Class notes - MT tutorial 4 1 Commuting operators ∑ ∑

S.R. Clark
Teaching notes
Class notes - MT tutorial 4
This week we shall mainly focus on solving PDE’s by examining a couple of the
questions on the MM4 problem sheet. However, before we do this we should consider
in more detail what the implications are of two operators A and B commuting or not.
1 Commuting operators
A common and subtle question about commutators is whether [A,B] ≠ 0 implies that
[A,B]|ψ〉 ≠ 0 for any state |ψ〉. The answer is no. A lack of commutivity between A
and B implies only that there is at least one 1 state |ψ〉 for which [A,B]|ψ〉 ≠ 0, while
all other states may in fact give [A,B]|ψ〉 = 0, despite the operators not commuting. To
see this let’s construct a diagonal representation of A as
A =
We can then build an operator B as
n
∑
j=1
B =B +
a j
∣ ∣aj
〉〈
a j
∣ ∣.
n
∑
j=3
b j
∣ ∣aj
〉〈
a j
∣ ∣,
whereB = ∑ 2 j=1 ∑ 2 k=1 b 〉
∣
jk a j 〈ak | and acts only in the 2D subspace spanned by {|a 1 〉,|a 2 〉}.
Thus outside of this subspace B shares all of its eigenstates with A. The commutator
[A,B] then equals
⎛
[A,B] =
⎜
⎝
0 b 12 (a 1 − a 2 ) 0 ··· 0
b ∗ 12 (a 1 − a 2 ) 0 0 ··· 0
0 0 0 ··· 0
.
.
.
. .. . 0 0 0 ··· 0
and is therefore potentially non-zero in the 2D subspace. Yet for the vast majority of
states, i.e. all those outside this subspace, A and B effectively commute. This is the
most benign example of non-commutativity possible. Note that it is not possible to
construct hermitian operators A and B so they share all but one eigenstate. Orthonormality
of the eigenstates implies that if they share n − 1 eigenstates then they must
share all n.
Another common question about commutators is what [A,B] = 0 actually implies.
Formally it means that a common set of eigenstates can be found for A and B. The
“can be found” part of that statement is easy to miss. Remember that when an operator
possesses degeneracies its eigenstates in the degenerate subspace are not unique. Take
the example above. We can make [A,B] = 0 by either settingb 12 = 0, so B is diagonal,
or by making a 1 = a 2 = a so A is degenerate. If this was the case then any superposition
α|a 1 〉+β|a 2 〉 is also an eigenstate of A with eigenvalue a. We can then simply
diagonalize B in this subspace and the resulting pair of orthogonal state (superpositions
1 In fact the minimum number of states for which the commutator is non-zero can never just be one, for
reasons which are clear from the example given above.
⎞
,
⎟
⎠
1

S.R. Clark
Teaching notes
of |a 1 〉 and |a 2 〉) are guaranteed to be eigenstates of A while also being eigenstates of
B by construction. Let’s make this concrete with an example. Take A and B to be
⎛
A = 1 ⎝ −1 0 5 ⎞ ⎛
0 4 0 ⎠ and B = 1 ⎝ 2 5 −4 ⎞
5 −7 5 ⎠.
2
3
5 0 −1
−4 5 2
A short calculation will reveal that [A,B] = 0. After a slightly longer calculation we
can diagonalize A and B to find eigenvalues a j = {2,2,−3} with corresponding eigenvectors
⎧⎛
{ ∣ 〉 ⎨ a j } = ⎝ 0 ⎞ ⎛
1
1 ⎠, √ ⎝ 1 ⎞ ⎛
1
0 ⎠, √ ⎝ 1 ⎞⎫
⎬
0 ⎠
⎩
0 2 1 2 ⎭ ,
−1
and eigenvalues b j = {−4,1,2} with corresponding eigenvectors
⎧ ⎛
{ ∣ 〉 ⎨ 1
b j } = √ ⎝ −1
⎞ ⎛
1
2 ⎠, √ ⎝ 1 ⎞ ⎛
1
1 ⎠, √ ⎝ 1 ⎞⎫
⎬
0 ⎠
⎩ 6 1 3 1 2 ⎭ .
