Lecture 27 Supplement 1: Angular Momentum Eigenvalues

Supplement #1 to Lecture #27

Angular Momentum Eigenvalues (from lecture notes by Professor Dud- ley Herschbach)

Consider any Hermitian operator J∼ whose components satisfy the following commutation rules

[Jx,Jy] = i}Jz and the cyclic permutations thereof. Equivalently, the rules may be written as

J∼ × J∼ = i}J∼ or as X [J`,Jm] = i} ε`mnJn n where

ε`mn = +1 if `, m, n are in cyclic order = −1 if `, m, n are in anti-cyclic order = 0 if any two of `, m, n are the same.

2 Seek to ﬁnd eigenvalues λ for J and µ for Jz such that

J 2 |λµi = λ |λµi

Jz |λµi = µ |λµi .

2 Since J and Jz are Hermitian, λ and µ are real, and |λµi are the simultaneous 2 eigenvectors which render J and Jz simultaneously diagonal. First show λ ≥ µ2 2 2 2 Proof: hλµ |J − Jz | λµi = λ − µ But 2 2 2 2 �2 �2 J − Jz = Jx+ J y + �Jz − �J z

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2 X 0 0 0 0 λµ J λµ = hλµ |Jx| λ µ i hλ µ |Jx| λµi x λ0µ 0 | {z } † 0 0 † λµ Jx λ µ and Jx = Jx

X 0 0 2 2 = |hλµ |Jx| λ µ i| → 0 and similarly for Jy term. λ0µ 0 So 2 2 2 λµ Jx + Jy λµ = λ − µ ≥ 0 Q. E. D. Since µ2 ≥ 0 this also implies λ ≥ 0. It is convenient to use the non-Hermitian operators

† † J± = Jx ± iJy Note J+ = J−,J− = J+.

These satisfy

[Jz,J±] = ±}J± since [Jz,Jx ± iJy] = i}Jy ± i(−i}Jx)

= }(Jx ± iJy) = }J±.

Apply this to |λµi and ﬁnd

(JzJ± − J±Jz) |λµi = ±}J± |λµi or

Jz(J± |λµi) = (J±Jz ± }J±) |λµi

= (µ ± })(J± |λµi) since Jz |λµi = µ |λµi .

Thus J± |λµi is an eigenvector of Jz with eigenvalue µ±}. Hence J+ “raises” the eigenvalue of µ to µ + } and J− “lowers” the eigenvalue of µ to µ − }. Now note 2 [J ,J±] = 0

2 since J commutes with its components Jx and Jy. Thus

2 2 J (J± |λµi) = J± J |λµi = λ(J± |λµi). | {z } λ |λµi

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2 Thus J± |λµi remains an eigenvector of J with the same eigenvalue λ as |λµi.

By repeated application of J+ we can get eigenvectors with Jz eigenvalues of µ + }, µ + 2}, . . . but the same eigenvalue λ of J 2 . Since µ2 ≥ λ, for a given

λ there must be some highest value of µ, call it µh, such that J+ |λµhi = 0 rather than generating a new eigenvector of still higher Jz–eigenvalue. Simi- larly, repeated application of J− gives µ − }, µ − 2}, . . . but would eventually 2 violate µ ≤ λ unless there is some lowest value of µ, call it µ`, such that

J− |λµ`i = 0.

Now we use these conditions to show µn = −µ`. Consider applying J− to

J+ |λµhi = 0. Note the identity:

J−J+ = (Jx − iJy)(Jx + iJy) 2 2 = Jx + Jy + i[Jx,Jy] 2 2 = J − Jz − }Jz. Thus 2 J−J+ |λµi = (λ − µ h − }µh) |λµhi = 0.

Taking the matrix element with hλµh| we ﬁnd

2 λ − µ h − }µh = 0. Similarly, 2 2 J+J− |λµ`i = (J − Jz + }Jz) |λµ`i leads to 2 λ − µ` + }µ` = 0. Hence

λ = µh(µh + }) = µ`(µ` − }) | {z }

Two solutions: µh = −µ` or µh = µ` − } but this second solution must be rejected since µh was assumed to be larger than µ`.

