J. Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 1 Solutions
1. (a)
(A × B) · (C × D) = εijk Aj B k εilm Cl Dm
= (δjl δkm − δjm δkl )Aj B k Cl Dm
= Aj B k Cj Dk − Aj B k Ck Dj
= (A · C)(B · D) − (A · D)(B · C) .
(b)
(∇ × ∇f )i = εijk ∂ j ∂ k f = εijk ∂ k ∂ j f
(since ∂ j ∂ k = ∂ k ∂ j )
= −εikj ∂ k ∂ j f
(since εijk = −εikj )
= −εijk ∂ j ∂ k f
(relabel j ↔ k)
= −(∇ × ∇f )i
and therefore ∇ × ∇f = 0.
(c)
∇ · (A × B) = εijk ∂ i (Aj B k ) = εijk [(∂ i Aj )B k + Aj (∂ i B k )]
= (∇ × A)k B k − Aj (∇ × B)j
(since εijk = −εjik )
= B · (∇ × A) − A · (∇ × B) .
(d)
[∇ × (A × B)]i = εijk εklm ∂ j (Al Bm )
= (δil δjm − δim δjl )[(∂ j Al )Bm + Al ∂ j Bm ]
= (∂ j Ai )Bj − (∂ j Aj )Bi + Ai ∂ j Bj − Aj ∂ j Bi
= (B · ∇)Ai − Bi (∇ · A) + Ai (∇ · B) − (A · ∇)Bi .
Therefore
∇ × (A × B) = A(∇ · B) − B(∇ · A) + (B · ∇)A − (A · ∇)B .
1
2. Since C is a constant vector we have
Z
Z
Z
3
3
C · ∇ × A d x = C · (∇ × A) d x = Ci εijk ∂j Ak d3 x
=
Z
∂j (−εjik Ci Ak ) d3 x = −
=−
Z
=−
Z
=
Z
(C × A) · n̂ da
j
Z
∇ · (C × A) d3 x
(by the divergence theorem)
Z
C j εjik ni Ak da
Z
n̂ × A da .
k i
εijk C A n da = +
C · (n̂ × A) da = C ·
But C was arbitrary, so we conclude that
Z
Z
∇ × A d3 x = n̂ × A da .
3. The definition of determinant is det(aij ) = εijk a1i a2j a3k and hence with


1
2
3


3
(aij ) =  4 −2
2
5 −1
we have
det(aij ) = ε123 a11 a22 a33 + ε132 a11 a23 a32 + ε312 a13 a21 a32
+ ε321 a13 a22 a31 + ε231 a12 a23 a31 + ε213 a12 a21 a33
= (1)(1)(−2)(−1) + (−1)(1)(3)(5) + (1)(3)(4)(5)
(−1)(3)(−2)(2) + (1)(2)(3)(2) + (−1)(2)(4)(−1)
= 79 .
4. We have

x − a11

 −a21

xI − A = 
..

.

−a12
···
−a1n
x − a22
···

−a2n 

.
..

.

