763312A QUANTUM MECHANICS I - Solution set 3 - Autumn 2016
1. Show that
a) for the normalizable solutions of the time-dependent Schrödinger
equation, the constant E has to be real.
b) if V (x) is even (or symmetric) i.e. V (−x) = V (x) then ψ(x) can be
taken to be either even or odd. (Hint: If ψ(x) is a solution of the timeindependent Schrödinger equation for a given E, then also is ψ(−x),
and hence also even and odd linear combinations ψ(x) ± ψ(−x). This
does not mean that every solution to time-independent Schrödinger
equation is even or odd; what it says is that if you’ve got one that is
not, it can always be expressed as a linear combination of solutions
(with the same energy) that are. So you might as well stick to ψ’s
that are even or odd.)
Solution
a) Let ψ(x, t) be a normalizable solution to the time-dependent S.E. Let us
consider the case when ψ is a stationary state i.e.
ψ(x, t) = φ(x)e−iEt/~ ,
(1.1)
where φ(x) is a normalizable solution to the time-independent S.E. and
E is the corresponding energy. Now suppose E is complex, E = E0 + i.
Then we would have
Z ∞
Z ∞
Z ∞
|ψ(x, t)|2 dx =
|φ(x)|2 e2t~ dx = e2t~
|φ(x)|2 dx.
(1.2)
−∞
−∞
−∞
Where the last equality holds since the exponential term doesn’t
R ∞ depend on
position x. Because ψ(x, t) was normalizable, the integral −∞ |ψ(x, t)|2 dx
R∞
is time-independent, and also −∞ |φ(x)|2 dx doesn’t depend on time. Thus
we must have that e2t~ is also time-independent, which is possible only
if = 0, i.e. if E = E0 is real.
A general solution, Ψ(x, t), to the time-dependent S.E. can be expressed
as a linear combination of stationary states and, thus, the statement holds
in general.
b) Let ψ(x) be an arbitrary solution to the time-independent S.E. with an
even potential function, V (x) = V (−x) ∀x, with energy E
−
~2 d2 ψ
(x) + V (x)ψ(x) = Eψ(x).
2m dx2
1
(1.3)
Let us make a change of variables from x to −x to the S.E., we get
−
~2 d2 ψ
(−x) + V (−x)ψ(−x) = Eψ(−x).
2m d(−x)2
(1.4)
Now d2 /d(−x)2 = d2 /dx2 and we assumed an even V (x), so we obtain
−
~2 d2 ψ
(−x) + V (x)ψ(−x) = Eψ(−x).
2m dx2
(1.5)
That is, also ψ(−x) is a solution to the S.E. with energy E. Since S.E. is
a linear differential equation, also linear combinations of ψ(x) and ψ(−x)
are solutions. Especially
ψ± (x) = ψ(x) ± ψ(−x)
(1.6)
are solutions. Now we can write
ψ(x)
=
=
=
1
1
(ψ(x) + ψ(x)) + (ψ(−x) − ψ(−x))
2
2
1
1
(ψ(x) + ψ(−x)) + (ψ(x) − ψ(−x))
2
2
1
(ψ+ (x) + ψ− (x)) .
2
(1.7)
(1.8)
(1.9)
Thus, every solution to the S.E. with an even potential can be written as
a sum of even and odd solutions. Therefore we can stick to just even and
odd solutions.
2
2. Let us consider a particle in an infinite potential well. Calculate the
expectation value of its
a) kinetic energy
b) potential energy.
Solution
a) In the lecture notes, we solved the time-independent Schrödinger equation
for a particle in an infinite potential well. We found out that the energy
eigenfunctions are
( q
2
nπ
a sin a x 0 < x < a,
un (x) =
(2.1)
0
otherwise,
where n ∈ {1, 2, 3, · · · }. Let us now calculate the expectation value of
kinetic energy for the nth energy eigenfunction. We obtain
Z ∞
hun |T |un i =
u∗n (x)T un (x) dx
(2.2)
−∞
Z a
=
u∗n (x)T un (x) dx,
0
where the last equality follows from the fact that un (x) vanishes if x ∈
/
(0, a). On the other hand, if x ∈ (0, a),
!
r
nπ
2
~2 ∂ 2
sin
x
T un (x) = −
2m ∂x2
a
a
r
~2 n2 π 2 2
nπ
=
sin
x
2m a2
a
a
~2 n2 π 2
=
un (x).
