Modern Quantum Mechanics, by Sakurai and Napoletano

Some notes on “Modern Quantum Mechanics,” by Sakuraiand Napoletano

Maarten Golterman,

Department of Physics and Astronomy, San Francisco State University,San Francisco, CA 94132, USA

These notes spell out some details of the arguments in the textbookRef. [1]. They do not replace the textbook, so they should be studiedalong with the textbook.

1. Section 2.7: quantum mechanics and electromagnetism

1.1. Gauge transformations in Electromagnetism

Let us start with the hamiltonian for a free particle,

H =~p2

2m. (1.1)

The claim is that if we wish to couple this particle to an EM field with ~E = −~∇φand ~B = ∇× ~A, the correct hamiltonian describing this is

H =1

2m

(~p− e

c~A)2

+ eφ (1.2)

=1

2m

(~p2 − e

c(~p · ~A+ ~A · ~p) +

e2

c2~A2

)+ eφ . (1.3)

Note that ~p and ~A(~x) do not commute, so that the cross terms cannot be simplified.Note also that, with this definition, H is hermitian.

We will now derive the Heisenberg equations of motion, and show that they leadto the Lorentz force law on our particle, thus validating our choice of hamiltonian.First, using the Heisenberg commutation rules, it is straightforward to show that

dxidt

=1

ih[xi, H] =

1

m

(pi −

e

cAi

)≡ Πi

m, (1.4)

where we defined the kinematical momentum Πi. Let us work this out in more detail.To find the commutator of xi with H, we only need the terms in H which depend onthe components pj of the canonical momentum ~p:

1

2m[xi, p

2j −

e

c(pjAj + pjAj)] =

1

2m

(−2

h

ipi + 2

h

i

e

cAi

)= − h

mi

(pi −

e

cAi

), (1.5)

1

and, multiplying this by 1/(ih), we find the right-hand side of Eq. (1.4).Because Πi depends on both pi and the coordinates xj (through ~A), the Πi do

not commute with themselves; using [pi, Aj] = (h/i)∂iAj (∂i ≡ ∂/∂xi),

[Πi,Πj] = [pi −e

cAi, pj −

e

cAj] (1.6)

=he

ic(−∂iAj + ∂jAi) (1.7)

=ihe

cεijkBk , (1.8)

where we use the convention that a repeated index is summed over. Using this, wefind that

dΠi

dt=

1

ih[Πi, H] (1.9)

=1

ih

1

2m[Πi,ΠjΠj] +

e

ih[Πi, φ] (1.10)

=1

2ihm

ihe

cεijk(ΠjBk +BkΠj) +

e

ih

h

i∂iφ (1.11)

=e

2mcεijk(ΠjBk +BkΠj)− e∂iφ . (1.12)

Combining the two Heisenberg equations, we find that

d2xidt2

=e

2cεijk

(dxjdtBk −Bj

dxkdt

)+ eEi , (1.13)

which is nothing else than the Lorentz force equation for a charged particle (withcharge e) in an EM field. Note that we switched the summed-over indices j and k inthe term BkΠj thus picking up a minus sign from the Levi–Civita tensor. Also, notethat, since dxj/dt = Πj does not commute with Bk, we cannot simplify this equationmore. We can do so in the classical limit, of course.

In the discussion above, we used that the commutation relations between opera-tors in the Heisenberg picture (if they are all taken at the same time) work the sameas in the Schrodinger picture. Suppose we have three operators X, Y and Z, with

A(H)(t) = U †(t)A(S)U(t) , (1.14)

and likewise for B and C, and suppose that

[A(S), B(S)] = C(S) , (1.15)

then

[A(H)(t), B(H)(t)] = [U(t)†A(S)U(t), U †(t)B(S)U(t)] (1.16)

= U †(t)A(S)U(t)U †(t)B(S)U(t)− U †(t)B(S)U(t)U †(t)A(S)U(t)

= U †(t)(A(S)B(S) −B(S)a(S)

)U(t)

= U †(t)C(S)U(t) = C(H)(t) ,

2

which is the same commutation rule as Eq. (1.15) with all Heisenberg operators atthe same time t.

1.2. Continuity equation with EM fields

In the 2nd problem of the 5th problem set, you will verify Eq. (2.7.30) of Ref. [1].Here, let us look at the expression in Eq. (2.7.33). Writing

ψ =√ρeiS/h , (1.17)

which just expresses the complex function in terms of its absolute value√ρ and phase

S/h. It should be straightforward to derive Eq. (2.7.33) from this. Furthermore, byusing Eq. (2.7.55), one sees that

√ρ does not change under a gauge transformation

(and, since ρ is the probability density, it should not!), but the phase, S, does; ittransforms as

iS/h→ iS/h+ ieΛ/(hc) → S → S + eΛ/c . (1.18)

Using now also Eq. (2.7.36), we conclude that ∇S − e ~A/c is gauge invariant, andthus that ~j defined in Eq. (2.7.33) is gauge invariant. Note that, since ~j has to begauge invariant, and is given by Eq. (2.7.33), we need to have S transform as givenin Eq. (1.15) above. This gives us thus an alternative way to derive Eq. (2.7.55) inRef. [1]!

1.3. Aharonov–Bohm effect

Here are some details about the magnetic field for the Aharonov–Bohm effect. Withρa being the radius of the cylinder, we define

~A =

{12B0z × ~r , ρ < ρa

12B0

ρ2aρϕ , ρ > ρa

, (1.19)

where we use cylindrical coordinates ρ, ϕ and z. Remember that

ρ = cosϕ x+ sinϕ y , (1.20)

ϕ = − sinϕ x+ cosϕ y , (1.21)

so that (~r = ~ρ+ zz)z × ~r = z × ~ρ = ρϕ . (1.22)

Inside the cylinder, we have that Ax = −12B0y, Ay = 1

2B0x and Az = 0. This leads to

Bx = By = 0m, Bz = B0. (Check these claims!) Outside the cylinder, we have that

Ax = −1

2B0ρ2a

ρsinϕ = −1

2B0

ρ2ay

x2 + y2, (1.23)

Ay =1

2B0ρ2a

ρcosϕ =

1

2B0

ρ2ax

x2 + y2, (1.24)

Az = 0 , (1.25)

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hence (check!)~B = 0 (outside cylinder) . (1.26)

2. Section 3.1: rotations

2.1. The rotation group

Rotations leave the inner product ~r1 · ~r2 between two (position) vectors ~r1 and ~r2

invariant.1 Using summation convention for repeated indices, the components of avector ~r transform under a rotation as

r′i = Rijrj , (2.1)

where Rij are the components of a 3× 3 matrix representing the rotation and r′i arethe transformed components. Requiring the invariance of the inner product:

~r′1 · ~r′2 = (R~r1) ·R~r2 = ~r1 ·RTR~r2 = ~r1 · ~r2 , (2.2)

in index notation:

r′1ir′2i = (Rijr1j)(Rikr2k = r1jR

TjiRikr2k = r1jr2j , (2.3)

and using that this has to be true for any pair ~r1, ~r2, we find that

RTR = 1 , (2.4)

with 1 the 3 × 3 unit matrix. In index notation, we need that RTjiRik = δjk, which

is Eq. (2.4) in index notation. Equation (2.4) implies that R is orthogonal (unitaryand real). This group is denoted as O(3): the group of orthogonal 3× 3 matrices. Ifwe also require, as we will do in this chapter, that det(R) = 1, it is the group SO(3)of “special” orthogonal 3× 3 matrices.

