angular momentum.pdf

L11 ANGULAR MOMENTUM

UCSC, PHYS 139A, SPRING 2008

GEY-HONG SAM GWEON

1. Orbital Angular Momentum

From CM, we are familiar with the following denition of the angular momentum

~L = ~r ~p

In QM, we dene angular momentum operators accordingly

(1.1) Lx = ypz zpy, Ly = zpx xpz , Lz = xpy ypxThe rst thing to note is the following canonical commutation relations for momentum and position

components:

(1.2) pa = i~ qa

, [qa, pb] = i~ab, [pa, pb] = 0, [qa, qb] = 0

where a = 1, 2, 3 standing for x, y, z, and q1, q2, q3 = x, y, z respectively. When going from CMto QM, there is one thing that needs to be always checked. Operators for observables must be

Hermitian. This is not always automatic because of operator algebra. In this case, we are ne since

Lx = pz y py z = pz y py z = ypz zpy = Lxwhere we made use of the fact that, because of [qa, pb] = i~ab, pa and qb for a 6= b can be swappedfreely. Now we calculate the commutation relations for the components of the angular momentum:

[Lx, Ly] = [ypz zpy, zpx xpz]= [ypz, zpx xpz] [zpy, zpx xpz]= [ypz, zpx] [ypz, xpz] [zpy, zpx] + [zpy, xpz]Using the rule [AB, C] = A[B, C] + [A, C]B and [A, BC] = B[A, C] + [A, B]C repeatedly, and usingEq. 1.2, we get

[Lx, Ly] = y[pz, z]px 0 0 + py[z, pz]x(note that x, py commute) = i~ypx + i~xpy

= i~Lz

Other commutators need not be calculated but they can be inferred by cyclic permutation of indices

(xyz) (zxy) (yzx)(1.3) [Lx, Ly] = i~Lz, [Ly, Lz] = i~Lx, [Lz, Lx] = i~Ly

The L2 operator is dened as

L2 = L2x + L2y + L

2z1

2 UCSC, PHYS 139A, SPRING 2008 GEY-HONG SAM GWEON

and the commuation relation between it and La is easy to see

[L2, Lx] = [L2x + L2y + L

2z, Lx]

= [L2y, Lx] + [L2z, Lx]

= Ly(i~Lz) i~LzLy + Lz(i~Ly) + i~LyLz= 0

Thus

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0

In the spherical coordinate system, what are the dierential operators for L? As it will turn out inthe next section, only Lz and L

2are necessary to know. We already know, from L9, what they are:

Lz = i~ (1.4)

L2 = ~2

sin

(sin

) ~

2

sin2 2

2(1.5)

We have stated then that the spherical harmonics {|lm = Ylm(, )} form the set of simulataneouseigenstates of Lz and L

2:

Lz|lm = m~|lmL2|lm = l(l + 1)~2|lm

l = 0, 1, 2, ....m = l, ..., 0, ..., lWe will see in the next section how this comes about.

2. General Angular Momentum Algebraic Approach

In this section, we take a very general approach to describe the angular momentum. This approach

is a little similar to the a, a+ description of the simple harmonic oscillator in that it is an operatorapproach, or an algebraic approach, rather than a dierential equation approach, or an analytical

approach. There is a fundamental dierence in the current case, though.

The operator, or algebraic, approach to describe angular momentum includes cases that simply

cannot be described within the analytical approach.

The reason for this is the spin angular momentum, which does not have any CM analogue, and so

we simply do not know how to set up a dierential equation for that!

So much for the motivation. Within the algebraic approach, the angular momentum operator is

dened as consisting of three observables Jx, Jy, Jz that satisfy the following fundamental relations1

(2.1) [Jx, Jy] = i~Jz, [Jy, Jz] = i~Jx, [Jz, Jx] = i~Jywhich are called fundamental commutation relations for angular momentum. Dene

(2.2) J = Jx iJyNote that J+J = J2x + J2y + i~JyJx i~JxJy = J2x + J2y + ~2Jz and so J2 = J+J+ J2z ~2Jz. JJ+can be treated similarly and we get

(2.3) J2 = JJ + J2z ~2Jz1

Here, we use J to denote general angular momentum. L is the standard notation for orbital angular momentumand S for spin angular momentum. J is for the general or the total angular momentum.