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Now we can conclude also that µh = µ` + n} where n is some integer. This follows since, if we start from |λµ`i and apply J+ repeatedly, we obtain the sequence of eigenvectors:

2 n |λµì,J+ |λµì,J+ |λµì, ... J+ |λµì = |λµni | {z } | {z } µ µ + | {z } | {z } ` ` } µ`+2} µ`+n}=µ} Thus

µh = −µ` = µ` + n} or n n µ = − , µ = + ` 2} h 2 } where n = 0, 1, 2,... is some integer (related to the value of λ). For convenience, we write

µ = m}, m = −j, −j + 1, ··· + j

n 1 3 where j = 2 , with j = 0,,2 1,, 2 2,... Then eigenvalues of Jz are −j}, (−j + 1)}, . . . j} | {z } 2j+1 diﬀerent values Eigenvalues of J 2 are given by

λ = µh(µh + }) = µ`(µ` − }) = j}(j} + }) = −j}(−j} − }) 2 λ = j(j + 1)}

Also, it is convenient to label the eigenvectors by j, m rather than λ, µ, so

2 2 J |jmi = j(j + 1)} |jmi

jz |jmi = m} |jmi . Comments We derived the above eigenvalues using only the commutation property and the Hermitian property. We ﬁnd that both integer and half-integer values of j and m are allowed.

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Actually, we have solved a much more general problem than that posed by the orbital angular momentum of a particle. Thus, for several particles in the same central force ﬁeld, the total angular momentum,

X (n) L∼ , n also satisﬁes these relations, even if the particles interact with each other. Spin angular momenta likewise satisfy these relations. For orbital angular momentum, L = q × p must require, in addition, that the ∼ ∼∼ system returns to its original state under a rotation by 2π . Such a rotation takes p → p and q → q so q × p → q × p and hence the eigenvectors of L2 ∼∼ ∼∼∼∼ ∼ ∼ and Lz must be unchanged:

e− i2πJz /} |jmi = e −i2πm |jmi e −i2πm = +1 if m is integer and hence integer eigenvalues are acceptable for 2 −i2πm L , Lz. Half-integer values give e = −1 and hence are not acceptable for orbital angular momentum. Half-integers do apply for spin angular momenta, which are not constructed from any q ×p and thus can take on both integer and half-integer eigenvalues. ∼ ∼ This illustrates the power of operator derivation. A more general case would not have been included if we had used wave mechanical methods and repre- sentations by diﬀerential operators. We have shown that, for a ﬁxed j value,

J+ |jmi = am |j, m + 1i and J− |jmi = bm |j, m − 1i , where am and bm are constants, possibly complex numbers. The proportionality constants are simply related to each other, since ⎛ ⎞∗ * +∗ Z a = hj, m + 1|J |jmi = jm |J † | j, m + 1 = ⎜ ψ∗ J ψ dτ⎟ m + + ⎝ jm − j,m+1 ⎠ |{z} | {z } J− bm+1ψj,m ∗ = b m+1

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Now, to evaluate am, consider the identity

2 2 J−J+ = J − Jz − }Jz.

Apply this to |jmi, then you have

2 J−J+ |jmi = amJ− |j, m + 1i = ambm+1 |jmi = |am| |jmi 2 2 2 2 J − Jz − }Jz |jmi = (j(j + 1) − m − m)} |jmi .

Hence

1/2 iφm am = [j(j + 1) − m(m + 1)] }e | {z } (j − m)(j + m + 1) ← another common way of writing it, where eiφm is an arbitrary phase factor. The usual convention is to take φ = 0; this ﬁxes the relative phases of the vectors |jmi having diﬀerent values of m but the same j.

The only non-vanishing matrix elements of J+ and J− are:

1/2 hj, m + 1|J+|jmi = hj, m|J−|j, m + 1i = [j(j + 1) − m(m + 1)] } 6 always the lower times the higher of the two m–values in Or you can write this alternatively as the matrix element

0 0 1/2 hj , m |J+|jmi = [j(j + 1) − m(m + 1)] }δj0,j δm 0 ,m+1 0 0 1/2 hj , m |J−|jmi = [j(j + 1) − m(m − 1)] }δj0,j δm0 ,m−1

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List of non-zero elements:

2 2 jm|J |jm = j(j + 1)} “add the bigger m to j and subtract the smaller” hjm|Jz|jmi = m} ? 1/2 z }| { 1/2 hj, m ± 1|J±|jmi = [j(j + 1) − m(m ± 1)] } = [(j ± m + 1)(j m)] } 1 1 J = (J + J ),J = (J − J ) x 2 + − y 2i + − 1 hj, m ± 1|J |jmi = [j(j + 1) − m(m ± 1)]1/2 x 2 } 1 hj, m ± 1|J |jmi = ± [j(j + 1) − m(m ± 1)]1/2 y 2i }