All elements of SO(3) can be written in the form

R(~ϕ) = eϕiTi , (2.5)

Tx =

0 0 0

0 0 −1

0 1 0

, Ty =

0 0 1

0 0 0

−1 0 0

, Tz =

0 −1 0

1 0 0

0 0 0

,

with Tx,y,z the generators of the group. Indeed, SO(3) is a group: it is associative(because matrix multiplication is associative), there is a unit element R(~ϕ = 0) = 1,each element R(~ϕ) has an inverse R(−~ϕ) = e−ϕiTi (rotate around the same axis bythe opposite angle), and it is closed under multiplication (this means that the productof two rotations is also a rotation, i.e., that the product of two elements of the group

1Of course, they also leave the inner product between two momentum vectors, etc., invariant.

4

is an element of the group). Closure is not trivial, and we will return to that below.First, let us work out an example:

Rz(ϕ) = eϕTz (2.6)

= 1 + ϕ

0 −1 0

1 0 0

0 0 0

+1

2ϕ2

−1 0 0

0 −1 0

0 0 0

+1

6ϕ3

0 1 0

−1 0 0

0 0 0

+ . . .

=

1− 1

2ϕ2 + . . . −ϕ+ 1

6ϕ3 + . . . 0

+ϕ− 16ϕ3 + . . . 1− 1

2ϕ2 + . . . 0

0 0 1

=

cosϕ − sinϕ 0

sinϕ cosϕ 0

0 0 1

,

which is a rotation around the z-axis by an angle ϕ. A general rotation with param-eters ~ϕ is a rotation around an axis with direction n = ϕ over an angle ϕ = |~ϕ|.

To prove closure, we need the following two ingredients. First, it is straightforwardto check that

[Tx, Ty] = Tz , [Ty, Tz] = Tx , [Tz, Tx] = Ty . (2.7)

(This can be written succinctly as [Ti, Tj] = εijkTk.) Second, we need the Baker–Campbell–Hausdorff formula:

eAeB = eA+B+ 12

[A,B]+ 112

([A,[A,B]]+[[A,B],B])+... . (2.8)

There is no closed-form expression, but one can prove that all terms in the exponent onthe right-hand side are “commutators of commutators,” like the ones explicitly shown.Now, if we apply this to A = ϕiTi and B = ψiTi for two rotations R(~ϕ) = eϕiTi andR(~ψ) = eψiTi , the exponent on the right-hand side of Eq. (2.8) becomes

(ϕi + ψi)Ti +1

2εijkϕiψjTk + . . . = (ϕi + ψi + εjkiϕjψk + . . .)Ti ≡ αi(~ϕ, ~ψ)Ti , (2.9)

and the product of R(~ϕ) and R(~ψ) is thus a rotation R(~α), with ~α a complicatedfunction of ~ϕ and ~ψ. We see that if we know the commutation rules of the generatorsTi, we know how elements of the group multiply. In other words, the algebra (2.7)defines the structure of the group SO(3). Note that, if [A,B] = 0, Eq. (2.8) collapsesto eAeB = eA+B. For rotations, this happens when we consider two rotations aroundthe same rotation axis. A group for which all generators commute is an abelian group.An example of an abelian group is the translation group: recall that translations aregenerated by the momenta pi, and we have that [pi, pj] = 0.

In fact, it is sufficient to consider only infinitesimal rotations, i.e., rotations overan infinitesimal angle ϕ = |~ϕ|. If we define ε = ϕ/N with N very large, so that ε isvery small, we have that

Rz(ϕ) = limN→∞

R(ε)N = limN→∞

(1 +

ϕ

NTz

)N= eϕTz . (2.10)

5

A group like this is called a Lie group, and the corresponding algebra (2.7) a Liealgebra.

2.2. Rotations in quantum mechanics

Now, let us see how this all works in quantum mechanics. In the discussion,note the similarity to the case of translations! The only thing that makes the caseof rotations more complicated (and more interesting!) is that rotations in general donot commute, while translations always do.

If a system is represented by a ket |α〉, the rotated system is represented by somedifferent ket

|α〉R = D(R)|α〉 , (2.11)

which defines a linear operator D(R) for each rotation R. Note that, of course, D(R)is represented by a matrix on the Hilbert space (all operators are), but the dimensionof D(R) is equal to the dimension of the Hilbert space, which is not (necessarily)equal to 3. The operators D(R) have to follow the same rules as the rotations Rthemselves; in jargon, the D(R) form a representation of the group SO(3). There isa homeomorphism R→ D(R) such that

R1R2 = R3 ⇒ D(R1)D(R2) = D(R3) . (2.12)

This implies that

|α〉R1R2 = D(R1)|α〉R2 = D(R1)D(R2)|α〉 . (2.13)

In words: if I first rotate a physical system, described by a ket |α〉, by R2, and then Irotate it again by R1, the ket describing this system should transform in accordancewith that.

Here is a trivial example: it could be that the physical system represented by |α〉does not change at all under rotations, which means that for such a system, for all R,D(R) = 1! Clearly, Eq. (2.12) is satisfied, and this one-dimensional representation iscalled the trivial representation. Referring to Eq. (3.1.15) of Ref. [1], for such a statewe have that ~J |α〉 = 0. In other words, such a state has zero angular momentum.A physical example is the hydrogen atom in the ground state: you may recall thatthe ”electron cloud” in the ground state looks spherically symmetric, so that indeedrotations have no effect.

3. Section 3.2: SU(2)

The simplest example of a non-trivial representation is described in great detailin Sec. 3.2 of Ref. [1]. Consider, in particular, Eq. (3.2.45) in Ref. [1], which worksout an explicit form of a 2-dimensional representation of the rotation group. Notethat the matrices on the right-hand side of this equation are 2 × 2 unitary matriceswith determinant equal to 1 (check this – you can do this by explicit calculation!).

6

These matrices form the group SU(2), i.e., the group of 2× 2 unitary matrices withdeterminant equal to 1. What we discover is that this group is homeomorphic withthe group SO(3), they are (almost) the same group, if we think about this groupin terms of their mathematical structure. They are not quite the same: clearly, forϕ = 2π, we have that R(2π) = 1, but if you substitute ϕ = 2π in Eq. (3.2.45), youwill find that

exp

(−iσ · n2π

2

)=

−1 0

0 −1

. (3.1)

In a sense, the group SU(2) is “twice as big” as SO(3), because the SU(2) matricesfor ϕ and ϕ + 2π are different: they differ by an oveall minus sign, i.e., D(n, ϕ) =D(n, ϕ+ 2π) (check this!), but get mapped onto the same element of SO(3). In thissense, in quantum mechanics, the group SU(2) is the more fundamental group.

Note that we already know that nature makes use of this representation: the spinof the electron transforms according to this representation! According to quantummechanics, the spin of the electron is thus a special example of angular momentum.

4. Section 3.5: j = 1 representation

In Sec. 3.5 of Ref. [1] we found a 3-dimensional representation of the rotationgroup, the j = 1 representation. An example of a particle with spin 1 is the Wmeson, which is electrically charged (in fact, W mesons with charges ±e exist). Onecan therefore imagine subjecting it to a Stern–Gerlach type of experiment, just likethe electron, and measure its possible polarizations along some axis. (In practice, aW meson decays pretty quickly, but that is an experimental problem!) Unlike thecase of the electron, the possible outcomes for the spin polarization are now +h, 0and −h. So, if we consider the spin of the W , it lives in a 3-dimensional Hilbertspace. Operators acting on kets in this space are represented by 3× 3 matrices.

