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Problems from Quantum Mechanics - Calculating Probabilities
Ahmed Sayeed
Department of Physics, University of Pune,
email: [email protected]
1. The energy eigenfunctions of a particle ψ1 (x) and ψ2 (x) are normalized and correspond to energies E1 6= E2 . ψ1 (x) = 0 outside the region R1 , and ψ2 (x) = 0 outside the region R2 .
(a)Supppose regions R1 and R2 do not overlap. Show that if the particle is in the region R1
√
it will stay there forever, and if the initial state is Ψ(x, 0) = [ψ1 (x) + ψ2 (x)]/ 2, show that
the probability density |Ψ(x, t)|2 is independent of time. (b) Show that if regions R1 and R2
overlap, the probability density |Ψ(x, t)|2 oscillates in time for the initial state given in (a).
Solution
(a) If the two regions are not overlapping and if the particle is in region R1 , that can only mean it
is in eigenstate ψ1 (x), because ψ2 (x) = 0 in this region. Therefore the time-dependent wavefunction is Ψ(x, t) = ψ1 (x)e−iE1 t/~ and the probability density distribution is |Ψ(x, t)|2 = |ψ1 (x)|2 ,
which vanishes outside R1 at all times.
For the given initial wavefuntion the time-dependent wavefunction is
i
1 h
Ψ(x, t) = √ ψ1 e−iE1 t/~ + ψ2 e−iE2 t/~
2
And the probability density distribution is
(1)
|Ψ(x, t)|2 = Ψ∗ (x, t)Ψ(x, t)
1
=
|ψ1 |2 + |ψ2 |2 + f (t)ψ1∗ ψ2 + g(t)ψ1 ψ2∗
2
Where f (t) = ei(E1 −E2 )t/~ and g(t) = e−i(E1 −E2 )t/~ . Now ψ1 (x) and ψ2 (x) do not overlap.
Which means when ψ1 (x) 6= 0 we have ψ2 (x) = 0 and vice versa. Therefore the products
ψ1∗ ψ2 = ψ1 ψ2∗ = 0 at all times. Thus we have
|Ψ(x, t)|2 =
1
|ψ1 |2 + |ψ2 |2
2
which is time independent.
(b) The bound energy eigenstates can always be chosen to be real. Therefore we can assume
ψ1 (x) and ψ2 (x) are real functions. And so we can set ψ1∗ ψ2 = ψ2∗ ψ1 = ψ1 ψ2 in eq. (1) and get
1
|ψ1 |2 + |ψ2 |2 + [f (t) + g(t)] ψ1 ψ2
2
1
(E1 − E2 )t
2
2
=
|ψ1 | + |ψ2 | + 2 cos
ψ1 ψ2
2
~
|Ψ(x, t)|2 =
which varies sinusoidally in time with angular frequency ω = (E1 − E2 )/~.
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2. In the ground state of a harmonic oscillator, what is the probability of finding the particle outside
the classically allowed region?
Solution
For a classical harmonic oscillator, as it moves awar from the equilibrium position (which we
take as x = 0) the potential energy V (x) = 12 mω 2 x2 (m is the mass and ω the angular frequency)
goes on increasing, till it becomes equal to total energy E, at position x = xM or x = −xM .
These two points are known as classical turning points, because the particle vecolcity changes
its direction at these points. Here x = xM is the amplitude, of course. Therefore we have
1
2 2
2 mω xM
= E ⇒ xM = (2E/mω 2 )1/2 . For the quantum harmonic oscillator in the
p
ground state (n = 0) E = E0 = ~ω/2 , which gives xM = ~/mω. Thus, the classically forbidVM (x) =
den region for the harmonic oscillator is x > xM and x < −xM , or in other words |x| > xM .