We can summarize elements of Jx, Jy, Jz by:

jm|J∼ |jm = zmˆ } 1 j, m ± 1|J|jm = (ˆx ± iyˆ) [j(j + 1) − m(m ± 1)]1/2 . ∼ 2 } Comment 2 Thus we have found all matrix elements of J with eigenvectors |jmi of J ,Jz. These eigenvectors and their properties are important, since any time we have a system of particles isolated in free space, their total angular momentum 2 J ,Jz commutes with the total Hamiltonian, no matter what kind of forces hold the system together (central or not). That is, the total angular momentum of an isolated system is a constant of the motion in quantum mechanics, just as in classical mechanics. Hence it is important to be able to take matrix elements of other operators in the angular momentum states which characterize an isolated system.

Examples

2 j = 0 : J+ = (0) J− = (0) Jz = (0) J = (0)

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1 1 3 0 1 0 0 2 0 2 4 0 j = : J+ = J− = Jz = 1 J = 3 2 0 0 1 0 0 − 2 0 4 j = 1: ⎛ √ ⎞ ⎛ ⎞ 0 2√ 0 1 0 0 J+ = ⎝0 0 2⎠ Jz = ⎝0 0 0 ⎠ 0 0 0 0 0 −1

⎛ ⎞ ⎛ ⎞ √0 0 0 2 0 0 2 J− = ⎝ 2√ 0 0⎠ J = ⎝0 2 0⎠ 0 2 0 0 0 2 j = 3 : 2 √ ⎛ ⎞ ⎛ 3 ⎞ 0 3 0 0 2 0 0 0 0 0 2 0 0 1 0 0 ⎜ √ ⎟ ⎜ 2 ⎟ J+ = ⎜ ⎟ Jz = ⎜ 1 ⎟ ⎝0 0 0 3 ⎠ ⎝0 0 − 2 0 ⎠ 3 0 0 0 0 0 0 0 − 2

⎛ 0 0 0 0⎞ ⎛ 15 0 0 0 ⎞ √ 4 15 ⎜ 3 0 0 0⎟ 2 ⎜ 0 4 0 0 ⎟ J− = ⎜ ⎟ J = ⎜ 15 ⎟ ⎝ 0 2 0 0⎠ ⎝ 0 0 4 0 ⎠ √ 15 0 0 3 0 0 0 0 4

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MOMENTA AS DISPLACEMENT OPERATORS: Geometrical Meaning of Commutation Rules Linear Momentum

Let |x1i be an eigenvector of the position operator X with eigenvalue x1, i.e.

X |x1i = x1 |x1i .

−iapx/} Consider the new state vector deﬁned by e |x1i; we might ask whether it is also an eigenvector of X. To ﬁnd out, evaluate

�−iapx/} −iapx/} −iapx/} X e |x1i = e (X |x1i) + X, e |x1i | {z } | {z } x1|x1i 6 ...... Now ...... d .. −iapx/} � −iapx/..} ... X, e = i e ... } ...... dpx ...... = ae −iap.... .x...... /...... }......

Thus

�−iapx/} −iapx/} −iapx/} X e |x1i = e (X + a) |x1i = e (x1 + a) |x1i

�−iapx/} = (x1 + a) e |x1i .

−iapx/} Hence e |x1i is indeed an eigenvector of X with eigenvalue x1+a instead −iapx/} of x1. The unitary operator e formed from the linear momentum operator px acts as a displacement operator for x position coordinates. Similarly, py generates displacements of the y coordinate and pz of the z coordinate. It is a geometrical fact that linear displacements of a point commute. For example:

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x2 y2 y ﬁrst move along x 6 6 - tb by a, then along y x1y1 −ibpy /} −iapx/} a by b |x2y2i = e e |x1y1i ⇒ t t- x

a - x y y 6 2 2 b t t reverse the −iapx/} −ibpy /} x1y1 order = e e |x1y1i ⇒ t - x

The same result is obtained by applying displacements in either order. This agrees with [px, py] = 0 (and {px, py} = 0).