However, even before thinking about quantum mechanics, we already knew a3× 3 matrix representation of the rotation group — it is how we defined the rotationgroup in the first place, as the group of rotations on vectors in 3-dimensional physicalspace leaving inner products of these vectors invariant. This raises the mathematicalquestion whether these two 3-dimensional representations are the same (or, moreprecisely, equivalent). Are there two different 3-dimensional representations, or isthere only one? In the systematic analyis of Sec. 3.5, only one j = 1 representation wasfound, so there is only one, and we would like to see how rotations in 3-dimensionalphysical space can be brought into the form of the j = 1 representation found inSec. 3.5.

Let us work this out with a simple example, considering a rotation around thez-axis. The generator we found in Sec. 3.5, Eq. (3.3.35b) is, writing out the matrixfor j = j′ = 1:

Jz = h

1 0 0

0 0 0

0 0 −1

. (4.1)

7

However, this does not coincide with (ih times) Tz, given in Eq. (2.5)! We note thatthe eigenvalues of Jz are h, 0, −h, while the eigenvalues of ihTz are also equal toh, 0, −h (check this!). So, these two matrices are in fact the “same” (more precisely,mathematically equivalent), but differ by a basis transformation. Below, we work outwhat this basis transformation is.

Start with a rotation over an angle ϕ generated by Tz; we have seen that thistakes the form ~r ′ = eTzϕ~r, or, in components (cf. Eq. (2.6)):

x′ = x cosϕ− y sinϕ , (4.2)

y′ = x sinϕ+ y cosϕ ,

z′ = z .

Now define new variables

u = (x+ iy)/√

2 , v = (x− iy)/√

2 . (4.3)

It is straightforward to work out that in terms of the variables u, v and z, Eq. (4.2)takes the form

u′ = ueiϕ , (4.4)

z′ = z .

v′ = ve−iϕ ,

and then that this transformation can be written asu′

z′

v′

= eiJzϕ/h

u

z

v

=

eiϕ 0 0

0 1 0

0 0 e−iϕ

u

z

v

. (4.5)

The expression on the right-hand side is just Eq. (4.4) in matrix form, but it is alsowhat you get if you work out eiJzϕ/h in matrix form, so the middle expression and theexpression on the right-hand side of Eq. (4.5) coincide! Therefore, in this form, therotation around the z-axis is indeed generated by iJz/h, so the generators iJz/h andTz only differ by a (unitary) basis transformation. This can be proven more generally(i.e., that a basis transformation exists that maps the Ti into the iJi/h), and thusthe two 3-dimensional representations are equivalent. Mathematically, we say thereis only one 3-dimensional representation. Of course, nothing prevents nature frommaking use of this representation in different physical applications, such as rotationsin physical 3-dimensional space, but also the existence of a 3-dimensional Hilbert spacecorresponding to the angular momentum states of a particle with angular momentumj = 1. Problem 3.14 of Ref. [1] asks you to work out the complete proof: a basistransformation exists that takes the set ihTx,y,z into the set Jx,y,z, respectively.

5. Section 3.6: orbital angular momentum

8

This section discusses a few technical points that arise in Sec. 3.6 of Ref. [1] insome more detail.

5.1. Derivation of Eq. (3.6.11)

Let us derive Eq. (3.6.11) of Ref. [1] in more detail, following the same logic that ledto Eq. (3.6.9). The idea is the same, but technically it is a little more involved. Thereason is that a rotation around the z-axis corresponds to a change only of the polarangle φ, while θ remains constant. With a rotation around the x-axis, both φ and θchange, and one thus expects terms involving both ∂/∂θ and ∂/∂φ.

Let us consider a rotation over an infinitesimal angle δϕ around the x-axis. Thisinfinitesimal rotation is represented by the operator 1− iδϕLx/h, and we have that

〈x′, y′, z′| (1− iδϕLx/h) |α〉 = 〈x′, y′, z′|(1− i

hδϕ(ypz − zpy)

)|α〉 (5.1)

=

(1− δϕy′ ∂

∂z′+ δϕz′

∂

∂y′

)〈x′, y′, z′|α〉

= 〈x′, y′ + z′δϕ, z′ − y′δϕ|α〉 .

Since rotations only change the polar angles, and not r, we can express the operatorLx in terms of derivatives with respect to φ and θ, instead of derivatives with respectto y′ and z′. Assume, therefore, that

Lx =h

i

(A∂

∂θ+B

∂

∂φ

). (5.2)

The task is to find A and B. For Lz this was done in Ref. [1], where it was found thatin that case A = 0, and B = 1. Now, because a rotation around the x-axis changesboth θ and φ, both A and B will be non-zero, and more complicated. Start from(

1− δϕA ∂

∂θ− δϕB ∂

∂φ

)〈r, θ, φ|α〉 = 〈r, θ − δϕA, φ− δϕB|α〉 , (5.3)

where we transitioned to write the wave function 〈x′, y′, z′|α〉 in polar coordinates.Comparing with Eq. (5.1), and remembering the relation between the cartesian andpolar coordinates, this gives us the equations

x′ ≡ r sin θ cosφ = r sin (θ − δϕA) cos (φ− δϕB) (5.4)

= r sin θ cosφ− r cos θ cosφ δϕA+ r sin θ sinφ δϕB ,

y′ + z′δϕ = r sin θ sinφ+ r cos θδϕ = r sin (θ − δϕA) sin (φ− δϕB)

= r sin θ sinφ− r cos θ sinφ δϕA− r sin θ cosφ δϕB ,

z′ − y′δϕ = r cos θ − r sin θ sinφδϕ = r cos (θ − δϕA)

= r cos θ − r sin θ δϕA .

From the last of these equations, we find that A = − sinφ, and substituting this intothe first or second equation, we then find that B = − cot θ cosφ (check both these

9

equations!). We thus arrive at

Lx =h

i

(− sinφ

∂

∂θ− cot θ cosφ

∂

∂φ

), (5.5)

which is Eq. (3.6.11) in position representation. A very similar calculation gives theexpression for Ly in Eq. (3.6.12).

5.2. Comment on Eq. (3.6.18)

In Eq. (3.6.18) of Ref. [1], we use the polar version of the gradient operator:

~∇ = r∂

∂r+ θ

1

r

∂

∂θ+ φ

1

r sin θ

∂

∂φ. (5.6)

With ~r = rr, we see that

~r · ~∇ = r∂

∂r, (5.7)

because, of course, r · θ = r · φ = 0.

5.3. Comment on Eq. (3.6.23)

For the discussion of spherical harmonics starting on page 202 of Ref. [1], it mayhelp to refer to the discussion on page 207, noting that there the energy eigenvalueis denoted by E, and not by n. Also note that this n has nothing to do with thedirectional unit vector n!

The directional eigenket |n〉 can also be written as

|n〉 = |θ, φ〉 . (5.8)

It makes sense to consider this ket, because for a spherically symmetric problem, thetotal energy and angular momentum eigen state (in position space) 〈~x ′|n, `,m〉 =〈r, θ, φ|n, `,m〉 in Eq. (3.6.22) factorizes into two parts, with one only depending onr, and the other only on the polar angles θ and φ.

