))))^(1/2))(Subscript[(OverHat[J]), -]^(j-m))(Ket[j,j]).

((SubPlus[(OverHat[J])])/(Sqrt[2j]))(Ket[j,-j+n])=(Sqrt[n+1])(Ket[j,-j+n+1])+𝒪((j^(-1/2))),
((SubMinus[(OverHat[J])])/(Sqrt[2j]))(Ket[j,-j+n])=(Sqrt[n])(Ket[j,-j+n-1])+𝒪((j^(-1/2))).

(Ket[j,-j+n])→(Ket[n]),
(((2j))^(-1/2))(SubMinus[(OverHat[J])])->a, (((2j))^(-1/2))(SubPlus[(OverHat[J])])→(SuperDagger[a]),

Summary

(OverHat[J])×(OverHat[J])=i (OverHat[J]).

The total angular momentum operator

((OverHat[J])^2)=(Subscript[(OverHat[J]), 1]^2)+(Subscript[(OverHat[J]), 2]^2)+(Subscript[(OverHat[J]), 3]^2).

The raising and lowering operators

(Subscript[(OverHat[J]), ±])=(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]).

((OverHat[J])^2)(Ket[j,m])=j(j+1)(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=m(Ket[j,m]),
(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Sqrt[j(j+1)-m(m±1)])(Ket[j,m±1]),

j=0,(1/2),1,(3/2),…,
m=-j,-j+1,…,j-1,j.

Representation Theory

Spin-1/2

Use Eq. () to derive the matrix representation for spin-1/2 angular momentum operators, and show that they are indeed related to Pauli matrices as indicated in Eq. ().

Block[{j=1/2,JJ,J3,Jp,Jm,J1,J2},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{Table[j+1-{m,m}->j(j+1),{m,-j,j}],Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
MatrixForm/@{J1,J2,J3}
]

MatrixForm/@PauliMatrix/@Range[3]

Basis: {(Ket[↑])=(Ket[1/2,+1/2]),(Ket[↓])=(Ket[1/2,-1/2])}.

Matrix representations:

(Subscript[(OverHat[S]), 1])≏(1/2)(({{0, 1}, {1, 0}})),(Subscript[(OverHat[S]), 2])≏(1/2)(({{0, -i}, {i, 0}})),(Subscript[(OverHat[S]), 3])≏(1/2)(({{1, 0}, {0, -1}})).

Eigenstates and eigenvalues

(Subscript[(OverHat[S]), 1]):(1/(Sqrt[2]))(({{1}, {1}}))->+(1/2),(1/(Sqrt[2]))(({{1}, {-1}}))->-(1/2),
(Subscript[(OverHat[S]), 2]):(1/(Sqrt[2]))(({{1}, {i}}))->+(1/2),(1/(Sqrt[2]))(({{1}, {-i}}))->-(1/2),
(Subscript[(OverHat[S]), 3]):(({{1}, {0}}))->+(1/2),(({{0}, {1}}))->-(1/2).

Spin-1

Basis: {(Ket[1,+1]),(Ket[1,0]),(Ket[1,-1])}.

Matrix representations (the same matrices also represent spin-1)

(Subscript[(OverHat[L]), 1])≏(1/(Sqrt[2]))(({{0, 1, 0}, {1, 0, 1}, {0, 1, 0}})),(Subscript[(OverHat[L]), 2])≏(1/((Sqrt[2])i))(({{0, 1, 0}, {-1, 0, 1}, {0, -1, 0}})),(Subscript[(OverHat[L]), 3])≏(({{1, 0, 0}, {0, 0, 0}, {0, 0, -1}})).

Use Eq. () to derive the matrix representation for spin-1 angular momentum operators in Eq. ().

Block[{j=1,JJ,J3,Jp,Jm,J1,J2},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{Table[j+1-{m,m}->j(j+1),{m,-j,j}],Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
MatrixForm/@{J1,J2,J3}
]

Basis transformation

(({{(Ket[1,+1])}, {(Ket[1,0])}, {(Ket[1,-1])}}))=(({{(i/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, -i}, {((-i)/(Sqrt[2])), (1/(Sqrt[2])), 0}}))(({{(Ket[x])}, {(Ket[y])}, {(Ket[z])}})).

Denote the transformation matrix by (OverHat[U]), the operators transform as (Subscript[(OverHat[L]), a])->(SuperDagger[(OverHat[U])])(Subscript[(OverHat[L]), a])(OverHat[U]),

(Subscript[(OverHat[L]), 1])≏(({{0, 0, 0}, {0, 0, -i}, {0, i, 0}})),(Subscript[(OverHat[L]), 2])≏(({{0, 0, i}, {0, 0, 0}, {-i, 0, 0}})),(Subscript[(OverHat[L]), 3])≏(({{0, -i, 0}, {i, 0, 0}, {0, 0, 0}})).

Implement the basis transformation Eq. () to transform Eq. () to Eq. ().

Block[{j=1,JJ,J3,Jp,Jm,J1,J2,U},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{Table[j+1-{m,m}->j(j+1),{m,-j,j}],Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
U={{I,1,0}/Sqrt[2],{0,0,-I},{-I,1,0}/Sqrt[2]};
MatrixForm[ConjugateTranspose[U].#.U]&/@{J1,J2,J3}
]

Rotation Operators

direction of θ: the direction of the rotational axis (following the right-handed rule),

magnitude of θ: the amount of angle to rotate.

Spin-1/2 representation, in the {(Ket[↑]),(Ket[↓])} basis,

(Subscript[(OverHat[R]), 1/2])(θ)≏(({{(e^(i θ/2)), 0}, {0, (e^(-i θ/2))}})).

Block[{j=1/2,JJ,J3,Jp,Jm,J1,J2,U},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{Table[j+1-{m,m}->j(j+1),{m,-j,j}],Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
MatrixForm[MatrixExp[I θ #]]&/@{J1,J2,J3}
]

Matrix Exp.

Spin-1 representation, in the {(Ket[x]),(Ket[y]),(Ket[z])} basis,

(Subscript[(OverHat[R]), 1])(θ)≏(({{cos θ, sin θ, 0}, {-sin θ, cos θ, 0}, {0, 0, 1}})).

Block[{j=1,JJ,J3,Jp,Jm,J1,J2,U},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{Table[j+1-{m,m}->j(j+1),{m,-j,j}],Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
U={{I,1,0}/Sqrt[2],{0,0,-I},{-I,1,0}/Sqrt[2]};
MatrixForm[MatrixExp[I θ ConjugateTranspose[U].#.U]]&/@{J1,J2,J3}
]

Matrix Exp.

This is indeed the case for spin-1 representation

(Subscript[(OverHat[R]), 1])(0,0,2π)≏(({{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}))≏𝟙.

But not for spin-1/2,

(Subscript[(OverHat[R]), 1/2])(0,0,2π)≏(({{-1, 0}, {0, -1}}))≏-𝟙,

The extra minus sign means that spin-1/2 is a projective representation of SO(3). Nevertheless, it is a linear representation of SU(2).

Fusion Category

Example: Fusion of Spins

States tensor product together:

(Subscript[{(Ket[↑]),(Ket[↓])}, A])⊗(Subscript[{(Ket[↑]),(Ket[↓])}, B])={(Ket[↑↑]),(Ket[↑↓]),(Ket[↓↑]),(Ket[↓↓])}.

Rotation operators multiply together:

(e^( i θ ·(Subscript[(OverHat[S]), A])))(e^( i θ ·(Subscript[(OverHat[S]), B])))=(Subscript[(OverHat[R]), A])(θ)(Subscript[(OverHat[R]), B])(θ):=(OverHat[R])(θ)=(e^( i θ ·(OverHat[S]))).

Angular momenta add together:

(Subscript[(OverHat[S]), A])=(1/2)(Subscript[(OverHat[σ]), A])⊗(Subscript[𝟙, B]),
(Subscript[(OverHat[S]), B])=(1/2)(Subscript[𝟙, A])⊗(Subscript[(OverHat[σ]), B]).

Write down the matrix representation for (Subscript[(OverHat[S]), A]) and (Subscript[(OverHat[S]), B]). [Hint: altogether six 4×4 matrices]

(Subscript[(OverHat[S]), A,1])=(1/2)(Subscript[(OverHat[σ]), A]^x)⊗(Subscript[𝟙, B])≏(1/2)(({{0, 1}, {1, 0}}))⊗(({{1, 0}, {0, 1}}))=(1/2)(({{0, 0, 1, 0}, {0, 0, 0, 1}, {1, 0, 0, 0}, {0, 1, 0, 0}})).

(Subscript[(OverHat[S]), A,2])=(1/2)(Subscript[(OverHat[σ]), A]^y)⊗(Subscript[𝟙, B])≏(1/2)(({{0, -i}, {i, 0}}))⊗(({{1, 0}, {0, 1}}))=(1/2)(({{0, 0, -i, 0}, {0, 0, 0, -i}, {i, 0, 0, 0}, {0, i, 0, 0}})).

(Subscript[(OverHat[S]), A,3])=(1/2)(Subscript[(OverHat[σ]), A]^z)⊗(Subscript[𝟙, B])≏(1/2)(({{1, 0}, {0, -1}}))⊗(({{1, 0}, {0, 1}}))=(1/2)(({{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, -1}})).