Angular momentum

Lx,Ly,Lz operators generate angular displacements or rotations; e.g.,

e −iφLx/} gives a rotation by angle φ about the x–axis, etc. However, geometrical rotations about diﬀerent axes do not commute. For example, 6z

...... consider a state representing a particle on the z–axis, |z0i. Now...... π π . t −i Ly /} −i Lx/} e 2 e 2 |z0i ⇒ particle on −y axis ? - y | {z } rotation by π/2 | {z } rotation by π/2 t about y–axis about x–axis x z But 6 ...... t π π . −i Lx/ −i Ly / .. 2 } 2 } .. e e |z i ⇒ particle on +x axis ... - 0 ... y ... | {z } | {z } R... rotation by π/2 rotation by π/2 about x–axis about y–axis x t The results of these two rotations taken in opposite order diﬀer by a rotation about the z–axis. Thus, because the rotations about diﬀerent axes don’t commute, we must expect the angular momentum operators, which generate these rotations, not to commute with each other. Indeed,

[Lx,Ly] = i}Lz

10 Lecture #27 Supplement #1 Page 27S(1)-11 corresponds to the above example, in which the commutator of rotations about the x and y axes depends on a z–axis rotation. Rotational Transformation Properties and Selection Rules The various observables of a dynamical system can be classiﬁed according to their transformation properties under rotations. This is of great value in determining the matrix elements of the corresponding operators and, in particular, leads to selection rules which limit the number of non–zero matrxix elements. i φ·J /} Under action of the rotation operator U = e ∼ ∼ an operator O is trans- formed according to 0 † O∼ = U∼ O∼ U∼ .

A scalar operator S∼ is one which is invariant to this transformation (e.g., the Hamiltonian of an isolated system). Hence, for a scalar operator

† U SU∼∼∼ = S∼ or

U∼ S∼ − S∼U∼ = 0 or [U ,S∼∼ ] = 0. Thus, a scalar commutes with every rotation operator. Consider, in particular, an inﬁnitesimal rotation dφ, for which ∼ i U = 1 + dφ · J. ∼ } ∼ ∼ Since the direction of dφ is arbitrary, S must commute with each component ∼ ∼ of J∼ , or [S∼,J∼ ] = 0. As shown below, this property leads to the selection rules Δj = 0, Δm = 0 for the non-zero matrix elements of a scalar operator.

A vector operator V∼ is one with three components, Vx,Vy,Vz which transform under rotations like the coordinates of a point. For an inﬁnitesimal rotation,

i i V 0 = 1 + dφ · J V 1 − dφ · J . ∼ } ∼ ∼∼ } ∼ ∼

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0 Now note that if the position vector ∼r is obtained from ∼r by rotation through a small angle dφ about an axis in the direction of the vector dφ, we have, to ∼ ﬁrst order in dφ, 0 . r = r + dφ × r :...... dφ ...... ∼∼ ∼ ...... ∼ ∼ ..... ..... .... ... .. . - . dφ X .. X .. X r .... X ∼ ..... X ..... X ..... and so X ...... XX...... r 0 Xz.... V 0 = V + dφ × V. ∼ ∼ ∼ ∼ ∼ Hence, if terms in (dφ)2 are neglected, we obtain i h i dφ × V = dφ · J V − V dφ · J . ∼ ∼ } ∼ ∼ ∼ ∼ ∼ ∼ Since dφ is arbitrary, this relation gives the commutator of V with any com- ∼ ∼ ponent of J. Thus, if dφ = εzˆ is a rotation about the z–axis, we ﬁnd ∼ ∼ i ε (zˆ × V ) = ε Jz V − V Jz ∼ } ∼ ∼ or Jz , V∼ = −i} (zˆ × V∼ ) or [Jz,Vx ] = −i}(−Vy) = i}Vy

[Jz,Vy] = −i}(−Vx) = −i}Vx

[Jz,Vz] = 0 etc. In this way we obtain a set of nine commutation rules:

[Jx,Vx] = 0 [Jy,Vx] = −i}Vz [Jz,Vx] = −i}Vy

[Jx,Vy] = i}Vz [Jy,Vy] = 0 [Jz,Vy] = −i}Vy

[Jx,Vz] = −i}Vy [Jy,Vz] = i}Vx [Jz,Vz] = 0 The selection rules for non-zero matrix elements of a vector operator, i.e. an operator which satisﬁes the above rules (e.g., position ∼r, linear momentum p, the angular momentum J itself) are shown below to be given by ∼ ∼

Δj = 0 and ± 1 for all components of V∼

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Δm = 0 for Vz

Δm = ±1 for V± = Vx ± iVy. Scalar Operators, S1 2 Deﬁned by [S∼,J∼] = 0, for all three components of J∼. Corollary is [S, J ] = 0 and [S, Jz] = 0. If we take the matrix elements, we have

j0 m0 |[S, J 2]|jm = 0 = j0 m 0|SJ 2 − J 2S|jm 2 0 0 0 0 = } hj m |Sj(j + 1) − j (j + 1)S|jmi 2 0 0 0 0 = } [j(j + 1) − j (j + 1)] hj m |S|jmi .