6. Sec. 3.8: addition of angular momentum

6.1. Definitions

Suppose we have a system with two different angular momentum operators, ~J1 and ~J2.Examples are the electron in the hydrogen atom, which has both an orbital angularmomentum ~J1 = ~L and a spin ~J2 = ~S, or a system with two electrons, each with theirown spin, ~J1 = ~S1 and a spin ~J2 = ~S2. Since these two angular momenta operatorsare unrelated to each other (i.e., they correspond to different, unrelated, observablesof the system), they commute:

[J1i, J2j] = 0 , (6.1)

10

for all components i and j of both operators. Formally, the operators ~J1 and ~J2 acton different Hilbert spaces. In the example of the hydrogen atom, ~L acts on the(angular part) of the spatial wave function, while ~S acts on the spin of the electron.To get a description of all properties of the electron, we can combine these two Hilbertspaces: using a position eigenbasis for the spatial properties of the electron, and theSz eigenkets |± 1

2〉 for the spin of the electron, a combined basis is |~x ′〉⊗ |± 1

2〉, where

hopefully the use of the “direct product” symbol ⊗ will become clear in what follows.If, instead of position eigenkets |~x ′〉, we only consider orbital angular momentumeigenkets in position space, we may use the orbital angular momentum eigenkets |`,m〉(here we are ignoring any quantum numbers not related to angular momentum, i.e.,quantum numbers which do not change under rotations), and a basis for the electron’s“combined” Hilbert space is

|`,m;±1

2〉 = |`,m〉 ⊗ | ± 1

2〉 , (6.2)

where we show two equivalent ways of writing states in this combined Hilbert space.(We say that the combined Hilbert space is the direct product of the two individualHilbert spaces.) In general, for a system with two different angular momenta ~J1 and~J2, using an eigenbasis for each, the combined eigenbasis of the product Hilbert spaceconsists of the eigenkets

|j1, j2;m1,m2〉 = |j1,m1〉 ⊗ |j2,m2〉 . (6.3)

Since ~J1 acts only on the states |j1,m1〉 and ~J2 only on the states |j2,m2〉, the notationon the right in Eq. (6.3) makes sense if instead of ~J1 we write ~J1 ⊗ 1, and instead of~J2 we write 1⊗ ~J2, with2

~J1|j1, j2;m1,m2〉 = ( ~J1 ⊗ 1)(|j1,m1〉 ⊗ |j2,m2〉) = ~J1|j1,m1〉 ⊗ |j2,m2〉 , (6.4)~J2|j1, j2;m1,m2〉 = (1⊗ ~J2)(|j1,m1〉 ⊗ |j2,m2〉) = |j1,m1〉 ⊗ ~J2|j2,m2〉 .

The notation with the ⊗ symbol exhibits the product structure of this Hilbert spacevery clearly, but often, in practice, the notation on the left-hand side of Eq. (6.4) isused. The notation with the ⊗ also makes it very clear that the two angular momentacommute:

[ ~J1 ⊗ 1,1⊗ ~J2] = [ ~J1,1]⊗ [1, ~J2] = 0 . (6.5)

With a little less mathematical precision we can also write that [ ~J1, ~J2] = 0, which ishow we think about it in physics, but the notation with the direct product symbolclarifies the mathematical structure of these operators more clearly. Note that thisis all just notation to give a mathematically precise description of the physics we areinterested in!

6.2. Total angular momentum

2For the states on the left of this equation I follow the notation of Ref. [1], in which first the j,and then the m quantum numbers are listed.

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We can now define a total angular momentum, which acts on the product Hilbertspace,

~J = ~J1 ⊗ 1 + 1⊗ ~J2 , (6.6)

or, more succinctly, ~J = ~J1 + ~J2. First, this is an angular momentum: becauseusing [J1i, J1j] = ihεijkJ1k and [J2i, J2j] = ihεijkJ2k, it is straightforward to showthat [Ji, Jj] = ihεijkJk (show this!). Since the components Ji satisfy the angularmomentum algebra, it is an angular momentum by definition. Of course, it alsomakes sense intuitively that the sum of two angular momenta is again an angularmomentum.

The basis |j1, j2;m1,m2〉 = |j1,m1〉⊗ |j2,m2〉 is composed of simultaneous eigen-kets of ~J2

1 , J1z, ~J22 and J2z. But, one can, of course, also try to find a basis composed

of eigenkets of ~J2 and Jz. However, such eigenkets are not the same as the eigenkets|j1, j2;m1,m2〉, because J1z and J2z do not commute with ~J2 (because they do notcommute with ~J1 · ~J2; show this!), so we cannot find simultaneous eigenkets of ~J2,Jz, J1z and J2z! In contrast, we can find simultaneous eigenkets of ~J2, Jz, ~J

21 and ~J2

2 ,because these operators all commute with each other (again, show this). We can thusconstruct a basis for our Hilbert space either using the eigenkets |j1, j2;m1,m2〉, orsimultaneous eigenkets |j1, j2; j,m〉 of the commuting set ~J2, Jz, ~J

21 and ~J2

2 , with

~J21 |j1, j2; j,m〉 = j1(j1 + 1)h2|j1, j2; j,m〉 , (6.7)~J22 |j1, j2; j,m〉 = j2(j2 + 1)h2|j1, j2; j,m〉 ,~J2|j1, j2; j,m〉 = j(j + 1)h2|j1, j2; j,m〉 ,Jz|j1, j2; j,m〉 = mh|j1, j2; j,m〉 .

Since these two bases span the same Hilbert space, there should be a unitary basistransformation relating the two, which is written in Eq. (3.8.33) of Ref. [1]. Theelements of this basis transformation are 〈j1, j2;m1,m2|j1, j1; j,m〉, and the task ofSec. 3.8 is the construction of these elements, the Clebsch–Gordan coefficients, giventwo values of j1 and j2.

An example where the new basis is useful is the case of a spin-orbit coupling ~L · ~S,where in this example ~L = ~J1 and ~S = ~J2. Since

~L · ~S =1

2~J2 − 1

2~L2 − 1

2~S2 =

1

2

(~J2 − ~J2

1 − ~J22

), (6.8)

the new basis eigenkets |j1, j2; j,m〉 are eigenstates of ~L · ~S = ~J1 · ~J2:

~J1 · ~J2|j1, j2; j,m〉 =1

2h2(j(j + 1)− j1(j1 + 1)− j2(j2 + 1)

)|j1, j2; j,m〉 . (6.9)

In contrast, the states |j1, j2;m1,m2〉 are not eigenstates of ~J1 · ~J2 = J1xJ2x+J1yJ2y +J1zJ2z. They are eigenstates of the last term in this sum, J1zJ2z, but not of the othertwo terms (convince yourself!).

6.3. Discussion of the addition of two angular momenta

12

Suppose we have a basis of simultaneous eigenkets |j1, j2;m1,m2〉 of two angularmomenta, ~J1 and ~J2:

~J21 |j1, j2;m1,m2〉 = j1(j1 + 1)h2|j1, j2;m1,m2〉 , (6.10)~J22 |j1, j2;m1,m2〉 = j2(j2 + 1)h2|j1, j2;m1,m2〉 ,

J1z|j1, j2;m1,m2〉 = m1h|j1, j2;m1,m2〉 ,J2z|j1, j2;m1,m2〉 = m2h|j1, j2;m1,m2〉 .

We want to find the basis transformation to another set of simultaneous eigenkets,now of the commuting set of operators ~J2

1 , ~J22 , ~J2 and Jz, where ~J = ~J1 + ~J2:

~J21 |j1, j2; j,m〉 = j1(j1 + 1)h2|j1, j2; j,m〉 , (6.11)~J22 |j1, j2; j,m〉 = j2(j2 + 1)h2|j1, j2; j,m〉 ,~J2|j1, j2; j,m〉 = j(j + 1)h|j1, j2; j,m〉 ,Jz|j1, j2; j,m〉 = mh|j1, j2; j,m〉 .