L11 ANGULAR MOMENTUM 3

The following commutation relations can be derived easily (the rst line was already derived in the

previous section, since only the fundamental commutation relations are used there):

[J2, Jx] = [J2, Jy] = [J2, Jz] = 0(2.4)

[J, Jz] = [Jx iJy, Jz] = i~Jy i(i~)Jx = ~J(2.5)

From the general formalism of QM, Eq. 2.4 means that J2 and any of Jx, Jy, Jz are compatibleobservables and so simulatneous eigenstates of the two of them can be found to form a natural

basis. By convention, we take simultaneous eigenstates of J2 and Jz, while in principle we couldjust as well choose Jx or Jy as the second observable instead of Jz. Two things are worthwhile to

note. (1) This example illustrates the basic point that just because an observable A is compatible

with two observables B, C does not mean that B, C are compatible with each other. (2) Once

a natural basis is chosen as simultaneous eigenstates of J2 and Jz, all other operators, e.g. Jxand Jy, are completely desribed using those basis states by denition. Let us call a simulatneous

eigenstate |jm, speccied by some value of the total angular momentum squared (J2) and the zcomponent

2

of the angular momentum (Jz). At the moment, we need to specify only the lattervalue, while we just specify the former as some nite number:

Jz|jm = m~|jmInvestigate the meaning of J, by applying the commutator of Eq. 2.5 on |jm:

[J, Jz]|jm = ~J|jmJm~|jm JzJ|jm = ~J|jmRe-arranging the terms, we get

(2.6) Jz

(J|jm

)= (m 1)~

(J|jm

)Thus, J+ increases the Jz value by 1, while J decreased it by 1. Namely, these operators areladder operators for angular momentum, in analogy with a that were ladder operators for theharmonic oscillator problem. There is an important dierence here, though: m~ values must bebounded both from below and from above, given a nite total angular momentum value. In fact,

it is the convention that the symbol j stands for the maximum value of m. In other words,

J+|jj = 0Now, apply J2 to |jj. From Eq. 2.3 (2nd form), we get

J2|jj = j(j + 1)~2|jjNow, consider operating J repeatedly until we reach mmin such that J|jmmin = 0. Note twothings. First, due to Eq. 2.3 (1st form), J2|jmmin = mmin(mmin 1)~2|jmmin. Second, sinceJ2 and J commute with each other, as can be seen from Eq. 2.4, if | is an eigenstate ofJ2 with eigenvalue , then J2J| = JJ2| = J|, i.e. J| is an eigenstate of J2 withthe same eigenvalue. Namely, in our case, any |jm has the same J2 eigenvalue, and especially,comparing the m = j and m = mmin cases, we have j(j + 1)~2 = mmin(mmin 1)~2. The solutionis mmin = j. Note that by the nature of J (or J+), Eq. 2.6, the dierence between m = j andm = j must be an integer, and so j can be any half integer. Since we dened j as the maximumJz value, it cannot be negative.

We summarize our ndings so far.

2

Standard notations for the z component of the angular momentum is m,ml,ms,mj , where the subscript maybe used whenever necessary to avoid ambiguity. Unfortunately, m is the same symbols as that for the electronmass. However, this is the standard practice and usually no confusion seems to arise.


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%

Jz|jm = m~|jmJ2|jm = j(j + 1)~2|jm

jm|jm = jjmmm = j, ..., 0, ..., jj = 0,

12, 1,

32, 2, ...

Note that the orthonormalization condition is imposed for |jm, possible since the group index jmis a discrete index (Denition 4.9 of L8). For a given j, the number of states with dierent mvalues is 2j + 1, which is often referred to as the multiplicity.

Denition 2.1. Spin, Fermion, Boson

These are the solutions for the angular momentum problem purely from the algebraic approach.

Note that there is a dierence from the solutions written for the orbital angular momentum in

that j can be now half integers as well as integers, while l was restricted to integers (previoussection and L9). There are two possibilities, (i) the algebraic solutions above are superuous and

really the half integer solutions are forbidden due to some reasons that we failed to consider,

and (ii) the algebraic solutions with half integer solutions correspond to some real physics that

we could not capture by dening QM angular momentum starting from CM angular momentum

as we did in Section 1. The correct one is (ii), and the angular momentum that has no CM

analogue is called the spin angular momentum. Spin is an intrinsic property of elementary particles.

In fact, elementary particles can be grouped into two kinds, one having half integer spin values

and the other integer spin values. The former is a group of so-called fermions and the latter

bosons. The spin angular momentum for all elementary fermions (udcstb quarks, and leptons

e,, ,e,, ; and their anti-particles),is j = 1/2, and so we call them spin 1/2 particles. The spinfor elementary bosons vary from 0 (Higgs boson not yet observed), 1 (photon), and 2 (graviton

not yet observed). The spin of composite particles is determined by the particular way the spins of

constituent particles are summed (see below for the topic of the angular momentum addition). For

instance pi mesons consist of a quark and an anti-quark in such a way that their spin is 0. So, pimesons themselves are bosons, although they are made up of two fermions.

Note 2.2. The Dimension of the Hilbert Space

The solutions presented in the box are all we need in order to compute any physical quantities.