(Subscript[(OverHat[S]), B,1])=(1/2)(Subscript[𝟙, A])⊗(Subscript[(OverHat[σ]), B]^x)≏(1/2)(({{1, 0}, {0, 1}}))⊗(({{0, 1}, {1, 0}}))=(1/2)(({{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}})).

(Subscript[(OverHat[S]), B,2])=(1/2)(Subscript[𝟙, A])⊗(Subscript[(OverHat[σ]), B]^y)≏(1/2)(({{1, 0}, {0, 1}}))⊗(({{0, -i}, {i, 0}}))=(1/2)(({{0, -i, 0, 0}, {i, 0, 0, 0}, {0, 0, 0, -i}, {0, 0, i, 0}})).

(Subscript[(OverHat[S]), B,3])=(1/2)(Subscript[𝟙, A])⊗(Subscript[(OverHat[σ]), B]^z)≏(1/2)(({{1, 0}, {0, 1}}))⊗(({{1, 0}, {0, -1}}))=(1/2)(({{1, 0, 0, 0}, {0, -1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, -1}})).

((OverHat[S])^2)=((((Subscript[(OverHat[S]), A])+(Subscript[(OverHat[S]), B])))^2)=(Subscript[(OverHat[S]), A]^2)+(Subscript[(OverHat[S]), B]^2)+2(Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B])
=(3/2)𝟙+2(Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B]).

(Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B])≏(1/4)(({{1, 0, 0, 0}, {0, -1, 2, 0}, {0, 2, -1, 0}, {0, 0, 0, 1}})).

Verify the matrix representation in Eq. ().

(Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B]):=(Subscript[(OverHat[S]), A,1])(Subscript[(OverHat[S]), B,1])+(Subscript[(OverHat[S]), A,2])(Subscript[(OverHat[S]), B,2])+(Subscript[(OverHat[S]), A,3])(Subscript[(OverHat[S]), B,3])
=(1/2)(({{0, 0, 1, 0}, {0, 0, 0, 1}, {1, 0, 0, 0}, {0, 1, 0, 0}}))(1/2)(({{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}}))+(1/2)(({{0, 0, -i, 0}, {0, 0, 0, -i}, {i, 0, 0, 0}, {0, i, 0, 0}}))(1/2)(({{0, -i, 0, 0}, {i, 0, 0, 0}, {0, 0, 0, -i}, {0, 0, i, 0}}))+(1/2)(({{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, -1}}))(1/2)(({{1, 0, 0, 0}, {0, -1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, -1}}))
=(1/4)(({{0, 0, 0, 1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {1, 0, 0, 0}}))+(1/4)(({{0, 0, 0, -1}, {0, 0, 1, 0}, {0, 1, 0, 0}, {-1, 0, 0, 0}}))+(1/4)(({{1, 0, 0, 0}, {0, -1, 0, 0}, {0, 0, -1, 0}, {0, 0, 0, 1}}))
=(1/4)(({{1, 0, 0, 0}, {0, -1, 2, 0}, {0, 2, -1, 0}, {0, 0, 0, 1}})).

Spin singlet state, (Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B]) =-3/4 => ((OverHat[S])^2)=0 (s=0):

Spin triplet states, (Subscript[(OverHat[S]), A])·(Subscript[(OverHat[S]), B]) =1/4 => ((OverHat[S])^2)=2 (s=1):

Block[{S,H,U},
S=Represent@(σ[#,0]+σ[0,#])/2&/@Range[3];
H=Total[#.#&/@S]-Last@S/10;
U=Transpose[Normalize/@Last/@SortBy[First]@Thread@Eigensystem[H]];
Style[Grid[Join[{{"space:","\!\(\(1\/2\)⊗\(1\/2\)\)","\!\(0⊕1\)"},{"basis:","\!\((Ket[↑↑]),(Ket[↑↓]),(Ket[↓↑]),(Ket[↓↓])\)","\!\((Ket[0,0]);(Ket[1,+1]),(Ket[1,0]),(Ket[1,-1])\)"}},{"\!\("<>#1<>"≏\)",MatrixForm[#2],MatrixForm[{{Grid[ConjugateTranspose[U].#2.U,Dividers->{2->Dashed,2->Dashed}]}}]}&@@@Thread@{{"S\&^\_1","S\&^\_2","S\&^\_3","\(S\)\&^\^2"},Append[S,tTr[{S,S},{{1,4},{3,5}}]]}]],FontSize->14,SingleLetterItalics->True,FontFamily->"CMU Serif"]]

The original basis {(Ket[↑↑]),(Ket[↑↓]),(Ket[↓↑]),(Ket[↓↓])} form a reducible representation of SU(2).

The new basis {(Ket[0,0])} and {(Ket[1,+1]),(Ket[1,0]),(Ket[1,-1])} respectively form irreducible representations of SU(2).

1/2⊗1/2→0 channel:

((Ket[0,0]))=(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}}))(({{(Ket[↑↑])}, {(Ket[↑↓])}, {(Ket[↓↑])}, {(Ket[↓↓])}})).

1/2⊗1/2→1 channel:

(({{(Ket[1,+1])}, {(Ket[1,0])}, {(Ket[1,-1])}}))=(({{1, 0, 0, 0}, {0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, 0, 1}}))(({{(Ket[↑↑])}, {(Ket[↑↓])}, {(Ket[↓↑])}, {(Ket[↓↓])}})).

Fusion Rules

Irreducible representations are simple objects in the category, labeled by their spin quantum numbers j=0,1/2,1,…

Other reducible representations are not simple. They can be considered as spin systems.

Semisimple property: each reducible representation (spin system) can split in to direct sum of irreducible representations (spins).

Matrix representation of group element g∈ SU(2) in representation j can be depicted as

Graphics[{{Red,Line[{{-1,0},{1,0}}],Text["\!\( j\)",{-0.8,0},{0,1}],Text["\!\( j\)",{0.8,0},{0,1}]},Text[Style["\!\( g\)",Background->White],{0,0}]},PlotRange->{-0.8,0.5},ImageSize->60]

Example: rotation θ=(0,0,θ), c.f. Eq. () and Eq. ()

=(({{(e^(i θ/2)), 0}, {0, (e^(-i θ/2))}})),
=(({{cos θ, sin θ, 0}, {-sin θ, cos θ, 0}, {0, 0, 1}})).

Row[{Graphics[{{Red,Line[{{-1,0},{1,0}}],Text["\!\(j\_1\)",{0,0},{0,1}]}},ImageSize->30],Style[" ⊗ ",14,FontFamily->"CMU"],Graphics[{{Blue,Line[{{-1,0},{1,0}}],Text["\!\(j\_2\)",{0,0},{0,1}]}},ImageSize->30],Style[" = ",14,FontFamily->"CMU"],Graphics[{{Red,Line[{{-1,0.5},{1,0.5}}],Text["\!\(j\_1\)",{0,0.5},{0,-1}]},
{Blue,Line[{{-1,-0.5},{1,-0.5}}],Text["\!\(j\_2\)",{0,-0.5},{0,1}]}},ImageSize->30]}]

The tensor structure specifies the fusion rule:

Composition of a spin-(Subscript[j, 1]) and a spin-(Subscript[j, 2]) systems is a spin system that contains irreducible representations ranging from |(Subscript[j, 1])-(Subscript[j, 2])| to (Subscript[j, 1])+(Subscript[j, 2]) (each of them appear once).

Example: 1/2⊗1/2→0⊕1.

Notice that the representation dimensions add up correctly

dim (Subscript[ℋ, (Subscript[j, 1])⊗(Subscript[j, 2])])=(∑_((Subscript[j, 3])=|(Subscript[j, 1])-(Subscript[j, 2])|)^((Subscript[j, 1])+(Subscript[j, 2])))(2(Subscript[j, 3])+1)=(2(Subscript[j, 1])+1)(2(Subscript[j, 2])+1).

The decomposition is implemented by the fusion vertex (a projective morphism), which can be viewed as three-leg tensors preserving the SU(2) symmetry.

