PHY4605 Problem Set 3 Solutions 1. Matrix algebra drill
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1
PHY4605
Problem Set 3
Solutions
1. Matrix algebra drill
a) Eigenvalues + eigenvectors
There are many related techniques for finding eigenvalues and eigenvectors systematically.
Eigenvalues: Define a characteristic matrix
• c = A − λ |{z}
1
identity
• det c = 0, solve for λ
A)
c(λ) =
det c =
3
5
− 45
−λ
4
5
3
5
(1)
−λ
¡3
¢2 ¡ 4 ¢2 ¡ 3
4 ¢¡ 3
4 ¢
−λ +
=
−λ+ i
−λ− i
5
5
5
5 5
5
3 4
λ= ± i
5 5
(2)
(3)
B)
1 − λ 1 −1
c(λ) = −1 3 − λ −1
−1
2 −λ
¡
¢¡
¢
¡
¢2 ¡
¢
det c = 1 − λ λ2 − 3λ + 2 = − λ − 1 λ − 2 ⇒
λ = 1 (multiplicity 2), λ = 2.
(4)
(5)
2
C)
−1 − λ 2
2
c(λ) = 2
2−λ
2
−3
−6 −6 − λ
¡
¢¡
¢
det c = −λ3 − 5λ2 − 6λ = −λ λ + 3 λ + 2 ⇒
λ = 0, λ = −3, λ = −2
(6)
(7)
Eigenvectors
Systematic procedure: first put characteristic matrix c(λ) in Hermite normal form for each
λ. Hermite normal form (HNF) means multiply rows of matrix by constants, add rows
together to get matrix in form with elements nonzero only on diagonal and above, with
first nonzero element to be equal to 1 on the diagonal. This method is equivalent to the
“sledgehammer” technique of plugging in for λ writing for each eigenvaector
¡
¢
vλ = a, b, c, · · · and solving Mvλ = λvλ for all a, b, c, · · ·
For matrix A compare methods:
A “Sledgehammer”
4
4
− i −5
i
3 4
= −4
λ = + i, c(λ) = 5
4
5 5
5 −1
− 54 i
5
i 1
a
4
= 0 ⇒
−
5 −1 i
b
ai + b = 0
a − ib = 0
⇒ a = ib = 0
1
i
(8)
(9)
3
e.g. a = 1, b = −i ⇒ v1 =
√1
2
1
−i
λ=
3 4
− i, c(λ) =
5 5
4
i
5
4
5
ai − b = 0
a + ib = 0
− 45
4
i
5
4 i −1
=
⇒
5 1 i
⇒ a = −ib
1
1
a = 1, b = i, v2 = √
2 i
(10)
(11)
“HNF”
i 1
3 4
4
λ = + i, c(λ) = −
5 5
5 −1 i
(12)
multiply by − 54 , first row by −i and add to the second row:
1 −i
⇒ a − ib = 0, etc.
c −→
0 0
3 4
4 i −1
λ = − i, c(λ) =
5 5
5 1 i
(13)
(14)
multiply by 54 , first row by −i and subtract from second row:
1 +i
⇒ a + ib = 0, etc.
c −→
0 0
(15)
(16)
(HNF not obviously useful for smaller matrices)
4
B
1 − λ 1 −1
c(λ) = −1 3 − λ −1
−1
2 −λ
0 1 −1
−1 2 −1
λ = 1, c −→ −1 2 −1 −→ 0 1 −1 −→
−1 2 −1
−1 2 −1
exchange rows 1,2
1 −2 1
0 1 −1
0 0 0
subtract row 1 from 3
1
triangular”
a=b
a − 2b + c = 0
1
−→
HNF ⇒
⇒
⇒ v1 = v2 = √ 1
(17)
| {z }
3
b=c
b−c=0
multiplicity 2
1
| {z }
normalization
−1 1 −1
1 −1 1
1 −1 1
λ = 2 c(λ) = −1 1 −1 −→
−→ −1 2 −2
−1 2 −2
−1 2 −2
−1 1 −1
0 0 0
now “upper
multiply row 1 by -1
exchange rows 2 and 3
0
1 −1 1
a−b+c=0
a=c
1
−→ 0 1 −1 ⇒
⇒
, v3 = √ 1
2
b−c=0
b=c
1
0 0 0
(18)
5
C
−1 2 2
1 2 2
1 2 2
λ = 0, c(0) = 2 2 2 −→ 1 1 1 −→ 0 1 1
−3 −6 −6
−1 −2 −2
0 −1 −1
1
2
2
0
a
+
2b
+
2c
=
0
1
⇒ a = 0 v1 = √ 1
−→ 0 1 1 ⇒
(19)
2
b+c=0
0 0 0
−1
1 1 1
1 1 1
2 2 2
λ = −3, c(−3) = 2 5 2 −→ 1 5/2 1 −→ 0 1/2 0
0 −1 0
−1 −2 −1
−3 −6 −3
1 1 1
−1
a+b+c=0
1
⇒ a − c v2 = √ 0
−→ 0 1 0 ⇒
(20)
2
b=0
0 0 0
1
1 2 2
1 2
2
1 2 2
λ = −2, c(−2) = 2 4 2 −→ 1 2
1 −→ 0 0 1 ⇒
−3 −6 −4
−1 −2 −4/3
0 0 −1/3
−2
a + 2b + 2c = 0
1
(21)
⇒ a = −2b, v3 = √ 1
5
c=0
0
b) det A =
¡ 3 ¢¡ 3 ¢
5
5
¡
¢¡ ¢
− − 45 45 = 1.
det B = 1(0 − (−2)) − 1(0 − 1) − 1(−2 − (−3)) = 2 + 1 − 1 = 2
det C = −1(−12 − (−12)) − 2(−12 − (−6)) + 2(−12 − (−6)) = 0 singular.
c) Similarity transformation
λ 0
1
contains eigenvalues 0 λ2
.. ..
