Fall 2019 Midterm1 Solution

Zewail City of Science and Technology–Fall 2019.
PEU 346: Mathematical Physics 1–Midterm 1 Solution
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(1) (a) Construct the most general 2 × 2 anti-Hermitian matrix.
(b) If M is anti-Hermitian
(i) Show that U = eiM is Hermitian
(ii) Show that the determinant of U is real.
(c) Two n × n matrices A and B are such that Tr(A) = 0 and Tr(B) 6= 0. Show
that A and B are linearly independent.
(d) Show that any n × n Hermitian matrix H can be written as
H = S + iP,
where S is a real symmetric matrix and P is a real anti-symmetric matrix.
(e) If T is an n × n anti-symmetric real matrix and S = eT
(i) Show that S is orthogonal
(ii) What is the determinant of S in this case?
(iii) Show that if n is odd then the determinant of T is zero.
Solution
(a) The most general 2 × 2 matrix is
A=
!
a b
,
c d
where a, b, c, d are complex numbers. Now we impose A† = −A
!
!
a b
a∗ c ∗
−
=
c d
b∗ d ∗
which gives
a∗ = −a,
d∗ = −d,
2
c∗ = −b.
Hence
a = ix1 ,
d = ix2 ,
b = x3 + ix4 ,
c = −b∗ = −x3 + ix4 ,
where x1 , x2 , x3 , x4 are real numbers. Now the general form of A is
!
ix1
x3 + ix4
A=
.
−x3 + ix4
ix2
Notice that Tr(A) = i(x1 + x2 ), i.e., pure imaginary.
(b)(i)
†
U † = e−iM = eiM = U
(ii)
det(U ) = det(eiM ) = eiTr(M ) .
From exercise (a), we know that Tr(M ) = ic, where c is some real number. Hence
det(U ) = e−c .
(c) To prove the linear independence we need to show that
aA + bB = 0
implies a = 0, b = 0. By taking the trace of the above equation and using Tr(A) = 0
and Tr(B) 6= 0
bTr(B) = 0,
b=0
which means
aA = 0.
But since A 6= 0, then a = 0.
(d) Since H † = H, then the matrix elements Hij satisfy
Hij = Hji∗ .
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The matrix elements Hij are complex numbers and as such we can write them as
Hij = xij + iyij ,
where xij , yij are real numbers. Now using Hij = Hji∗
xij + iyij = xji − iyji
which gives
xij = xji ,
yij = −yji .
Hence xij are the matrix elements of a real symmetric matrix S and yij are the
matrix elements of a real anti-symmetric matrix P ,
H = S + iP.
(e)(i)
S̃ = eT̃ = e−T = S −1
(ii)
det(S) = det(eT ) = eTr(T ) = e0 = 1,
where we used the fact that the diagonal elements of an anti-symmetric matrix are
zero (Tii = −Tii ).
(iii) By taking the determinant of the two sides
T̃ = −T
then
det(T̃ ) = (−)n det(T ).
If n is odd then det(T ) must be zero.
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(2) (a) Consider the following matrices:
Σi = σi ⊗ In ,
i = 1, 2, 3,
where In is an n × n identity matrix and σi are the Pauli matrices which satisfy
X
[σi , σj ] = 2i
ijk σk ,
{σi , σj } = 2δij I2 .
k
(i) Show that
[Σi , Σj ] = 2i
X
ijk Σk ,
{Σi , Σj } = 2δij I2n ,
k
where I2n is an 2n × 2n identity matrix.
(ii) Show that Tr(Σi ) = 0, i = 1, 2, 3.
(b) The matrix B satisfies B 2 = −I, show that
eiθB = I cosh θ + iB sinh θ.
(c) A general 2 × 2 Hermitian matrix H can be written in terms of Pauli matrices
as:
H = c0 I 2 + c1 σ 1 + c2 σ 2 + c3 σ 3 .
Please answer the following questions without using any explicit form of the Pauli
matrices.
(i) Show that c0 , c1 , c2 , c3 are real.
(ii) If Tr(H) = 0, show that c0 = 0.
Solution
(a) (i) We will use the following property of the direct products
(A ⊗ B)(C ⊗ D) = AC ⊗ BD.