In what follows, to avoid cluttered notation, we will use units in which h = 1. SinceJz = J1z + J2z,

Jz|j1, j2;m1,m2〉 = (m1 +m2)|j1, j2;m1,m2〉 , (6.12)

and therefore, that

|j1, j2; j, j〉 = |j1, j2; j,m = j = j1 + j2〉 = |j1, j2;m1 = j1,m2 = j2〉 . (6.13)

This state, with j = j1+j2 is the unique state with the maximal Jz eigenvalue possible,and thus is the highest m-value component of a multiplet with angular momentumj = j1+j2. The largest possible values of j and m are thus j = j1+j2 and m = j1+j2.

We find the other componets of this multiplet by acting with the lowering operatorJ− = J1− + J2−:

1√2jJ−|j1, j2; j, j〉 =

1√2j

(J1− + J2−)|j1, j2; j, j〉 (6.14)

=1√2j

(J1− + J2−)|j1, j2; j1, j2〉

=1√2j

(√2j1|j1, j2; j1 − 1, j2〉+

√2j2|j1, j2; j1, j2 − 1〉

),

where we used that

J±|j,m〉 =√

(j ∓m)(j ±m+ 1)|j,m± 1〉 . (6.15)

We thus find that

|j1, j2; j, j − 1〉 =1√2j

(√2j1|j1, j2; j1 − 1, j2〉+

√2j2|j1, j2; j1, j2 − 1〉

). (6.16)

Note that this state is normalized because (√

2j1/√

2j)2+(√

2j2/√

2j)2 = (j1+j2)/j =1. Also note that this state is a linear combination of the two possible states with

13

m = j1 + j2 − 1, namely the states |j1, j2; j1 − 1, j2〉 and |j1, j2; j1, j2 − 1〉. Up to anoverall phase, there is a unique state orthogonal to this state:

|j1, j2; j − 1, j − 1〉 ≡ 1√2j

(√2j2|j1, j2; j1 − 1, j2〉 −

√2j1|j1, j2; j1, j2 − 1〉

). (6.17)

Excercise: check that the states on the right-hand sides of Eqs. (6.16) and (6.17) areindeed orthogonal. Moreover, if we act with J+ on the latter state, we find

J+|j1, j2; j − 1, j − 1〉 = (J1+ + J2+)|j1, j2; j − 1, j − 1〉 (6.18)

=1√2j

(J1+ + J2+)(√

2j2|j1, j2; j1 − 1, j2〉 −√

2j1|j1, j2; j1, j2 − 1〉)

=1√2j

(√2j1

√2j2|j1, j2; j1, j2〉 −

√2j2

√2j1|j1, j2; j1, j2〉

)= 0 .

This justifies the name we gave this state in Eq. (6.17): it is the component withmaximal m-value of a multiplet with angular momentum j1 + j2 − 1.

So far, we have thus identified two total angular momentum multiplets: one withj = j1 +j2, and one with j = j1 +j2−1. Of course, we can find the other components(states) in these two multiplets by “laddering down,” i.e., applying the operator J−repeatedly.

Let us carry out one more step of this exercise by applying the operator J− tothe state we found in Eq. (6.16) (omitting the overall normalization, which is notrelevant for the rest of the argument):

J−|j1, j2; j, j − 1〉 = (J1− + J2−)1√2j

(√2j1|j1, j2; j1 − 1, j2〉+

√2j2|j1, j2; j1, j2 − 1〉

)=

1√2j

(√2j1

√2(2j1 − 1)|j1, j2; j1 − 2, j2〉 (6.19)

+ 2√

2j1

√2j2|j1, j2; j1 − 1, j2 − 1〉

+√

2j2

√2(2j2 − 1)|j1, j2; j1, j2 − 2〉

).

Acting with J− on the state (6.17) will also give a linear combination of the three states|j1, j2; j1−2, j2〉, |j1, j2; j1−1, j2−1〉 and |j1, j2; j1, j2−2〉, but it will be a different linearcombination, orthogonal to the state (6.19); this is the state |j1, j2; j − 1,m = j − 2〉.Excercise: Why? Work this out! But, since these two states are linear combinationsof three states, there is a third independent linear combination of these states, whichis orthogonal to both |j1, j2; j, j − 2〉 and |j1, j2; j − 1, j − 2〉. This new state will bethe maximal m-value component of a multiplet with j = j1 + j2 − 2! It is thus thestate |j1, j2; j − 2, j − 2〉. Exercise: Convince yourself ! Of course, the minimal valueto which we can lower m1 is −j1, and the minimal value to which we can lower m2 is−j2; acting with Ji− on a state with mi = −ji will give zero (i = 1, 2).

We see that m1 or m2 get lowered each time we apply J− = J1−+J2−; in Eq. (6.19)to minimally m1 = j1− 2 or m2 = j2− 2. Continuing with the tower of states we are

14

obtaining starting with Eq. (6.13), i.e., our first multiplet, we find, ultimately, thestate

|j1, j2; j,−j〉 = |j1, j2; j,m = −j = −j1 − j2〉 = |j1, j2;m1 = −j1,m2 = −j2〉 . (6.20)

One way to see that this is where we have to end up is that we could have started withthis state, and obtain the same tower of states by acting with the raising operatorJ+ = J1+ + J2+.

Collecting what we know thus far, we find ~J2, Jz multiplets with j = j1 + j2,j = j1 + j2 − 1, j = j1 + j2 − 2, etc. Where does this end, i.e., what is the lowest jvalue we will find, carrying out this procedure to the end? In constructing our first(j = j1 + j2) multiplet, we started with one state from the collection in Eq. (6.10) (cf.Eq. (6.13)), we then found a linear combination of two of them (cf. Eq. (6.16)), andthen a linear combination of three of them (cf. Eq. (6.19)). Each time we laddereddown one step we found one more state to show up in the linear combination, andeach time we also found a new multiplet: this is how we found the multiplets withj = j1 + j2 − 1 and j = j1 + j2 − 2. Now, without loss of generality, take j1 ≥ j2.Then, the number of states in these “growing” linear combinations stops growingonce we have encountered the state |j1, j2;m1 = j1,m2 = −j2〉, because states withm2 = −j2−1 do not exist. (Note how the m2 quantum number in the last state in thelinear combinations in Eqs. (6.16) and (6.19) gets lowered by one unit in each step.)That means that the “latest” new multiplet we found, by systematically carrying outour procedure is the multiplet with j = j1 − j2 (which is non-negative, because wechose j1 ≥ j2). When m2 = −j2 is reached in our lowering process, no new statesfrom the collection (6.10) will show up in our linear combinations anymore, so allstates with m1 and m2 such that m = m1 + m2 = j1 − j2 will have been used tocomplete multiplets we already were constructing. Therefore, the possible j valuesfor the total angular momentum ~J = ~J1 + ~J2 are

j = j1 + j2 , j1 + j2 − 1 , . . . , |j1 − j2| , (6.21)

where we dropped the restriction j1 ≥ j2. Since the two bases (6.10) and (6.11) haveto consist of an equal number of states, a simple counting argument, given in Ref. [1],confirms that we found all states in the new basis (6.11). Problem 3.24 of Ref. [1]asks you to work this out for the case j1 = 1, j2 = 1.