(2.3)
2m a2
Thus
hun |T |un i =
~2 n 2 π 2
2m a2
Z
a
u∗n (x)un (x) dx.
(2.4)
0
We recall that we have normalized the functions un (x) (cf. lecture notes).
Thus the integral at the right-hand side is equal to unity. It immediately
follows that
n2 π 2 ~2
.
(2.5)
hun |T |un i =
2ma2
3
b) The expectation value of potential energy is
Z ∞
hun |V |un i =
u∗n (x)V (x)un (x) dx
−∞
0
Z
u∗n (x)V (x)un (x) dx +
=
−∞
Z ∞
+
Z
a
u∗n (x)V (x)un (x) dx
0
u∗n (x)V
(x)un (x) dx.
(2.6)
0
The second integral vanishes because V (x) = 0 when x ∈ (0, a). However,
both the first and the last integral are undefined because the integrands
are undefined (the integrands are of the form 0 · ∞ · 0). However, when we
consider the infinite potential well as a limiting case of the finite potential
well, we obtain the result that there is no contribution to the expectation
value of potential energy from the outside of the well. Therefore we set
the first and the last integral equal to zero. We obtain the result that the
expectation value of potential energy vanishes. That is,
hun |V |un i = 0.
1
4
(2.7)
3. Let us consider the eigenvalue equation
d2
,
dx2
under the boundary condition U (−a) = u(a) = 0 (a is a positive real
number). Determine those eigenfunctions u that correspond to a positive
eigenvalue λ.
 = −
Âu = λu,
Solution
Our eigenvalue equation
d2
u = λu
(3.1)
dx2
is a second order homogeneous equation with constant coefficients. The
associated auxiliary equation reads
−
r2 = −λ.
(3.2)
The roots of the auxiliary equation are
√
r1 = i λ,
√
r2 = −i λ.
(3.3)
The general solution
u(x)
= C1 er1 x + C2 er2 x
√
√
= C1 ei λx + C2 e−i λx
√ √ √
√
= C1 cos λx + i sin λx + C2 cos λx − i sin λx
√
√
= A cos λx + B sin λx.
(3.4)
Let us next consider the solutions we obtain for certain values of A and
B.
A = 0 and B = 0
We obtain the zero function, which is not considered to be an eigenfunction.
A ∈ C arbitrary and B = 0
For the resulting eigenfunctions u(x) = A cos
tions read
√
cos λa = 0.
√
λx the boundary condi(3.5)
This implies that
√
λa =
π
+ nπ
2
⇐⇒
5
λ=
π2
2
(1 + 2n) .
4a2
(3.6)
A = 0 and B ∈ C arbitrary
For the resulting eigenfunctions u(x) = B sin
tions read
√
sin λa = 0.
√
λx the boundary condi(3.7)
This implies that
√
⇐⇒
λa = nπ
λ=
π2 2
n .
a2
(3.8)
A 6= 0 and B 6= 0
It is only a minor task to show that the eigenvalues must now be of the
form (3.6) and of the form (3.8). Therefore we have to find out whether
there are n1 , n2 ∈ {1, 2, 3, · · · } satisfying
π2
π2 2
2
(1
+
2n
)
=
n
1
4a2
a2 2
⇐⇒
1 + 2n1 = 2n2 .
(3.9)
We see that the left-hand side is odd for all n1 , whereas the right-hand
side is even for all n2 . Therefore there are no eigenvalues that are of the
form (3.6) and of the form (3.8).
