HOMEWORK ASSIGNMENT 3: Solutions Fundamentals of Quantum Mechanics

PHYS851 Quantum Mechanics I, Fall 2009
HOMEWORK ASSIGNMENT 3: Solutions
Fundamentals of Quantum Mechanics
1. [10pts] The trace of an operator is defined as T r{A} =
set.
P
m hm|A|mi,
where {|mi} is a suitable basis
(a) Prove that the trace is independent of the choice of basis.
Answer:
Let {|mi} and {|em i}
Pbe two independent
P basis sets for our Hilbert space.
We must show that m hem |A|em i = m hm|A|mi.
Proof:
X
hem |A|em i =
m
=
X
mm′ m′′
X
mm′ m′′
=
X
m′ m′′
=
X
m′ m′′
=
X
m′
=
X
m
hem |m′ ihm′ |A|m′′ ihm′′ |em i
(1)
hm′′ |em ihem |m′ ihm′ |A|m′′ i
(2)
hm′′ |m′ ihm′ |A|m′′ i
(3)
δm′ m′′ hm′ |A|m′′ i
(4)
hm′ |A|m′ i
(5)
hm|A|mi
(6)
(b) Prove the linearity of the trace operation by proving T r{aA + bB} = aT r{A} + bT r{B}.
Answer:
T r{aA + bB} =
X
hm|aA + bB|mi
(7)
m
=
X
(ahm|A|mi + bhm|B|mi)
(8)
X
X
hm|A|mi + b
hm|B|mi
(9)
m
= a
m
m
= a T r{A} + b T r{B}
1
(10)
(c) Prove the cyclic property of the trace by proving T r{ABC} = T r{BCA} = T r{CAB}.
Answer:
First, if T r{ABC} = T r{BCA} then it follows that T r{BCA} = T r{CAB}, so we need only
prove the first identity.
T r{ABC} =
X
hm|ABC|mi
(11)
m
=
X
mm′ m′′
=
X
mm′ m′′
hm|A|m′ ihm′ |B|m′′ ihm′′ |C|mi
(12)
hm′′ |C|mihm|A|m′ ihm′ |B|m′′ i
(13)
= T r{CAB}
2
(14)
2. Consider the system with three physical states {|1i, |2i, |3i}. In this basis, the Hamiltonian matrix
is:


1 2i 1
(15)
H =  −2i 2 −2i 
1 2i 1
Find the eigenvalues {ω1 , ω2 , ω3 } and eigenvectors {|ω1 i, |ω2 i, |ω3 i} of H. Assume that the initial
state of the system is |ψ(0)i = |1i. Find the three components h1|ψ(t)i, h2|ψ(t)i, and h3|ψ(t)i. Give
all of your answers in proper Dirac notation.
Answer:
The eigenvalues are solutions to
det |H − ~ωI| = 0
(16)
Taking the determinate in Mathematica gives
4ω + 4ω 2 − ω 3 = 0
(17)
ω(ω 2 − 4ω − 4) = 0
(18)
which factorizes as
which has as its solutions
ω1 = 2(1 −
√
2)
ω2 = 0
√
ω3 = 2(1 + 2)
(19)
(20)
(21)
the corresponding eigenvectors are
√
1
|ω1 i = (|1i + 2i|2i + |3i)
2
1
|ω2 i = √ (−|1i + |3i)
2
√
1
|ω3 i = (|1i − 2i|2i + |3i)
2
(22)
(23)
(24)
The components of |ψ(t)i are found via |ψ(t)i = e−iHt |ψ(0)|i, giving
√
√ 1
2 + e−i2(1− 2)t + e−i2(1+ 2)t
4
√
√ i h2|ψ(t)i = √ e−i2(1− 2)t − e−i2(1+ 2)t
2 2
√
√ 1
h3|ψ(t)i =
−2 + e−i2(1− 2)t + e−i2(1+ 2)t
4
h1|ψ(t)i =
The mathematic script I used to work this problem is on the following page:
3
(25)
(26)
(27)
4
3. Cohen-Tannoudji: pp 203-206: problems 2.2, 2.6, 2.7
2.2 (a) The operator σy is hermitian:
† ∗ 0 −i
0 i
0 −i
†
σy =
=
=
= σy
i 0
−i 0
i 0
We find the eigenvalues via det |σy − ωI| = 0:
−ω −i
det i −ω
= ω2 − 1 = 0
(28)
(29)
The solutions are ω = 1 and ω = −1.
