CHEM 301: Homework assignment #3
Solutions
1. Show that the position, momentum, and Hamiltonian operators are linear.
(10%)
Solution:
By definition, an operator is linear if it satisfies the following condition:
 [c1 f1 (x) + c2 f2 (x)] = c1 Âf1 (x) + c2 Âf2 (x) ,
for any functions f1 (x) and f2 (x) and arbitrary real or complex constants
c1 and c2 .
• The position operator x̂ = x:
x [c1 f1 (x) + c2 f2 (x)] = xc1 f1 (x) + xc2 f2 (x) =
= c1 xf1 (x) + c2 xf2 (x) .
The linearity condition is satisfied, so the operator is linear.
d
:
• The momentum operator p̂ = −ih̄ dx
d
d
d
−ih̄ dx
[c1 f1 (x) + c2 f2 (x)] = −ih̄ dx
[c1 f1 (x)] − ih̄ dx
[c2 f2 (x)] =
d
d
f2 (x) .
= c1 −ih̄ dx f1 (x) + c2 −ih̄ dx
The linearity condition is satisfied, so the operator is linear.
2
2
h̄ d
• The Hamiltonian operator Ĥ = − 2m
dx2 + V (x):
−
=
=
h̄2 d2
2m dx2 + V (x) [c1 f1 (x) + c2 f2 (x)] =
h̄2 d2
h̄2 d2
− 2m
dx2 [c1 f1 (x)] + V (x) [c1 f1 (x)] − 2m dx2
h
[c2 f2 (x)] + V (x) [c2 f2 (x)] =
i
(x) + c1 [V (x) f1 (x)] − c2
(x) + c2 [V (x) f2 (x)] =
i
h 2 2
i
h̄ d
+ V (x) f1 (x) + c2 − 2m
+
V
(x)
f2 (x)
2
dx
h̄2 d2
c1 − 2m
dx2 f1
h 2 2
h̄ d
= c1 − 2m
dx2
i
h
h̄2 d2
2m dx2 f2
The linearity condition is satisfied, so the operator is linear.
2. Check whether the following pairs of operators commute:
How to approach the problem: By definition, two operators commute
if
ÂB̂f (x) = B̂Âf (x) ,
where we first act on the function f (x) with the operator that is closer to
it and then act on the result with the operator that is further to the left.
1
•
2
d
d
and dx
2 + 2 dx (15%)
Solution:
d
dx
d
dx
|{z}
Â
d2
d
d d2 f (x)
df (x)
+2
f (x) =
+2
=
dx2
dx
dx
dx2
dx
{z
}
|
B̂
d2 f (x)
d3 f (x)
+
2
=
dx3
dx2
2
d
d
d
d df (x)
d2
+
2
f
(x)
=
+
2
=
dx2
dx |{z}
dx
dx2
dx
dx
{z
}
|
B̂
3
=
Â
2
d f (x)
d f (x)
+2
dx3
dx2
The expressions that we get in both cases are the same, so the two
operators commute.
d
• x and dx
(15%):
Solution:
d
x
f (x) =
|{z}
dx
|{z}
Â
B̂
df (x)
=x
dx


d 
x f (x) =
dx |{z}
|{z}
Â
B̂
= f (x) + x
df (x)
dx
The expressions that we get when we reverse the order of operators
are not the same, so the two operators do not commute.
• SQR(x) (take the square of x) and SQRT(x) (take the square root
of x) (15%):
Solution:
This problem is formulated slightly ambiguously and can be interpreted two-fold.
2
2
a) The more
√ straightforward interpretation is to assume that  = x
and B̂ = x. Then,
√
5
x2
x f (x) = x 2 f (x)
|{z}
|{z}
Â
B̂
√
5
x x2 f (x) = x 2 f (x)
|{z} |{z}
B̂
Â
Since applying the operators in any order gives the same result, the
operators commute.
b) It could also be supposed that the operators are just SQR (take
the square) and SQRT (take the square root), and x is an arbitrary
function (say, x (t)). Then we can write:


p
2


SQR SQRT x (t) =
x (t) = x (t) ,
| {z } | {z }
Â
B̂
because by definition the square root of x is a number the square of
which is x. Applying the operators in reverse order is less straightforward. First, we must agree that SQRT denotes the principal square
root (that is, it has a positive value if x (t) is positive, a positive
imaginary value if x (t) is negative, and the definition is somewhat
more complicated for complex numbers). For example, the principal
square root of 4 is 2, not −2, and the principal square root of −4 is 2i,
not −2i. For simplicity, we will also assume that x (t) is a real-valued
function. Then,