Also,

0 0 hj m |[S, Jz]|jmi = 0 0 0 = hj m |SJz − JzS|jmi 0 0 0 = } hj m |Sm − m S|jmi 0 0 0 = }(m − m ) hj m |S|jmi .

Therefore, hj0m0|S|jmi must vanish unless j = j0 and m0 = m0 . “Selection rules” for non-zero elements are: Δj = 0 and Δm = 0.

Let sjm ≡ hjm|S|jmi denote the non-vanishing element. Since this is the only non-zero matrix element, |jmi is an eigenvector of S, i.e. S |jmi = sjm |jmi. Now we can show that the eigenvalues of the scalar operator S don’t depend on m. Since S commutes with J± = Jx ± iJy, we have

S (J+ |jmi) = J+S |jmi = sjm (J+ |jmi) .

But J+ |jmi is proportional to |j, m + 1i and still has some eigenvalue sjm 2 of S. We could continue this with J+ → m + 2,... and with J− → m − 1,

1These notes were prepared by Professor Dudley Herschbach of Harvard University

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2 J− → m−2, etc., and would get the same eigenvalue sjm of S for all m states of a given j. Hence we would obtain

0 0 hj m |S|jmi = hjkSkji δjj0 δmm0 where hjkSkji is called a reduced matrix element, a number that does not depend on m. The above equation only describes the properties of S which are associated with its scalar character. In general, the states of the system will depend upon other quantum numbers in addition to j and m. If these are denoted collectively by α, the scalar operator need not be diagonal in α, so the general statement becomes

0 0 0 0 hα j m |S|αjmi = hα jkSkαji δjj0 δmm0 for

[S, J∼ ] = 0. Vector Operators, V

Deﬁnition: A vector operator V∼ with respect to the angular momentum J∼ is any set of three operators Vx,Vy,Vz that satisfy the following commutation rules: X [Ji,Vj ] = i} εijkVk εijk = 1, ijk cyclic k = −1, ijk anti-cyclic = 0, any two subscripts the same This is shorthand for

[Jx,Vx] = 0 [Jy,Vx] = −i}Vz [Jz,Vx] = i}Vy

[Jx,Vy] = i}Vz [Jy,Vy] = 0 [Jz,Vy] = −i}Vx

[Jx,Vz] = −i}Vy [Jy,Vz] = i}Vx [Jz,Vz] = 0. It is convenient to use 1 1 V = V ± iV V = (V + V ); V = (V − V ). ± x y x 2 + − y 2i + −

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Selection Rules for m

Consider the commutators involving Jz, take matrix elements of the commutators:

a) [Jz,Vz] = 0

0 0 0 0 0 0 0 0 hj m |JzVz − VzJz|jmi = hj m |m }Vz − Vzm}|jmi = }(m −m) hj m |Vz|jmi

0 0 0 Thus hj m |Vz|jmi = 0 unless m = m, Δm = 0

b) [Jz,V+] = [Jz,Vx + iVy] = i}Vy + i(−i}Vx) = }V+

0 0 0 0 hj m |JzV+ − V+Jz|jmi = } hj m |V+|jmi.

0 0 0 }(m − m − 1) hj m |V+|jmi = 0

0 0 0 hj m |V+|jmi = 0 unless m − m = +1, Δm = +1

c) Similarly, [Jz,V−] = −}V− and

0 0 0 hj m |V−|jmi = 0 unless m − m = −1, Δm = −1

Selection Rules for j

To ﬁnd the selection rules for j, we want to examine commutators of V∼ with 2 J∼ . For this, some vector identities are useful. First we show

(1) J∼ × V ∼ + V ∼ × J ∼ = 2i}V∼ . This relation is another way to deﬁne a vector operator. It states that, because of the non-commuting algebra of quantum mechanics, J∼ × V ∼ =6

−V∼ × J ∼ as would hold for ordinary vectors.