7. Section 3.11: tensor operators

7.1. Details of Eq. (3.11.13)

Let us check that the three terms on the right-hand side of Eq. (3.11.13) indeed onlytransform among themselves, and not into each other. Start with the first term. First,~U · ~V is the inner product of the vectors ~U and ~V , so, it is invariant under rotations.Furthermore

δij →∑i′j′

Rii′Rjj′δi′j′ =∑i′Rii′Rji′ =

∑i′Rii′R

Ti′j = δij , (7.1)

15

so also δij is invariant under rotations, and thus the whole first term is invariant underrotations. In more technical terms, this part transforms in the j = 0 representation ofthe rotation group (i.e., it does not transform at all, just like the angular momentumeigenstate |j = 0,m = 0〉!). Note that δij, a two-index object, does not transform atall either — such a tensor is called an invariant tensor.

Now let us consider the second term. Applying a rotation to it:

UiVj − UjVi = UiVj − ViUj →∑i′j′

(Rii′Rjj′Ui′Vj′ −Rii′Rjj′Vi′Uj′) (7.2)

=∑i′j′

Rii′Rjj′ (Ui′Vj′ − Vi′Uj′) .

We see that the anti-symmetrized product of Ui and Vj transform among each other,and not into the other two terms in Eq. (3.11.13). In fact, we recognize the compo-nents of the cross product ~U × ~V , which has three components, and transforms like avector. Thus, the three anti-symmetrized products in the second term of Eq. (3.11.13)transform in the j = 1 representation of the rotation group.

Likewise, the symmetrized products in the third term of Eq. (3.11.13) also trans-form among themselves:

UiVj + UjVi = UiVj + ViUj →∑i′j′

(Rii′Rjj′Ui′Vj′ +Rii′Rjj′Vi′Uj′) (7.3)

=∑i′j′

Rii′Rjj′ (Ui′Vj′ + Vi′Uj′) .

Now, note that if we set i = j in the third term, and we sum over i, we get∑i

(1

2(UiVi + UiVi)−

1

3~U · ~V δii

)=∑i

UiVi = 0 , (7.4)

so the third term is traceless, and the “33” component of this combination is thusequal to minus the sum of the “11” and “22” components. Therefore, there are onlyfive independent components in the third term of Eq. (3.11.13). They all transforminto each other, in a five dimensional representation of the rotation group; thus, theytransform in the j = 2 representation. These j = 0, j = 1 and j = 2 tensors areexamples of spherical tensors.

7.2. Elements of the derivation of Eq. (3.11.25)

First, recall that the ket |n〉 is just the angular part of the position eigenket. We canexpress the position eigenket |~x ′〉 in terms of polar coordinates: |r, θ, φ〉. If we onlyconsider rotations, the coordinate r does not play a role, and we can just considerthe “abbreviated” ket |n〉 = |θ, φ〉. The Y m

` (θ, φ) are nothing else than the (orbital)angular momentum eigenstates |`m〉 expressed on a position eigenbasis:

Y m` (θ, φ) = 〈θ, φ|`m〉 . (7.5)

The steps leading from Eq. (3.11.18) to Eq. (3.11.22b) in Ref. [1] should be quitestraightforward; note that Eq. (3.11.18) is the same equation as Eq. (3.5.49), but now

16

we consider orbital angular momentum, so we relabel j → `, and instead of takingthe rotation to be R, we take it to be R−1 (which is also a rotation of course, so weare free to do that).

Note that Eq. (3.11.22a) is a definition, inspired by Eq. (3.11.21): it defines aspherical tensor T (k)

q by requiring it to transform in the same way as Y qk . Note the

(annoying!) choice of making the label k ↔ ` the superscript on T , even thoughit is the subscript on Y , and likewise, the label q ↔ m is a subscript on T , but asuperscript on Y !

Then, since according to Eq. (3.11.4)

D(R) = 1− i εh~J · n , (7.6)

the signs in Eq. (3.11.23) would be different if you derive it directly from Eq. (3.11.22a):

(1− i ε

h~J · n

)T (k)q

(1 + i

ε

h~J · n

)=

k∑q′=−k

T(k)q′ 〈kq′|

(1− i ε

h~J · n

)|kq〉 . (7.7)

This still leads to Eq. (3.11.24), as it should, of course. From there it should also bestraightforward to derive Eq.(3.11.25), following the “directions” given in Ref. [1], ifin Eqs. (3.5.35b) and (3.5.41) you substitute j → k and m→ k.

7.3. Solving the recursion relation for Clebsch–Gordan coefficients: anexample

Clebsch–Gordan coefficients satisfiy the recursion relation (3.8.49) of Ref. [1]:√(j ∓m)(j ±m+ 1)〈j1j2;m1,m2|j1j2; j,m± 1〉 (7.8)

=√

(j1 ±m1)(j1 ∓m1 + 1)〈j1j2;m1 ∓ 1,m2|j1j2; j,m〉

+√

(j2 ±m2)(j2 ∓m2 + 1)〈〈j1j2;m1,m2 ∓ 1|j1j2; j,m〉 .

Here, we work out an example. We take j1 = j2 = 12, so that j can take the values 0

and 1. Taking the lower sign everywhere in Eq. (7.8) for j = 1, m = 0, m1 = 12

andm2 = 1

2yields:

√2

⟨1

2

1

2;1

2,1

2

∣∣∣∣∣12 , 1

2; 1, 1

⟩=√

1

⟨1

2

1

2;−1

2,1

2

∣∣∣∣∣12 1

2; 1, 0

⟩+√

1

⟨1

2

1

2;1

2,−1

2

∣∣∣∣∣12 1

2; 1, 0

⟩,

(7.9)while taking the lower sign, j = 1, m = 1, m1 = 1

2and m2 = −1

2yields:

√2

⟨1

2

1

2;1

2,−1

2

∣∣∣∣∣12 , 1

2; 1, 0

⟩=√

0

⟨1

2

1

2;3

2,−1

2

∣∣∣∣∣12 1

2; 1, 1

⟩+√

1

⟨1

2

1

2;1

2,1

2

∣∣∣∣∣12 1

2; 1, 1

⟩.

(7.10)Note that, of course, m1 cannot be equal to 3/2, and indeed, that coefficient ismultiplied by zero in Eq. (7.10). Now take⟨

1

2

1

2;1

2,1

2

∣∣∣∣∣12 1

2; 1, 1

⟩= 1 . (7.11)

17

This happens to be the correct value (why?), but since the recursion relation (7.8) ishomogeneous, we have to fix one coefficient to find all the others. Substituting thisin Eq. (7.10), we find ⟨

1

2

1

2;1

2,−1

2

∣∣∣∣∣12 , 1

2; 1, 0

⟩=

1√2, (7.12)

and substituting both Eq. (7.11) and Eq. (7.12) into Eq. (7.9) then yields⟨1

2

1

2;−1

2,1

2

∣∣∣∣∣12 , 1

2; 1, 0

⟩=

1√2. (7.13)

Note that Eq. (7.11) is equivalent to∣∣∣∣∣12 1

2; 1, 1

⟩=

∣∣∣∣∣12 1

2;1

2,1

2

⟩, (7.14)

while Eqs. (7.12) and (7.13) are equivalent to∣∣∣∣∣12 , 1

2; 1, 0

⟩=

1√2

(∣∣∣∣∣ 1

2

1

2;1

2,−1

2

⟩+

∣∣∣∣∣ 1

2

1

2;−1

2,1

2

⟩). (7.15)

These are just equations (3.8.15a) and (3.8.15b) of Ref. [1], in more “pedantic” no-tation. It is straightforward to find other Clebsch–Gordan coefficients with j = 1,by choosing other values of m, m1 and m2 in Eq. (7.8) (note that we always shouldchoose m = m1 + m2 − 1 for the “upper” equation, and m = m1 + m2 + 1 for the“lower” equation—why?).