Summary
The eigenvalues λn and the corresponding eigenfunctions un :
λn
λn
=
=
π2
4a22 (1 +
π
2
a2 n :
2
2n) :
un (x)
un (x)
6
√
= A cos λn x,
√
= B sin λn x.
4. Consider a particle in an infinite potential well with the initial condition
Ψ(x, 0) = A ψ1 (x) + eiφ ψ2 (x) ,
(4.1)
where φ is some constant. Find Ψ(x, t), |Ψ(x, t)|2 and hxi. Study the
special cases φ = 0, π/2, π. Comment on your results.
Solution
The states ψ1 and ψ2 are the ground and first excited state of the infinite
potential well, they are given in the lecture notes to be
r
πx 2
(4.2)
sin
ψ1 (x) =
a
a
r
2
2πx
ψ2 (x) =
sin
,
(4.3)
a
a
where a is the width of the potential well. Ψ(x, t) can be found in the
same manner as in Ex. 2.4, we obtain
r πx −iE1 t/~
2
2πx −iE2 t/~
sin
e
, (4.4)
e
+ eiφ sin
Ψ(x, t) = A
a
a
a
where En =
~2 π 2 n2
2ma2
and n ∈ Z+ .
Let us normalize Ψ(x, t), this can be done by normalizing Ψ(x, 0), since
the normalization condition does not depend on time. We obtain
Z a
1 =
|Ψ(x, 0)|2 dx
(4.5)
0
Z a
= |A|2
(|ψ1 (x)|2 + |ψ2 (x)|2 + eiφ ψ1∗ (x)ψ2 (x)
(4.6)
0
+e−iφ ψ1 (x)ψ2∗ (x))dx
Z a
Z a
= |A|2
|ψ1 (x)|2 dx +
|ψ2 (x)|2 dx
0
0
Z a
Z a
+eiφ
ψ1∗ (x)ψ2 (x)dx + e−iφ
ψ1 (x)ψ2∗ (x)dx .
0
(4.7)
(4.8)
(4.9)
0
Eigenfunctions ψn (x) are orthonormal, thus, the first two integrals are
equal to unity and the last two go to zero. That is
1 = 2|A|2
⇔
1
A= √ .
2
(4.10)
We chose A to be real and positive, this is an arbitrary choice and has
no physical meaning. Any other phase factor is just as good. Thus,
normalized Ψ(x, t) reads
r 1
πx −iE1 t/~
2πx −iE2 t/~
iφ
Ψ(x, t) =
sin
e
+ e sin
e
. (4.11)
a
a
a
7
Next we want to calculate the probability density.
|Ψ(x, t)|2
Ψ∗ (x, t)Ψ(x, t)
(4.12)
πx πx 1
2πx
=
+ sin
sin
sin2
e−i(E2 −E1 )t~+iφ
a
a
a
a
πx 2πx i(E2 −E1 )t~−iφ
2πx
2
sin
+ sin
e
+ sin
a
a
a
=
Let us define ω =
we obtain
|Ψ(x, t)|
~π 2
2ma2 .
With this we can write (E2 − E1 )/~ = 3ω and
2πx
1
2
2 πx
+ sin
sin
=
a
a
a
πx 2πx
+2 sin
sin
cos (3ωt − φ) ,
a
a
2
(4.13)
(4.14)
where we have used Euler’s formula to write the complex exponentials as
a cosine function.
Next in line is the expectation value of x.
Z a
Z a
hxi =
Ψ∗ (x, t)xΨ(x, t)dx =
x|Ψ(x, t)|2 dx
0
0
Z
1 a
2πx
2 πx
2
=
x sin
+ sin
a 0
a
a
πx 2πx
cos (3ωt − φ) dx
+2 sin
sin
a
a
(4.15)
(4.16)
(4.17)
Let us go through the integrals one by one.