..
.
−an1
−an2
..
.
···

x − ann
Now, by definition, the determinant of this matrix is
det(xI − A) = εi1 ··· in (xI − A)i1 1 · · · (xI − A)in n
= εi1 ··· in (xδi1 1 − ai1 1 ) · · · (xδin n − ain n ) .
2
Note that this is a sum of products, where each product consists of one of the
two terms within each pair of parenthesis. Then
det(xI − A) = εi1 ··· in (xn δi1 1 · · · δin n − ai1 1 xn−1 δi1 2 · · · δin n
− ai2 2 xn−1 δi1 1 δi3 3 · · · δin n − · · · − ain n xn−1 δi1 1 · · · δin−1 n−1
+ terms of lower degree in x + (−1)n ai1 1 · · · ain n )
= xn ε1···n − xn−1 (ai1 1 εi1 2···n + ai2 2 ε1i2 3···n + · · · + ain n ε1···(n−1)in )
+ terms of lower degree in x + (−1)n εi1 ··· in ai1 1 · · · ain n
= xn − xn−1 (a11 + a22 + · · · ann )
+ terms of lower degree + (−1)n det A
= xn − (tr A)xn−1 + terms of lower degree + (−1)n det A .
5. (a) The kinetic
energy (which in this case is the total energy) is given by
P
2
m
v
T =
i i /2. Take the origin of our coordinate system to be the
i
center of mass. Then for a rotating object we have vi = ωri where ω = θ̇,
and hence
1X
1
T =
mi ri2 ω 2 = Iω 2
2 i
2
where I is the moment of inertia. But we also have
X
X
mi ri × vi
ri × pi =
L=
i
i
so that (since v is perpendicular to r)
X
X
mi ri2 ω = Iω .
mi vi ri =
L=
i
i
Eliminating ω we have
1 L2
2 I
and hence the Hamiltonian is given by
T =
H=
1 2
L
2I
where now L2 is the usual angular momentum operator.
(b) The eigenfunctions of L2 are the spherical harmonics Ylm , and the eigenvalues are simply ~2 l(l+1), and hence the energy eigenvalues of the rotator
are given by
~2
El =
l(l + 1) .
2I
Note that these energy levels are (2l + 1)-fold degenerate.
3
6. (a) We have
ψ(θ, φ, 0) =
3Y11 + 4Y73 + Y71
√
26
and hence a measurement of ψ will necessarily result in one of three possibilities:
p
√
√
i. L = ~ 1(1 + 1) = 2~ and Lz = ~ with a probability of (3/ 26)2 =
0.346154.
p
√
√
ii. L = ~ 7(7 + 1) = 56~ and Lz = 3~ with a probability of (4/ 26)2 =
0.615385.
p
√
√
iii. L = ~ 7(7 + 1) = 56~ and Lz = ~ with a probability of (1/ 26)2 =
0.0384615.
(b) By Problem 5 we know that the energy levels of D2 are given by
El =
~2
l(l + 1) .
2I
From
ψl (x, t) = ψl (x, 0)e−iEl t/~
we have
4
3
1
ψ(θ, φ, t) = √ Y11 e−iE1 t/~ + √ Y73 e−iE7 t/~ + √ Y71 e−iE7 t/~
26
26
26
where
~2
28~2
and
E7 =
.
I
I
(c) The expectation value of E is defined by
E1 =
hEi = hψ(θ, φ, t)|Hψ(θ, φ, t)i .
But
1 2
L ψ(θ, φ, t)
2I
~2 3 · 2 1 −iE1 t/~ 4 · 56 3 −iE7 t/~ 1 · 56 1 −iE7 t/~
√ Y1 e
+ √ Y7 e
+ √ Y7 e
=
2I
26
26
26
Hψ(θ, φ, t) =
so using the orthonormality of the Ylm we easily find
hEi = hψ(θ, φ, t)|Hψ(θ, φ, t)i
~2 9 · 2 16 · 56 1 · 56
=
+
+
2I 26
26
26
= 18.6538
4
~2
.
I
But we are given that ~/4πIc = 30.4 cm−1 , and hence using π = 3.1416,