If j is xed for a given problem, then the Hilbert space is spanned by {|jm|m = j, ..., 0, ..., j}and so its dimension is 2j + 1. This is the case of spin, which is xed for a given particle. Forthe electron, spin is 1/2, and so the Hilbert space is two dimensional, i.e. there are two basisvectors corresponding to m = 1/2. For the orbital angular momentum problem for the Hydrogenproblem, j is not xed, but covers all integers 0, 1, 2, ..... In that case, the dimension of the Hilbertspace is innite dimensional, which comes from the basic fact that in the real space, the Hilbert

space is a space of functions, and thus innite dimensional.

Note 2.3. Wave Function

You may ask what is the wave function for |jm? In the case of the orbital angular momentum, Land |lm, we have already noted how L2 and Lz are represented by dierential operators in and, and how |lm is the spherical harmonic function, Ylm(, ). This is all well, but actually we donot need to know what this function looks like in order to gure out all physical quantities. It is

quite the same situation as in the harmonic oscillator problem. Life was much easier when we used

the a algebra on eigenstates of a+a for the simple harmonic oscillator problem than dealingwith the explicit functions of x (or p). Similarly, in the case of the angular momentum, it is much

easier to use the L algebra on simulataneous eigenstates of L2, Lz rather than dealing with the


explicit functions , . Besides, in the case of spin, we do not even know3 how to write the wavefunction in terms of , . This is related to the fact that we do not know what the microscopicmechanism for the spin 1/2 angular momentum is and we know the existence of the spin onlythrough its observed properties. In that case, the best we can do is to deal with the abstract

states |jm. However, the structure of science is such that that is all we need for now.

Let us close this section by expressing the action of ladder operators more denitely. From

Eq. 2.6, we can dene a number Cjm, by J|jm = Cjm,|jm 1. Noting that J = J andjm|J = j,m 1|Cjm,, we have4

jm|JJ|jm =[jm|J

] [J|jm

]= |Cjm,|2

On the other hand, from Eq. 2.3, we have

jm|JJ|jm =[j(j + 1)m2 m] ~2

=[j2 m2 + j m] ~2

= (j m)(j m+ 1)~2

Equating these two results, we can determine Cjm, up to an arbitrary phase. But, the states|jm themselves are arbitrary up to a phase and so we have some freedom. By convention, relativephases of |jm are chosen so that Cjm, values are real and positive. Thus we get(2.7) J|jm = ~

(j m)(j m+ 1)|j,m 1

3. Conservation of Angular Momentum

The angular momentum is conserved when the system is invariant under rotation. For a spherically

symmetric system, the angular momentum is conserved since by denition the system is isotropic.

A less symmetric problem can occur, e.g. if a

~B eld is applied along the z direction. In that case,the system becomes only cylindrically symmetric. Lz is conserved, but not Lx, Ly.

When the angular momentum is conserved, according to Denition 8.3 of L8, it should be possible

to take simultaneous eigenstates of H and the angular momentum. In the previous section, wesaw that the eigenstates of the angular momentum is characterized by J2, Jz eigenvalues. Thus,it is no surprise that in the spherically symmetric potential problem we found that eigenstates

can be written as |nlm = Rnl(r)|lm, where |lm (Ylm) are the eigenstates of the orbital angularmomentum. This remains true when the problem becomes only cylindrically symmetric. The

dierence is that in the cylindrically symmetric problem the degeneracy between states with dierent

m values is lifted, and that Lx, Ly are no longer conserved.

The following table summarizes this situation in general.

3

This might sound impossibly ridiculous at rst. You may ask, what is wrong if I just dene the angular

momentum of spin as Sz = i~/s, where s is the angle in the spin space and just solve the eigenvalueequation for it? Also, similarly for S2, why can't we just use Eq. 1.5 with s and s? The problem isthat if you solve those two dierential equations Sz|1/2,1/2 = ~2 |1/2,1/2 and S2|1/2,1/2 = 3~

2

4|1/2,1/2in terms of s and s dened this way, then you get results for the two wave functions, corresponding to|1/2,1/2, that do not obey the general angular momentum properties, Eq. 2.6. There are also other mathematicalproblems. In short, this way of naively justifying the spin angular momentum is futile. See more discussions in

Modern Quantum Mechanics by J. J. Sakurai.

4

A comment on the notation. Griths uses expressions such as O| and |O. These expressions are notstandard, while it is perhaps understandable why they were chosen for pedagogy. In the standard notation, these

two are written as, |O and O|. As we convered in the formalism, (O|) = |O. In my opinion, thestandard notation is better, since it is more powerful and cleaner.