Let the corresponding eigenvectors be |+i and |−i, so that
σy |±i = ±|±i.
(30)
Hit this equation with the bra h1| and insert the projector onto the {|1i, |2i} basis:
(−h1|σy |1i ± 1) h1|±i − h1|σy |2ih2|±i = 0
(31)
inserting the values of the matrix elements of σy then gives:
±h1|±i + ih2|±i = 0
(32)
a non-normalized solution is then
h1|±i = i
h2|±i = ∓1
(33)
(34)
the normalized eigenvectors are then given, up to arbitrary overall phase-factors, by
|+i =
|−i =
1
√ (i|1i − |2i)
2
1
√ (i|1i + |2i)
2
(35)
(36)
(b) The projectors are given by I± = |±ih±|. In matrix form, in the {|1i, |2i} basis, these are
h1|±ih±|1i h1|±ih±|2i
I± =
(37)
h2|±ih±|1i h2|±ih±|2i
!
i ∓1
i −i
=
=
√ √
√ √
2 2
2 2
∓1
−i
∓1 √
∓1
√ √
√
2 2
2 2
1
i
∓
2
2
1
± 2i
2
2 = I and I + I = I:
we need to show that I±
±
+
−
1 ∓i 1 ∓i 2
2
2
2
2
I±
=
±i
±i
1
1
2
2
21 12
i
i
+
∓
∓
4
4
4
4
=
1
1
+
± 4i ± 4i
4 4
1
i
∓
2
2
=
1
± 2i
2
= I±
5
(38)
(39)
(40)
(41)
(42)
(43)
I+ + I− =
1
2
+ 2i
− 2i
1
2
+ 21
− 2i
1 0
=
0 1
= I
=
1
2
+ 2i
(c) The results for M and Ly are attached.
6
+
1
2
− 2i
i
− 2i + 2
1
1
2 + 2
+ 2i
1
2
(44)
(45)
(46)
(47)
2_2M.nb
1
H* Enter the matrix M *L
M = 882, I Sqrt@2D<, 8-I Sqrt@2D, 3<<;
MatrixForm@MD
In[190]:=
i
2
j
j
j
j
è!!!
k -Â 2
è!!!
2 y
z
z
z
z
3 {
Out[191]//MatrixForm=
Â
H* now compute the Hermitian conjugate and make sure it is self-adjoint *L
In[192]:=
MatrixForm@Conjugate@Transpose@MDDD
i
2
j
j
j
j
è!!!
-Â
2
k
è!!!
2 y
z
z
z
z
3 {
Out[192]//MatrixForm=
Â
H* Now find the eigenvalues *L
In[198]:=
y = Solve@Det@M - w IdentityMatrix@2DD ã 0D;
H* extract the eigenvalues from y *L
w1 = w ê. y@@1, 1DD
In[213]:=
Out[213]=
1
w2 = w ê. y@@2, 1DD
In[214]:=
Out[214]=
4
H* now we find the eigenvectors the old-fashioned way*L
H* e1 and e2 will be the normalized eigenvectors *L
In[227]:=
y = Solve@HHM - w1 [email protected], c<L@@1DD ã 0, cD
Out[227]=
Â
99c Ø ÅÅÅÅÅÅÅÅÅ
è!!! ==
2
81, c ê. y@@1DD<
e1 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
Sqrt@Conjugate@81, c ê. y@@1DD<D.81, c ê. y@@1DD<D
In[229]:=
Out[229]=
2
Â
9$%%%%%%
ÅÅÅÅ , ÅÅÅÅÅÅÅÅÅ
è!!! =
3
3
2_2M.nb
2
In[230]:=
y = Solve@HHM - w2 [email protected], c<L@@1DD ã 0, cD
88c Ø -Â
Out[230]=
è!!!
2 <<
81, c ê. y@@1DD<
e2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
Sqrt@Conjugate@81, c ê. y@@1DD<D.81, c ê. y@@1DD<D
In[231]:=
Out[231]=
1
2
9 ÅÅÅÅÅÅÅÅÅ
ÅÅÅÅ =
è!!! , -Â $%%%%%%
3
3
H* we can form the projectors via the outer-product function *L
In[243]:=
I1 = Outer@Times, Conjugate@e1D, e1D;
MatrixForm@I1D
2
i
ÅÅÅ
j
3
j
j
j
è!!!!
j
j - ÅÅÅÅÅÅÅÅÅ
 2
Å
k
3
 2 y
ÅÅÅÅÅÅÅÅÅ
Å z
3
z
z
z
z
z
1
ÅÅÅ
3
{
è!!!!