 p
SQRT SQR x (t) = x2 (t) = |x (t)| .
| {z } | {z }
B̂
Â
Thus, if we postulate that all functions to which the operators SQR
and SQRT can be applied only have non-negative values, then x =
|x|, and the two operators commute. If x (t) can become negative for
any values of t, the operators SQR and SQRT do not commute. For
instance, if
x (t) = −4,
then
but
√
−4
2
2
= (2i) = −4,
q
√
2
(−4) = 16 = 4.
3
Thus, determining whether two operators commute may require the
knowledge of what functions they are allowed to take as their arguments. For instance, SQR and SQRT commute if we restrict their
allowed arguments to real, non-negative valued functions, but generally do not commute.
(This is just an illustration to make you aware of one of the subtleties
that you might run into when determining if two operators commute.
No such ambiguous problems will be on the test!)
2
d
d
and dx
• x2 dx
2 (15%):
Solution:


3
2

d 
 d f (x) = x2 d f (x)
x2


2
dx} |{z}
dx
dx3
| {z
Â


d2 
x2 d f (x) = d

 dx
2
dx | {z
dx}
|{z}
B̂
B̂

d
df (x)
x2
=
dx
dx
Â
2
2
df (x)
df (x)
d
d
d
2 d f (x)
2 d f (x)
2x
+x
2x
+
x
=
=
=
dx
dx
dx2
dx
dx
dx
dx2
3
3
df (x)
d2 f (x)
d2 f (x)
df (x)
d2 f (x)
2 d f (x)
2 d f (x)
=2
+ 2x
+
2x
+
x
=
2
+
4x
+
x
dx
dx2
dx2
dx3
dx
dx2
dx3
The expressions that we get when we reverse the order of operators
are not the same, so the two operators do not commute.
3. The wavefunction ψ (x) of a paricle is written as a linear combination of
eigenfunctions ϕn (x), n = 1, 2, 3, of a linear operator  that corresponds
to observable a:
ψ (x) = (0.3 + 0.4i) ϕ1 (x) + (0.4 − 0.5i) ϕ2 (x) + (0.5 + 0.3i) ϕ3 (x) .
The eigenvalue corresponding to eigenfunction ϕn (x) is an , n = 1, 2, 3.
• Calculate the probability of obtaining the result an , n = 1, 2, 3, in a
measurement of the observable a. (10%)
Solution:
The probability of getting the result an , n = 1, 2, 3, in a measurement
of the observable a is given by
– the square of the absolute value of the coefficient cn in front of
the function ϕn
– in the expansion of the wavefunction ψ (x)
4
– into as a linear combination of eigenfunctions of the operator Â
that corresponds to observable a.
The expansion in terms of eigenfunctions of  is:
ψ (x) = (0.3 + 0.4i) ϕ1 (x) + (0.4 − 0.5i) ϕ2 (x) + (0.5 + 0.3i) ϕ3 (x) .
|
|
|
{z
}
{z
}
{z
}
c1
c2
c3
Thus, the probabilities of obtainig the results a1 , a2 , a3 in a measurement of the observable a are given by:
a1 :
∗
2
|(0.3 + 0.4i)| = (0.3 + 0.4i) (0.3 + 0.4i) = (0.3 − 0.4i) (0.3 + 0.4i) =
= 0.09 + 0.12i − 0.12i − 0.16i2 = 0.09 + 0.16 = 0.25
a2 :
∗
2
|0.4 − 0.5i| = (0.4 − 0.5i) (0.4 − 0.5i) = (0.4 + 0.5i) (0.4 − 0.5i) =
