Physics 70007, Fall 2009
Solutions to HW #4
November 2009
1. (Sakurai 2.11)
Consider a particle subject to a one-dimensional simple harmonic oscillator potential. Suppose at
t = 0 the state vector is given by
exp
−ipa
~
|0i
where p is the momentum operator and a is some number with dimension of length. Using the
Heisenberg picture, evaluate the expectation value hxi for t ≥ 0.
Recall that in the Heisenberg picture, the state kets/bras stay xed, while the operators evolve
in time. At time t = 0, Heisenberg-picture operators equal their Schrodinger-picture counterparts
(this species their initial value), so we will write e.g. x(0) ≡ x̂. Then
ip̂a
−ip̂a hx (t)i = 0exp
x (t) exp
0
~
~
We can express x (t) in terms of Schrodinger operators by solving the Heisenberg equations of
motion for x (t) and p (t). These are coupled rst-order linear ordinary dierential equations, and
the method of solution is straightforward. Sakurai carries this through in section 2.3; here we will
only quote the nal result (Sakurai 2.3.45a):
sin ωt
x (t) = (cos ωt) x (0) +
p (0)
mω
sin ωt
= (cos ωt) x̂ +
p̂
mω
Using this,
hx (t)i =
=
ip̂a
sin ωt
−ip̂a 0exp
(cos ωt) x̂ +
p̂ exp
0
~
mω
~
ip̂a
−ip̂a sin ωt
(cos ωt) 0exp
x̂ exp
h0|p̂|0i
0 +
~
~
mω
In the rst matrix element, we need to commute x̂ with one of the exponentials. We learned how
to do this in a previous homework (Sakurai problem 1.29a): we found
i.e.
−ip̂a
x̂, exp
=
~
−ip̂a
−ip̂a
x̂ exp
− exp
x̂ =
~
~
∂
−ip̂a
−ip̂a
i~ exp
= a exp
∂ p̂
~
~
−ip̂a
a exp
~
so
ip̂a
−ip̂a 0exp
x̂ exp
=
0
~
~
=
=
ip̂a
−ip̂a
−ip̂a
0exp
a exp
+ exp
x̂ 0
~
~
~
a h0|0i + h0|x̂|0i
a + h0|x̂|0i
1
and thus
x (t) = (cos ωt) (a + h0|x̂|0i) +
sin ωt
mω
h0|p̂|0i
Later on in this homework, we will see that the needed matrix elements are both zero: x̂ and p̂ are
non-diagonal in the SHO energy eigenbasis (#3 in this homework: Sakurai problem 2.13). Here we
take this result as given; thus
hx (t)i = a cos ωt.
2. (Sakurai 2.12)
(a) Write down the wave function (in coordinate space) for the state specied in Problem 11 at
t = 0. You may use
"
0
hx |0i =
1
−1/2
π − /4 x0
1
exp −
2
x0
x0
2 #
x0 ≡
,
The state in question from the previous problem is |ψi ≡ exp
space wave function for this state is
ψ (x) =
−ip̂a
~
~
mω
1/2 !
|0i; so the desired coordinate-
−ip̂a xexp
0
~
In order to use the wave function given to us in the problem statement, we insert the identity
strategically into our matrix element:
ˆ
ψ (x) =
−ip̂a 0
0
dx xexp
x hx |0i
~
0
The quickest way to proceed at this point is to remember, from a previous homework (Sakurai problem 1.28c), that the exponential is the spatial translation operator and so operates upon position
eigenkets in a very simple way:
−ip̂a
~
exp
|x0 i = |x0 + ai
(note the opposite signs attached to a on the left-hand and right-hand sides of this relation!) Then
we have
ˆ
ψ (x)
dx0 hx|x0 + ai hx0 |0i
=
ˆ
dx0 δ (x0 + a − x) hx0 |0i
=
ˆ
=
dx0 δ (x0 − (x − a)) hx0 |0i
= hx − a|0i
"
=
1
−1/2
π − /4 x0
1
exp −
2
x−a
x0
2 #
(b) Obtain a simple expression for the probability that the state is found in the ground state at t = 0.
Does this probability change for t > 0?
The probability that a measurement of the energy gives the ground state energy, is given by the
squared amplitude of the overlap matrix element between our state and the ground state:
2
Prob (ground state) = |h0|ψi|2
2
ˆ
= dx0 h0|x0 i hx0 |ψi
2
ˆ
∗
0
0
0
= dx hx |0i hx |ψi
ˆ
= dx0
"
2
=
The needed integral is
1
exp −
2
ˆ
2
02
0
x
ax dx0 e− x20 + x20 1
−1/2
π − /4 x0
1 − xa2
e 0
πx20
ˆ
∞
x0
x0
c2
2
4c1
2 #!
"
1
−1/2
π − /4 x0
1
exp −
2
x0 − a
x0
2 #!2
r
π
c
1
−∞
√
´∞
2
(this can be obtained from the known result −∞ dx exp −x = π by completing the square in
dx e
−c1 x2 +c2 x
=e
·
the exponent). We obtain
1
2
2
exp −a /x0 · exp
Prob (ground state) =
πx20
2
1a
= exp − 2 .