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0.0.1 Proof:

X re-label.. via j ↔ k (J × V + V × J)i = (εijkJjVk + εijkVjJk) ..... ∼∼∼ ∼ ...... 6...... 6. 66 ...... j,k Then use ε = −ε X ijk ijk = εijk(Jj Vk − VkJj ) j,k X X using the deﬁnition = ε [J , V ] = i ε ε V ijk j k } ijk jk` ` of a vector operator j,k jk`

Note εijkεjk` = εijkε`jk as a cyclic permutation of subscripts leaves εijk unchanged. Then factor 2 appears because both odd-odd X εijk ε`j k = 2δi` and even-even permutations give a j,k contribution So X (J∼ × V∼ + V∼ × J∼)i = 2i} δi`V` = 2i}Vi Q.E.D. ` Now we show 2 (2) [J∼ , V∼ ] = i} (V∼ × J − J × V∼∼∼ ).

Proof:

2 X 2 [J , Vj ] = [Ji , Vj ] i X = {Ji[Ji,Vj ] + [Ji,Vj ]Ji} i ⎧ ⎫ X ⎨ ⎬ [J 2,V ] = i J ε V + ε V J j } i ijk k ijk k i i,k ⎩ |{z} |{z} ⎭

= i} ( −J∼ × V∼ + V∼ × J∼)j Q.E.D.

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It is convenient to deﬁne the operator 1 K ≡ (V × J − J × V ). ∼ 2 ∼ ∼∼ ∼

This is Hermitian (since V∼ and J∼ are) and is a vector operator if V∼ is. Then Equation (2) states 2 [J∼ ,V∼ ] = 2i}K∼ . (1)

However, we can’t yet use this commutator to get selection rules on V∼ , since the matrix elements of the commutator K∼ would seem to bear no simple relation to those of V∼ . We will ﬁnd that selection rules can be obtained from an identity involving the double commutator, 2 2 2 2 2 (3) [J∼ , [J∼ ,V∼ ]] = 2} {J V∼∼ − 2(J∼ · V∼ )J∼ + V∼ J∼ }.

This can be proven by examining further the properties of K∼ .

2 2 2 [J , [J ,V∼∼ ]] = 2i} [J∼ ,K∼ ].

Since K∼ is a vector operator, we have from (2) that

2 [J∼ ,K∼ ] = i} ( K∼ × J∼ − J∼ × K∼ ).

Also, from Equation (1) we have

J × K∼∼ + K∼ × J∼ = 2i}K∼ .

Hence

J∼ × K∼ − K∼ × J∼ = J∼ × K∼ − (2i}K − J∼ × K∼ )

= 2J∼ × K∼ − 2i}K∼ .

Also, from equation (1) 1 K ≡ (V × J − J × V ) = V × J − i V . ∼ 2 ∼ ∼ ∼ ∼ ∼ ∼ } ∼

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Thus,

J∼ × K∼ = J × (V∼ × J∼ ) − i} (J∼ × V∼ ) X (J∼ × K∼ )i = εijk Jj (V∼ × J∼ )k −i} (J∼ × V∼ )i j,k | {z }

X shift to ε = ε since cyclic εijkJj ε k`mV`Jm ..... `mk k`m ...... 6 ...... jk`m ...... permutation of subscripts leaves ε ...... unchanged

X X |(δi`δjm − δimδj`)Jj V`Jm = (Jj ViJj − Jj Vj Ji) j`m j X = (Jj Jj Vi − Jj [Jj ,Vi] − Jj Vj Ji) j

2 X = J Vi − Jj i}εji`V` −(J − V )Ji ∼ ∼ replace by −ε , j` 6 ij` | {z } a non-cyclic X permutation + i} Jj εij`V` = i} (J∼ × V∼ )i. j`

So we ﬁnd

2 (J × K∼∼ )i = J∼ Vi − (J∼ · V∼ )Ji + i}(J × V )i − i}(J × V )i or 2 J × K∼∼ = J∼ V∼ − (J∼ · V∼ )J∼. Now we can use these results to simplify the double commutator,

2 2 [J∼ , [J,V∼∼ ]] = 2i}[J∼ ,K∼ ] = (2i})(i})(−1)(2J∼ × K∼ − 2i}K∼ ) 2 = 2} (2J∼ × K∼ − 2i}K∼ ) 2 2 2 = 2} {2J V∼ − 2(J · V∼∼ )J∼ − [J∼ , V∼ ] | {z }

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2 2 J V∼ − V∼ J and ﬁnally,

2 2 2 2 [J , [J∼∼ ,V∼ ]] = 2} {J V∼ − 2(J · V )J∼∼∼ + V∼ J } Q.E.D.