Graphics[{{Red,BSplineCurve[{{-1.5,0.5},{-1,0.5},{-0.5,0.5},{-0.2,0.2},{0,0}}],Text["\!\(j\_1\)",{-1.2,0.5},{0,1}]},
{Green,BSplineCurve[{{-1.5,-0.5},{-1,-0.5},{-0.5,-0.5},{-0.2,-0.2},{0,0}}],Text["\!\(j\_2\)",{-1.2,-0.5},{0,1}]},
{Blue,Line[{{0,0},{1.5,0}}],Text["\!\(j\_3\)",{1.2,0},{0,1}]},
{EdgeForm[Black],White,Disk[{0,0},0.35]},
Text["\!\( π\)",{0,0}]},BaselinePosition->{0,0},ImageSize->85]

such that ∀g∈ SU(2):

Graphics[{Translate[{{Red,BSplineCurve[{{-1.8,0.5},{-1,0.5},{-0.5,0.5},{-0.2,0.2},{0,0}}],Text["\!\(j\_1\)",{-1.5,0.5},{0,1}]},
Text[Style["\!\( g\)",Background->White],{-1,0.5}],
{Green,BSplineCurve[{{-1.8,-0.5},{-1,-0.5},{-0.5,-0.5},{-0.2,-0.2},{0,0}}],Text["\!\(j\_2\)",{-1.5,-0.5},{0,1}]},
Text[Style["\!\( g\)",Background->White],{-1,-0.5}],
{Blue,Line[{{0,0},{1,0}}],Text["\!\(j\_3\)",{0.8,0},{0,1}]},
{EdgeForm[Black],White,Disk[{0,0},0.35]},
Text["\!\( π\)",{0,0}]},{-1.5,0}],
Text["=",{0,0}],Translate[{{Red,BSplineCurve[{{-1.5,0.5},{-1,0.5},{-0.5,0.5},{-0.2,0.2},{0,0}}],Text["\!\(j\_1\)",{-1.2,0.5},{0,1}]},
{Green,BSplineCurve[{{-1.5,-0.5},{-1,-0.5},{-0.5,-0.5},{-0.2,-0.2},{0,0}}],Text["\!\(j\_2\)",{-1.2,-0.5},{0,1}]},
{Blue,Line[{{0,0},{1.8,0}}],Text["\!\(j\_3\)",{1.5,0},{0,1}]},
Text[Style["\!\( g\)",Background->White],{1,0}],
{EdgeForm[Black],White,Disk[{0,0},0.35]},
Text["\!\( π\)",{0,0}]},{2,0}]},BaselinePosition->{0,0},ImageSize->200]

Example: fusion of two spin-1/2s, c.f. Eq. () and Eq. ()

(Subscript[π, 0]):1/2⊗1/2->0,
(Subscript[π, 0])≏(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}})),
(Subscript[π, 1]):1/2⊗1/2->1,
(Subscript[π, 1])≏(({{1, 0, 0, 0}, {0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, 0, 1}})).

(({{(Subscript[j, 1]), (Subscript[j, 2]), (Subscript[j, 3])}, {(Subscript[m, 1]), (Subscript[m, 2]), (Subscript[m, 3])}}))=(BraKet[(Subscript[j, 3]),(Subscript[m, 3]), (Subscript[j, 1]),(Subscript[m, 1]);(Subscript[j, 2]),(Subscript[m, 2])]).

π((Subscript[j, 1])⊗(Subscript[j, 2])→(Subscript[j, 3]))=(∑_((Subscript[m, 1]),(Subscript[m, 2]),(Subscript[m, 3])))(Ket[(Subscript[j, 3]),(Subscript[m, 3])])(({{(Subscript[j, 1]), (Subscript[j, 2]), (Subscript[j, 3])}, {(Subscript[m, 1]), (Subscript[m, 2]), (Subscript[m, 3])}}))(Bra[(Subscript[j, 1]),(Subscript[m, 1]);(Subscript[j, 2]),(Subscript[m, 2])]).

either look up in Mathematica:

ClebschGordan[{1/2,-1/2},{1/2,1/2},{1,0}]

or calculate as eigenbasis of the Casimir operator ((OverHat[J])^2):

Graphics[{Translate[{{Red,Line[{{-1,0.4},{1,0.4}}],Text["\!\(j\_1\)",{-0.8,0.4},{0,-1}],
Text["\!\(j\_1\)",{0.8,0.4},{0,-1}]},
{Blue,Line[{{-1,-0.4},{1,-0.4}}],Text["\!\(j\_2\)",{-0.8,-0.4},{0,1}],
Text["\!\(j\_2\)",{0.8,-0.4},{0,1}]},
{{EdgeForm[Black],White,Rectangle[{-0.4,-0.6},{0.4,0.6},RoundingRadius->0.2]},Text["\!\(J\&^\^2\)",{0,0}]}},{-1.8,0}],Text["\!\(=\*UnderscriptBox[\(∑\), \(j\_3\)]\)",{0,0},{0,0.4}],
Translate[{{Green,Line[{{-1,0},{1,0}}],Text["\!\(j\_3\)",{-0.5,0},{0,1}],
Text["\!\(j\_3\)",{0.5,0},{0,1}]},
{EdgeForm[Black],White,Polygon@CirclePoints[{0.35,0},4]},
Text["\!\(λ\)"],Translate[{Scale[{{Red,BSplineCurve[{{-1,0.4},{-0.8,0.4},{-0.4,0.4},{-0.2,0.2},{0,0}}]},
{Blue,BSplineCurve[{{-1,-0.4},{-0.8,-0.4},{-0.4,-0.4},{-0.2,-0.2},{0,0}}]}},{#1,1},{0,0}],
{Red,Text["\!\(j\_1\)",{-#1 0.8,0.4},{0,-1}],
Blue,Text["\!\(j\_2\)",{-#1 0.8,-0.5},{0,1}]},
{EdgeForm[Black],White,Disk[{0,0},0.35]},
Text["\!\( π"<>#2<>"\)",{0,0}]},{-#1,0}]&@@@{{1,"\^*"},{-1,""}}},{2.8,0}]},ImageSize->220]

Represent ((OverHat[J])^2) in the (Subscript[j, 1])⊗(Subscript[j, 2]) space.

Diagonalize ((OverHat[J])^2), group eigenvectors by eigenvalues.

Within each degenerated subspace: the eigenvalue (Subscript[λ, (Subscript[j, 3])])=(Subscript[j, 3])((Subscript[j, 3])+1) or the degeneracy 2(Subscript[j, 3])+1 infers the spin quantum number (Subscript[j, 3]), the eigenvectors form the fusion vertex π((Subscript[j, 1])⊗(Subscript[j, 2])→(Subscript[j, 3])).

Compute the fusion vertex of π(1/2⊗1->1/2) and π(1⊗1/2->1/2).

Jvec[j_]:=Block[{JJ,J3,Jp,Jm,J1,J2},
{JJ,J3,Jp}=SparseArray[#,{#,#}&@2j+1]&/@{
Table[j+1-{m,m}->j(j+1),{m,-j,j}],
Table[j+1-{m,m}->m,{m,-j,j}],
Table[j+1-{m+1,m}->Sqrt[j(j+1)-m(m+1)],{m,-j,j-1}]};
Jm=Transpose@Jp;
{J1,J2}={Jp+Jm,-I Jp+I Jm}/2;
{J1,J2,J3}];
MatrixForm/@Jvec[1]

j1=1/2;j2=1;
JA=KroneckerProduct[#,IdentityMatrix[2j2+1]]&/@Jvec[j1];
JB=KroneckerProduct[IdentityMatrix[2j1+1],#]&/@Jvec[j2];
J=JA+JB;
Jsqr=Total[#.#&/@J];
MatrixForm@Jsqr

Eigenvalues@Jsqr

({{(Subscript[λ, j])=j(j+1), deg.=2j+1, j}, {3/4, 2, 1/2}, {15/4, 4, 3/2}})

pi=Orthogonalize/@GroupBy[Thread@Eigensystem@Jsqr,First->Last];
MatrixForm/@pi

MatrixForm[pi[3/4].J[[3]].Transpose[pi[3/4]]]

π(1/2⊗1->1/2)=(({{0, -(1/(Sqrt[3])), 0, (Sqrt[(2/3)]), 0, 0}, {0, 0, -(Sqrt[(2/3)]), 0, (1/(Sqrt[3])), 0}})).

π(1⊗1/2->1/2)=(({{0, (Sqrt[(2/3)]), -(1/(Sqrt[3])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[3])), -(Sqrt[(2/3)]), 0}})),

({{1/2⊗1 basis, 1⊗1/2 basis}, {(Ket[1/2,+1/2;1,+1]), (Ket[1,+1;1/2,+1/2])}, {(Ket[1/2,+1/2;1,0]), (Ket[1,+1;1/2,-1/2])}, {(Ket[1/2,+1/2;1,-1]), (Ket[1,0;1/2,+1/2])}, {(Ket[1/2,-1/2;1,+1]), (Ket[1,0;1/2,-1/2])}, {(Ket[1/2,-1/2;1,0]), (Ket[1,-1;1/2,+1/2])}, {(Ket[1/2,-1/2;1,+1]), (Ket[1,-1;1/2,-1/2])}})

F-Symbols

Deploy[Graphics[GraphicsComplex[{{1,0},{0,1},{-1,2},{1,2},{-2,3},{0,3},{2,3}},{Line[{{1,5},{3,6},{2,7}}],{EdgeForm[Black],White,Disk[#,0.4]&/@{2,3}},
Text["\!\(π\)",#]&/@{2,3},
Text@@@{{"\!\(j\_1\)",5,{0,-1}},{"\!\(j\_2\)",6,{0,-1}},{"\!\(j\_3\)",7,{0,-1}},{"\!\(j\_4\)",3,{-0.5,1.5}},{"\!\(j\_6\)",2,{-0.5,1.5}}}}],ImageSize->80]]

The composition follows from the tensor network contraction.

For (Subscript[j, 4])=0,

(Subscript[α, 0])=π(0⊗1/2->1/2)(π(1/2⊗1/2->0)⊗π(1/2->1/2))
≏(({{0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0}})).