Hamiltonian Symmetry Conserved Angular Momentum Stationary States

(angular momentum part)

p2

2m + V (r) Spherical J2, Jx, Jy, Jz |jm

ddtJa = ddtJ2 = 0 for a = x, y, z Ejm is independent of mJ can be L or S or the total degeneracy at least 2j + 1

p2

2m + V (r) +AJz Cylindrical J2, Jz only |jm

ddtJz = ddtJ2 = 0 but ddtJx,y 6= 0 Ejm is dependent on mNote 3.1. Orbital and Spin Subspaces

In the above table, the nature of J is left ambiguous on purpose. Generally, a particle's angularmometum comprises of the orbital part and the spin part. In the above table, J can be the orbitalangular momentum L, the spin angular momentum S, or the sum of the two. When we learn howto add angular momenta, we will update the above table so that stationary states are written

more completely in terms of all relevant quantum numbers. For now, consider J above as meaningonly one of L, S and the sum of the two.

The above discussion makes it clear that in the spherical symmetric case [H, Ja] = 0 for a = x, y, zand in the cylindrically symmetric case [H, J2] = [H, Jz] = 0 but [H, Jx,y] 6= 0. For any givenHamiltonian, this can be explicitly checked. The following two examples illustrate this point, and

along the way, we derive some useful algebra, such as Eqs. 3.1,3.2, which are actually specic

examples of very general properties of operator algebra in the angular momentum space. Other

than that, how much satisfaction one draws from the following two examples depends on how much

one appreciates the rst sentence of this section that it is the rotational invariance that leads to

the consevation of the angular momentum.

Example3.2. SphericallySymmetricPotentialandOrbitalAngularMomentumConservation

First of all, in this and the next example, the Einstein summation convention (any repeating indices

imply summation) and the Levi-Civita symbol

5 abc will be used for the sake of economy. Also, qameans x, y, z for a = 1, 2, 3 respectively. The orbital angular momentum (~r ~p) is then expressedas La = abcqbpc.

Take the case

H =p2

2m+ V (r)

(Reminder: r is the operator for r, not the unit vector along the radial direction!) From Eqs. 1.1,1.2,the following can be seen

[La, qb] = [acdqcpd, qb]= acdqc[pd, qb]= acdqcdb(i~)= i~abcqc(3.1)

Similarly,

[La, pb] = [acdqcpd, pb]= acd[qc, pb]pd= acdpdcb(i~)= i~abcpc(3.2)

5

The denition of this symbol is as follows. (i) 123 = 1. (ii) abc is invariant under a cyclic permutation. (iii)abc = bac (anti-symmetry). (iv) abc = 0 if a = b. From this denition it follows that abc is zero if any of thetwo indices are the same, abc acquires a minus sign if any of the two indices are swapped.


where in the last step, c = b is set rst using cb and then the dummy index d is changed to c.These imply

[La, r2] = [La, qbqb]

= qb[La, qb] + [La, qb]qb= qbabcqc + abcqcqb

bc are dummy summation indices and can be redened as cb = (abc + acb)qbqcanti-symmetry of the Levi-Civita symbol = 0

The proof that

[La, p2] = 0is totally unnecessary, since it is precisely the same as the one just given if one substitutes p forq. Thus, La commutes with any function of p

2and r2 and thus with a spherically symmetric

Hamiltonian,

p2

2m + V (r).[H, L2] = [H, Lx,y,z] = 0

Example 3.3. Cylindrically Symmetric Potential and Orbital Angular Momentum

Conservation

Let us say that the symmetry is broken so that the system is symmetric on rotation around the zaxis, but not symmetric on other rotations. We can write

H =p2

2m+ V (r, )

(Reminder: r, are operators not unit vectors.) In this case, the Hamiltonian commutes with

Lz = i~/, since H is independent of , but not with Lx, Ly, since Lx, Ly necessarily involvepz, which involves a dierentiation on . Another way to see this latter point is that H is notinvariant when changes: when = 0 and changes that is the process in which Lx drives themotion, when = pi/2 it is Ly, so just by considering these two cases it is obvious that Lx,y cannotbe conserved in general if V is a function of . You can do an explicit evaluation of Lx, Ly in

terms of and as well (Section 4.3.2 of Griths), if you are unsatised at this point. Lastly, L2

commutes with H, since [L2, cos()] = [L2, zr ] = 0 because L2commutes with both z and r. Thus,

L2 commutes with any function of .

[H, L2] = [H, Lz] = 0, [H, Lx,y] 6= 0

4. Spin 1/2

Who Has Seen the Wind?

Christina Rossetti

Who has seen the wind?

Neither I nor you:

But when the leaves hang trembling

The wind is passing thro'.

Who has seen the wind?

Neither you nor I:

But when the trees bow down their heads

The wind is passing by.