a
2
Z a
a2 cos 2πx
ax sin 2πx
x
2 πx
a
a
dx =
−
−
x sin
2
a
4
8π
4π
0
0
a2
4
a
2
a2 cos 4πx
ax sin 4πx
x
a
a
=
−
−
2
4
32π
8π
0
=
Z
a
x sin2
0
2πx
a
dx
=
Z
2
a
x sin
0
πx a
sin
2πx
a
=
=
a2
4Z
(4.18)
(4.19)
a
πx πx 4
x cos
− cos3
dx
a
a
0
a
πx a
3πx
3 πx
12πx sin
+ 9a cos
− a cos
9π 2
a
a
a
0
= −
8
16a2
9π 2
(4.20)
All of the integrals need to be solved by integration by parts (derivate x
and integrate the trigonometric function). In the last integral one can use
formulae
sin 2α
=
2 sin α cos α,
(4.21)
sin2 α
=
1 − cos2 α
(4.22)
to get rid of the product of trigonometric functions. Substituting the
results to the expression for hxi, we obtain
1 a2
a2
16a2
hxi =
cos (3ωt − φ)
(4.23)
+
−
a 4
4
9π 2
a
32
=
1 − 2 cos (3ωt − φ) .
(4.24)
2
9π
Case φ = 0:
r πx −iE1 t/~
1
2πx −iE2 t/~
sin
e
(4.25)
Ψ(x, t) =
e
+ sin
a
a
a
πx 1
2πx
|Ψ(x, t)|2 =
sin2
(4.26)
+ sin2
a
a
a
πx 2πx
+2 sin
sin
cos (3ωt)
(4.27)
a
a
32
a
1 − 2 cos (3ωt)
(4.28)
hxi =
2
9π
32
At time t = 0 hxi = a2 1 − 9π
2 .
Case φ =
π
2:
r πx −iE1 t/~
1
2πx −iE2 t/~
Ψ(x, t) =
sin
e
(4.29)
e
+ i sin
a
a
a
πx 1
2πx
|Ψ(x, t)|2 =
sin2
+ sin2
(4.30)
a
a
a
πx 2πx
+2 sin
sin
sin (3ωt)
(4.31)
a
a
a
32
hxi =
1 − 2 sin (3ωt)
(4.32)
2
9π
At time t = 0 hxi = a2 .
9
Case φ = π:
r 1
πx −iE1 t/~
2πx −iE2 t/~
Ψ(x, t) =
e
sin
− sin
e
(4.33)
a
a
a
πx 2πx
1
+ sin2
|Ψ(x, t)|2 =
sin2
(4.34)
a
a
a
πx 2πx
sin
−2 sin
cos (3ωt)
(4.35)
a
a
a
32
hxi =
(4.36)
1 + 2 cos (3ωt)
2
9π
32
At time t = 0 hxi = a2 1 + 9π
2 .
Physically the phase difference φ between the states ψ1 and ψ2 amounts
to shifting the t = 0 point.
10
5. Show that E must exceed the minimum value of V (x) for every normalizable solution to the time-independent Schrödinger equation.
Solution
Start with the time-independent S.E.
−
~2 d2 ψ
(x) + V (x)ψ(x) = Eψ(x),
2m dx2
(5.1)
where ψ(x) is a normalizable solution. Reorganizing the terms gives
d2 ψ
2m
(x) = 2 [V (x) − E] ψ(x).
2
dx
~
00
(5.2)
Now, if E < Vmin , ψ and ψ have always the same sign. This means
that ψ always curves away from the axis. However ψ has to go to zero as
x → −∞, since it is normalizable. At some point ψ has to differ from zero
(If not, then it is zero everywhere, and hence not normalizable.). Let’s
assume that it happens to the positive direction. Now the slope of ψ is
positive and increasing, so it gets bigger and bigger as x increases. It can’t
ever start heading back to zero, because then the second derivative would
be negative. If ψ starts going negative, by the same arguments as above,
it just gets more and more negative. In both cases it is impossible for ψ
to return to zero as x → ∞, and hence it is not normalizable. Thus, we
conclude E > Vmin for normalizable solutions to the S.E..
11