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Physics 4617/5617: Quantum Physics

Course Lecture Notes

Dr. Donald G. Luttermoser East Tennessee State University

Edition 5.

Abstract

These class notes are designed for use of the instructor and students of the course Physics 4617/5617: Quantum Physics. This edition was last modified for the Fall 2006 semester.

−z

∂y

( z ∂f ∂x

)

• z

∂y

( x ∂f ∂z

) − z

∂x

( y ∂f ∂z

)

+z

∂x

( z ∂f ∂y

)

• x

∂z

( y ∂f ∂z

) − x

∂z

( z ∂f ∂y

)}

( (^) ¯h i

) 2  y ∂f ∂x

• yz ∂^2 f ∂z ∂x − yx ∂^2 f ∂z^2 − z^2 ∂^2 f ∂y ∂x +zx ∂

(^2) f ∂y ∂z − zy ∂

(^2) f ∂x ∂z

• z^2 ∂

(^2) f ∂x ∂y

• xy ∂

(^2) f ∂z^2 −x ∂f ∂y −^ xz

∂^2 f ∂z ∂y

  (^).

2. All the terms cancel in pairs (by virtue of the equality of cross- derivatives) except two:

[Lx, Ly] f =

( (^) ¯h i

) 2 ( y

∂x − x

∂y

) f = i¯hLzf,

and we conclude (dropping the test function) that

[Lx, Ly] = i¯hLz. (VI-6)

3. By cyclic permutation of the indices, it follows that

[Ly, Lz] = i¯hLx and [Lz, Lx] = i¯hLy. (VI-7)

Exercise: Prove Eq. (VI-7) by the technique shown in §VI.B.1.

4. Since Lx, Ly, and Lz do not commute, they are incompatible ob- servables. Hence, according to the generalized uncertainty prin- ciple (Eq. IV-100), we can write

σ^2 Lx σ^2 Ly ≥

2 i 〈ihL¯ z〉

) 2

¯h^2 4 〈Lz〉^2 ,

or σLx σLy ≥ ¯h 2 |〈Lz 〉|. (VI-8)

5. It would therefore be futile to look for states that are simultane- ously eigenfunctions of Lx and of Ly.
6. On the other hand, the square of the total angular momentum does commute with (separately) Lx, Ly, and Lz: a) The square of the total angular momentum is

L^2 ≡ L^2 x + L^2 y + L^2 z. (VI-9)

b) Using the definition of the commutator (Eq. IV-42) along with Eqs. (IV-43), (VI-6), (VI-7), and the following com- mutator relations, [ A,^ ˆ Bˆ + Cˆ

]

[ A,^ ˆ Bˆ

]

[ A,^ ˆ Cˆ

] (VI-10) [ A,^ ˆ Aˆ

] = 0 , (VI-11)

we can write [ L^2 , Lx

]

[ L^2 x, Lx

]

[ L^2 y, Lx

]

[ L^2 z, Lx

]

= Lx [Lx, Lx] + [Lx, Lx] Lx + Ly [Ly, Lx] + [Ly, Lx] Ly +Lz [Lz, Lx] + [Lz, Lx] Lz = 0 + 0 + Ly(−i¯hLz) + (−ihL¯ z)Ly + Lz(i¯hLy) +(i¯hLy)Lz = 0.

Likewise, the same proof can be made for the Ly and Lz components giving [ L^2 , Lx

] = 0,

[ L^2 , Ly

] = 0,

[ L^2 , Lz

] = 0, (VI-12)

or more compactly, [ L^2 , L

] = 0. (VI-13)

and then from the first equation of Eq. (VI-14), we get

L^2 (L±f) = λ(L±f), (VI-18)

so L±f is an eigenfunction of L^2 with the same eigenvalue λ.

8. Now let’s take a look at the second eigenfunction equation of Eq. (VI-14). For this equation, let’s replace f with L±f and write

Lz (L±f) = LzL±f − L±Lzf + L±Lzf = [Lz, L±] f + L±(Lzf) = ±¯hL±f + L±(μf) = (μ ± ¯h)(L±f), (VI-19)

so L±f is an eigenfunction of Lz with the new eigenvalues μ ± ¯h.

9. L+ is called the “raising” operator because it increases the eigen- value of Lz by ¯h.
10. L− is called the “lowering” operator because it decreases the eigenvalue of Lz by ¯h.
11. We also can investigate various relationships with these angular momentum ladder operators.