Now we can obtain selection rules by taking matrix elements of this relation. Consider two cases:

Case I: Elements diagonal in j: Wigner-Eckart Theorem

jm0|[J 2,A]|jm = hjm0|j(j + 1)A − Aj(j + 1)|jmi = 0 for any operator A. Thus,

0 2 2 2 0 2 2 jm |[J , [J ,V ]]|jm = 0 = 2} jm |J V∼ − 2(J∼ · V∼ )J∼ + V∼ J |jm or

2 0 0 } j(j + 1) jm |V∼ |jm − jm |(J∼ · V∼ )J∼ |jm = 0 | {z }

X 0 00 00 00 00 = jm |(J∼ · V∼ |j m j m |J∼ |jm . j00 m 00

The operator (J · V∼∼ ) is a scalar with respect to J∼ and therefore diagonal in both m and j, so that j00 = j and m00 = m0 , and its matrix elements are independent of m. Hence we ﬁnd

j|J · V |j jm0|V |jm = ∼∼ jm0|V |jm ∼ }2j(j + 1) ∼

This is the Wigner-Eckart theorem for a vector operator.

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Powell & Craseman, page 371 j|J · V |j jm 0|V |jm = ∼ ∼ jm 0|J|jm ∼ j(j + 1)}2 ∼

6 Suppose V precesses around J. The time averaged J∼ ∼ ∼

...... value of the component normal to J is zero. The ...... ∼ ...... time average of V is therefore parallel to J and has ...... ∼ ∼ ...... magnitude . . . - . . . . . J · V ...... .. .. ...... ... |J| ...... ...... Hence, on this model, the average is J · V V∼ ∼ ∼ |J| J · V (J · V ) J - V = ∼∼ J = ∼ ∼ J 2 ∼ |J| |J|

The theorem is very useful, as it states that, for any vector operator V∼ , the matrix elements diagonal in j are simply proportional to the corresponding matrix elements of J∼ itself. The proportionality constant, c0(j) = j |(J · V∼∼ )|j (}j(j + 1)) is the same for all m–states. Therefore, we have via the Wigner-Eckart Theorem:

1/2 hj, m + 1|V+|jmi = c0(j)[j(j + 1) − m(m + 1)]

hjm|Vz|jmi = c0(j)m 1/2 hj, m − 1|V−|jmi = c0(j)[j(j + 1) − m(m − 1)]

0 with c0(j) = hα jkV kαji a reduced matrix element. In particular, we note that all matrix elements of V∼ between j = 0 states vanish.

Case II: Elements non-diagonal in j

Now consider j0 6= j, again take matrix elements of Equation (3). LHS gives

j0 m 0|[J 2 , [J 2,V ]]|jm = j0 m0 |J 2(J 2V − VJ 2) − (J 2V − VJ 2)J 2|jm = {j02(j0 + 1)2 − 2j(j + 1)j(j + 1) + j2(j + 1)2} hj0 m0 |V |jmi .

20 Lecture #27 Supplement #1 Page 27S(1)-21

RHS gives * + 2 0 0 2 2 2 0 0 0 0 2h j m |J V∼ − 2(J∼ · V ∼ )J ∼ +VJ ∼ |jm = 2} {j (j + 1) + j(j + 1)} j m |V∼ |jm | {z } 0 0 0 drops out as j m |J∼ |jm = 0, because j =6 j.

Equating LHS = RHS and rearranging gives

0 2 0 2 0 0 {(j − j) − 1}{(j + 1 + 1) −1} j m |V∼ |jm = 0 | {z } This factor > 0 since j0 =6 j and j0 ≥ 0, j ≥ 0

Therefore 0 0 j m |V∼ |jm = 0 unless (j0 − j)2 − 1 = 0 or j0 = j ± 1. The complete selection rules for any vector operator thus are:

0 0 j m |V∼ |jm = 0 unless j0 = j =6 0 or j0 = j ± 1 and, for any j0 , j m 0 = m or m0 = m ± 1.

We have already found (page 19) the matrix element for j0 = j. Now we will do j0 = j + 1.

Consider [J+,V+] = [Jx + iJy,Vx + iVy]

= [Jx,Vx] + i[Jx,Vy] + i[Jy,Vx] − [Jy,Vy] ↓ ↓ ↓ ↓

0 i}Vz − i}Vz 0 = 0.

