Chem 5314 homework #1 solutions, out of 58 marks
Problem 1 – 12 marks
a – 4 marks) Show that u(x, t) = sin(kx − ωt) satisfies the classical wave equation by
using the trigonometric identity sin(A − B) = sin A cos B − cos A sin B.
Solution:
The wave equation is
1 ∂ 2u
∂ 2u
=
∂x2
v 2 ∂t2
(1)
On the left hand side, for the given u(x, t), we have
∂ 2u
∂ 2 [sin(kx) cos(ωt) − cos(kx) sin(ωt)]
∂
[k cos(kx) cos(ωt) + k sin(kx) sin(ωt)] (2)
=
=
∂x2
∂x2
∂x
= −k 2 sin(kx) cos(ωt) + k 2 cos(kx) sin(ωt) = −k 2 u(x, t)
(3)
On the right hand side, we have
1 ∂ 2 [sin(kx) cos(ωt) − cos(kx) sin(ωt)]
1 ∂
= 2 [−ω sin(kx) sin(ωt) − ω cos(kx) cos(ωt)] (4)
2
2
v
∂t
v ∂t
=
1
1
[−ω 2 sin(kx) cos(ωt) + ω 2 cos(kx) sin(ωt)] = − 2 ω 2 u(x, t)
2
v
v
(5)
Then, using that v = ω/k, we can see that the left hand side and the right hand side are
equal, and thus we have shown that this u(x, t) satisfies the wave equation.
b – 4 marks) Show that u(x, t) = sin(kx − ωt) satisfies the classical wave equation by
directly using the function sin(kx − ωt) in the wave equation.
Solution:
For this problem we need the chain rule, which states, in general, that
d
f (g(x)) = f 0 (g(x))g 0 (x)
dx
(6)
On the left hand side of Eq. (1) we have
∂2
∂
∂(kx − ωt)
∂
sin(kx − ωt) =
[cos(kx − ωt)
]=
[k cos(kx − ωt)]
2
∂x
∂x
∂x
∂x
(7)
Taking one more partial gives
= −k 2 sin(kx − ωt) = −k 2 u(x, t)
1
(8)
On the right hand side of Eq. (1) we have
∂(kx − ωt)
1 ∂
1 ∂2
1 ∂
] = 2 [−ω cos(kx − ωt)]
sin(kx − ωt) = 2 [cos(kx − ωt)
2
2
v ∂t
v ∂t
∂t
v ∂t
(9)
Taking one more partial gives
=−
1 2
1
ω sin(kx − ωt) = − 2 ω 2 u(x, t)
2
v
v
(10)
Then, using that v = ω/k, we can see that the left hand side and the right hand side are
equal, and thus we have shown that this u(x, t) satisfies the wave equation.
c – 4 marks) Show that u(x, t) = ei(kx−ωt) satisfies the classical wave equation.
Solution:
We can use the Euler identity eiθ = cos θ + i sin θ or we can directly differentiating the
function ei(kx−ωt) . In the latter case, we need the chain rule. On the left hand side of Eq. (1)
we have
∂ i(kx−ωt) ∂(i(kx − ωt))
∂
∂ 2 i(kx−ωt)
e
=
[e
]=
[ikei(kx−ωt) ] = −k 2 ei(kx−ωt)
2
∂x
∂x
∂x
∂x
(11)
On the right hand side of Eq. (1) we have
1 ∂ 2 i(kx−ωt)
1 ∂
1
e
= 2 [−iωei(kx−ωt) ] = − 2 ω 2 ei(kx−ωt)
2
2
v ∂t
v ∂t
v
(12)
Then, using that v = ω/k, we can see that the left hand side and the right hand side are
equal, and thus we have shown that this u(x, t) satisfies the wave equation.
Problem 2 – 11 marks
An operator B̂ is said to be linear if, for every pair of functions f and g and scalar c,
B̂(f + g) = B̂f + B̂g and B̂(cf ) = cB̂f .
a – 8 marks) Prove that the operator
 =
d
dx
(13)
is a linear operator by using the definition of a derivative:
d
f (x + h) − f (x)
f (x) = lim
h→0
dx
h
(14)
Solution:
We must begin with the definition of a derivative at the point x = x0 :
f (x0 + h) − f (x0 )
h→0
h
f 0 (x0 ) ≡ lim
2
(15)
To show linearity, we will establish that
Â(af + bg) = aÂf + bÂg
(16)
where  = d/dx, and where a,b are constants, and f,g are functions f(x), g(x). This combines
the two properties of linearity in a single equation. Consider the point x = x0 .