L+L− = (Lx + iLy)(Lx − iLy) = L^2 x − iLxLy + iLyLx + L^2 y = L^2 x + L^2 y − i[Lx, Ly], (VI-20)

L−L+ = (Lx − iLy)(Lx + iLy) = L^2 x + iLxLy − iLyLx + L^2 y = L^2 x + L^2 y + i[Lx, Ly], (VI-21)

and the commutator of L+ and L− results from subtracting Eq. (VI-21) from Eq. (VI-20) which gives

[L+, L−] = L+L− − L−L+ = − 2 i[Lx, Ly] = 2¯hLz. (VI-22)

Exercise: Verify the following commutator relations:

[Lz, L+] = ¯hL+ (VI-23) [Lz, L−] = −¯hL− (VI-24) [ L^2 , L+

]

[ L^2 , L−

] = 0. (VI-25)

If we add Eqs. (VI-20) and (VI-21) we get

L+L− + L−L+ = 2(L^2 x + L^2 y) = 2(L^2 − L^2 z). (VI-26)

Finally, if we use Eq. (VI-26) along with Eq. (VI-9) we get

L^2 = L^2 x + L^2 y + L^2 z

(L+L− + L−L+) + L^2 z = L+L− − ¯hLz + L^2 z (VI-27) = L−L+ + ¯hLz + L^2 z. (VI-28)

Exercise: Prove Eqs. (VI-27) and (VI-28).

12. For a given value of λ in Eq. (VI-14), a “ladder” of states is obtained with each “rung” separated from its neighbors by one unit of ¯h in the eigenvalue of Lz (see Figure VI-1). a) To ascend the ladder, we apply the raising operator.

b) To descend the ladder, we apply the lowering operator.

c) This procedure cannot go on forever — eventually a state is reached for which the z-component exceeds the total angular momentum which is impossible (see below).

c) Using Eq. (VI-29), it is obvious that L−L+ft = L−(L+ft) = L−(0) = 0. (VI-32)

d) Now, using the second equation of Eq. (VI-30) in conjunc- tion with the first equation of Eq. (VI-30) along with Eq. (VI-28), we can write L^2 ft = (L−L+ + ¯hLz + L^2 z)ft = (0 + ¯h^2 +^2 ¯h^2 )ft = (¯h^2 +^2 ¯h^2 )ft = ( + 1)¯h^2 ft = λft (VI-33)

14. Similarly, there must exist a “bottom rung,” fb, such that

L−fb = 0. (VI-34) Let ¯h be the eigenvalue of Lz at this bottom rung, then Lzfb =¯hfb; L^2 fb = λfb. (VI-35) a) Using Eq. (VI-34), we see that L+L−fb = L+(L−fb) = L+(0) = 0. (VI-36)

b) Now, using the second equation of Eq. (VI-35) in conjunc- tion with the first equation of Eq. (VI-35) along with Eq. (VI-27), we can write L^2 fb = (L+L− − ¯hLz + L^2 z)fb = (0 − ¯h^2 +^2 ¯h^2 )fb = (−¯h^2 +^2 ¯h^2 )fb = ( − 1)¯h^2 fb = λfb (VI-37)

15. It is clear from the comparison of Eq. (VI-33) with Eq. (VI-37) that λ ¯h^2

= ( + 1) = ( − 1). (VI-38)

a) This implies that either = − or = + 1. We can reject the second solution since is defined to be the top most rung and this second equation would place above `.

b) Let N be the number of steps one must make from the bottom () state to get to the top () state, then N = − = +, or ` =

N

2.^ (VI-39)

c) At this point, let’s note that L^2 (the angular momentum squared) has the same units as ¯h squared (remember that ` is either an integer or half-integer as shown in Eq. VI- 39), hence the angular momentum has units of ¯h. As such, let’s rewrite the eigenvalue of the Lz operator as μ = m¯h.

d) From these last two points, we see that the maximum value of m is (see Eq. VI-30) and the minimum value of m is − (see Eq. VI-35), where ` can be an integer or half-integer.

e) From this formalism, it is clear that there are 2+ 1 “rungs” (i.e., values of m) in the angular momentum lad- der:, − 1 , − 2 , ..., 1 , 0 , − 1 , ..., −` as shown in Figure (VI-2).

16. Since the eigenvalues are composed of `’s and m’s, so too must be the eigenfunctions. As such, we can now write the two angular momentum eigenvalue equations as

L^2 fm =(+ 1)¯h^2 fm ; Lzfm = m¯hfm , (VI-40) where ` = 0,

2 ,^1 ,^

2 , ...;^ m^ =^ −,^ −^ + 1, ..., ^ −^1 ,.^ (VI-41) This is the reason why we chose (+1) as our differential equation constant for the Schr¨odinger equation in Eqs. (V-14) and (V-15).

