Problem 1:

The Hamiltonian operator for a two state system is given by
H = a(|1><1|-|2><2|+|1><2|+|2><1|),
where a is a number with the dimensions of energy.
(a)  Find the eigenvalues of H and the corresponding eigenkets |ψ1> and |ψ2> (as linear combinations of |1> and |2>).
(b)  A unitary transformation maps the {|1>, |2>} basis onto the {|ψ1>, |ψ2>} basis.  We have U|i> = |ψi>.  Write down the matrix of U and the matrix of U in the {|1>, |2>} basis.

Solution:

(a)  In the {|1>, |2>} basis the matrix of H is
.
The eigenvalues of H are found from

For the eigenvectors we find

(b)  The matrix of U has the eigenvectors as its columns.

Problem 2:

Let P and Q be two linear operators and let [P,Q] = -iħ.  Find
(a)  [Q,P],
(b)  [Q,Pn],
(c)  [P,Qn] .

Solution:

• Concepts:
Commutator algebra
• Reasoning:
We are asked to find several commutators.
The product of two linear operators A and B, written AB, is defined by
AB|Ψ> = A(B|Ψ>).
The order of the operators is important.
The commutator [A,B] is by definition [A,B] = AB - BA.
Two useful identities using commutators are
[A,BC] = B[A,C] + [A,B]C and [AB,C] = A[B,C] + [A,C]B.
Proof: [A,BC] = ABC - BCA + (BAC - BAC) = ABC + B[A,C] - BAC = B[A,C] + [A,B]C.
• Details of the calculation:
(a)  [Q,P] = iħ,  [Q,P2] = P[Q,P] + [Q,P]P = 2iħP.

(b)  [Q,Pn+1] = [Q,PPn] = P[Q,Pn] + [Q,P]Pn.
If [Q,Pn]  =iħnPn-1, then [Q,Pn+1] = PiħnPn-1 + iħPn = iħ(n+1)Pn.
[Q,Pn] = iħnPn-1 holds for n = 1 and n = 2.  Therefore it holds for all n.

(c)  [P,Q] = -iħ,  [P,Q2] = Q[P,Q] + [P,Q]Q = -2iħQ.
[P,Qn+1] = [P,QQn] = Q[P,Qn] + [P,Q]Qn.
If [P,Qn] = -iħnQn-1, then [P,Qn+1] = -QiħnQn-1 - iħQn =- iħ(n+1)Qn.
[P,Qn] = -iħnQn-1 holds for n = 1 and n = 2.  Therefore it holds for all n.

Problem 3:

Suppose |i> and |j> are eigenkets of some Hermitian operator A.  Under what conditions can we conclude that |i> + |j> is also an eigenket?  Justify your answer.

Solution:

• Concepts:
Eigenvalues of operators
• Reasoning:
An operator operating on the elements of the vector space V has certain kets, called eigenkets, on which its action is simply that of rescaling.  Ω|V> = ω|V>.  |V> is an eigenket (eigenvector) of Ω, ω is the corresponding eigenvalue.
• Details of the calculation:
|i> and |j> are eigenkets of A.
A|i> = ai|i>,  A|j> = aj|j>.
A(|i> + |j>) = ai|i> + aj|j> ≠ aij(|i> + |j>) unless ai = aj, i.e.  unless |i> and |j> have the same eigenvalue.
Problem 4:

Using the rules of bra-ket algebra, prove or evaluate the following:
(a)  tr(XY) = tr(YX), where X and Y are operators.
(b)  (XY)= YX
(c)  exp(i f(A)) in ket-bra form, where A is a Hermitian operator whose eigenvalues are known.
(d)  ∑a'ψ*a'(r')ψa'(r''), where ψa'(r') = <r'|a'>, with {|a'>} being a basis.

Solution

(a)  Let {|i>} denote a basis for the vector space.
tr(XY) = ∑i<i|XY|i> = ∑ij<i|X|j><j|Y|i> =  ∑ij<j|Y|i><i|X|j> =  ∑j<j|YX|j> = tr(YX)
(<i|X|j> and <j|Y|i> are numbers and we can reverse their order when multiplying.)

(b)  <i|(XY)|j> = <XYi|j> = <j|XY|i>* = <Xj|Y|i>* = <Yi|X|j >= <i|YX|j>
for any |i> and |j>.

(c) Let {|j>} be the eigenbasis of A, A|j> = aj|i>.
exp(i f(A))|j> = [∑0 (i f(A))n/n!]|j> = [∑0 (if(aj))n/n!]|j> = exp(i f(aj))|j>
<k|exp(i f(A))|j> = = [∑0(if(aj))n/n!]|δkj = exp(i f(aj))δkj.

For an arbitrary vector |ψ> = ∑jcj|j> we have
<ψ|exp(i f(A))|ψ> = ∑j|cj|2exp(i f(aj))

(d)  ∑a'ψ*a'(r')ψa'(r'') =  ∑a'<a'|r'><r''|a'> = ∑a'<r''|a'><a'|r'>= <r''|r'> = δ(r'' - r')
if {|a’>} is a basis.

Problem 5:

Assume {|1>, |2>, |3>, ... , |n>} forms an orthonormal basis for the vector space V.  Let Ωij be the matrix elements of the Hermitian operator Ω in this basis.  Assume that the set {|1>, |2>, |3>, ... , |n>} is not an eigenbasis of Ω, but that the unitary transformation U|i> = |ωi> changes this basis to the eigenbasis {|ω1>, |ω2>, |ω3>, ... , |ωn>} of Ω.  We have Ω|ωi> = ωii>.  In this basis the matrix of Ω is diagonal.  Let Uij denote the matrix elements of the unitary operator in the {|1>, |2>, |3>, ... , |n>} basis, Uij = <i|U|j> = <i|ωi>.  The matrix elements of Ω in the {|ωi>} basis are the same as the matrix elements of UΩU in the {|i>} basis.  The matrix of UΩU in the {|i>} basis is diagonal, <i|UΩU|j> = 0 if i ≠ j.
<i|UΩU|j> = ∑kl<i|U|k><k|Ω|l><l|U|j> =  ∑klUikΩklUlj.
We can therefore diagonalize the matrix Ω by multiplying it from the left by U and from the right by U.
Consider the matrix

in the {|i>} basis.

(a)  Is this matrix Hermitian?
(b)  Find its eigenvalues ωi and eigenvectors |ωi>.
(c)  Find the matrix of U in the {|i>} basis.
(d)  Find the matrix of U in the {|i>} basis. (U changes {|i>} into {|ωi>}.)
(e)  Verify that UU= UU = I.
(f)  Verify that the matrix UΩU in the {|i>} basis is diagonal.

Solution

(a)  Ω is Hermitian, Ωij = Ωji*.
(b)> det(ωI - Ω) = 0,  ω = 1, 0, -1.
ω1 = 1:  -c1 + c3 = 0,  c2 = 0,
ω2 = 0:  c1 = c3 = 0,  c2 = arbitrary,
ω3 = -1:  c1+c3 = 0,  c2 = 0.

are normalized eigenvectors.

(c)  Uij = <i|ωj>

The matrix has the eigenvectors as its columns.

(d)  Uij = Uij*

(e)  UU= UU = I.

(f)  UΩ U is a diagonal matrix with the eigenvalues along its diagonal.