Let us consider also the case j = 0, which is a separate set of recursion relations,because rotations do not mix different j values. If j = 0 then also m = 0, and theleft-hand side of Eq. (7.8) will now vanish. Choosing m1 = 1

2and m2 = 1

2in the

“upper” equation thus yields:

√0 =√

1

⟨1

2

1

2;−1

2,1

2

∣∣∣∣∣12 1

2; 0, 0

⟩+√

1

⟨1

2

1

2;1

2,−1

2

∣∣∣∣∣12 1

2; 0, 0

⟩, (7.16)

and thus ⟨1

2

1

2;−1

2,1

2

∣∣∣∣∣12 1

2; 0, 0

⟩= −

⟨1

2

1

2;1

2,−1

2

∣∣∣∣∣12 1

2; 0, 0

⟩. (7.17)

We can choose the righ-hand side of this equation to be equal to a constant c, andthat fixes the other j = 0 Clebsch–Gordan coefficient. We thus find that∣∣∣∣∣12 1

2; 0, 0

⟩= c

(∣∣∣∣∣12 1

2;1

2,−1

2

⟩−∣∣∣∣∣12 1

2;−1

2,1

2

⟩), (7.18)

which is nothing else than Eq. (3.8.15d) of Ref. [1] if we choose c = 1/√

2. Try findingthe remaining equation, Eq. (3.8.15c) from the recursion relations!

8. Section 4.2: parity

18

Here is another proof of Theorem 4.1 (which is just an application of Theorem1.2). Suppose that H commutes with π, and that H is not degenerate. It followsthat, for two energy eigenkets |Em〉 and |En〉

0 = 〈En|[H, π]|Em〉 = (En − Em)〈En|π|Em〉 . (8.1)

Thus, if n 6= m we find that 〈En|π|Em〉 = 0, because then Em 6= En. Now, takem = n and define pn = 〈En|π|En〉. We have that

π|En〉 =∑m

|Em〉〈Em|π|En〉 = pn|En〉 , (8.2)

because only the term m = n in the sum contributes. Thus, |En〉 is a parity eigenstate,with eigenvalue pn. Since the only possible eigenvalues of π are ±1, we thus concludethat energy eigenstates are parity even (pn = +1), or parity odd (pn = −1). Anexample is the one-dimensional harmonic oscillator. Note that non-degeneracy of thespectrum of H is important!

9. Section 5.1: non-degenerate perturbation theory

Consider a hamiltonianH = H0 + λV , (9.1)

, where we have solved the eigenvalue problem for H0, i.e., we know the eigenvaluesE(0)n and eigenkets |n(0)〉, with

H0|n(0)〉 = E(0)n |n(0)〉 . (9.2)

We do not know the eigenvalues En and eigenkets |n〉 of the full hamiltonian, and wewant to approximate them in perturbation theory in λ, which we will assume to be asmall parameter.

We expand

En = E(0)n + λ∆(1)

n + λ2∆(2)n + . . . , (9.3)

|n〉 = |n(0)〉+ λ|n(1)〉+ λ2|n(2)〉+ . . . ,

and substitute this into Eq. (9.1):

(H0 + λV )(|n(0)〉+ λ|n(1)〉+ λ2|n(2)〉+ . . .) (9.4)

= (E(0)n + λ∆(1)

n + λ2∆(2)n + . . .)(|n(0)〉+ λ|n(1)〉+ λ2|n(2)〉+ . . .) .

Terms of order λk should be equal on both sides of this equation. At order λ0 = 1,we recover Eq. (9.2), as one would expect. At order λ, we find:

V |n(0)〉+H0|n(1)〉 = ∆(1)n |n(0)〉+ E(0)

n |n(1)〉 , (9.5)

where we already divided by λ. This is a vector equation, with the vectors kets in ourHilbert space. Therefore, let us consider the components of this vector equation on

19

the basis of normalized energy eigenkets of the unperturbed hamiltonian H0, {|k(0)〉}.Taking first |k(0)〉 = |n(0)〉, we find, taking the inner product with |n(0)〉,

〈n(0)|V |n(0)〉+ 〈n(0)|H0|n(1)〉 = ∆(1)n + E(0)

n 〈n(0)|n(1)〉 . (9.6)

Using that 〈n(0)|H0 = 〈n(0)|E(0)n two terms cancel between the left- and right-hand

sides, and we find a very simple expression for the first-order energy shift:

∆(1)n = 〈n(0)|V |n(0)〉 . (9.7)

Now let us take the inner product with |k(0)〉 6= |n(0)〉:

〈k(0)|V |n(0)〉+ 〈k(0)|H0|n(1)〉 = E(0)n 〈k(0)|n(1)〉 , (9.8)

where we used that 〈k(0)|n(0)〉 = 0. Using that 〈k(0)|H0 = 〈n(0)|E(0)k , we find

〈k(0)|n(1)〉 =1

E(0)n − E(0)

k

〈k(0)|V |n(0)〉 , (9.9)

or

|n〉 = cn(λ)|n(0)〉+ λ∑k 6=n

1

E(0)n − E(0)

k

|k(0)〉〈k(0)|V |n(0)〉+O(λ2) . (9.10)

Let us pause to consider what we learned. First, our calculation does not provideus with a solution for the component of |n(1)〉 in the |n(0)〉 direction. We thus havean unknown constant cn(λ) multiplying |n(0)〉 in Eq. (9.10); all we know is thatcn(λ = 0) = 1. For now, we will arbitrarily set cn(λ) = 1 for all λ, but we will returnto this issue below. Second, we observe that the result is singular if unperturbedeigenkets |k(0)〉 6= |n(0)〉 exist with E

(0)k = E(0)

n , i.e., if the n-th level is degenerate! Fornow, we will assume that the n-th level is not degenerate, and postpone a discussionof the degenerate case till later.

Let us also consider what happens at order λ2. From Eq. (9.4) we obtain

V |n(1)〉+H0|n(2)〉 = ∆(2)n |n(0)〉+ ∆(1)

n |n(1)〉+ E(0)n |n(2)〉 , (9.11)

where we divided by λ2. Proceeding as we did before, we first take the inner productwith |n(0)〉; using again that 〈n(0)|H0 = 〈n(0)|E(0)

n the terms with |n(2)〉 cancel again,and we find

∆(2)n = 〈n(0)|V |n(1)〉 . (9.12)

Here we used that, with choosing cn(λ) = 1, |n(1)〉 does not have a component along|n(0)〉, i.e., 〈n(0)|n(1)〉 = 0 (so we made a clever choice for cn(λ)!). Substituting the sumin Eq. (9.10), which equals |n(1)〉 into this, we get our final result for the second-orderenergy shift:

∆(2)n =

∑k 6=n

1

E(0)n − E(0)

k

〈n(0)|V |k(0)〉〈k(0)|V |n(0)〉 . (9.13)

Defining a short-hand Vkn = 〈k(0)|V |n(0)〉, we can also write this as

∆(2)n =

∑k 6=n

|Vkn|2

E(0)n − E(0)

k

, (9.14)

20

if we recall that Vnk = V ∗kn. Again, this result works only if the n-th level is notdegenerate. Finally, take the inner product of Eq. (9.11) with |k(0)〉 6= |n(0)〉:

〈k(0)|V |n(1)〉+ 〈k(0)|H0|n(2)〉 = ∆(1)n 〈k(0)|n(1)〉+ E(0)

n 〈k(0)|n(2)〉 , (9.15)

and using again that 〈k(0)|H0 = 〈n(0)|E(0)k and 〈k(0)|n(0)〉 = 0 we obtain

〈k(0)|n(2)〉 =1

E(0)n − E(0)

k

(〈k(0)|V |n(1)〉 −∆(1)

n 〈k(0)|n(1)〉)

(9.16)

=1

E(0)n − E(0)

k

∑`6=n

1

E(0)n − E(0)

`

〈k(0)|V |`(0)〉〈`(0)|V |n(0)〉

− 1

(E(0)n − E(0)

k )2〈n(0)|V |n(0)〉〈k(0)|V |n(0)〉

=1

E(0)n − E(0)

k

∑`6=n

Vk`V`n

E(0)n − E(0)

`

− VknVnn

(E(0)n − E(0)

k )2.