−1
Thus the first two eigenvectors of A are degenerate, while B is non-degenerate. Notice
also that as computed above A and B share precisely their third eigenvector. What this
example illustrates is that because A and B commute, and B is non-degenerate, eigenvectors
of B will automatically be eigenvectors of A and are the unique choice for a
common eigenbasis for the two operators. However, since A is degenerate, the converse
is not true, eigenvectors of A are not necessarily eigenvectors of B. The choice made
above for |a 1 〉 and |a 2 〉 is precisely an example of this since neither is an eigenvector
of B, whereas a simple calculation can confirm that |b 1 〉 and |b 2 〉 are eigenvectors of
A. It is simple to see that we can form |b 1 〉 and |b 2 〉 as superpositions of |a 1 〉 and |a 2 〉.
2 Troublesome problems explained
MM question 4.5: Find all separable solutions of Laplace’s equation ∇ 2 V = 0 in 2D
polar coordinates?
We start by assuming that V(r,θ,φ) = R(r)S(θ) and insert this into the equation for the
Laplacian in 2D polars
[
∇ 2 V(r,θ) = r ∂ (
r ∂ ] )+ ∂2
∂r ∂r ∂θ 2 V(r,θ) = 0,
which gives
S(θ)r ∂ ∂r
(
r ∂ )R(r)+R(r) ∂2
S(θ) = 0.
∂r
∂θ2 We then rearrange and introduce a separation constant m 2 to give
(
r ∂
r ∂ )
R(r) = m 2 ,
R(r) ∂r ∂r
∂ 2
m 2 = − 1
S(θ) ∂θ 2 S(θ).
2

S.R. Clark
Teaching notes
We can immediately solve the S(θ) equation
∂ 2
∂θ 2 S(θ) = −m2 S(θ), (1)
to get solutions S(θ) = C m e imθ = D m e −imθ with the constraint that since θ is an angular
variable satisfying 0 ≤ θ ≤ 2π, then S(θ) = S(θ+2π) we have that m must be an integer
or zero. This leaves an equation for R(r) of the form
r ∂ (r ∂ )
∂r ∂r R(r) = m 2 R(r). (2)
We first solve this equation for the case when m = 0 which is equivalent to solving
∂
∂r R(r) = A 0
r , (3)
where A 0 is a constant, yielding a solution R(r) = A 0 ln(r)+B 0 . For m ≠ 0 we solve
the equation using the ansatz R(r) = r k giving k 2 = m 2 implying that k = ±m. Thus we
have solutions R(r) of the form
{
Am r m + B m r −m m > 0
R(r) =
A 0 ln(r)+B 0 m = 0 . (4)
Linear combinations of the separable solutions R(r)S(θ) are capable of describing the
most general solution because the equation is linear and homogeneous. In the question
we are given the following:
(i) Given V(r = a,θ) = V 0 (1+cos(θ)), and that V is finite as r → 0, find V(r,θ) in
the region r ≤ a. (ii) Given, separately, that V(r = b,θ) = 2V 0 sin 2 (θ), and V is finite
as r → ∞, find V(r,θ) for r ≥ b.
For (i) we see that neither ln(r) nor the r −m solutions can appear so A 0 = 0 and
B m = 0 for m > 0. We can then identify that m = 1 with A 1 =
a 1, C 1 = D 1 =
2 1V 0 and
B 0 = V 0 , so
r
V(r,θ) = V 0 +V 0 cos(θ), (5)
a
for r ≤ a. For (ii) we can similarly reject r m and ln(r) solutions so A m = 0 for all m.
The boundary condition at r = b is V(r = b,θ) = 1 2 V 0(2 − e 2iθ − e −2iθ ), from which we
identify B 0 = V 0 , and C 2 = D 2 = − 1 2 V 0 with B 2 = b 2 . Thus the solution is
for r ≥ b.