Out[244]//MatrixForm=
In[245]:=
I2 = Outer@Times, Conjugate@e2D, e2D;
MatrixForm@I2D
1
i ÅÅÅ
j
3
j
j
j
è!!!!
j
j ÅÅÅÅÅÅÅÅÅ
 2
k 3 Å
 2 y
- ÅÅÅÅÅÅÅÅÅ
Å z
3
z
z
z
z
z
2
ÅÅÅ
3
{
è!!!!
Out[246]//MatrixForm=
H* To verify the othogonality relation, we compute [email protected]
In[251]:=
[email protected]
Out[251]=
0
H* Lastly, we verify the closure relation *L
In[252]:=
MatrixForm@I1 + I2D
Out[252]//MatrixForm=
J
1 0
N
0 1
*L
2_2Ly.nb • 9/28/09
1
H* First we enter the matrix Ly *L
—
Ly = ÅÅÅÅÅÅÅÅ 880, Sqrt@2D, 0<, 8-Sqrt@2D, 0, Sqrt@2D<, 80, -Sqrt@2D, 0<<;
2I
MatrixForm@LyD
0
i
j
j
j
j
j
j
—
j
j
ÅÅÅÅÅÅÅ
è!!!!Å
j
j
2
j
j
j
j
j
j 0
k
—
- ÅÅÅÅÅÅÅ
è!!!!Å
2
0
—
ÅÅÅÅÅÅÅ
è!!!!Å
2
y
z
z
z
z
z
z
Â
—
- ÅÅÅÅÅÅÅ
z
è!!!!Å z
z
2 z
z
z
z
z
z
0 z
{
0
H* Now we take the transpose and complex conjugate *L
MatrixForm@Conjugate@Transpose@LyDDD ê. Conjugate@—D Ø —
i
0
j
j
j
j
j
j
—
j
j
ÅÅÅÅÅÅÅ
è!!!!Å
j
j
2
j
j
j
j
j 0
k
—
- ÅÅÅÅÅÅÅ
è!!!!Å
2
0
—
ÅÅÅÅÅÅÅ
è!!!!Å
2
y
z
z
z
z
z
z
Â
—
- ÅÅÅÅÅÅÅ
z
è!!!!Å z
z
2 z
z
z
z
z
0 z
{
0
H* By comparison, we see it is Hermitian *L
H* This the the matrix whose determinant gives the spectrum of Ly *L
MatrixForm@Ly - w IdentityMatrix@3DD
i -w
j
j
j
j
j
j
—
j
j
ÅÅÅÅÅÅÅ
è!!!!Å
j
j
2
j
j
j
j
j 0
k
—
- ÅÅÅÅÅÅÅ
è!!!!Å
2
-w
—
ÅÅÅÅÅÅÅ
è!!!!Å
2
y
z
z
z
z
z
z
Â
—
- ÅÅÅÅÅÅÅ
z
è!!!!Å z
z
2 z
z
z
z
z
-w z
{
0
H* Here we let mathematica solve the characteristic polynomial *L,
y = Solve@Det@Ly - w IdentityMatrix@3DD ã 0D;
Solve::svars : Equations may not give solutions for all "solve" variables.