(af + bg)(x0 + h) − (af + bg)(x0 )
d
(af + bg) = lim
h→∞
dx
h
1
= lim (af (x0 + h) + bg(x0 + h) − af (x0 ) − bg(x0 ))
h→∞ h
1
= lim (af (x0 + h) − af (x0 ) + bg(x0 + h) − bg(x0 ))
h→∞ h
f (x0 + h) − f (x0 )
g(x0 + h) − g(x0 )
= a lim
+ b lim
h→∞
h→∞
h
h
0
0
= af (x0 ) + bg (x0 )
(17)
(18)
(19)
(20)
(21)
which completes the proof.
b – 3 marks) List two non-linear operators and demonstate that they are non-linear
with some specific examples.
Solution: There are many possibilities here. One good example is the affine transformation “add 2”, where the operator  takes a function and adds the number 2 to each
function value. For example, with the function f (x) = x2 , Âf (x) = x2 + 2. With this
choice, Â(cf ) = cx2 + 2 and cÂf = cx2 + 2c. So for c 6= 1 we do not have linearity.
Another example is “take the logarithm”, where Âf (x) = log(f (x)). It is easy to show
that this is nonlinear.
Problem 3 – 8 marks
For (a), (b), and (c) you can look up the answers using any resource.
a – 1 mark) Write down, up to (and including) 7th powers of x, the Taylor series for
sin x.
Solution:
sin x = x −
x 3 x5 x7
+
−
+ ···
3!
5!
7!
(22)
b – 1 mark) Write down, up to 7th powers of x, the Taylor series for cos x.
Solution:
cos x = 1 −
x2 x4 x6
+
−
+ ···
2!
4!
6!
3
(23)
c – 1 mark) Write down, up to 7th powers of x, the Taylor series for ex .
Solution:
ex = 1 + x +
x2 x3 x4 x5 x6 x7
+
+
+
+
+
+ ···
2!
3!
4!
5!
6!
7!
(24)
d – 2 marks) Write down, up to 7th powers of x, the Taylor series for eix by using your
answer (c).
Solution:
eix = 1 + ix +
i2 x2 i3 x3 i4 x4 i5 x5 i6 x6 i7 x7
+
+
+
+
+
+ ···
2!
3!
4!
5!
6!
7!
(25)
Now, we have the relations i2 = −1, i3 = −i, i4 = 1, i5 = i, i6 = −1, and i7 = −i, giving
x3 x4
x5 x6
x7
x2
−i +
+i −
− i + ···
2!
3!
4!
5!
6!
7!
(26)
x2 x4 x6
x3 x5 x7
+
−
+i x−
+
−
e = 1−
2!
4!
6!
3!
5!
7!
(27)
eix = 1 + ix −
or, rearranging,
ix
e – 3 marks) By comparing your answer (d) to the Euler formula eiθ = cos θ + i sin θ
show how you could identify the sin x and cos x Taylor series (assuming you didn’t know
them).
Solution: By looking at Equation (27) and using the Euler formula eiθ = cos θ + i sin θ,
we can immediately identify the Taylor series for sin x and cos x in Equations (22) and (23).
Problem 4 – 9 marks
In class we used the relationship
∇V · dr = dV
(28)
where r = (x, y, z) and V is the potential energy.
a – 3 marks) Explain this relationship at a mathematical level, using the following hint.
Hint: In thermodynamics, if we think of the pressure P as a function of the temperature
T and the volume V , we can write
∂P
∂P
dP =
dT +
dV
∂T V
∂V T
Solution:
4
∇V · dr = (
∂V ∂V ∂V
∂V
∂V
∂V
,
,
) · (dx, dy, dz) =
dx +
dy +
dz = dV
∂x ∂y ∂z
∂x
∂y
∂z
(29)
b – 3 marks) Explain your answer to part a) at a conceptual level (no equations allowed).
Solution:
The way to think about this is that V depends on three independent variables, namely
x, y, z. The total change in V can be decomposed into separate contributions from each independent variable, with the other variables held constant (this is the way a partial derivative
is defined).
c – 3 marks) There is another way to think about the relationship ∇V ·dr = dV , namely
as a relationship between vectors using the general formula a · b = |a| |b| cos θ where |a| is
the length of vector a and θ is the angle between the vectors a and b. Also, recall from
calculus that ∇V points in the direction of steepest descent of V . If ∇V · dr = 0 what can
you say about the direction dr in which the particle has moved?