The spin 1/2 angular momentum is fundamental in descrbing elementary particles, since all elementary

fermions have spin 1/2. It is dened as a dynamical variable in a two dimensional Hilbert space

satisfying the general angular momentum algebra, Eq. 2.1. Unfortunately, [in the standard view]

we do not really yet understand how spin 1/2 arises microscopically (for instance, you are welcome

to do Prob. 4.25 and see the hint of why this is), so this is as concrete as we can get.


Then, how do we ever know what the spin is? The angular momentum, orbital or spin, gives rise to

the magnetic moment. We know the existence of spin 1/2, because when we apply magnetic eld to

align magnetic moments in atoms, we discover that some magnetic moment splits into two dierent

energy levels, which is impossible to understand if we use orbital angular momentum alone. In the

same vein, the Stern-Gerlach experiment Example 4.4 of Griths on certain atomic beams, like

an Ag beam, splits the incoming beam into two distinct outgoing beams, which can be understood

only if we invoke an angular momentum with degeneracy 2, that is spin 1/2. Thus, the above poem.

The Stern-Gerlach set up is shown below (http://www.modelofreality.org). The essence of the

Stern-Gerlach set up is an inhomogeneous magnetic eld,

~B = x~i+ (B0 +z)~k, where ~i,~k are unitvectors along the x axis and z axis, respectivley (see gure). Then, ~F = (~ ~B) = z~k x~i.Note that this classical thinking

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does not include the quantum eect,the Larmor precession which

will be discussed later in this note. Here it suces to note that this precession can result in some

net spin along the x direction and this may result in some force in the x direction. The result is alip like image on the lm, where the central part corresponds to net force along the x directionaveraging out to zero, while the edge of the lip corresponding to the case when the net force

along the x direction is nite [Question for readers: why is the vertical separation reduced to zeroat the far, left or right, edge of the lip?].

Then, spin 1/2 is, by denition,

Sa =~2a

where a, a = x, y, z or equivalently a = 1, 2, 3, are the Pauli matrices.

x =[

0 11 0

], y =

[0 ii 0

], z =

[1 00 1

]Note that I do not use for Pauli matrices, but they are obviously QM operators, being matrices.Since they cannot be confused with numbers, I will take hats from them to be ecient. I will still

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Here, the classical approximation is justied since we are dealing with atoms. For Ag atoms, the typical energy

when it is hot is say, 100 meV, which translates to the momentum p =

2mE 2 108 940 106 0.1 eV/c 1.4 105 eV/c, where c is the speed of light. In contrast, if the uncertainty in position is say, 1 micron, thenp ~/1 micron 2 101 eV/c. And so, we see that p p, justifying the classical approximation. So, that isthat for Ag atoms. How about if one tries Stern-Gerlach for an electron? This argument alone remains valid for

that case as well, if the energy is high enough. However, an electron is charged, and so, not only there is force

due to (~ ~B), but also is there the Lorenz force q~v ~B. Since there is a small component of ~B along the xdirection, there will be a Lorenz force in the z direction, Fz evxx due to the uncertainty of the wave packetalong the x direction. This force better be smaller than z due to (~ ~B), for the Stern-Gerlach eect to beobservable. The ratio is R = evxx/z. Plugging in z = e~/(2m) (see Eq. 4.3) and vx px/m, one sees thatR 2xpx/~ 1! So, the Stern-Gerlach eect is not possible for an electron. What if Ag atom is ionized in theoven? That is perfectly ne, due to the fact that vx px/MAg while z = e~/(2m), and so R m/M 104.


keep on spin though. Many goody propertis of spin physics have been explored in Homework7.For instance, they are Hermitian operators, as they should be, and also satisfy the commutation

relations [a, b] = 2iabcc, so that

[Sa, Sb] = i~abcScsatisfying the fundamental angular momentum commutation relations. Also, note that if we take

|z |1/2, 1/2 =[

10

], |z |1/2,1/2 =

[01

]then the phase convention of Eq. 2.7 is satised, since

S+ =~2

(x + iy)

= ~[

0 10 0

]S = ~

[0 01 0

]

S|1/2, 1/2 = ~|1/2,1/2S+|1/2,1/2 = ~|1/2, 1/2In the above, the notation is such that

(4.1) |z = |+ = + = |1/2, 1/2and

(4.2) |z = | = = |1/2,1/2

The spin wave functions, in whatever symbol they are represented by, mean column vectors in the

spin Hilbert space, and are generally referred to as spinors.