(b) As it turns out (see Equations VI-40 and VI-41), the square of the eigenvalue of Lz never even equals the eigenvalue of L^2 (except in the special case ` = m = 0). Comment on the implications of this result. Show that it is enforced by the uncertainty principle (see Eq. VI-8), and explain how the special case gets away with it.

Implication : There is no state in which the angular momentum points in a specific direction. If there were, then we would simultaneously know Lx, Ly, and Lz — but these are incompatible observables — if you know Lx precisely, you cannot know Ly and Lz precisely. The uncertainty principle of Eq. (VI-8) says that Lx, Ly, and Lz cannot be simultaneously determined, except in the special case 〈Lx〉 = 〈Ly〉 = 〈Lz 〉 = 0, which is to say when ` = 0 and m = 0.

C. Eigenfunctions of Angular Momentum.

1. We need to rewrite Lx, Ly, and Lz (see Eqs. VI-3, 4, 5) in spher- ical coordinates. a) The angular momentum operator in vector notation is

L = h¯ i (r × ∇), (VI-42)

where the gradient operator in spherical coordinates is

∇ = ˆr

∂r

• θˆ

r

∂θ

• φˆ

r sin θ

∂φ

, (VI-43)

and r = rˆr, so

L = ¯h i

[ r(ˆr × rˆ)

∂r

• (ˆr × θˆ)

∂θ

• (ˆr × φˆ)

sin θ

∂φ

] . (VI-44)

But (ˆr × rˆ) = 0, (ˆr × θˆ) = φˆ, and (ˆr × φˆ) = −θˆ, and hence

L = ¯h i

( φ^ ˆ ∂ ∂θ − θˆ

sin θ

∂φ

)

. (VI-45)

b) We now need to express the unit vectors θˆ and φˆ in their Cartesian components: θˆ = (cos θ cos φ) ˆx + (cos θ sin φ) ˆy − (sin θ) ˆz φ^ ˆ = −(sin φ) ˆx + (cos φ) ˆy.

c) Using these unit vector relations in Eq. (VI-45) gives us

L = ¯h i

[ (− sin φ ˆx + cos φ yˆ)

∂θ

(VI-46)

− (cos θ cos φ xˆ + cos θ sin φ yˆ − sin θ zˆ)

sin θ

∂φ

] .

Looking at the Cartesian unit vector components for com- parison with L = Lx xˆ + Ly yˆ + Lz zˆ, we see that

Lx = ¯h i

( − sin φ

∂θ −^ cos^ φ^ cot^ θ

∂φ

) , (VI-47)

Ly = ¯h i

(

• cos φ

∂θ − sin φ cot θ

∂φ

) , (VI-48)

and Lz = ¯h i

∂φ.^ (VI-49)

d) We shall also need the raising and lowering operators:

L± = Lx ± iLy = h¯ i

[ (− sin φ ± i cos φ) ∂ ∂θ − (cos φ ± i sin φ) cot θ ∂ ∂φ

] ,

but cos φ ± i sin φ = e±iφ, so handling the raising and lowering operators separately we have

L+ = ¯h

[ (+ cos φ + i sin φ)

∂θ

• i(cos φ + i sin φ) cot θ

∂φ

]

= +¯he+iφ

∂θ

• i cot θ

∂φ

)

L^2 f`m =

( L+L− + L^2 z − ¯hLz

) f`m = he¯ iφ

(∂fm ` ∂θ

• i cot θ ∂f`m ∂φ

) ( −¯he−iφ

)

×

( (^) ∂fm ` ∂θ −^ i^ cot^ θ

∂f`m ∂φ

) − ¯h^2 ∂^2 f`m ∂φ^2 −^

¯h^2 i

∂fm ∂φ = h¯^2(+ 1) fm. (VI-53)

i) At this point, let’s make use of Eq. (VI-51) in Eq. (VI-53). Taking the θ and φ derivatives of Eq. (VI- 51), we get: ∂f`m ∂θ =^ e

imφ dg dθ (VI-54) ∂f`m ∂φ = imeimφ^ g. (VI-55)

ii) Using these equations in Eq. (VI-53) gives

−eiφ

∂θ

• i cot θ ∂ ∂φ

) ( ei(m−1)φ

) (^ ∂g ∂θ

• mg cot θ

)