c = 2.9979 × 1010 cm/sec, and ~ = 1.0546 × 10−27 erg-sec we have
hEi = 18.6538(4πc~)(30.4 cm−1 ) = 2.253 × 10−13 erg
= (2.253 × 10−13 erg)(0.624 × 1012 eV/erg)
= 0.141 eV .
Another more general way to look at this is to note that using Hψn =
En ψn and hψm |ψn i = δmn we have
X
DX
E
cm ψm e−iEm t/~ H
hEi = hψ|Hψi =
cn ψn e−iEn t/~
m
=
e−i(En −Em )t/~ c∗m cn En hψm |ψn i
X
|cn | En .
m,n
=
n
X
n
2
2
This is just the usual statistical definition of average value where |cn | is
the probability of having the energy En . Using the results of parts (a)
and (b) we can then immediately write
hEi = 0.346154(~2/I) + 0.615385(28~2/I) + 0.0384615(28~2/I)
= 18.6539(~2/I)
which is the same as above (up to roundoff error).
7. (a) By direct calculation we have, for example,
0 1
0 −i
1
0
σ1 σ2 =
=i
= iσ3 = iε123 σ3 .
1 0
i
0
0 −1
And in general we find that (for i 6= j)
σi σj = iεijk σk
which also implies that for i 6= j we have σj σi = −σi σj . Therefore, for
i 6= j it follows that
σi σj + σj σi = 0 .
Also by direct calculation we can verify that
σi 2 = I .
Combining these last two results we have
σi σj + σj σi = 2Iδij .
5
For example,
−i
0
+
0
i
i
0
−i 0
=
+
0 −i
0 i
σ1 σ2 + σ2 σ1 =
0 1
1 0
0
i
−i
0
0
1
1
0
= 0.
Then adding the equations
σi σj − σj σi = 2iεijk σk
and
σi σj + σj σi = 2Iδij
it is easy to see that
2σi σj = 2Iδij + 2iεijk σk
or simply
σi σj = Iδij + iεijk σk .
(b) Using the last result from part (a) we have (where there is no difference
between upper and lower indices since we are in R3 with a Cartesian
coordinate system)
(a · σ)(b · σ) = ai σi bj σj = ai bj (Iδij + iεijk σk )
= ai bi I + iai bj εijk σk = (a · b)I + i(a × b)k σk
= (a · b)I + i(a × b) · σ .
(c) Expanding the exponential we have
e
−iθ·σ/2
2
3
1
1
θ·σ
2 θ·σ
3 θ·σ
+ (−i)
+ ···
+ (−i)
= I + (−i)
2
2!
2
3!
2
2
3
4
θ·σ
1 θ·σ
i θ·σ
1 θ·σ
=I −i
−
+
+
+ ··· .
2
2!
2
3!
2
4!
2
From part (b) we see that (for any unit vector â)
(â · σ)2 = (â · â)I = I
(â · σ)3 = â · σ
(â · σ)4 = I .
6
Therefore, letting a · σ = θ · σ/2 = (θ/2)(θ̂ · σ) we obtain
e
−iθ·σ/2
2
3
4
θ
1 θ
1 θ
1 θ
= I − i(θ̂ · σ) −
I + i(θ̂ · σ)
+
I + ···
2 2! 2
3! 2
4! 2
2
4
3
1 θ
1 θ
θ
1 θ
=I 1−
+
+ · · · − i(θ̂ · σ) −
+ ···
2! 2
4! 2
2 3! 2
= I cos
θ
θ
− i(θ̂ · σ) sin .
2
2
Using the explicit form of each σi we have
θ̂ · σ = θ̂1 σ1 + θ̂2 σ2 + θ̂3 σ3
"
# "
# "
0 θ̂1
0 −iθ̂2
θ̂3
=
+
+
θ̂1 0
iθ̂2
0
0
#
"
θ̂1 − iθ̂2
θ̂3
=
θ̂1 + iθ̂2
−θ̂3
#
"
θ̂−
θ̂3
=
θ̂+ −θ̂3
0
−θ̂3
#
and therefore
e−iθ·σ/2 = I cos
=
"
θ
θ
− i(θ̂ · σ) sin
2
2
cos θ/2 − iθ̂3 sin θ/2
−iθ̂− sin θ/2
−iθ̂+ sin θ/2
cos θ/2 + iθ̂3 sin θ/2
#
.
(d) Since Si = (~/2)σi and (for i 6= j) σi σj = iεijk σk , we have (again for
i 6= j)
i~
Si Sj = εijk Sk .
2