Verify Eq. ().

(Subscript[α, 0])=π(0⊗1/2->1/2)(π(1/2⊗1/2->0)⊗π(1/2->1/2))
≏(({{1, 0}, {0, 1}}))[(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}}))⊗(({{1, 0}, {0, 1}}))]
=(({{1, 0}, {0, 1}}))(({{0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0}}))
=(({{0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[2])), 0, -(1/(Sqrt[2])), 0, 0}})).

For (Subscript[j, 4])=1,

(Subscript[α, 1])=π(1⊗1/2->1/2)(π(1/2⊗1/2->1)⊗π(1/2->1/2))
≏(({{0, (Sqrt[(2/3)]), -(1/(Sqrt[6])), 0, -(1/(Sqrt[6])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[6])), 0, (1/(Sqrt[6])), -(Sqrt[(2/3)]), 0}})).

Verify Eq. ().

(Subscript[α, 1])=π(1⊗1/2->1/2)(π(1/2⊗1/2->1)⊗π(1/2->1/2))
≏(({{0, (Sqrt[(2/3)]), -(1/(Sqrt[3])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[3])), -(Sqrt[(2/3)]), 0}}))[(({{1, 0, 0, 0}, {0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, 0, 1}}))⊗(({{1, 0}, {0, 1}}))]
=(({{0, (Sqrt[(2/3)]), -(1/(Sqrt[3])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[3])), -(Sqrt[(2/3)]), 0}}))(({{1, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, (1/(Sqrt[2])), 0, (1/(Sqrt[2])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[2])), 0, (1/(Sqrt[2])), 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 1}}))
=(({{0, (Sqrt[(2/3)]), -(1/(Sqrt[6])), 0, -(1/(Sqrt[6])), 0, 0, 0}, {0, 0, 0, (1/(Sqrt[6])), 0, (1/(Sqrt[6])), -(Sqrt[(2/3)]), 0}})).

Deploy[Graphics[GraphicsComplex[{{1,0},{0,1},{-1,2},{1,2},{-2,3},{0,3},{2,3}},{Line[{{1,5},{4,6},{2,7}}],{EdgeForm[Black],White,Disk[#,0.4]&/@{2,4}},
Text["\!\(π\)",#]&/@{2,4},
Text@@@{{"\!\(j\_1\)",5,{0,-1}},{"\!\(j\_2\)",6,{0,-1}},{"\!\(j\_3\)",7,{0,-1}},{"\!\(j\_5\)",4,{0.5,1.5}},{"\!\(j\_6\)",2,{-0.5,1.5}}}}],ImageSize->80]]

For (Subscript[j, 5])=0,

(Subscript[β, 0])=π(1/2⊗0->1/2)(π(1/2->1/2)⊗π(1/2⊗1/2->0))
≏(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}})).

Verify Eq. ().

(Subscript[β, 0])=π(1/2⊗0->1/2)(π(1/2->1/2)⊗π(1/2⊗1/2->0))
≏(({{1, 0}, {0, 1}}))[(({{1, 0}, {0, 1}}))⊗(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}}))]
=(({{1, 0}, {0, 1}}))(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}}))
=(({{0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, (1/(Sqrt[2])), -(1/(Sqrt[2])), 0}})).

For (Subscript[j, 5])=1,

(Subscript[β, 1])=π(1/2⊗1->1/2)(π(1/2->1/2)⊗π(1/2⊗1/2->1))
≏(({{0, -(1/(Sqrt[6])), -(1/(Sqrt[6])), 0, (Sqrt[(2/3)]), 0, 0, 0}, {0, 0, 0, -(Sqrt[(2/3)]), 0, (1/(Sqrt[6])), (1/(Sqrt[6])), 0}})).

Verify Eq. ().

(Subscript[β, 1])=π(1/2⊗1->1/2)(π(1/2->1/2)⊗π(1/2⊗1/2->1))
≏(({{0, -(1/(Sqrt[3])), 0, (Sqrt[(2/3)]), 0, 0}, {0, 0, -(Sqrt[(2/3)]), 0, (1/(Sqrt[3])), 0}}))[(({{1, 0}, {0, 1}}))⊗(({{1, 0, 0, 0}, {0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, 0, 1}}))]
=(({{0, -(1/(Sqrt[3])), 0, (Sqrt[(2/3)]), 0, 0}, {0, 0, -(Sqrt[(2/3)]), 0, (1/(Sqrt[3])), 0}}))(({{1, 0, 0, 0, 0, 0, 0, 0}, {0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, (1/(Sqrt[2])), (1/(Sqrt[2])), 0}, {0, 0, 0, 0, 0, 0, 0, 1}}))
=(({{0, -(1/(Sqrt[6])), -(1/(Sqrt[6])), 0, (Sqrt[(2/3)]), 0, 0, 0}, {0, 0, 0, -(Sqrt[(2/3)]), 0, (1/(Sqrt[6])), (1/(Sqrt[6])), 0}})).

=(∑_(Subscript[j, 5]))(Subscript[((Subscript[F, (Subscript[j, 6])]^((Subscript[j, 1])(Subscript[j, 2])(Subscript[j, 3])))), (Subscript[j, 5])]^(Subscript[j, 4]))

(({{(Subscript[α, 0])}, {(Subscript[α, 1])}}))=(({{-(1/2), -((Sqrt[3])/2)}, {((Sqrt[3])/2), -(1/2)}}))(({{(Subscript[β, 0])}, {(Subscript[β, 1])}})).

The matrix (Subscript[F, (Subscript[j, 6])]^((Subscript[j, 1])(Subscript[j, 2])(Subscript[j, 3]))) is called F-matrix, whose elements are called F-symbols. The F-matrix is unitary.

F-symbol is related to Wigner 6j-symbol by a normalization factor

(Subscript[((Subscript[F, (Subscript[j, 6])]^((Subscript[j, 1])(Subscript[j, 2])(Subscript[j, 3])))), (Subscript[j, 5])]^(Subscript[j, 4]))=(((-))^((Subscript[j, 1])+(Subscript[j, 2])+(Subscript[j, 3])+(Subscript[j, 6])))(Sqrt[(2(Subscript[j, 4])+1)(2(Subscript[j, 5])+1)]){({{(Subscript[j, 1]), (Subscript[j, 2]), (Subscript[j, 4])}, {(Subscript[j, 3]), (Subscript[j, 6]), (Subscript[j, 5])}})}.

SixJSymbol[{1/2,1/2,0},{1/2,1/2,0}]

Graphics[{MapThread[Translate[#1,#2]&,{{GraphicsComplex[{{-3,0},{-1,0},{1,0},{3,0},{-2,-1},{0,-1},{2,-1},{-1,-2},{1,-2},{0,-3},{1,-4}},
{Line[{{1,11},{2,9},{3,6},{4,10}}],
Text@@@{{"\!\(j\_1\)",1,{0,-1}},{"\!\(j\_2\)",2,{0,-1}},{"\!\(j\_3\)",3,{0,-1}},{"\!\(j\_4\)",4,{0,-1}},{"\!\(j\_5\)",11,{2,0}},{"\!\(j\_10\)",9,{1.5,0}},{"\!\(j\_9\)",10,{-2,0}}}}],
GraphicsComplex[{{-3,0},{-1,0},{1,0},{3,0},{-2,-1},{0,-1},{2,-1},{-1,-2},{1,-2},{0,-3},{1,-4}},
{Line[{{1,11},{2,9},{3,7},{4,10}}],
Text@@@{{"\!\(j\_1\)",1,{0,-1}},{"\!\(j\_2\)",2,{0,-1}},{"\!\(j\_3\)",3,{0,-1}},{"\!\(j\_4\)",4,{0,-1}},{"\!\(j\_5\)",11,{2,0}},{"\!\(j\_9\)",10,{-2,0}},{"\!\(j\_8\)",9,{-2,0}}}}],
GraphicsComplex[{{-3,0},{-1,0},{1,0},{3,0},{-2,-1},{0,-1},{2,-1},{-1,-2},{1,-2},{0,-3},{1,-4}},
{Line[{{1,11},{2,5},{3,7},{4,10}}],
Text@@@{{"\!\(j\_1\)",1,{0,-1}},{"\!\(j\_2\)",2,{0,-1}},{"\!\(j\_3\)",3,{0,-1}},{"\!\(j\_4\)",4,{0,-1}},{"\!\(j\_5\)",11,{2,0}},{"\!\(j\_6\)",8,{2,0}},{"\!\(j\_8\)",9,{-2,0}}}}],
GraphicsComplex[{{-3,0},{-1,0},{1,0},{3,0},{-2,-1},{0,-1},{2,-1},{-1,-2},{1,-2},{0,-3},{1,-4}},
{Line[{{1,11},{2,5},{3,8},{4,10}}],
Text@@@{{"\!\(j\_1\)",1,{0,-1}},{"\!\(j\_2\)",2,{0,-1}},{"\!\(j\_3\)",3,{0,-1}},{"\!\(j\_4\)",4,{0,-1}},{"\!\(j\_5\)",11,{2,0}},{"\!\(j\_6\)",8,{2,0}},{"\!\(j\_7\)",10,{2,0}}}}],
GraphicsComplex[{{-3,0},{-1,0},{1,0},{3,0},{-2,-1},{0,-1},{2,-1},{-1,-2},{1,-2},{0,-3},{1,-4}},
{Line[{{1,11},{2,6},{3,8},{4,10}}],
Text@@@{{"\!\(j\_1\)",1,{0,-1}},{"\!\(j\_2\)",2,{0,-1}},{"\!\(j\_3\)",3,{0,-1}},{"\!\(j\_4\)",4,{0,-1}},{"\!\(j\_5\)",11,{2,0}},{"\!\(j\_10\)",8,{-1.5,0}},{"\!\(j\_7\)",10,{2,0}}}}]},
8CirclePoints[5]}],
Block[{arr},
arr[{p1_,p2_}]:={{GrayLevel[0.7],AbsoluteThickness[5],Arrow[{(p1+p2)/2+0.12(p1-p2),(p1+p2)/2-0.12(p1-p2)}]},Text["\!\( F\)",(p1+p2)/2,1.5{{0,1},{-1,0}}.Normalize[p2-p1]]};
Translate[{Arrowheads[0.08],arr[10CirclePoints[5][[#]]]&/@{{4,3},{3,2},{4,5},{5,1},{1,2}}},{0,-1}]]},
ImageSize->240]