The notable things about the spin 1/2 space is that wave functions do not return to itself on

rotation by 2pi, but only on rotation by 4pi. (Homework 07)

The exact relationship between the spin 1/2 and the magnetic moment is

~ = ~S

where the gyromagnetic ratio (more standard notation is gB), is pretty close to 2B/~, forthe electron, where B is the Bohr magneton

(4.3) B =e~2m

B = 9.271024 J/T(SI) or 5.79109 eV/gauss. So, for an electron in a ~B eld, the Hamiltonianis

H = ~ ~B= B~ ~B(dene z axis as pararllel to

~B) = BBzz= Lz

L BBz is the freqeuncy at which the spin precesses in this problem, and this is called theLarmor precession. as we have seen in Homework 7. Note an important fact.

10 UCSC, PHYS 139A, SPRING 2008 GEY-HONG SAM GWEON'

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In the Larmor precession problem for spin 1/2, expectation values such as Sx and Sy rotatewith an angular velocity L. When Lt = 2pi, the wave function does not return to itself. Rather

|(t = 2pi/L) = |(t = 0)This is a fundamental property of spin 1/2 you rotate the particle by 360

and the wave function

acquires the negative sign. It is as though the spin space is double the size of the ordinary

space as only 4pi rotation brings the wave function to itself!

Note 4.1. The Complete Wave Function for the Hydrogen Problem

When we discussed the periodic table, we mentioned the spin angular momentum already. Now we

are at a potision to write down the total wave function, including the spin.

nlmlms(~x) = Rnl(r)Ylml(, )sgn(ms)= |n, l|l,ml|ms |n, l,ml,mswhere is dened as in Eqs. 4.1,4.2, |n, l is the ket dened in the radial wave function space,|l,ml is the ket dened in angular function space, representing Ylml , and |ms is the short-handfor |s = 1/2,ms used in Eqs. 4.1,4.2. A general wave function is then

(~x) =

n,l,ml,ms

CnlmlmsRnl(r)Ylml(, )sgn(ms)

| =

n,l,ml,ms

Cnlmlms |n, l,ml,ms

In the pure-Coulomb problem that we have been dealing with so far, energy eigen-values depend

only on n and so we have

|(t) =

n,l,ml,ms

exp(iEnt/~)Cnlmlms |n, l,ml,ms

How can one view this complete wave function which is an amalgam of the spatial part and the

spin part? One can think of | as a ~x dependent function whose value at each ~x is a columnvector (spinor). Or, one can think of it as a spinor, whose component is a function of ~x. Thesetwo views are totally equivalent.

5. Addition of Angular momenta

Let us consider two angular momenta

~J1 and ~J2. Let us be very general and treat these two asany angular momenta. By denition, they are assumed to describe completely dierent degrees of

freedom, which means

(5.1) [J1,a, J2,b] = 0

for any a = x, y, z and b = x, y, z. An example of such two dierent angular momenta can beorbital angular momentum and spin angular momentum of a given particle. Or, they could be

spin angular momenta of two particles of the same kind or dierent kinds. We consider the sum

~J ~J1 + ~J2. The z component is(5.2) Jz = J1,z + J2,z

and the raising/lowering operators are

(5.3) J = J1, + J2,


and the length squared operator is, noting that

~J1 and ~J2 represent separate degrees of freedomand so they commute,

J2 = J21 + J22 + 2 ~J1 ~J2(5.4)

= J21 + J22 + 2J1,xJ2,x + 2J1,yJ2,y + 2J1,zJ2,z(5.5)

Noting that J1,x = 12(J1,+ + J1,

)and J1,y = 12i

(J1,+ J1,

)and so on, we see that

2J1,xJ2,x + 2J1,yJ2,y = J1,+J2, + J1,J2,+and so

(5.6) J2 = J21 + J22 + J1,+J2, + J1,J2,+ + 2J1,zJ2,zQuestions arise. When we form the sum of two angular momenta, does the sum also satisfy the

fundamental commutation relations of the angular momentum, Eq. 2.1? The answer is yes. Let us

see:

[Ja, Jb] = [J1,a + J2,a, J1,b + J2,b]

(Eq. 5.1) = [J1,a, J1,b] + [J2,a, J2,b]

= abcJ1,c + abcJ2,c= abcJcand thus

[Ja, Jb] = abcJcwhich is the same as Eq. 2.1 (the notation abc is explained in Example 3.2). Given these fundamental

commutation relations, it follows immediately that Jz commutes with J2. How about J1,z? From

Eq. 5.5, it can be noted that the two terms in the middle has a nite commutator with J1,z andthat is

[J2, J1,z] = [2J1,xJ2,x + 2J1,yJ2,y, J1,z]

= 2[J1,x, J1,z]J2,x + 2[J1,y, J1,z]J2,y= 2J1,yJ2,x + 2J1,xJ2,yThus, J1,z does not commute with J

2and nor does J2,z by symmetry. Only the total component

Jz = J1,z + J2,z commutes with J2. How about J21 ? Applying Eq. 2.4 to J1, one notices easilythat J2 as expressed by Eq. 5.4 or Eq. 5.5 commute with J21 . By symmetry, it also commutes

with J22 . Let us summarize. Before we do it, just one note on notation. Before the angularmomentum is added, one can consider all possible combinations of m1 and m2 and write the stateas |j1,m1, j2,m2 |j1,m1|j2,m2. This juxtaposition of two states corresponding to two dierentdegrees of freedom technically means that we are considering a tensor product, or a direct product,

Hilber space [cf. Denition 4.21 of L8].#

"

!