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This gives
Σi Σj − Σj Σi = σi ⊗ In σj ⊗ In − σj ⊗ In σi ⊗ In
= σi σj ⊗ In − σj σi ⊗ In
= σi , σj ⊗ In
X
= 2i
ijk σk ⊗ In
(1)
k
= 2i
X
ijk Σk .
k
Similarly,
Σi Σj + Σj Σi = σi ⊗ In σj ⊗ In + σj ⊗ In σi ⊗ In
= σi , σj ⊗ In
= 2δij In ⊗ In
(2)
= 2δij I2n .
(ii) Since every Σi is the commutator of two other Σ’s, for example,
Σ1 =
Then
1
[Σ2 , Σ3 ].
2i
1
Tr Σ1 = Tr [Σ2 , Σ3 ]) = 0.
2i
The same for Σ2 and Σ3 .
(b) One can expand eiθB and then collect the even and odd powers of B. However,
we can just define A = iB and now A2 = I which enables us to write
eiθB = ei(−iθ)A = eiφA
where φ = −iθ. Using (proved in Arfken)
eiφA = I cos φ + iA sin φ.
Then
eiφA = eiθB = I cos(−iθ) + i(iB) sin(−iθ) = I cosh θ + iB sinh θ.
(c)(i) Since H = H † , then
H = c0 I2 + c1 σ1 + c2 σ2 + c3 σ3 = H † = c∗0 I2 + c∗0 σ1 + c∗0 σ2 + c∗0 σ3 .
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But the σi ’s are linearly independent, hence
c∗µ = cµ ,
µ = 0, 1, 2, 3.
(ii) Since Tr(H) = 0 and Tr(σi ) = 0, then
0 = 2c0 ,
c0 = 0.
(3) (a) Show that
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X
γ µ γ ν Pµ Pν = P02 − P12 − P22 − P32 I4 ,
µ,ν=0
µ
where γ are the Dirac matrices which satisfy
µ ν
γ , γ = 2η µν I4 .
(b) The matrix γ 5 is defined as
γ 5 = iγ 0 γ 1 γ 2 γ 3 .
Please answer the following questions without using any explicit form of γ 5 .
(i) Show that γ 5 is Hermitian.
(ii) Show (γ 5 )2 = I4 .
(iii) Show that {γ 5 , γ 1 } = 0.
(iv) Show that Tr(γ 5 ) = 0.
Solution
(a) Since Pµ Pν = Pν Pµ , i.e., symmetric in µ, ν, then we can decompose γ µ γ ν as
γ µγ ν =
1 µ ν 1 µ ν
γ ,γ + γ ,γ .
2
2
Only the symmetric part will contribute to the sum
3
X
µ,ν=0
γ µ γ ν Pµ Pν =
3
3
X
1 X µ ν
γ , γ Pµ Pν =
η µν Pµ Pν .
2 µ,ν=0
µ,ν=0
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†
†
(b) (i) Using γ 0 = γ 0 , γ i = −γ i
(iγ 0 γ 1 γ 2 γ 3 )† = iγ 3 γ 2 γ 1 γ 0 .
Now we just need to reorder the γ’s taking care of the minus signs one picks along
the way we can show that
iγ 3 γ 2 γ 1 γ 0 = −iγ 0 γ 3 γ 2 γ 1 = −iγ 0 γ 1 γ 3 γ 2 = iγ 0 γ 1 γ 2 γ 3 .
(ii)
(γ 5 )2 = −(γ 0 γ 1 γ 2 γ 3 )(γ 0 γ 1 γ 2 γ 3 ).
The rest is very similar to (i).
(iii)
γ 5 γ 1 = iγ 0 γ 1 γ 2 γ 3 γ 1 = −iγ 1 γ 0 γ 1 γ 2 γ 3 = −γ 1 γ 5 ,
where we passed γ 1 from right to left and three minus signs were accumulated.
(iv)
Tr(γ 5 ) = iTr(γ 0 γ 1 γ 2 γ 3 ) = iTr(γ 3 γ 0 γ 1 γ 2 ) = −iTr(γ 0 γ 1 γ 2 γ 3 ) = −Tr(γ 5 ),
where in the second equality we used the cyclic property of the trace and in the
third equality we reordered γ 3 .
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