. .
U−1MU = M0 which diagonalizes M (i.e. M0 is diagonal) and
···
· · · may be constructued directly from eignevectors of M:
..
.
h
i
U = v1 v2 · · ·
(22)
meaning arrange vectors in adjacent columns. To find inverse, use formula
µ
¶T
1
−1
M =
cof M
det M
(23)
6
where cof M is “cofactor matrix” consisting of replacing every element with the
determinant of the smaller matrix obtained by crossing out row and column of the element
in question, multiplied by (−1)P , where P is the number of steps from upper left element
to the one in question.
i
1 1
, cof U =
A U=
−i i
−1
righthand corner we replace
i
, det U = 2, since e.g. for the element in the upper
1
it with det{−i} = −i, multiplied by (−1) since it is 1
step away from upper left corner.
T
i i
i −1
1
= 1
=
2i −1 1
2i i 1
U−1
(24)
check
UU−1
1
i
−1
1
=
=
−i i 2i i 1
0
1 1
0
(25)
1
and that
1 i −1 3/5 −4/5 1
U−1 AU =
2i i 1
4/5 3/5
−i
1
i
=
3
4
+
0
4
i
5
0
3
5
−
4
i
5
(26)
C
0 −2 −1
1 1 −1
U = −1 1 0 , det U = −1, cof U = 2 1 −2
1 0 1
1 1 −2
1 2 1
−1 −2 −1
1
U−1 =
1 1 1 = −1 −1 −1
(−1)
−1 −2 −2
1 2 2
|
{z
}
(28)
cof U T
−1 −2 −1 −1 2 2 0 −2 −1 0 0 0
U−1 CU = −1 −1 −1 2 2 2 −1 1 0 = 0 −2 0
0 0 3
1 0 1
−3 −6 −6
1 2 2
(27)
(29)
7
d) Inverses of A and B
B−1
3/5
4/5
1
A−1 = cof AT =
1
−4/5 3/5
T
2 1 1
1 −1 1
1
1
T
= cof B = −2 −1 −3 = 1/2 −1/2 1
2
2
2 2 4
1/2 −3/2 2
(30)
(31)
can’t invert C since det C = 0.
e)
³
´
M = a0 + a · σ, a = a1 a2 a3
(32)
claim:
a0 − a · σ
a20 − a · a
¡ 2 ¡
¢2 ¢
1
= 2
a0 − a · σ
a0 − a · a
M−1 =
MM−1
(33)
(34)
(cross terms cancel since a0 = a0 σ0 , σ0 commutes with everything!) But (a · σ)2 = ai aj σi σj
(Einstein’s sum convention assumed). This sum includes terms like
a1 a2 σ1 σ2 + a2 a1 + σ2 σ1 = 0
(35)
σi σj + σj σi = 0 fori 6= j, i, j = 1, 2, 3
(36)
since
so only terms which survive are
a21 σ12 + a22 σ22 + a23 σ32 = a21 + a22 + a23 = a · a QED
(37)
2. Phase shift in a vector potential.
1. Since the vector potential should wind around the solenoid, let’s take it to be
Φ/2πρθ̂ outside the solenoid. Check to make sure field is indeed zero. One way is to
express the curl in cylindrical coordinates:
8
µ
∇×A =
1 ∂Az ∂Aθ
−
ρ ∂θ
∂z
¶
µ
ρ̂ +
∂Aρ ∂Az
−
∂z
∂ρ
¶
µ
θ̂ +
∂Aθ Aθ 1 ∂Aρ
+
−
∂ρ
ρ
ρ ∂θ
¶
ẑ.
Only the Aθ term is finite, and the two terms involving it proportional to ẑ cancel,
giving B = 0 outside the solenoid. Inside (ρ < a), Aθ = Φρ/(2πa2 ) and the two terms
in the curl add, giving B = Φ/(πa2 )ẑ = B0 ẑ.
2.
µ
ψA (r, t) = exp
ie
~
Z
r
¶
A · ds ψA=0 (r, t) ≡ eieS/~ ψ0
r0
Note
ieS/~
∇(e
µZ r
¶
ie
ψ0 ) = ∇
A · ds eieS/~ ψ0 + eieS/~ ∇ψ0
~
r0
ie
= AeieS/~ ψ0 + eieS/~ ∇ψ0
~
and since A is time independent
i~
∂ ieS/~
∂ψ0
(e
ψ0 ) = i~eieS/~
∂t
∂t
Apply Π ≡ p − eA) to ψA :
µ
(−i~∇ − eA)e
ieS/~
ψ0 = −i~
¶
ie ieS/~
ieS/~
Ae
ψ0 + e
∇ψ0 − eAeieS/~ ψ0
~
= −i~ieS/~ ∇ψ0 = eieS/~ pψ0
and again
(−i~∇ − eA)2 eieS/~ ψ0 = (−i~∇ − eA)(−i~eieS/~ ∇ψ0 ) = eieS/~ p2 ψ0
So
1
p2
[p − eA(r)]2 ψA = eieS/~
ψ0 and
2m
2m
∂ψ0
∂ψA
= i~eieS/~
,
i~
∂t
∂t
so dividing by eieS/~ we recover original Schrödinger equation in A = 0.