Therefore, also using Si2 = ~2 /4, we have
Sx Sy Sx Sz Sy =
i~
~2
i~3
~4
Sz Sx Sz Sy = − Sy Sz Sy = −
Sx Sy =
Sz .
2
4
8
16
Similarly,
Sx Sz Sx Sy Sx = −
~2
i~3 2 i~5
i~
Sy Sx Sy Sx = − Sz Sy Sx =
S =
.
2
4
8 x
32
7
(e) It is clear from part (d) that any product of the Si ’s can be reduced to
either a constant or a single Sj (times a constant). Thus the most general
spin-dependent operator A must be expressible as the sum of a constant
plus terms linear in each Sj , i.e.,
A = A0 + A1 Sx + A2 Sy + A3 Sz .
8. (a) We want to express the spin state
1
χ= √
13
" #
2
3
as a linear combination of the eigenstates of Sy . We already know that
the eigenvalues are going to be ±~/2, but we can easily prove it:
−λ −i~/2 det(Sy − λI) = = λ2 − ~2 /4 = 0
i~/2
−λ and hence λ± = ±~/2 as claimed.
To find the eigenvector corresponding to λ+ = +~/2 we solve
"
#" #
"
#
1 i
a
a + ib
~
~
(Sy − λ+ I)v+ = −
=−
= 0.
2 −i 1
2 −ia + b
b
Hence b = ia so the eigenvector is of the form
" # " #
a
a
v+ =
=
.
b
ia
Normalizing
v+ says that 2 |a|2 = 1, so taking the phase to be 1 we have
√
a = 1/ 2. Thus we write the normalized eigenvector as
" #
1
1
(y)
.
χ+ = √
2 i
Similarly, for λ− = −~/2 we have (there is no relation between the a, b in
this equation and the a, b in the equation for v+ )
"
#" #
"
#
a
~ 1 −i
~ a − ib
(Sy − λ− I)v− =
=
=0
2 i
2 ia + b
1
b
so that b = −ia and the normalized eigenvector is
"
#
1
1
(y)
.
χ− = √
2 −i
8
We now write (again, these a, b are completely different)
" #
" #
"
#
2
1
1
a 1
b
(y)
(y)
χ= √
= aχ+ + bχ− = √
+√
13 3
2 i
2 −i
so that equating components on both sides we have
p
2 2/13 = a + b
and
Solving these yields
p
−i3 2/13 = a − b .
a=
r
b=
r
and
2
3
− i√
13
26
3
2
+ i√ .
13
26
Thus the probability that a measurement of Sy will yield +~/2 is
2
|a| =
2
9
1
+
=
13 26
2
and the probability that a measurement of Sy will yield −~/2 is also
|b|2 =
9
1
2
+
= .
13 26
2
Note that these are independent of the original state χ as long as both
components of χ are real.
(b) What happens if the components of χ are complex? Let us write
" #
α
χ=
β
2
2
where |α| + |β| = 1. Now we have
" #
" #
"
#
α
1
a 1
b
(y)
(y)
χ=
= aχ+ + bχ− = √
+√
2 i
2 −i
β
so that
1
α = √ (a + b)
2
and
i
β = √ (a − b)
2
1
a = √ (α − iβ)
2
and
1
b = √ (α + iβ) .
2
and hence
9
Now the probability of obtaining +~/2 when you measure the spin in the
y-direction is given by
2
|a| =
=
1
1
2
2
(α − iβ)(α∗ + iβ ∗ ) = [|α| + |β| + i(αβ ∗ − α∗ β)]
2
2
1
[1 − 2 Im(αβ ∗ )] .
2
Similarly, the probability of obtaining −~/2 is
|b|2 =
=
1
1
(α + iβ)(α∗ − iβ ∗ ) = [|α|2 + |β|2 − i(αβ ∗ − α∗ β)]
2
2
1
[1 + 2 Im(αβ ∗ )] .
2
Note that in the case where α and β are both real, these probabilities
reduce to simply 1/2 as in part (a).
10