The probability that the particle is found outside the classically allowed region is the probability
that the particle is found within the forbidden allowed region. Which is
∞
Z
ψ02 dx
P (|x| > xM ) =
Z
xM
Z
−∞
+
−xM
ψ02 dx
∞
ψ02 dx
=2
(because ψ02 is even)
xM
2
The ground state wavefunction for the harmonic oscillator is ψ0 (x) = αe−ξ /2 , where α =
p
p
√
(mω/π~)1/4 , ξ = mω/~x. In the following we also use c = mω 2 /~ = πα2 . With this we
have
P (|x| > xM ) = 2α
=
2
2α2
Z
∞
xM
∞
Z
2
e−ξ dx
2
e−ξ dξ
(because ξ = cxM =
c ξ=1
Z ∞
2
2
e−ξ dξ
=√
π ξ=1
√
2
π
=√ ·
erfc(1) = erfc(1) ≈ 0.1573
π 2
p
p
mω/~ · ~/mω = 1)
(erfc(x) = 1 − erf(x))
3. A charged harmonic oscillator is placed in a constant electric field (That is, in addition to harmonic potential it also experiences a potential due to electric field). Find the energy eigenvalues
and eigenfunctions.
Solution
Hamiltionian for the charged harmonic occillator is
H=
1 2 mω 2 x2
p +
− qEx
2m
2
where q is the charge and E is the electric field. We use the change of variable y = x − qE/mω 2
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to get
1 2 mω 2 y 2
q2E 2
p +
−
2m
2
2mω 2
= H0 − C
H=
where H0 = p2 /2m + mω 2 y 2 /2, and C = q 2 E 2 /2mω 2 is a constant term. We can see that H0
has the form of a simple harmonic oscillator Hamiltonian with a potential function V (y). Now
we have
H0 = H + C
H0 ψn = (H + C)ψn
= En ψn
(En and ψn are eigenvalues and the eigenfunctions of H0 )
⇒ Hψn = En ψn − Cψn
= (En − C)ψn
Thus, ψn are also eigenfunctions of H with corresponding eigenvalues (En − C). That is, the
eigenvalues (allowed energy values) of the charged harmonic oscillator are
En −
q2E 2
1
q2E 2
= (n + )~ω −
2
2mω
2
2mω 2
and eignefunctions are
qE
ψn (y) = ψn x −
mω 2
4. The initial state of a harmonic oscillator is a linear combination of the ground state and the first
excited state: ψ(x) = 3ψ0 (x) + 4ψ1 (x). (a) Normalize this state. (b) Write the time-dependent
wavefunction. (c) Find the time-dependent probability density distribution. (d) Find the expectation value of position as a function of time. Notice that it oscillates sinusoidally. Find
the amplitude and angular frequency of the oscillation. (e) Using the result of (d) to find
hpi (expectation value of momentum) and check that Ehrenfest theorum holds. (f ) What are
the probabilities that an energy measurement on state Ψ(x, t) of (b) gives energy values ~ω/2,
3~ω/2, 5~ω/2, 2~ω? Note: Here we are talking about energy values obtained in individual measurements, and not any kind of average of a number of measurements. (g) Sketch |Ψ(t)|2 at
t = 0, π/ω, 2π/ω, 3π/ω, 4π/ω.
Solution
Let A be the normalization constant, so that ψ(x) = A[3ψ0 (x) + 4ψ1 (x)] is normalized. In the
following we use harmonic oscillator energies E0 = ~ω/2 and E1 = 3~ω/2, and wavefunctions
√
2
2
ψ0 (x) = αe−ξ /2 and ψ1 (x) = 2αξe−ξ /2 . α, c and ξ are defined in the solution to problem 2.
All the integral limits are from −∞ to +∞.