V(r,θ) = V 0 −V 0
b 2
cos(2θ), (6)
r2 MM question 4.8: Perform a separation of variables on Laplace’s equation ∇ 2 V = 0
in spherical polar coordinates?
We start by assuming that V(r,θ,φ) = R(r)T(θ)P(φ) and insert this into the equation
for the Laplacian in spherical polars
[ ( 1 ∂
∇V(r,θ,φ) =
r 2 r 2 ∂ )
+
∂r ∂r
∂ 2
1
r 2 sin 2 (θ) ∂φ 2 + 1 ∂
r 2 sin(θ) ∂θ
(
sin(θ) ∂ )]
V(r,θ,φ),
∂θ
3

S.R. Clark
Teaching notes
which then gives
[ T(θ)P(φ) ∂
(r 2 ∂ )
r 2 ∂r ∂r R(r) + R(r)T(θ) ∂ 2 R(r)P(φ) ∂
r 2 sin 2 P(φ)+
(θ) ∂φ2 r 2 sin(θ) ∂θ
We now separate the equation for φ first with a separation constant of −m 2 as
1
P(φ) ∂φ 2 P(φ) = −m2 ,
[
−m 2 = − sin2 (θ) ∂
(r 2 ∂ )
R(r) ∂r ∂r R(r) − sin(θ)
T(θ)
∂ 2
leaving the P(φ) equation
∂
∂θ
(
sin(θ) ∂
∂θ T(θ) )]
= 0.
(
sin(θ) ∂
∂θ T(θ) )]
,
∂ 2
∂φ 2 P(φ) = −m2 P(φ). (7)
We then separate the remaining equation once more with a separation constant λ as
1 ∂
(r 2 ∂ )
R(r) ∂r ∂r R(r) = λ,
(
1 ∂
λ = −
sin(θ) ∂ )
sin(θ)T(θ) ∂θ ∂θ T(θ) + m2
sin 2 (θ) ,
giving the R(r) and T(θ) equations
∂
(r 2 ∂ )
∂r ∂r R(r) = λR(r), (8)
− 1 (
∂
sin(θ) ∂ )
sin(θ) ∂θ ∂θ T(θ) + m2
sin 2 T(θ) = λT(θ). (9)
(θ)
We are now in a position to answer following sub-questions:
(i) Why is m an integer or zero, (ii) Verify that for m = 0 the function T(θ) = cos(θ)
is a solution for a certain value of λ, and find that value, (iii) Find solutions for R(r)
corresponding to the particular value of λ determined in (ii), and (iv) What is the most
general solution for V(r,θ,φ) having this dependence?
For (i) we see that P(φ) = e imφ is a solution and since φ is an angular variable
satisfying 0 ≤ φ ≤ 2π, then P(φ) = P(φ+2π) which immediately implies that m must
be integer or zero. For (ii) we can set m = 0 and substitute T(θ) = cos(θ) into Eq. (9)
and identify λ = 2. For (iii) we insert λ = 2 into Eq. (8) and find solutions by trying
the ansatz R(r) = r m . This gives m(m + 1)r m = λr m , and has solutions R(r) = r and
R(r) = 1 . Finally for (iv) the most general solution with this θ and φ dependence is
r 2 V(r,θ,φ) = cos(θ)
[Ar+ B ]
r 2 , (10)
where A and B are constants. Using this solution we can answer the final part of the
question:
An uncharged conducting sphere of radius a is placed at the origin in an initially uniform
electrostatic field (0,0,E). Show that it behaves an electric dipole.
4

S.R. Clark
Teaching notes
The potential V(r,θ) satisfies the boundary condition V(r = a,θ) = 0 for all θ on
r = a, since it is a conductor. At very large distances from the sphere the field must
essentially be the same as if it was absent entirely, giving a boundary condition V(r →
∞,θ) = −Ez = −Er cos(θ). We can now use these boundary conditions to determine
the two constants A and B: firstly V(r,θ) = −Er cos(θ) at r → ∞ implies that A = −E
while V(r,θ) = 0 at r = a implies B = Ea 3 giving
]
V(r,θ) = −E cos(θ)
[r − a3
r 2 . (11)
Thus V(r,θ) behaves like an electric dipole as the deviation from the potential from a
uniform field −Ez is ∝ cos(θ)
r 2 .