= 0.16 − 0.20i + 0.20i − 0.25i2 = 0.16 + 0.25 = 0.41
a3 :
∗
2
|0.5 + 0.3i| = (0.5 + 0.3i) (0.5 + 0.3i) = (0.5 − 0.3i) (0.5 + 0.3i) =
= 0.25 + 0.15i − 0.15i − 0.09i2 = 0.25 + 0.09 = 0.34
• Assume that we performed a measurement of a on the system and
obtained the value a2 . What is the probability of obtaining the value
a3 in a subsequent measurement of a, if no other measurements were
performed on the system? (10%)
Solution:
Remember that performing a measurement on a system changes its
wavefunction. In particular, if we obtained a value of a2 in a measurement (and did not do anything else to further change the state of
the system), then we know the value of the observable a with absolute
certainty. Thus, the wavefunction of the system must now be ϕ2 . We
can write it as
ψnew (x) = |{z}
0 ϕ1 (x) + |{z}
1 ϕ2 (x) + |{z}
0 ϕ3 (x) .
c1
c2
c3
• Check that the wavefunction ψ (x) is normalized. (The eigenfunctions
of the linear operator  are all normalized.) (10%)
Solution:
We will use the fact that eigenfunctions of an operator corresponding
to different eigenvalues are orthogonal, that is that:
+∞
Z
ϕ ∗n ϕm dx = 0
−∞
5
for m 6= n. We will also use the condition that any wavefunction,
including an eigenfunction of an operator, must be normalized :
+∞
Z
ϕ ∗n ϕn dx = 1.
−∞
We now write the normalization condition and expand it using the
definition of ψ (x):
+∞
+∞
Z
Z
2
|ψ (x)| dx =
ψ ∗ (x) ψ (x) dx =
−∞
−∞
+∞
Z
=
∗
[c1 ϕ1 (x) + c2 ϕ2 (x) + c3 ϕ3 (x)] [c1 ϕ1 (x) + c2 ϕ2 (x) + c3 ϕ3 (x)] dx =
−∞
+∞
Z
=
[c∗1 ϕ∗1 (x) + c∗2 ϕ∗2 (x) + c∗3 ϕ∗3 (x)] [c1 ϕ1 (x) + c2 ϕ2 (x) + c3 ϕ3 (x)] dx =
−∞
+∞
Z
=
[c∗1 ϕ∗1 (x) c1 ϕ1 (x) + c∗2 ϕ∗2 (x) c1 ϕ1 (x) + c∗3 ϕ∗3 (x) c1 ϕ1 (x) +
−∞
+ c∗1 ϕ∗1 (x) c2 ϕ2 (x) + c∗2 ϕ∗2 (x) c2 ϕ2 (x) + c∗3 ϕ∗3 (x) c2 ϕ2 (x) +
+c∗1 ϕ∗1 (x) c3 ϕ3 (x) + c∗2 ϕ∗2 (x) c3 ϕ3 (x) + c∗3 ϕ∗3 (x) c3 ϕ3 (x)] dx
+∞
+∞
+∞
Z
Z
Z
c∗1 ϕ∗1 (x) c1 ϕ1 (x) dx +
=
−∞
c∗2 ϕ∗2 (x) c1 ϕ1 (x) dx +
−∞
=
c∗3 ϕ∗3 (x) c1 ϕ1 (x) dx+
−∞
+∞
+∞
+∞
Z
Z
Z
∗ ∗
∗ ∗
+
c1 ϕ1 (x) c2 ϕ2 (x) dx +
c2 ϕ2 (x) c2 ϕ2 (x) dx +
c∗3 ϕ∗3 (x) c2 ϕ2 (x) dx+
−∞
−∞
−∞
+∞
+∞
+∞
Z
Z
Z
∗ ∗
∗ ∗
+
c1 ϕ1 (x) c3 ϕ3 (x) dx +
c2 ϕ2 (x) c3 ϕ3 (x) dx +
c∗3 ϕ∗3 (x) c3 ϕ3 (x) dx =
−∞
−∞
6
−∞
=
+∞
+∞
+∞
Z
Z
Z
∗
∗
∗
∗
ϕ2 (x) ϕ1 (x) dx +c3 c1
ϕ∗3 (x) ϕ1 (x) dx +
ϕ1 (x) ϕ1 (x) dx +c2 c1
c∗1 c1
−∞
−∞
|
+
{z
}
=1
|
=0
}
|
{z
=0
}
+∞
+∞
+∞
Z
Z
Z
∗
∗
∗
∗
ϕ1 (x) ϕ2 (x) dx +c2 c2
ϕ2 (x) ϕ2 (x) dx +c3 c2
ϕ∗3 (x) ϕ2 (x) dx +
c∗1 c2
−∞
−∞
{z
|
+
−∞
{z
}
=0
|
−∞
{z
=1
}
|
{z
=0
}
+∞
+∞
+∞
Z
Z
Z
∗
∗
∗
∗
ϕ1 (x) ϕ3 (x) dx +c2 c3
ϕ2 (x) ϕ3 (x) dx +c3 c3
ϕ∗3 (x) ϕ3 (x) dx =
c∗1 c3
−∞
−∞
|
2
{z
2
|
}
=0
2
−∞
{z
=0
= |c1 | + |c2 | + |c3 | = 0.25 + 0.41 + 0.34 = 1.
Thus, the wavefunction is normalized.
7
}
|
{z
=1
}