(a)
Z
2
|Ψ(x, 0)| dx = A
2
Z
(9|ψ0 |2 + 16|ψ1 |2 + 12ψ0∗ ψ1 + 12ψ1∗ ψ0 )dx
= A2 (9 + 16 + 0 + 0) = 25A2 = 1 ⇒ A =
1
5
(Using orthonomality of ψ0 and ψ1 )
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(b)
Ψ(x, t) =
i 1h
i
1h
3ψ0 (x)e−iE0 t/~ + 4ψ1 (x)e−iE1 t/~ =
3ψ0 (x)e−iωt/2 + 4ψ1 (x)e−3iωt/2
5
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(c)
1 2
9ψ0 + 16ψ12 + 12ψ0 ψ1 e−iωt + 12ψ1 ψ0 eiωt
25
1 =
9ψ0 (x)2 + 16ψ1 (x)2 + 24ψ0 ψ1 cos(ωt)
25
|Ψ(x, t)|2 =
(d)
Z
x|Ψ(x, t)|2
Z
Z
Z
1
2
2
9 xψ0 dx + 16 xψ1 dx + 24 xψ0 ψ1 cos(ωt)dx
(from (c))
=
25
Z
1
=
24 xψ0 ψ1 cos(ωt)dx
(first two integrals vanish as integrands are odd)
25
Z
24
=
cos(ωt)
xψ0 ψ1 dx
25
√ Z
2
24
2
cos(ωt) ·
ξ 2 e−ξ
=
2
25
πα
√
( substituting wavefunctions ψ0 and ψ1 , and using x = cξ, c = πα2 )
r
r
24
~
~
24
=
cos(ωt)
=
cos(ωt)
25
2mω
25 2mω
q
~
Thus the amplitude is 24
25
2mω and angular frequency frequency is ω. This frequency is same as
hxi =
the classical angular frequency, but this is just an accident. If we choose a linear combination of
eignefunctions ψ1 and ψ2 we will get an angular frequency 2ω, as can be seen from the inspection
of result in (c).
(e) In the following V is the potential energy. We have to check if the Ehrenfest relation
∂V dhpi
− ∂x = dt holds in this case.
r
d
24 mω~
hpi = hxi = −
sin(ωt)
dt
25
2
r
dhpi
24 mω~
=−
ω cos(ωt)
⇒
dt
25
2
mω 2 x2
∂V
We have V =
⇒
= mω 2 x
2
∂x
∂V
Therefore −
= −mω 2 hxi
∂x
r
24
~
= −mω 2 ·
cos(ωt)
25 2mω
r
24 mω~
dhpi
=−
ω cos(ωt) =
25
2
dt
from (d)
from (d)
(f ) The energy value ~ω/2 is the eigenvalue for eigenstate ψ0 . Therefore, from (b) the probability
of getting this value is (3/5)2 = 9/25. Similarly the probability of getting the energy value 3~ω/2
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is (4/5)2 = 16/25. The energy value 5~ω/2 is the eigenvalue of eigenstate ψ2 , which does not
appear in the linear combination we are considering. Therefore, the probability of getting this
value is zero. The energy value 2~ω is not an allowed energy value for a quantum harmonic
oscillator (no eigenstate corresponding to this energy), and so the probability of getting this
value is zero.
(g) This is shown in Figure 1. We see that the probability density distribution oscillates with
time. We can compare it to oscillation of probability density of a classical harmonic ocillator,
which would be simply a delta function oscillating between the two turning points mentioned in
solution to problem 2.
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t = 0.0АΩ
Ρ
40
30
20
10
-2
0
-1
1
2
1
2
1
2
1
2
1
2
x
t = 0.2АΩ
Ρ
40
30
20
10
-2
0
-1
x
t = 0.5АΩ
Ρ
40
30
20
10
-2
0
-1
x
t = 0.7АΩ
Ρ
40
30
20
10
-2
0
-1
x
t = 1.0АΩ
Ρ
40
30
20
10
-2
-1
0
x
Figure 1: Oscillation of probability density ρ(x) = |Ψ(x, t)|2 with time (Problem 4(g)). Only half the
cycle is shown. You get the sequence of the other half if you move from bottom to top.
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