21 Lecture #27 Supplement #1 Page 27S(1)-22

1/2 Take matrix element and use hj, m + 1|J+|j, mi = }[j(j + 1) − m(m + 1)] 1/2 = }[(j + m + 1)(j − 1)]

0 = hj + 1, m + 1|(J+V+ − V+J+|j, m − 1i

= hj + 1, m + 1|J+|j + 1, mi hj + 1, m|V+|j, m − 1i

− hj + 1, m + 1|V+|jmi hjm|J+|j, m − 1i where we use the Δm = +1 selection rule for V+ and J+. This provides a recurrence relation for the matrix elements.

hj + 1, m + 1|J+|j + 1, mi hj + 1, m|V+|j, m − 1i = hj + 1, m + 1|V+|jmi

× hjm|J+|j, m − 1i ((( 1/2 (((( 1/2 }([(j(−(m( + 1)(j + m + 2)] }([(j( −( m + 1)(j + m)]

So hj + 1, m|V |j, m − 1i hj + 1, m + 1|V |j, mi + = + . (j + m)1/2 (j + m + 2)1/2 This takes on a simple pattern if we divide both sides by (j + m + 1)1/2: hj + 1, m|V |j, m − 1i hj + 1, m + 1|V |j, mi −c (j, m) ≡ + = + + [(j + m + 1)(j + m)]1/2 [(j + m + 2)(j + m + 1)]1/2

= −c+(j, m + 1).

Since m was arbitrary, c+(j, m) = c+(j, m + 1) = c+(j, any other m) so the ratio c+(j) must be independent of m. The m–independence of the matrix element is therefore given by

1/2 hj + 1, m + 1|V+|j, mi = −c+(j)[(j + m + 2)(j + m + 1)] ,

0 with c+(j) = α , j + 1kV∼ kα, j a reduced matrix element that depends on the detailed nature of V∼ , not merely on its vector character. However, it can be evaluated if the matrix element of V∼ can be evaluated for any single m value, e.g., m = j or m = 0, for which the evaluation is often simpler than in the general case.

22 Lecture #27 Supplement #1 Page 27S(1)-23

0 Now determine the j = j + 1 elements of VZ using the above result for V+. Start with

−2}VZ = [J−,V+] which expresses VZ in terms of J− and V+, whose matrix elements we now know.

−2} hj + 1, m|VZ |j, mi = hj + 1, m|J−|j + 1, m + 1i hj + 1, m + 1|V+|j, mi

− hj + 1, m|V+|j, m − 1i hj, m − 1|J−|j, mi 1/2 1/2 = }[(j + m + 2)(j − m + 1)] (−c+(j))[(j + m + 2)(j + m + 1)] 1/2 1/2 − }(−c+(j))[(j + m + 1)(j + m)] [(j + m)(j − m + 1)] 1/2 = −}c+(j)[(j +m + 2) − (j+ m)][( j + m + 1)(j − m + 1)] 1/2 = −2}c+(j)[(j + m + 1)(j − m + 1)]

Thus, 1/2 hj + 1, m|VZ |j, mi = c+(j)[(j + m + 1)(j − m + 1)] . Similarly, from

}V− = [J−,VZ ] we ﬁnd

1/2 hj + 1, m − 1|V−|j, mi = c+(j)[(j − m + 2)(j − m + 1)] .

Results for j0 = j − 1 are derived in analogous fashion and involve a third re- 0 duced matrix element, c−(j) = hα , j − 1kV kα, ji. Hence the m–dependence of a scalar or vector operator follows from its scalar or vector character only. Classiﬁcation of operators by their transformation properties under rotation can be extended to tensors of any rank. In each case the form of the matrix elements is determined except for factors that depend on α and j.

23 Lecture #27 Supplement #1 Page 27S(1)-24

SUMMARY: Non-zero Matrix Elements of a Vector Op- erator, V∼

1/2 Δj = +1 hj + 1, m ± 1|V±|jmi = c+(j)[(j ± m + 2)(j ± m + 1)] 1/2 hj + 1, m|VZ |jmi = c+(j)[(j + m + 1)(j − m + 1)]

1/2 Δj = 0 hj, m ± 1|V±|jmi = c0(j)[(j ± m + 1)(j m)]

hjm|VZ |jmi = c0(j)m

1/2 Δj = −1 hj − 1, m ± 1|V±|jmi = ±c−(j)[(j m)(j m − 1)] 1/2 hj − 1, m|VZ |jmi = c−(j)[(j − m)(j + m)]

24 MIT OpenCourseWare https://ocw.mit.edu/

5.73 Quantum Mechanics I Fall 2018

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