When one adds two angular momenta

~J1 and ~J2, the total angular momentum ~J ~J1 + ~J2 ischaracterized by J2 and Jz. J

21 and J

22 are compatible with J

2, but J1,z and J2,z (or any otherseparate components along x or y) are not. In terms of states labeled with quantum numbers

|j1,m1, j2,m2 add the two angular momenta |j,m, j1, j2

Note that there are (2j1 + 1)(2j2 + 1) distinct states, before angular momenta are added up. Afterthey are added, there should be the same number of distinct states. It is easy to notice that there

is one state on the left hand side with m1 = j1 and m2 = j2, giving m = j1 + j2. This is themaximum value that m can ever have, and therefore correspond to j = j1 + j2. For this j value, wecan keep applying J and nd all states |j = j1 + j2,m, j1, j2. Now consider m1 +m2 = j1 + j2 1states on the left hand side. There are two such states, since m1 or m2, but not both, canbe lowered by 1 from their respective maximum values to give this m1 +m2 value. Notice that


j = j1 + j2 state on the right hand side consumed one linear combination of m1 +m2 = j1 + j2 1state on the left hand side, and so there is one state left. This state must be the maximum mstate for the j = j1 + j2 1 state. Continuing this way, we can nd all j values. What is theminimum j value? This is determined by the condition that the number of distinct states shouldbe the same before and after angular momenta are added:

(2j1 + 1)(2j2 + 1) =j1+j2j=jmin

(2j + 1)

= (j1 + j2)(j1 + j2 + 1) jmin(jmin 1) + (j1 + j2 jmin + 1)which gives a quadratic equation

j2min = (j1 + j2)2 + 2(j1 + j2) 4jj2 2(j1 + j2)

= (j1 j2)2

which means

jmin = |j1 j2|So, here is another important nding.

(5.7) j1, j2 add the two angular momenta j = |j1 j2|, |j1 j2|+ 1, ..., j1 + j2 1, j1 + j2The discussion above that leads to the identication of jmin actually indicates how |j,m, j1, j2 canbe constructed. When we are done with that construction, we get a general relation such as

(5.8) |j,m, j1, j2 =

m1+m2=m

Cj1j2jm1m2m|j1,m1, j2,m2

The coecients Cj1j2jm1m2m are called Clebsh-Gordan coecients. Some simple cases are tabulated inTable 4.8 of Gritths, as a column vector. The reverse relationship also exists

|j1,m1, j2,m2 =j

Cj1j2jm1m2m|j,m = m1 +m2, j1, j2

These coecients can be read o from Table 4.8 of Griths as a row vector. These two sets

of vectors are both natural basis sets for the subspace where j1 and j2 are xed, and thus thefollowing identities hold within that subspace.

m1,m2

|j1,m1, j2,m2j1,m1, j2,m2| = 1

j1,m1, j2,m2|j1,m1, j2,m2 = m1m1m2m2j,m

|j,m, j1, j2j,m, j1, j2| = 1

j,m, j1, j2|j,m, j1, j2 = jjmmExample 5.1. Sum of Two 1/2 Spins

Two electrons. One electron plus a proton. A quark plus an anti-quark to form a meson. All of

these examples correspond to this important example. Let us use the following notation, which is

quite standard.

| |j1 = 1/2, m1 = 1/2, j2 = 1/2, m2 = 1/2| |j1 = 1/2, m1 = 1/2, j2 = 1/2, m2 = 1/2| |j1 = 1/2, m1 = 1/2, j2 = 1/2, m2 = 1/2| |j1 = 1/2, m1 = 1/2, j2 = 1/2, m2 = 1/2


Also, it is standard, when it is not ambiguous, to leave out j1, j2 in |j,m, j1, j2. Simply by lookingup Table 4.8, we can write

|11 = | |10 = 1

2(| + | )

|1 1 = | These form the triplet states for the total spin 1 case. According to Eq. 5.7, adding j1 = 1/2 andj2 = 1/2 should result in total j = 1 and j = 0. In the spin context, we say s1 = 1/2 and s2 = 1/2give s = 1 and s = 0. The s = 0 state is a singlet (again from Table 4.8)

|00 = 12

(| | )

One qualitatively says that s = 1 state is where the two spins are parallel, while s = 0 state is wherethe two spins are anti-parallel. As you can see from the wave functions, there is more-than-half

truth to this statement, and so this qualitative statement is made quite often.