(Subscript[((Subscript[F, (Subscript[j, 5])]^((Subscript[j, 6])(Subscript[j, 3])(Subscript[j, 4])))), (Subscript[j, 8])]^(Subscript[j, 7]))(Subscript[((Subscript[F, (Subscript[j, 5])]^((Subscript[j, 1])(Subscript[j, 2])(Subscript[j, 8])))), (Subscript[j, 9])]^(Subscript[j, 6]))=(∑_(Subscript[j, 10]))(Subscript[((Subscript[F, (Subscript[j, 7])]^((Subscript[j, 1])(Subscript[j, 2])(Subscript[j, 3])))), (Subscript[j, 10])]^(Subscript[j, 6]))(Subscript[((Subscript[F, (Subscript[j, 5])]^((Subscript[j, 1])(Subscript[j, 10])(Subscript[j, 4])))), (Subscript[j, 9])]^(Subscript[j, 7]))(Subscript[((Subscript[F, (Subscript[j, 9])]^((Subscript[j, 2])(Subscript[j, 3])(Subscript[j, 4])))), (Subscript[j, 8])]^(Subscript[j, 10])).

Hydrogen Atom

Background: Classical Mechanics

Model Hamiltonian

The proton is much heavier than the electron (mass ratio (Subscript[m, p])/(Subscript[m, e])≈1836). So the proton is essentially pinned at the center of mass (i.e. we can ignore its motion).

Consider electron moving in the electric static potential created by the proton.

V(r)=-(α/r),

where α=(e^2)/(4π (Subscript[ϵ, 0])) is the fine-structure constant, describing the strength of the electrostatic Coulomb force.

H=((p^2)/(2m))-(α/r),

r is the displacement from the proton and r=|r| is the electron-proton distance.

p=m (OverDot[r]) is the momentum.

m is the mass of electron (or more precisely the reduced mass of the electron in this two-body problem).

r→(r/(m α)), p→m α p,

H->m (α^2) H.

Every quantity is dimensionless now. The energy m (α^2)/2 is called the Rydberg unit of energy.

The classical equation of motion is given by the Hamiltonian equation

((dr)/(dt))=((∂H)/(∂p)),((dp)/(dt))=-((∂H)/(∂r)).

Substitute in Eq. (),

((dr)/(dt))=p,((dp)/(dt))=-(r/(r^3)).

Angular Momentum

((dL)/(dt))=0.

Check Eq. ().

((dL)/(dt))=((dr)/(dt))×p+r×((dp)/(dt))
=p×p-r×(r/(r^3))=0.

Laplace-Runge-Lenz (LRL) Vector

Manipulate[Graphics[{Arrowheads[{{0.001,1,Arrowhead["Alignment"->0]}}],{Cases[PolarPlot[1/(1+A Cos[α]),{α,-π,π}],_Line,∞],
If[A>0.,{Red,Arrow[{{0,0},{A,0}}]},Nothing]},{Arrow[{{0,0},#}],Text["\!\(r \)",#/2,Reverse[{1,-1}Normalize[#]]],Translate[{{Green,Arrow[{{0,0},Normalize[#]}],Text["\!\(r\&^\)",Normalize[#]/2,1.2Sign[θ]Reverse[{-1,1}Normalize[#]]]},{Blue,Arrow[{{0,0},Normalize[#]+{A,0}}],Text["\!\(p × L\)",(Normalize[#]+{A,0})/2,1.2Sign[θ]Reverse[{1,-1}Normalize[(Normalize[#]+{A,0})]]]},{Red,Arrow[{Normalize[#],Normalize[#]+{A,0}}],Text["\!\(A \)",Normalize[#]+{A,0}/2,{0,-Sign[θ]+0.1}]}},#]}&@(1/(1+A Cos[θ]){Cos[θ],Sin[θ]}),{Red,Point[{0,0}],AbsoluteThickness[1],Circle[{0,0},Offset@{6,6},{0,θ}],Text["θ",Offset[6{Cos[θ/2],Sin[θ/2]}],-0.7{Cos[θ/2],Sin[θ/2]}]}},PlotRange->{{-1/(1-A)-1,1/(1+A)+1+A},{-1/Sqrt[1-A^2]-1,1/Sqrt[1-A^2]+1}},PlotRangePadding->Scaled[0.05],ImageSize->{Automatic,180}],{{A,0.7},0.01,0.9,ImageSize->Small},{{θ,π-ArcTan[Sqrt[1-A^2]/A]},-π,π,ImageSize->Small},Paneled->False,LabelStyle->"InlineFormula"]

((dA)/(dt))=0.

Check Eq. ().

((dA)/(dt))=((dp)/(dt))×L-(1/r)((dr)/(dt))-r(d/(dt))((1/r))
=-(r/(r^3))×L-(1/r)p+(r/(r^2))((dr)/(dt))
=-(1/(r^3))r×(r×p)-(1/r)p+(r/(2(r^3)))((2rdr)/(dt))
=-(1/(r^3))(r (r·p)-(r^2)p)-(1/r)p+(r/(2(r^3)))((d(r^2))/(dt))
=-r (1/(r^3))(r·((dr)/(dt)))+(r/(2(r^3)))((dr·r)/(dt))
=(r/(2(r^3)))(-((dr·r)/(dt))+((dr·r)/(dt)))
=0.

Let θ be the angle between r and A,

A·r=A r cos θ=r·(p×L)-r.

Permuting the scalar triple product

r·(p×L)=L·(r×p)=(L^2),

Eq. () => the orbit equation

This describes an elliptical orbit, with

a=((L^2)/(1-(A^2))),c=(((L^2)A)/(1-(A^2))).

Block[{a=1,e=0.8,b,c,ah},
c=a e;
b=Sqrt[a^2-c^2];
ah=Arrowheads[{{-0.001,0,#},{0.001,1,#}}]&@Graphics[{Line[Offset/@{{-6,-2},{0,0},{-6,2}}]}];
Graphics[{{AbsoluteThickness[1],Dashed,Line[{a #,0}&/@{-1.2,1.2}],Line[{0,b #}&/@{-1.2,1.2}]},{AbsoluteThickness[1.25],ah,{Red,Arrow[{{c,0},{0,b}}],Text["\!\( a\)",{c/2,b/2},{-1,-0.5}],Arrow[{{-a,0},{0,0}}],Text["\!\( a\)",{-a/2,0},{0,0.8}]},{Green,Arrow[{{0,0},{0,b}}],Text["\!\( b\)",{0,b/2},{2,0}]},{Blue,Arrow[{{0,0},{c,0}}],Text["\!\( c\)",{c/2,0},{0,0.8}]}},{Text["\!\( F\)",{c,0},{0,1.2}],Text["\!\( F\^′\)",{-c,0},{0,-0.8}]},Circle[{0,0},{a,b}],{AbsolutePointSize[5],Point[{c#,0}&/@{-1,1}]}}]]

Conservation of the magnitude A=|A| => conservation of the eccentricity of the orbit (in fact e≡c/a=A).

Conservation of the direction of A => conservation of the orientation of the orbit in the orbital plane: the periapsis (the point of closest approach) does not move with respect to the force center (focal point) in time.

SO(4) Symmetry

Operator Definition

Coordinate operator: (OverHat[r])=((Subscript[(OverHat[r]), 1]),(Subscript[(OverHat[r]), 2]),(Subscript[(OverHat[r]), 3])). Further define

(OverHat[r]):=(Sqrt[(OverHat[r])·(OverHat[r])])=(Sqrt[(Subscript[(OverHat[r]), 1]^2)+(Subscript[(OverHat[r]), 2]^2)+(Subscript[(OverHat[r]), 3]^2)]).

Momentum operator: (OverHat[p])=((Subscript[(OverHat[p]), 1]),(Subscript[(OverHat[p]), 2]),(Subscript[(OverHat[p]), 3])).