Again, we see that the component 〈n(0)|n(2)〉 does not follow from our calculation,but, recall that we chose 〈n(0)|n〉 = 1 for the full solution, to all orders in λ.

Let us now address this latter point. We want the state |n〉 to be normalized,but the state we constructed actually is not normalized! So, let us define

|n〉N =√Zn|n〉 , (9.17)

for some real number Zn > 0 such that |n〉N is normalized, N〈n|n〉N = 1. Observingthat 〈n(0)|n〉 = 1, we see that

√Zn = 〈n(0)|n〉N , so that the probability to find |n〉N

in the state |n(0)〉 is equal to Zn.Using Eq. (9.17) we can calculate Zn from 1 = N〈n|n〉N = Zn〈n|n〉:

Z−1n = 〈n|n〉 (9.18)

=(〈n(0)|+ λ〈n(1)|+ λ2〈n(2)|+ . . .

) (|n(0)〉+ λ|n(1)〉+ λ2|n(2)〉+ . . .

)= 1 + λ2〈n(1)|n(1)〉+O(λ3) ,

where in the last step we used that |n(1)〉 and |n(2)〉 were cleverly chosen to be or-thogonal to |n(0)〉. Substituting |n(1)〉 into this, we find (prove this!)

Zn = 1− λ2∑k 6=n

|Vkn|2

(E(0)n − E(0)

k )2+O(λ3) . (9.19)

10. Sec. 5.2: degenerate perturbation theory

Here we review a of of the technical details of Sec. 5.2 of Ref. [1], considering firstthe projectors needed to project on the degenerate subspace D and its “complement”(which contains all states orthogonal to any of the states inside D), and then a fewcalculational details of the example of the linear Stark effect.

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10.1. Projectors

In Sec. 5.2 of Ref. [1], the projector P0 onto the subspace D spanned by the degenerateH0 eigenstates is defined as

P0 =g∑i=1

|m(0)i 〉〈m

(0)i | , (10.1)

where we use a shorthand notation

|m(0)i 〉 ≡ |E

(0)D , i〉 , (10.2)

where E(0)D is the common energy eigenvalue, and the index i labels the g degenerate

eigenstates. The projector P1 onto the space orthogonal to D is defined as

P1 = 1− P0 (10.3)

=∑k

|E(0)k 〉〈E

(0)k | −

g∑i=1

|m(0)i 〉〈m

(0)i |

=∑

k/∈{m1,...,mg}|E(0)

k 〉〈E(0)k | . (10.4)

We have that

P 20 =

g∑i=1

g∑j=1

|m(0)i 〉〈m

(0)i |m

(0)j 〉〈m

(0)j | (10.5)

=g∑i=1

g∑j=1

|m(0)i 〉δij〈m

(0)j | =

g∑i=1

g∑j=1

|m(0)i 〉〈m

(0)i | = P0 ,

P 21 = (1− P0)2 = 1− 2P0 + P 2

0 = 1− P0 = P1 , (10.6)

andP0P1 = P1P0 = 0 , (10.7)

because 〈m(0)j |E

(0)k 〉 = 0 if k /∈ {m1, . . . ,mg}, and thus E

(0)k 6= E

(0)D .

10.2. Linear Stark effect

Here are a few more details on the linear Stark effect. Let us derive Eq. (5.2.19)of Ref. [1] in more detail. First, notation. Historically, instead of labeling hydrogenenergy eigenstates with `, the labels s (` = 0), p (` = 1), d (` = 2), etc. were used,so

|2s〉 ≡ |n = 2, ` = 0,m = 0〉 , |2p,m = 0〉 ≡ |n = 2, ` = 1,m = 0〉 . (10.8)

The two matrix elements in Eq. (5.2.19) are complex conjugates of each other (sincethe operator V is hermitian), so we need to calculate only one of them:

〈2s|V |2p,m = 0〉 = −e| ~E|〈2s|z|2p,m = 0〉 (10.9)

22

= −e| ~E|〈2s|r cos θ|2p,m = 0〉

= −e| ~E|∫ ∞

0drr2

∫ 1

−1d(cos θ)

∫ 2π

0dφ

×[R20(r)(Y 0

0 )(θ, φ)]∗r cos θ

[R21(r)Y 0

1 (θ, φ)]

= −2πe| ~E|∫ ∞

0drr3R∗20(r)R21(r)

∫ 1

−1d(cos θ)(Y 0

0 )∗(θφ) cos θY 01 (θ, φ)

= −e| ~E|∫ ∞

0drr3 1

8a30

(2− r

a0

)r√3a0

e−r/a0

× 2π∫ 1

−1d(cos θ)

1√4π

√3

4πcos2 θ

= −e| ~E|(3√

3a0

)×(

1√3

)= 3e| ~E|a0 ,

where a0 is the Bohr radius, we used Eqs. (B.5.7) and (B.6.7) of Ref. [1], and we usedthat ∫ ∞

0dx xne−x = Γ(n+ 1) = n! (10.10)

after substituting y = r/a0. Therefore, the matrix (5.2.18) of Ref. [1] is equal to

P0V P0 = 3e| ~E|a0

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

, (10.11)

on the basis of the four degenerate H0 eigenstates|2s〉

|2p,m = 0〉|2p,m = 1〉|2p,m = −1〉

. (10.12)

The 4× 4 matrix P0V P0 has eigenvalues

+3e| ~E|a0 , −3e| ~E|a0 , 0 , 0 . (10.13)

The energy levels of the perturbed p-wave states thus become

E+2 = − e2

8a0

+ 3e| ~E|a0 , 1 state , (10.14)

E−2 = − e2

8a0

− 3e| ~E|a0 , 1 state ,

E2 = − e2

8a0

, 2 states .

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The eigenstates for the first two eigenvalues are, respectively,

1√2

(|2s〉+ |2p,m = 0〉) ,1√2

(|2s〉 − |2p,m = 0〉) . (10.15)

(Check this!) See Eqs. (5.2.20) and (5.2.21), as well as Fig. 5.1 in Ref. [1].For perturbation theory to work, we want the energy shifts to be small compared

to the differences between energy levels of the unperturbed states. The smallestenergy difference for the n = 2 energy level is that with the n = 3 energy level

E3 − E2 = − e2

18a0

+e2

8a0

=5e2

72a0

. (10.16)

We thus want 3e| ~E|a0 � 5e2

72a0, or | ~E| � 5e2

216a20≈ 2× 10−19 V/m.

Appendix A. Typos in Ref. [1]

Eq. (3.5.4b) should readJz|a, b〉 = b|a, b〉 . (A.1)

Eq. (3.5.33) should read

m = −j,−j + 1, . . . , j − 1, j︸ ︷︷ ︸ .2j + 1 states

References

[1] J. J. Sakurai and J. Napoletano, Modern Quantum Mechanics, 2nd edition,Addison-Wesley, 2011

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