+m^2 geimφ^ − mgeimφ = eimφ

[ − d dθ

( (^) dg dθ

• mg cot θ

)

• (m − 1) cot θ

(dg dθ

• mg cot θ

)

+m(m − 1)g ] = ( + 1)geimφ.

iii) Canceling eimφ^ gives

−d

(^2) g dθ^2 − m dg dθ cot θ + mg csc^2 θ + (m − 1) cot θ dg dθ +m(m − 1) (1 + cot^2 θ) g = − d^2 g dθ^2 −^ cot^ θ

dg dθ +^ m

(^2) g csc (^2) θ = ( + 1)g.

iv) Now, multiplying each term by − sin θ produces

sin^2 θ d^2 g dθ^2 +sin^ θ^ cos^ θ

dg dθ −m

(^2) g = −(+1)g sin (^2) θ.

v) This is a differential equation for g(θ), but we can write it in the more familiar form of

sin θ d dθ

( sin θ dg dθ

)

• [( + 1) sin^2 θ − m^2 ] g = 0. (VI-56)

2. Note that Eq. (VI-56) is precisely the equation we determined for the θ component of the three dimensional Schr¨odinger equation (see Eq. V-24): g(θ) = Θ(θ). Using this realization in Eq. (VI-52) means that fm = Θ(θ) Φ(φ) = Y (^)m. (VI-57)

a) As such, spherical harmonics are precisely the (nor- malized) eigenfunctions of L^2 and Lz.

b) When we solved the Schr¨odinger equation by separation of variables in §V.A of the notes, we were inadvertently con- structing simultaneous eigenfunctions of the three com- muting operators H, L^2 , and Lz:

Hψ = Eψ, L^2 ψ = (+1)¯h^2 ψ , (^) zψL = m¯hψ. (VI-58)

c) There is a slight inconsistency here: i) The algebraic theory of angular momentum per- mits ` (and hence also m) to take on half -integer values (see Eq. VI-41), whereas the analytic method yielded eigenfunctions only for integer values (see Eq. V-28).

so L^2 = a^2 M^2 v^2 and thus H =

L^2

Ma^2

But from quantum mechanics, we know the eigenvalues of L^2 are ¯h^2 ( + 1), or, since we usually label energies with n and E is the eigenvalue for H, we get En = ¯h

(^2) n(n + 1) Ma^2 (n = 0, 1 , 2 , ...).

(b) What are the normalized eigenfunctions for this system? What is the degeneracy of the nth energy level?

Since there is no radial component to this problem, the eigenfunction is just going to be described by spherical harmonics:

ψnm(θ, φ) = Y (^) nm (θ, φ).

The degeneracy of the nth energy level is just the number of m-values for a given n which is given in §VI.B.15.e as 2 n + 1.

D. Spin and Particle Families

1. In classical mechanics, a rigid body admits two kinds of angular momentum: a) Orbital: L = r × p, associated with the motion of the center of mass. Such motion is referred to as a revolution about the center of mass.

b) Spin: S = Iω, associated with the motion about the center of mass. Such motion is referred to as a rotation about an axis.

c) One then talks about the total angular momentum: orbital (L) + spin (S) angular momenta.

2. By analogy, we have the same description on the microscopic level for quantum mechanics: a) Orbital: The motion of an electron about the nucleus of an atom as described by spherical harmonics with the orbital angular momentum quantum number ` and the magnetic (or azimuthal) quantum number m. This is sometimes referred to as the extrinsic angular momentum (L).

b) Spin: Unlike the classical case, this isn’t the spin of the electron about an axis, since the electron is a point particle (even though we describe electron spin about an axis in elementary physics). Here “spin” is nothing more than the intrinsic angular momentum (S) of the electron. Since this is intrinsic, spin angular momentum is independent of spatial coordinates (r, θ, φ).

3. In quantum mechanics, the algebraic theory of spin is just a car- bon copy of the theory of orbital angular momentum. a) It obeys the following fundamental commutation relations:

[Sx, Sy] = i¯hSz, [Sy, Sz ] = ihS¯ x, [Sz, Sx] = ihS¯ y. (VI-59)

b) The eigenvectors S^2 and Sz satisfy:

S^2 |s ms〉 = s(s + 1)¯h^2 |s ms〉; Sz|s ms〉 = ms¯h|s ms〉. (VI-60) Note that the L^2 and Lz eigenstates are eigenfunctions whereas the S^2 and Sz eigenstates are eigenvectors. As