MM question 4.9: The temperature T(x,t) of a one-dimensional bar, whose sides are
perfectly insulated, obeys the heat flow equation
∂
∂2
T(x,t) = κ T(x,t), (12)
∂t ∂x2 where κ is a positive constant. The bar extends from x = 0 to x = L and is perfectly
insulated at x = L. At times t < 0 the temperature is 0 ◦ C throughout the bar and at
t = 0 the uninsulated end at x = 0 is placed in contact with a heat bath at 100 ◦ C.
Determine a solution for the temperature of the bar at subsequent times and estimate
the time taken for the average temperature of the bar to reach 90 ◦ C if L = 1m and
κ = 0.001m 2 s −1 .
We start by performing separation of variables of the form T(x,t) = F(t)X(x) with a
separation constant −q 2 as
giving the F(t) and X(x) equations
1 ∂
κF(t) ∂t F(t) = −q2 ,
−q 2 =
∂ 2
1
X(x) ∂x 2 X(x),
∂
∂t F(t) = −κq2 F(t),
∂ 2
∂x 2 X(x) = −q2 X(x).
These equations have solutions of the form F(t) = e −κq2t and X(x) = e iqx , and clearly
our choice of separation constant was made to ensure decay in time and oscillatory
solutions in space, as intuition expects. In principle q could be complex, however since
T(x,t) must be real we try q being real. This gives a general solution of the form
T(x,t) = T 0 +C 0 x+ ∑ e −κq2t [A q sin(qx)+B q cos(qx)], (13)
q>0
where A q and B q are constants, while C 0 and T 0 are constants accounting for a possible
q = 0 term. When t → ∞ then we know the temperature must be in thermal equilibrium
with the heat bath with a temperature of 100 ◦ C throughout. This implies that T 0 =
5

S.R. Clark
Teaching notes
100 ◦ C. At x = 0 we have that T(0,t) = T 0 for all t > 0 which gives B q = 0. The perfect
insulation at x = L means that there is no heat flow at the boundary so temperature
gradient must vanish there as
∂
∂x T(x,t) ∣
∣∣∣x=L
= 0. (14)
This implies that C 0 = 0 and −∑ q>0 qe −κq2t A q cos(qL) = 0 for all q so
q n = π L (n+ 1 2
), (15)
with the index n an integer. The general solution for T(x,t) is then of the form
T(x,t) = T 0 +
∞
∑
n=0
e −κq2t A n sin(q n x). (16)
Finally, we have the boundary condition that at t = 0 we have T(0 < x < L,0) = 0 ◦ C.
To use this to fix the coefficients A n we employ the orthonormality relation
Z
2 L
sin(q n x)sin(q m x)dx = δ nm , (17)
L 0
since we can multiply through by sin(q n x) and integrate to get
A n + 2T 0
L
Z L
0
sin(q n x)dx = A n + 2T 0
q n L . (18)
Putting this together we get the full solution for T(x,t) as
∞
( ) [ ( )
T(x,t)
100 = 1 − 4 (2n+1)πx (2n+1)π 2
∑
n=0
(2n+1)π sin exp −κ
t]
. (19)
2L
2L
We can now compute the average temperature of the bar 〈T(t)〉 as
〈T(t)〉 = 1 L
Z L
0
T(x,t)dx = T 0 −
∞
∑
n=0
2T 0
q 2 n L2 e−κq2nt . (20)
An estimate for the time required to heat the bar to 90 ◦ C can be made by keeping only
the slowest decaying term in the summation q 0 = π
2L
which gives
T 0 − 〈T(t)〉
≤ 8 ( −κπ 2 )
T 0 π 2 exp t
4L 2 . (21)
Thus we have an estimate t c = − 4L2 ln( π2
κπ 2 80
) ≈ 14 minutes.
6