Even if one did not look up Table 4.8, this case is simple enough so that we can follow the

procedure described just before Eq. 5.7. First, we can identify that

|11 = | Apply S = S1,+ S2, to this. First note that, from what we obtained in Section 4, S1,| 1 = ~| 1and S1,+| 1 = ~| 1, and similarly for S2,. Thus, we get

S|11 = ~ (| + | )This should be by denition ~

(1 + 1)(1 1 + 1)|10 = ~2|10, using Eq. 2.7 for the total spin

S. Therefore, we get

|10 = 12

(| + | )

Applying S on this, we get

2~| , which must be equal to ~(1 + 0)(1 0 + 1)|11 = 2~|11.So we get nally

|1 1 = | For the |00 state, we need to make it a linear combination of | , | and also make it orthogonalto |11. Thus we get (with a sign convention),

|00 = 12

(| | )

For your condence, you can apply Eq. 5.6 to these states and double check that they are what

you think they are in terms of S. For instance,

S2|10 =(S21 + S

22 + 2S1,zS2,z + S1,+S2, + S1,S2,+

) 12

(| + | )

=(S21 + S

22 + 2S1,zS2,z

) 12

(| + | ) +(S1,+S2, + S1,S2,+

) 12

(| + | )

= ~2(

34

+34 21

4

)|10+ ~2 1

2(| + | )

= ~22|10= 1(1 + 1)~2|10Example 5.2. Strong Zeeman Eect

Cylindrical symmetry results if an external

~B eld is applied. In this case, (see Eq. 6.71 of Griths)

H Z =e

2m(~L+ 2 ~S) ~B


must be added to the Hamiltonian. Note that H Z commutes with Lz and Sz but not withJz = Lz + Sz. Also, H Z commutes with L

2and S2 but not with J2, since it is in the form of

H Z ( ~J + ~S) ~B. So the following table can be obtained, which is an update of the table of Section1.

Hamiltonian Symmetry Conserved Angular Momentum Angular Momentum Part of

(

~J ~L+ ~S) Stationary State~L: orbital, ~S: spin, ~J : total (Dm: minimum degeneracy)

p2

2m + V (r) Spherical J2, L2, S2, Jz, Lz, Sz |jmj or |mlms

H0 ddt Ix,y,z = ddt I2 = 0 E is independent of mj ,ml,ms(vanilla Hydrogen) for any I = L, S, J Dm = (2l + 1)(2s+ 1)

H0 + H Z L, S: cylindrical L2, S2, Lz, Sz but not J

2, Jz |mlms but not |jmjH Z (~L+ 2 ~S) ~B J : none ddt Iz = ddt I2 = 0 E is dependent on ml,ms(strong Zeeman) for I = L, S only

H0 + H so J : spherical J2, Jz, L

2, S2 but not Lz, Sz |jmjH so ~L ~S ddt Jx,y,z = 0 E is independent of mj(spin orbit)

ddt I2 = 0 for any I = L, S, J Dm = 2j + 1Example 5.3. Spin Orbit Interaction

The spin orbit interaction occurs for an electron with non-zero angular momentum. Qualitatively, it

is easy to understand why it occurs. From the electron's reference frame, the nucleus is going around

the electron, with an angular momentum which is opposite to the electron's angular momentum

around the nucleus. So, the nucleus moves in a current loop and thus has a magnetic dipole

moment. The electron at rest at origin of this reference frame has a magnetic dipole moment as

well due to spin, ~ = 2B ~S. Two dipole moments interact via dipole-dipole interaction. So, thespin orbit interaction has the following form:

H so = C~L ~Swhere C is a positive constant, and ~L, ~S are the orbital angular momentum and the spin angularmomentum of the electron. [For details of C, look it up in any QM book, including Griths. It iseasy to gure out that C gets large for small r and large Z, the atomic number.] In this case,

note that the Hamiltonian breaks the spherical symmetry for both

~L and ~S. Moreover, since bothof these vectors, aligning each other, are dynamic, the problem does not even have a cylindrical

symmetry for

~L and ~S. Another way of saying/seeing this is that H so does not commute witheither Lz or Sz. On the other hand, as we can see from Eq. 5.4, ~L ~S = 12

(J2 L2 S2

)and so

Jz is a good quantum number, since

[Jz, ~L ~S] = 12[Jz, J2 L2 S2]

=12

([Jz, J2] [Jz, L2] [Jz, S2]

)(S and L commute as they belong in dierent Hilbert spaces) =

12

(0 [Lz, L2] [Sz, S2]

)= 0

This situation is summarized in the last row of the above table.

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