2 x0
2
(a/x20 )
4 · 1/x20
!r
π
1/x2
0
2
The probability at later times is most easily evaluated in the Schrodinger picture. Our state |ψi
evolves via the time-evolution operator:
iĤt
|ψ (t)i = exp −
~
!
|ψi
while the ground state base ket |0i does not change (the operator Ĥ does not change with time, so
its eigenkets also stay the same). So
Prob (ground state at t > 0) = |h0|ψ (t)i|2
* ! +2
iĤt = 0exp −
ψ ~ Now what we can do is apply the time-evolution operator to the left, which is easy because we know
its eect upon an energy eigenstate:
Prob (ground state at t > 0)
2
!
i 21 ~ω t
= exp −
h0|ψi
~
2
1
= exp − iωt h0|ψi
2
But this extra factor is just a phase factor (a complex number with magnitude 1). When we take
the magnitude squared, it simply becomes 1 and drops out of the calculation. Thus we get the
same probability as we did at t = 0!
3
3. (Sakurai 2.13)
Consider a one-dimensional simple harmonic oscillator.
(a) Using
√
ip
n |n − 1i
a |ni
= √
,
a† |ni
n + 1 |n + 1i ,
mω
evaluate hm|x|ni, hm|p|ni, hm|{x, p}|ni, mx2 n , and mp2 n .
a
a†
r
=
mω
2~
x±
First we solve for x and p in terms of a and a† :
r
x=
r
~
a + a† ,
2mω
p=i
~mω †
a −a
2
Then
r
~
2mω
r
~
=
2mω
r
~
=
2mω
hm|x|ni =
Similarly
ma + a† n
hm|
√
√
n |n − 1i +
n δm,n−1 +
√
√
n + 1 |n + 1i
n + 1 δm,n+1
r
√
~mω √
n + 1 δm,n+1 − n δm,n−1
2
†
i~
†
Next {x, p} = xp + px = 2 a + a a − a + a† − a a† + a = i~ a†2 − a2 , so
p
p
(n + 1) (n + 2) δm,n+2 − n (n − 1) δm,n−2
hm|{x, p}|ni = i~
hm|p|ni = i
Then x2 =
1), so
~
2mω
a + a†
2 m x n =
Finally p2 = − ~mω
2
2
=
~
2mω
a2 + aa† + a† a + a†2 =
~
2mω
a†2 + 2a† a − 1 + a2 (using a, a† =
p
~ p
(n + 1) (n + 2) δm,n+2 + (2n + 1) δm,n + n (n − 1) δm,n−2
2mω
2
a† − a = ~mω
−a2 + aa† + a† a − a†2 = ~mω
−a†2 + 2a† a − 1 − a2 , so
2
2
p
2 ~mω p
mp n =
− (n + 1) (n + 2) δm,n+2 + (2n + 1) δm,n − n (n − 1) δm,n−2
2
(b) Check that the virial theorem holds for the expectation values of the kinetic and the potential
energy taken with respect to an energy eigenstate.
The virial theorem in one dimension takes the form (see e.g. Bransden & Joachain, Intro to QM,
1st ed., p.227):
2 hT i =
x
∂V
∂x
where the expectation values are evaluated in an energy eigenstate. For our potential V = 12 mω 2 x2 ,
x
∂V
= x · mω 2 x = 2V
∂x
so the virial theorem reduces to hT i = hV i. We check that this holds:
hT i ≡ hn|T |ni =
1 2 1 ~mω
~ω
np n =
(2n + 1) =
(2n + 1)
2m
2m 2
4
4
hV i ≡ hn|V |ni =
mω 2 2 mω 2 ~
~ω
nx n =
(2n + 1) =
(2n + 1)
2
2 2mω
4
so the theorem is satised.
4. (Sakurai 2.15)
Consider a function, known as the correlation function , dened by
C (t) = hx (t) x (0)i ,
where x (t) is the position operator in the Heisenberg picture. Evaluate the correlation function
explicitly for the ground state of a one-dimensional simple harmonic oscillator.
As in the rst problem, we use Sakurai 2.3.45a for the simple harmonic oscillator:
x (t) = (cos ωt) x (0) +
sin ωt
mω
p (0)
Then writing x (0) and p (0) as x̂ and p̂, we have
C (t) = (cos ωt) 0x̂2 0 +
sin ωt
mω
h0|p̂x̂|0i
Now we use the results of the previous problem to evaluate the matrix elements. To do the second
one, we rewrite p̂x̂:
p̂x̂ =
1
1
({x̂, p̂} − [x̂, p̂]) = ({x̂, p̂} − i~)
2
2
Then
C (t)
=
=
=
~
sin ωt
1
(cos ωt) ·
(1) +
· (0 − i~)
2mω
mω
2
~
(cos ωt − i sin ωt)
2mω
~e−iωt
.
2mω
5