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 Mis anti-Hermitian
(i) Show that U=eiM is Hermitian
(ii) Show that the determinant of Uis real.
(c) Two n×nmatrices Aand Bare such that Tr(A) = 0 and Tr(B)6= 0. Show
that Aand Bare linearly independent.
(d) Show that any n×nHermitian matrix Hcan be written as
H=S+iP,
where Sis a real symmetric matrix and Pis a real anti-symmetric matrix.
(e) If Tis an n×nanti-symmetric real matrix and S=eT
(i) Show that Sis orthogonal
(ii) What is the determinant of Sin this case?
(iii) Show that if nis odd then the determinant of Tis 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
c d!= a∗c∗ b∗d∗!
which gives
a∗=−a, d∗=−d, c∗=−b.
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Hence
a=ix1, d =ix2, b =x3+ix4, c =−b∗=−x3+ix4,
where x1, x2, x3, x4are real numbers. Now the general form of Ais
A= ix1x3+ix4
−x3+ix4ix2!.
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 cis 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 A6= 0, then a= 0.
(d) Since H†=H, then the matrix elements Hij satisfy
Hij =H∗
ji.
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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 =H∗
ji
xij +iyij =xji −iyji
which gives
xij =xji, yij =−yji.
Hence xij are the matrix elements of a real symmetric matrix Sand yij are the
matrix elements of a real anti-symmetric matrix P,
H=S+iP.
(e)(i)
˜
S=e˜
T=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
˜
T=−T
then
det( ˜ T) = (−)ndet(T).
If nis 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 Inis an n×nidentity matrix and σiare the Pauli matrices which satisfy
[σi, σj] = 2iX
k
ijkσk,{σi, σj}= 2δij I2.
(i) Show that
[Σi,Σj]=2iX
k
ijkΣk,{Σi,Σj}= 2δij I2n,
where I2nis an 2n×2nidentity matrix.
(ii) Show that Tr(Σi) = 0, i= 1,2,3.
(b) The matrix Bsatisfies B2=−I, show that
eiθB =Icosh θ+iB sinh θ.
(c) A general 2 ×2 Hermitian matrix Hcan be written in terms of Pauli matrices
as:
H=c0I2+c1σ1+c2σ2+c3σ3.
Please answer the following questions without using any explicit form of the Pauli
matrices.
(i) Show that c0, c1, c2, c3are 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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