Hamiltonian operator:

Angular momentum operator: (OverHat[L])=((Subscript[(OverHat[L]), 1]),(Subscript[(OverHat[L]), 2]),(Subscript[(OverHat[L]), 3])), where

or in component form as

(Subscript[(OverHat[L]), a])=(Subscript[ϵ, a b c])(Subscript[(OverHat[r]), b])(Subscript[(OverHat[p]), c]).

LRL vector operator: (OverHat[A])=((Subscript[(OverHat[A]), 1]),(Subscript[(OverHat[A]), 2]),(Subscript[(OverHat[A]), 3])), where

or in component form as

(Subscript[(OverHat[A]), a])=(1/2)(Subscript[ϵ, a b c])((Subscript[(OverHat[p]), b])(Subscript[(OverHat[L]), c])-(Subscript[(OverHat[L]), b])(Subscript[(OverHat[p]), c]))-((Subscript[(OverHat[r]), a])/(OverHat[r])).

The expression is modified from the classical definition Eq. () to ensure the Hermiticity of the LRL vector as quantum operators.

Operator Algebra

[(Subscript[(OverHat[r]), a]),(Subscript[(OverHat[p]), b])]=i (Subscript[δ, a b])𝟙.

[(OverHat[H]),(Subscript[(OverHat[L]), a])]=0,
[(OverHat[H]),(Subscript[(OverHat[A]), a])]=0,
[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]),
[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[A]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[A]), c]),
[(Subscript[(OverHat[A]), a]),(Subscript[(OverHat[A]), b])]=-2 i  (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c])(OverHat[H]),

Verify Eq. (). [Hint: try writing a computer program to help.]

Notations:

comm[A,B] - represents the commutator of operators A and B,

A**B - represents the non-commutative multiplication of A and B (like matrix multiplication),

A⋀B -  represent the cross product A×B of two vector operators A and B (the symbol × has been used to denote normal multiplication).

ϵ - Levi-Civita tensor.

rules={};

Rules to dispatch non-commutative multiplication or commutator over a list of operators

rules=Join[rules,{
As_List**B_:>(#**B&/@As),
A_**Bs_List:>(A**#&/@Bs),
comm[As_List,B_]:>(comm[#,B]&/@As),
comm[A_,Bs_List]:>(comm[A,#]&/@Bs)
}];

For example,

comm[{a,b},{c,d}]//.rules

Rules regarding non-commutative multiplications

rules=Join[rules,{
(A_+B_)**C_:>A**C+B**C,
A_**(B_+C_):>A**B+A**C,
(x_?NumericQ A_)**B_:>x A**B,
A_**(x_?NumericQ B_):>x A**B,
_**0:>0,
0**_:>0,
A_**𝟙:>A,
𝟙**A_:>A
}];

For example, applying these rules can expand expressions involving non-commutative multiplications

(a+2b)**(3 c+4d)//.rules

Rules regarding commutation relations

General properties

rules=Join[rules,{
comm[A_+B_,C_]:>comm[A,C]+comm[B,C],
comm[A_,B_+C_]:>comm[A,B]+comm[A,C],
comm[x_?NumericQ A_,B_]:>x comm[A,B],
comm[A_,x_?NumericQ B_]:>x comm[A,B],
comm[A_**B_,C_]:>A**comm[B,C]+comm[A,C]**B,
comm[A_,B_**C_]:>B**comm[A,C]+comm[A,B]**C,
comm[A_,𝟙]:>0,
comm[𝟙,A_]:>0
}];

For example, applying these rules can expand the following commutators

comm[a+2b,3c+4d]//.rules
comm[a**b,c**d]//.rules

Operator specific properties

[(Subscript[(OverHat[r]), a]),(Subscript[(OverHat[p]), b])]=i (Subscript[δ, a b])𝟙,
[((OverHat[r])^n),(Subscript[(OverHat[p]), a])]=i n ((OverHat[r])^(n-2))(Subscript[(OverHat[r]), a])
[(Subscript[(OverHat[r]), a]),((OverHat[p])^n)]=i n ((OverHat[p])^(n-2))(Subscript[(OverHat[p]), a])

rules=Join[rules,{
comm[(Subscript[r, a_]),(Subscript[p, b_])]:>I KroneckerDelta[a,b]𝟙,
comm[(Subscript[p, b_]),(Subscript[r, a_])]:>-comm[(Subscript[r, a]),(Subscript[p, b])],
comm[r^(n_:1),(Subscript[p, a_])]:>I n If[n==2,𝟙,r^(n-2)]**(Subscript[r, a]),
comm[(Subscript[p, a_]),r^(n_:1)]:>-comm[r^n,(Subscript[p, a])],
comm[(Subscript[r, a_]),p^(n_:1)]:>I n If[n==2,𝟙,p^(n-2)]**(Subscript[p, a]),
comm[p^(n_:1),(Subscript[r, a_])]:>-comm[(Subscript[r, a]),p^n],
comm[r^(n_:1),p^(m_:1)]:>comm[r^n,If[m==2,𝟙,p^(m-2)]**Sum[(Subscript[p, a])**(Subscript[p, a]),{a,3}]],
comm[p^(m_:1),r^(n_:1)]:>-comm[r^n,p^m],
comm[(Subscript[o_, a_]),(Subscript[o_, b_])]:>0,
comm[(Subscript[o_, a_]),o_^(n_:1)]:>0,
comm[o_^(n_:1),(Subscript[o_, a_])]:>0,
comm[o_^(n_:1),o_^(m_:1)]/;AtomQ[o]:>0
}];

For example, applying these rules can break down commutation relations

comm[(Subscript[r, 1]),(Subscript[p, 2])]//.rules
comm[p^2,(Subscript[r, 3])]//.rules
comm[(Subscript[p, 1]),1/r]//.rules

Rules regarding normal ordering (to rearrange the operators in a product in a canonical order by using commutation relations)

rules=Join[rules,{
(Subscript[r, a_])**(Subscript[r, b_])/;a>b:>(Subscript[r, b])**(Subscript[r, a]),
(Subscript[p, a_])**(Subscript[p, b_])/;a>b:>(Subscript[p, b])**(Subscript[p, a]),
(Subscript[p, a_])**(Subscript[r, b_]):>(Subscript[r, b])**(Subscript[p, a])+comm[(Subscript[p, a]),(Subscript[r, b])],
(Subscript[p, a_])**r^(n_:1):>r^n**(Subscript[p, a])+comm[(Subscript[p, a]),r^n],
p^(n_:1)**(Subscript[r, a_]):>(Subscript[r, a])**p^n+comm[p^n,(Subscript[r, a])],
(Subscript[o_, a_])**o_^(n_:1):>o^n**(Subscript[o, a]),
r^(n_:1)**r^(m_:1):>If[n+m==0,𝟙,r^(n+m)],
p^(n_:1)**p^(m_:1):>If[n+m==0,𝟙,p^(n+m)],
p^(n_:1)**r^(m_:1):>If[n==2,𝟙,p^(n-2)]**r^m**p^2-m(m+1)If[m==2,𝟙,r^(m-2)]
}];

Rules to define cross product of operators

First define the Levi-Civita tensor as ϵ

ϵ=Normal@LeviCivitaTensor[3]

Then use tensor contraction to define cross product

(Subscript[((OverHat[A])⋀(OverHat[B])), a]):=(Subscript[ϵ, a b c])(Subscript[(OverHat[A]), b])(Subscript[(OverHat[B]), c])

rules=Join[rules,{
As_List⋀Bs_List:>Activate@TensorContract[Inactive[ϵ⊗Outer[NonCommutativeMultiply,As,Bs]],{{2,4},{3,5}}]
}];

For example

{a,b,c}⋀{d,e,f}//.rules

A special rule to help packing (Subscript[(OverHat[o]), 1]^2)+(Subscript[(OverHat[o]), 2]^2)+(Subscript[(OverHat[o]), 3]^2) back to ((OverHat[o])^2) for o=r,p.

rule2=(Subscript[o_, 3])**(Subscript[o_, 3]):>o^2-(Subscript[o, 1])**(Subscript[o, 1])-(Subscript[o, 2])**(Subscript[o, 2]);

Starting with (OverHat[r]) and (OverHat[p])

𝕣={(Subscript[r, 1]),(Subscript[r, 2]),(Subscript[r, 3])};
𝕡={(Subscript[p, 1]),(Subscript[p, 2]),(Subscript[p, 3])};

Hamiltonian

H=(((OverHat[p])^2)/2)-(1/(OverHat[r])).

H=p^2/2-1/r

Angular momentum

(OverHat[L])=(OverHat[r])×(OverHat[p]).

𝕃=𝕣⋀𝕡//.rules

LRL vector

(OverHat[A])=(1/2)((OverHat[p])×(OverHat[L])-(OverHat[L])×(OverHat[p]))-((OverHat[r])/(OverHat[r])).