H* so the eigenvalues are: *L
More…
2_2Ly.nb • 9/28/09
w1 = w ê. y@@1, 1DD
w2 = w ê. y@@2, 1DD
w3 = w ê. y@@3, 1DD
0
-—
—
H* "Eigensystem" will give the eigenvalues and Hun-normalized eigenvectors *L
y = Eigensystem@LyD;
e1p = y@@2, 1DD;
e2p = y@@2, 2DD;
e3p = y@@2, 3DD;
H* To normalize them, we compute the normalization constants *L
n1 = Sqrt@[email protected];
n2 = Sqrt@[email protected];
n3 = Sqrt@[email protected];
H* The normalized eigenvectors are then: *L
e1 = e1p ê n1
e2 = e2p ê n2
e3 = e3p ê n3
1
1
9 ÅÅÅÅÅÅÅÅÅ
è!!! , 0, ÅÅÅÅÅÅÅÅÅ
è!!! =
2
2
1
Â
1
9- ÅÅÅÅ , ÅÅÅÅÅÅÅÅÅ
=
è!!! , ÅÅÅÅ
2
2
2
1
Â
1
9- ÅÅÅÅ , - ÅÅÅÅÅÅÅÅÅ
=
è!!! , ÅÅÅÅ
2
2
2
H* Here we form the projectors by using Mathematicas outer-product function *L
I1 = Outer@Times, Conjugate@e1D, e1D;
I2 = Outer@Times, Conjugate@e2D, e2D;
I3 = Outer@Times, Conjugate@e3D, e3D;
2
2_2Ly.nb • 9/28/09
3
MatrixForm@I1D
MatrixForm@I2D
MatrixForm@I3D
1
Å
i ÅÅÅ
j
2
j
j
j
j
j
0
j
j
j
j
1
j ÅÅÅ
k 2Å
0
0
0
1Å
ÅÅÅ
i
j
4
j
j
j
j
j
Â
j
j
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
j
j
2 2
j
j
j
j
1Å
j
j - ÅÅÅ
4
k
1Å
i
ÅÅÅ
j
4
j
j
j
j
j
Â
j
j
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
j
j
2 2
j
j
j
j
1Å
j - ÅÅÅ
4
k
1
ÅÅÅ
Å y
2 z
z
z
z
z
0 z
z
z
z
z
1
ÅÅÅÅ z
2 {
Â
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
2
1Å
ÅÅÅ
2
Â
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
2
y
z
z
z
z
z
z
Â
z
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å z
z
2 2 z
z
z
z
z
1
z
z
ÅÅÅÅ
4
{
1Å
- ÅÅÅ
4
1Å y
- ÅÅÅ
4 z
z
z
z
z
z
Â
z
ÅÅÅÅÅÅÅÅ
è!!!Å!Å z
z
2 2 z
z
z
z
z
1
ÅÅÅÅ z
4
{
Â
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
2
1Å
ÅÅÅ
2
Â
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
2
H* Here we square the matrices. By comparison we see that Ij 2 =Ij *L
MatrixForm@MatrixPower@I1, 2DD
MatrixForm@MatrixPower@I2, 2DD
MatrixForm@MatrixPower@I3, 2DD
1Å
ÅÅÅ
i
j
2
j
j
j
j
j
0
j
j
j
j
1
j ÅÅÅ
k 2Å
0
0
0
1Å
i ÅÅÅ
j
4
j
j
j
j
j
Â
j
j
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
j
j
2 2
j
j
j
j
1Å
j - ÅÅÅ
4
k
1Å
ÅÅÅ
i
j
4
j
j
j
j
j
Â
j
j
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
j
j
j
j 2 2
j
j
1Å
j
j - ÅÅÅ
4
k
1Å
ÅÅÅ
y
2 z
z
z
z
z
0 z
z
z
z
1Å z
z
ÅÅÅ
2 {
Â
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
1Å
- ÅÅÅ
4
2
1Å
ÅÅÅ
2
-
Â
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
1Å
ÅÅÅ
4
2
Â
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
2
1Å
ÅÅÅ
2
Â
- ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2
Â
ÅÅÅÅÅÅÅÅ
è!!!Å!Å
2 2
2
y
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
{
1Å y
- ÅÅÅ
4 z
z
z
z
z
z
Â
z
ÅÅÅÅÅÅÅÅ
è!!!Å!Å z
z
2 2 z
z
z
z
z
1Å
z
z
ÅÅÅ
4
{
H* Lastly, we sum the projectors to verify the closure relation *L
MatrixForm@I1 + I2 + I3D
1 0 0y
i
j
z
j
z
j
j
0 1 0z
z
j
z
j
z
0
0
1
k
{
2.6 Let
σx =
By definition we have
eiασx =
0 1
1 0
(48)
∞
X
(iα)m m
σx
m!
m=0
For the m = 0 term, we have σx0 = I, and for the
0
2
σx =
1
1
=
0
= I
(49)
m = 1 term, σx1 = σx . For the m = 2 term, we find
1
0 1
(50)
0
1 0
0
(51)
1
(52)
From this it follows that σx3 = σx and σx4 = I, and so on. So we see that all odd powers give σx and
all even powers give I. Thus we can write
iασx
e
∞
∞
X
X
(iα)2m
(iα)2m+1
= I
+ σx
(2m)!
(2m + 1)!
m=0
m=0
∞
∞
X
X
(−1)m (α)2m+1
(−1)m (α)2m
+ iσx
= I
(2m)!
(2m + 1)!
(53)
(54)
m=0
m=0
= I cos(α) + iσx sin(α)
where the last step is possible because we recognize the series expansions for sin and cos.