Solution:
The gradient of V by definition is the vector of greatest change in potential energy. dr
by definition is the direction vector (of a particle that is moving through a potential energy
field). dV is the change in potential energy experienced by the particle. When you take
the dot product of the gradient vector ∇V and the direction vector dr you are finding the
length of the projection of dr onto the unit vector of the gradient, i.e. the component of
the direction vector that is in the same direction as the gradient. This scalar is equal to the
change in potential energy experienced by the particle as it moves through the field. This
makes sense when you consider the properties of the dot product. A particle moving in a
direction perpendicular to the gradient will not experience a potential energy change just as
when you walk along a level horizontal surface you don’t experience a change in gravitational
potential energy (this is the concept of a level set). Therefore if ∇V · dr = 0 we can say the
particle is moving on a level set of the potential energy function.
5
Problem 5 – 8 marks
The heat equation is:
∂u
= α∇2 u
∂t
(30)
where α is the thermal diffusivity of a substance, and u(x, y, z, t) decribes the temperature of
a substance as a function of space and time. Show that the separation of variables procedure
can be successfully applied to this partial differential equation to separate the space and time
variables. You should obtain, as your final answer, a series of ordinary differential equations.
Solution:
Postulate u(x, y, z, t) = S(x, y, z)T (t). Then LHS = SdT /dt and RHS = αT ∇2 S.
Divide through by u to get LHS = (dT /dt)/T and RHS = (α∇2 S)/S. For LHS = RHS
both sides must be independent of x, y, z, t. So to make it work put LHS = RHS = c. This
gives
dT (t)
= cT (t)
dt
(31)
and
∇2 S(x, y, z) =
c
S(x, y, z)
α
(32)
Now you can separate the 3 spatial variables in Eq. (32) since this is a linear partial
differential equation to get each variable by itself in a separate equation. To do this we put
S(x, y, z) = X(x)Y (y)Z(z), giving
X 00 Y Z + XY 00 Z + XY Z 00 =
c
XY Z
α
(33)
which, after dividing through by S(x, y, z) = X(x)Y (y)Z(z) becomes
X 00 Y 00 Z 00
c
+
+
=
X
Y
Z
α
(34)
Now we are in a position to argue that each of the three terms on the RHS must be a
constant. To see this for the first term, we move the other terms to the LHS, giving
X 00
c
Y 00 Z 00
= −
−
X
α
Y
Z
(35)
so that we can see that X 00 (x)/X(x) cannot, in fact, depend on x so that we can set it equal
to a constant d:
X 00
=d
X
⇒
X 00 (x) = dX(x)
6
(36)
Likewise, we can generate equations in y and z:
Y 00 (y) = eY (y)
and
Z 00 (z) = f Z(z)
(37)
which also results in the condition d + e + f = c/α.
Problem 6 – 6 marks
Suppose we have a wavefunction
ψ(x, t) = ψ1 (x)e−iE1 t/~ + ψ2 (x)e−iE2 t/~
(38)
with E1 6= E2 . Write down an expression for the probability density corresponding to this
wavefunction and reduce it to an explicitly real form containing only trigonometric functions
(i.e. get rid of all i’s). Based on this expression, describe the time dependence (i.e. find the
frequency) of the probability density.
Solution:
The only assumption I will make is that the energies E1 and E2 are real numbers.
ψ(x, t)ψ ∗ (x, t) = [ψ1 (x)e−iE1 t/~ + ψ2 (x)e−iE2 t/~ ][ψ1∗ (x)eiE1 t/~ + ψ2∗ (x)eiE2 t/~ ]
(39)
= ψ1 (x)ψ1∗ (x) + ψ2 (x)ψ2∗ (x) + ψ1 (x)ψ2∗ (x)ei∆Et/~ + ψ1∗ (x)ψ2 (x)e−i∆Et/~
(40)
where ∆E = E2 − E1 . Now, let us focus on the time dependence. The first and second term
do not depend on time. The third and fourth terms are complex conjugates of each other.
When we take the sum of terms that are complex conjugates of each other, the result we
obtain is twice the real part of either term (they have the same real part). Therefore, the
only term with time dependence looks like
2 Real[ψ1∗ (x)ψ2 (x)] Real[ei∆Et/~ ]
(41)
where Real[z] is the real part of the complex number z. Thus the time dependence is
cos(∆Et/~)
(42)
and therefore the angular frequency is ∆E/~, so that we conclude that the time dependence
depends on the energy difference ∆E between the two stationary states in which the particle
exists.
7
Problem 7 – 4 marks
Consider the operator
 = x
d
d
− x
dx dx
(43)
What does this operator do to a function f (x)? Based on your answer, express this operator
in a simpler form.
Solution:
d
d
Âf (x) = x − x f (x)
dx dx
d
= xf 0 (x) − (xf (x)) = xf 0 (x) − (f (x) + xf 0 (x)) = −f (x)
dx
(44)
from which we conclude that  = −1. Namely the operator A acts on a function by
multiplying the function by −1.
8