𝔸=(𝕡⋀𝕃-𝕃⋀𝕡)/2-(1/r)**𝕣//.rules

[(OverHat[H]),(Subscript[(OverHat[L]), a])]=0

comm[H,𝕃]//.rules

[(OverHat[H]),(Subscript[(OverHat[A]), a])]=0

comm[H,𝔸]//.rules//.rule2//.rules//Expand

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c])

(comm[𝕃,𝕃]//.rules)-I ϵ.𝕃//Expand

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[A]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[A]), c])

(comm[𝕃,𝔸]//.rules)-I ϵ.𝔸//Expand

[(Subscript[(OverHat[A]), a]),(Subscript[(OverHat[A]), b])]=-2 i  (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c])(OverHat[H])

(comm[𝔸,𝔸]//.rules)+(2I (ϵ.𝕃)**H//.rules)//.rule2//.rules//Expand

(OverHat[A])·(OverHat[L])=(OverHat[L])·(OverHat[A])=0,
2(OverHat[H])(((OverHat[L])^2)+𝟙)-((OverHat[A])^2)+𝟙=0.

Verify Eq. ().

(OverHat[A])·(OverHat[L])=0

Tr[𝔸**𝕃//.rules]//Expand

(OverHat[L])·(OverHat[A])=0

Tr[𝕃**𝔸//.rules]//Expand

2H(((OverHat[L])^2)+𝟙)-((OverHat[A])^2)+𝟙=0

(2H**(Tr[𝕃**𝕃//.rules]+𝟙)-Tr[𝔸**𝔸//.rules]+𝟙)//.rules//.rule2//.rules//Expand

The conservation laws of angular momentum and LRL vector are manifested by

[(OverHat[H]),(OverHat[L])]=[(OverHat[H]),(OverHat[A])]=0.

One might expect (OverHat[L]) and (OverHat[A]) to generate a symmetry group, however their algebra is not closed --- in particular, the [(OverHat[A]),(OverHat[A])] commutator leads to new operators like (OverHat[L])(OverHat[H]), which can not be written as a linear combination of (OverHat[L]) and (OverHat[A]) --- thus the algebra must be further extended.

Or ... maybe not. If we restrict our focus within each eigen subspace of (OverHat[H]) (the subspace spanned by eigenvectors of (OverHat[H]) with the same eigenvalue), then (OverHat[H]) is effectively reduced to a number (the eigenvalue) in the eigen subspace, and the algebra Eq. () is closed.

Emergent SO(4) Symmetry

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]),
[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[N]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[N]), c]),
[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[N]), b])]= i  (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]),

(OverHat[N])·(OverHat[L])=(OverHat[L])·(OverHat[N])=0,
2E(((OverHat[L])^2)+((OverHat[N])^2)+𝟙)+𝟙=0.

Eq. () further reduces to

[(Subscript[(OverHat[T]), a]),(Subscript[(OverHat[T]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[T]), c]),
[(Subscript[(OverHat[S]), a]),(Subscript[(OverHat[S]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[S]), c]),
[(Subscript[(OverHat[T]), a]),(Subscript[(OverHat[S]), b])]=0.

Derive Eq. () from Eq. ().

[(Subscript[(OverHat[T]), a]),(Subscript[(OverHat[T]), b])]=(1/4)[(Subscript[(OverHat[L]), a])+(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])+(Subscript[(OverHat[N]), b])]
=(1/4)([(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]+[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[N]), b])]+[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])]+[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[N]), b])])
=((i (Subscript[ϵ, a b c]))/4)((Subscript[(OverHat[L]), c])+(Subscript[(OverHat[N]), c])+(Subscript[(OverHat[N]), c])+(Subscript[(OverHat[L]), c]))
=i (Subscript[ϵ, a b c])(Subscript[(OverHat[T]), c]).

[(Subscript[(OverHat[S]), a]),(Subscript[(OverHat[S]), b])]=(1/4)[(Subscript[(OverHat[L]), a])-(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])-(Subscript[(OverHat[N]), b])]
=(1/4)([(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]-[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[N]), b])]-[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])]+[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[N]), b])])
=((i (Subscript[ϵ, a b c]))/4)((Subscript[(OverHat[L]), c])-(Subscript[(OverHat[N]), c])-(Subscript[(OverHat[N]), c])+(Subscript[(OverHat[L]), c]))
=i (Subscript[ϵ, a b c])(Subscript[(OverHat[S]), c]).

[(Subscript[(OverHat[T]), a]),(Subscript[(OverHat[S]), b])]=(1/4)[(Subscript[(OverHat[L]), a])+(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])-(Subscript[(OverHat[N]), b])]
=(1/4)([(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]-[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[N]), b])]+[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[L]), b])]-[(Subscript[(OverHat[N]), a]),(Subscript[(OverHat[N]), b])])
=((i (Subscript[ϵ, a b c]))/4)((Subscript[(OverHat[L]), c])-(Subscript[(OverHat[N]), c])+(Subscript[(OverHat[N]), c])-(Subscript[(OverHat[L]), c]))
=0.

This implies that (OverHat[T]) and (OverHat[S]) generates two independent SU(2) group, which combines into the SO(4) group (roughly speaking), see (HW ).

Eq. () reduces to the following constraints

((OverHat[T])^2)=((OverHat[S])^2),
2E (4((OverHat[T])^2)+𝟙)+𝟙=0.

Prove Eq. ().

((OverHat[T])^2)=(1/4)((((OverHat[L])+(OverHat[N])))^2)=(1/4)(((OverHat[L])^2)+(OverHat[L])·(OverHat[N])+(OverHat[N])·(OverHat[L])+((OverHat[N])^2)),
((OverHat[S])^2)=(1/4)((((OverHat[L])-(OverHat[N])))^2)=(1/4)(((OverHat[L])^2)-(OverHat[L])·(OverHat[N])-(OverHat[N])·(OverHat[L])+((OverHat[N])^2)).

((OverHat[T])^2)=((OverHat[S])^2)=(1/4)(((OverHat[L])^2)+((OverHat[N])^2)).

2E(((OverHat[L])^2)+((OverHat[N])^2)+𝟙)+𝟙=2E(4((OverHat[T])^2)+𝟙)+𝟙=0.

t=0,(1/2),1,(3/2),2,….

(Subscript[m, t]),(Subscript[m, s])=-t,-t+1,…,t-1,t.

For each fixed t, there are (((2t+1))^2) degenerated states.

Spectrum

Energy Levels

((OverHat[T])^2)=-(1/4)((1/(2E))+1)𝟙,

-(1/4)((1/(2E))+1)=t (t+1),

E=-(1/(2(4t (t+1)+1)))=-(1/(2(((2t+1))^2))).

Given t=0,(1/2),1,(3/2),2,…, one defines the principal quantum number

n:=2t+1=1,2,3,4,…

The energy levels are labeled by the positive integer n∈(SubPlus[ℤ])

Restoring the energy unit by undo the rescaling in Eq. (), the energy levels are given by

(Subscript[E, n])=-((m (α^2))/(2(n^2))).

The corresponding eigenstates (Ket[t,(Subscript[m, t]),(Subscript[m, s])]) (with t=(n-1)/2) are labeled by (Subscript[m, t]),(Subscript[m, s])=-t,-t+1,…,t-1,t => level degeneracy: (((2t+1))^2)=(n^2).

Ground State

The energy (Subscript[E, n])=-m (α^2)/(2(n^2)) is minimized when n=1 (or t=0).

When t=0, (Subscript[m, t]) and (Subscript[m, s]) has only one choice, i.e. (Subscript[m, t])=(Subscript[m, s])=0.

So the ground state has to be the (Ket[0,0,0]) state (also denoted as the 1,s orbital).

(OverHat[H]) (Subscript[ψ, 1,s])=(Subscript[E, 1])(Subscript[ψ, 1,s]),

((((OverHat[p])^2)/(2m))-(α/(OverHat[r])))(Subscript[ψ, 1,s])=-((m (α^2))/2)(Subscript[ψ, 1,s]).

(Subscript[ψ, 1,s])(r)∝(e^(-r/a)),

Verify that Eq. () is the solution of Eq. ().

((((OverHat[p])^2)/(2m))-(α/(OverHat[r]))+((m (α^2))/2))(Subscript[ψ, 1,s])(r)
=(-((Subscript[∇, r]^2)/(2m))-(α/r)+((m (α^2))/2))(Subscript[ψ, 1,s])(r)
=0.

(-((Subscript[∇, r]^2)/(2m))-(α/r)+((m (α^2))/2))(e^(-r/a))=0

Block[{r,ψ},
r=Sqrt[x^2+y^2+z^2];
ψ=Exp[-r/a];
Simplify[-Laplacian[ψ,{x,y,z}]/(2m)-(α/r)ψ+(m α^2/2) ψ]]

-1+a m α=0,

a=1/(m α) is called the Bohr radius.

The ground state energy is given by

as if the electron in orbiting the proton in a circular orbital of radius a (the classical picture).

In quantum mechanics, the electron does not move along a deterministic trajectory, instead, it distributes in a probability cloud, called the electron cloud.

p(r)=((|ψ(r)|)^2)∝(e^(-2r/a)).

Excited States

States with the same principal quantum numbers are degenerated in energy.

Within each degenerated sectors, the states (Ket[t,(Subscript[m, t]),(Subscript[m, s])]) form a complete set of basis (with (Subscript[m, t]),(Subscript[m, s])=-t,…,t), which can be viewed as a combined system of two “spins” of the same angular momentum t.