7
(55)
2.7 We can start by computing σy2 and seeing what
0
2
σy =
i
1
=
0
= I
happens
−i
0 −i
0
i 0
0
1
(56)
(57)
(58)
Since the previous derivation followed strictly from σx2 = I, then the same result must be valid for
σy .
eiασy = I cos(α) + iσy sin(α)
(59)
For σu = λσx + µσy , here we start by computing σu :
0
λ − iµ
σu =
λ + iµ
0
(60)
Computing the square gives
σu2 =
0
λ − iµ
λ + iµ
0
λ2 + µ 2
0
=
0
λ2 + µ 2
1 0
=
0 1
= I
0
λ − iµ
λ + iµ
0
(61)
(62)
(63)
(64)
Thus it follows immediately that
eiασu = I cos(α) + iσu sin(α)
iσx 2
e
−0.42 0.91i
0.91i −0.42
0.54 0.84i
= I cos(1) + iσx sin(1) =
0.84i 0.54
e2iσx = I cos(2) + iσx sin(2) =
eiσx
(65)
0.54 0.84i
0.84i 0.54
= =
−0.42 0.91i
=
0.91i −0.42
0.54 0.84i
0.84i 0.54
= e2iσx
0
i+1
ei(σx +σy ) = exp
i−1
0
0.16
0.70 + 0.70i
=
−0.70 + 0.70i
0.16
8
(66)
(67)
(68)
(69)
(70)
(71)
(72)
iσy
e
iσx iσy
e
e
=
0.54 0.84
−0.84 0.54
0.54 0.84i
0.84i 0.54
0.54 0.84
=
−0.84 0.54
0.29 − 0.71i 0.45 + 0.45i
=
−0.45 + 0.45i 0.29 + 0.71i
6= ei(σx +σy )
9
(73)
(74)
(75)
(76)
4. Cohen-Tannoudji ;pp341-350: problem 3.14
a. From inspection, the eigenvalues of H are ~ω0 and 2~ω0 , with the latter being doubly degenerate. The eigenstates are |u1 i, |u2 i, and |u3 i, with the first being the non-degenerate state. Thus
a measurement of the energy would yield ~ω0 with probability 1/2 and 2~ω0 with probability 1/2.
1
1
3
hHi = ~ω0 1 + 2
= ~ω0
2
2
2
1
5
1
= ~2 ω 2
hH 2 i = ~2 ω02 1 + 4
2
2
2
r
p
5 9
1
∆H = hH 2 i − hHi2 = ~ω0
− = ~ω0
2 4
2
b. The eigenvalues of A are a and −a, with eigenstates |a, 1i = |u1 i, |a, 2i =
|−ai =
c.
√1 (||u2 i
2
− |u3 i)
√1 (|u2 i
2
+ |u3 i), and
Clearly, the initial state is a superposition of |a, 1i and |a, 2i, so the probability to obtain a
is 1 and to obtain −a is 0. Thus after the measurement, the state will remain unchanged.
e−iω0 t
e−i2ω0 t
e−i2ω0 t
|ψ(t)i = √ |u1 i +
|u2 i +
|u3 i
2
2
2
d. Because the initial state is a superposition of degenerate eigenstates of A, we know that
hAi(t) = a
e−i2ω0 t
e−i2ω0 t
e−iω0 t
√ |u1 i +
|u2 i +
|u3 i
B|ψ(t)i = b (|u1 ihu2 | + |u2 ihu1 | + |u3 ihu3 |)
2
2
2
−i2ω0 t
e
e−iω0 t
e−i2ω0 t
= b
|u1 i + √ |u2 i +
|u3 i
2
2
2
so that
hψ(t)|B|ψ(t)i = b
e−iω0 t eiω0 t 1
√ + √ +
2 2
2 2 4
=b
1
1
+ √ cos(ω0 t)
4
2
e. From the previous problem, we can see that a measurement of A at time t will yield a. The
eigenvalues of B are b and −b, with eigenstates |b, 1i = |u3 i, |b, 2i = √12 (|u1 i + |u2 i), and
|−bi =
√1 (|u1 i −
2
|u2 i).
−i2ω t
e−iω0 t
+ e 2√20
2
√
2
3
8 + 4 cos(ω0 t)
Projecting |ψ(t)i onto |−bi gives h−b|ψ(t)i =
Taking the square modulus gives p(−b) =
By probability conservation, we have then p(b) = 1 − p(−b) =
10
5
8
−
√
2
4
cos(ω0 t)