(Ket[n,l,m])

n - principal quantum number (n=1,2,…),

l - angular quantum number (l=0,1,…,n-1),

m - magnetic quantum number (m=-l,-l+1,…,l-1,l).

Atomic Structure

(Subscript[ψ, n l m])(r).

Atomic orbitals do not depict the exact location of an electron in an atom, but rather provide a probability distribution ((|(Subscript[ψ, n l m])(r)|)^2)  of where the electron is likely to be, which can be visualized as electron clouds:

DynamicModule[{nmax=4,n=1,l=0,m=0,ψ,compile,name,cfun,legend,plot,orbit,head,cloud},
ψ[$n_Integer,$l_Integer,$m_Integer]/;$n>=1&&$l<$n&&-$l<=$m<=$l:=ψ[$n,$l,$m]=FullSimplify[Block[{ρ=2r},
Sqrt[($n-$l-1)!/(2$n($n+$l)!)]
Exp[-ρ/2] ρ^$l LaguerreL[$n-$l-1,2 $l+1,ρ]] Which[$m>0,Sqrt[2]Re[#],$m==0,#,$m<0,Sqrt[2]Im[#]]&[SphericalHarmonicY[$l,Abs[$m],θ,ϕ]],
{r>=0,θ∈Reals,ϕ∈Reals}];
compile=Compile[{{x,_Real},{y,_Real},{z,_Real}},
Block[{ρ=x^2+y^2,r,θ,ϕ},If[ρ>0,r=Sqrt[ρ+z^2];θ=ArcCos[z/r];ϕ=ArcTan[x,y],
r=Abs[z];θ=π/2 Sign[z];ϕ=0];
#],CompilationTarget->"C"]&;
name[$n_,$l_,$m_]:=Style[StringReplace[StringTemplate["\!\(`1`,`2`\_\(`3`\)\)"][$n,<|0->"s",1->"p",2->"d",3->"f"|>[$l],StringReplace[ToString[InputForm[Block[{poly=z^($l-Abs[$m])ComplexExpand[Which[$m>0,Re[#],$m==0,#,$m<0,Im[#]]&[(x+I y)^Abs[$m]]]},Replace[poly,a_?NumericQ b_:>b]]]],{" "->"","^"->"\^","*"->" "}]],{"s\_\(1\)"->"s"}],"InlineFormula"];
cfun=(Blend[{{0,RGBColor[0.,0.,0.,0.]},{1/256,RGBColor[0.14,0.75,0.95,0.02]},
{4/256,RGBColor[0.,0.6,0.1,0.04]},{16/256,RGBColor[1.,0.84,0.23,0.08]},{64/256,RGBColor[1.,0.6,0.,0.16]},{1,RGBColor[0.8,0.2,0.,0.32]}},#]&);
legend=Deploy@Column[{Style["\!\(p(r)\) [a.u.]",Black,"InlineFormula"],Graphics[{Raster[Transpose@{Table[{#1,#2,#3,Clip[16#4]}&@@cfun[10^lgx],{lgx,-3,0,0.1}]},{{0,-3},{1,0}}]},AspectRatio->8,Frame->True,FrameTicks->{{None,{#,If[#==0,"1",StringTemplate["\!\(10\^`1`\)"][#]]}&/@Range[-3,1]},{None,None}},ImageSize->{Automatic,100}]}];
plot[$n_,$l_,$m_]:=Block[{ps=Range[-8,8,0.25],nψ=compile[ψ[$n,$l,$m]],data,vol},data=Table[Abs[nψ[x,y,z]]^2,{z,ps},{y,ps},{x,ps}];
vol=RawArray["Byte",Round[255 Clip[data/0.01]]];
Show[{Image3D[vol,"Byte",Method->{"FastRendering"->True,"InterpolateValues"->True},ColorFunction->cfun,BoxRatios->1],Graphics3D[{AbsoluteThickness[1],Translate[Scale[{#1,Rotate[Line[{{#,0,0}&/@{-0.5,-0.25},{#,0,0}&/@{0.5,0.25}}],2π #2/3,{1,1,1},{0,0,0}]}&@@@{{Red,0},{Green,1},{Blue ,2}},Length[ps]-1,{0,0,0}],(Length[ps]-1){1,1,1}/2]}]},Background->Black,SphericalRegion->True,Boxed->False,ViewPoint->{5,3,2},ViewVertical->{0,0,1},ImageSize->180]];
cloud=<||>;
orbit[$n_,$l_,$m_]:=If[$l<$n,cloud[{$n,$l,$m}]=plot[$n,$l,$m];Item[Button[name[$n,$l,$m],{n,l,m}={$n,$l,$m},Appearance->"Frameless"],Background->Dynamic[If[{n,l,m}=={$n,$l,$m},LightGray,White]]],""];
head=Block[{ls,ms},
{ls,ms}=Thread@Flatten[Table[{l,m},{l,0,nmax-1},{m,SortBy[Range[-l,l],Abs]}],1];
ls=Flatten[Prepend[ConstantArray[SpanFromAbove,Length@#-1],Style[StringTemplate["\!\(l=`1`\)"][#[[1]]],"InlineFormula"]]&/@Split[ls]];
ms=Which[#==0,Style["\!\(m=0\)","InlineFormula"],#<0,Style[StringTemplate["\!\(m=±`1`\)"][Abs[#]],"InlineFormula"],#>0,SpanFromAbove]&/@ms;{Prepend[ls,""],Prepend[ms,SpanFromLeft]}];
Column[{Row[{Dynamic[cloud[{n,l,m}]],legend},Spacer[1]],Deploy[Grid[Transpose@Join[head,
Table[Prepend[Flatten@Table[orbit[$n,$l,$m],{$l,0,nmax-1},{$m,SortBy[Range[-$l,$l],Abs]}],Style[StringTemplate["\!\(n=`1`\)"][$n],"InlineFormula"]],{$n,1,nmax}]],Alignment->{Center,Center},Dividers->All,ColumnWidths->{2.2,3.5,2.5,2.5,3.5,4.2},FrameStyle->LightGray]]},Alignment->Center],SaveDefinitions->True]

n - labels energy shells, controls the radial excitation,

l - labels energy sub-shells, controls the angular excitation, and are given the following names

({{l=, 0, 1, 2, 3, 4, 5, …}, { , s, p, d, f, g, h, …}})

m - specifies the orientation of the orbital.

Block[{nmax=10,cut=15,x0=-5,x1=2,cols,name,shell,split},
cols=<|1->Red,2->Orange,3->RGBColor[0.9,0.75,0],4->Green,5->Cyan,6->Blue,7->Purple|>;
name[n_,l_]:=Style[StringTemplate["\!\(`1`,`2`\)"][n,Lookup[<|0->"s",1->"p",2->"d",3->"f",4->"g",5->"h",6->"i",7->"j"|>,l,ToString[l]]],"InlineFormula"];
shell=GroupBy[Thread[{#,Range@Length@#-1}&@SortBy[Flatten[Table[{n,l},{n,1,nmax},{l,0,n-1}],1],{#.{1,1},#{1,0}}&]],#[[1,1]]&];
split=Cases[Flatten[Values@shell,1],{{_,0},e_}/;e<=cut:>e-0.5];
Graphics[{{Lighter[Lookup[cols,#2],0.9],Rectangle[{0,#1[[1]]},{x1+5,#1[[2]]}]}&@@@Thread[{#,Range@Length@#}&@Thread@Through[{Most,Rest}@Append[split,cut+0.5]]],{GrayLevel[0.7],Dashed,AbsoluteThickness[1.2],Translate[Line[{#,0}&/@{0,x1+5}],{0,#}]&/@split},
{GrayLevel[0.3],Text[StringTemplate["Period `1`"][#2],{x1+5,#1},{1.2,0.2}]&@@@Thread[{#,Range@Length@#}&[Mean/@Thread@Through[{Most,Rest}@split]]]},Table[Block[{es=Last/@shell[n],e0,er,col},
e0=Mean[es];er=Abs[Subtract@@MinMax[es]];
col=Lookup[cols,n,Black];
{col,If[e0<=cut,{AbsoluteThickness[1.5+0.2(n-1)],Line[{{x0,e0},{-er/3,e0}}],Text[StringTemplate["\!\(n=`1`\)"][n],{x0,e0},{1.1,0}]},Nothing],If[#<=cut||e0<=cut,Line[{{-er/3,e0},{0,#}}],Nothing]&/@es,If[#2<=cut,Translate[{Line[{#,0}&/@{0,x1}],Text[name@@#1,{x1,0},{-1.5,0.2}]},{0,#2}],Nothing]&@@@shell[n]}],{n,Keys@shell}]},PlotRange->{{x0-2,x1+5},{-0.6,cut+0.5}},ImageSize->240]
]

As the atomic number increases, more electrons will fill the energy sub-shells from the bottom up. The electron configuration determines the chemical properties of the atom.

Periodic table in chemistry: The chemical properties of an element are dependent on its outermost electrons. Therefore, whenever a new s orbital begins to be filled, a new period of elemental properties commences.