**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 |ψ_{1}> and
|ψ_{2}> (as linear combinations of |1> and |2>).

(b) A unitary transformation maps the {|1>, |2>} basis onto the {|ψ_{1}>,
|ψ_{2}>}
basis. We have U|i> = |ψ_{i}>. Write
down the matrix of U and the matrix of U^{†} in the {|1>, |2>}
basis.

(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,P^{n}],

(c) [P,Q^{n}] .

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,P^{2}] = P[Q,P] + [Q,P]P = 2iħP.

(b) [Q,P^{n+1}] = [Q,PP^{n}] = P[Q,P^{n}] + [Q,P]P^{n}.

If [Q,P^{n}] =iħnP^{n-1}, then [Q,P^{n+1}] = PiħnP^{n-1 }+ iħP^{n }= iħ(n+1)P^{n}.

[Q,P^{n}] = iħnP^{n-1}holds for n = 1 and n = 2. Therefore it holds for all n.

(c) [P,Q] = -iħ, [P,Q^{2}] = Q[P,Q] + [P,Q]Q = -2iħQ.

[P,Q^{n+1}] = [P,QQ^{n}] = Q[P,Q^{n}] + [P,Q]Q^{n}.

If [P,Q^{n}] = -iħnQ^{n-1}, then [P,Q^{n+1}] = -QiħnQ^{n-1 }- iħQ^{n }=- iħ(n+1)Q^{n}.

[P,Q^{n}] = -iħnQ^{n-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> = a_{i}|i>, A|j> = a_{j}|j>.

A(|i> + |j>) = a_{i}|i> + a_{j}|j> ≠ a_{ij}(|i> + |j>) unless a_{i }= a_{j}, i.e. unless |i> and |j> have the same eigenvalue.

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)^{† }= Y^{†}X^{†}

(c) exp(i f(A)) in ket-bra form, where A is a Hermitian operator
whose eigenvalues are known.

(d) ∑_{a'}ψ*_{a'}(**r**')ψ_{a'}(**r**''),
where ψ_{a'}(**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> = ∑_{i}∑_{j}<i|X|j><j|Y|i> =
∑_{i}∑_{j}<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>^{*
}= <X^{†}j|Y|i>^{* }= <Yi|X^{†}|j >= <i|Y^{†}X^{†}|j>

for any |i> and |j>.

(c) Let {|j>} be the eigenbasis of A, A|j> = a_{j}|_{i}>.

exp(i f(A))|j> = [∑_{0}^{∞} (i f(A))^{n}/n!]|j> =
[∑_{0}^{∞} (if(a_{j}))^{n}/n!]|j>
= exp(i f(a_{j}))|j>

<k|exp(i f(A))|j> = =
[∑_{0}^{∞}(if(a_{j}))^{n}/n!]|δ_{kj}
= exp(i f(a_{j}))δ_{kj}.

For an arbitrary vector
|ψ> = ∑_{j}c_{j}|j>
we have

<ψ|exp(i f(A))|ψ> = ∑_{j}|c_{j}|^{2}exp(i f(a_{j}))

(d) ∑_{a'}ψ*_{a'}(**r**')ψ_{a'}(**r**'')
= ∑_{a'}<a'|**r**'><**r**''|a'> = ∑_{a'}<**r**''|a'><a'|**r**'>= <**r**''|**r**'>
= δ(r'' - r')

if {|a’>} is a basis.

Assume {|1>, |2>, |3>, ... , |n>} forms an orthonormal basis for the
vector space V. Let Ω_{ij} 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> = |ω_{i}>
changes this basis to the eigenbasis {|ω_{1}>, |ω_{2}>, |ω_{3}>,
... , |ω_{n}>} of Ω.
We have Ω|ω_{i}> = ω_{i}|ω_{i}>.
In
this basis the matrix of Ω is diagonal.
Let U_{ij}
denote the matrix elements of the unitary operator in the {|1>, |2>, |3>, ...
, |n>} basis, U_{ij }= <i|U|j> = <i|ω_{i}>.
The
matrix elements of Ω in the {|ω_{i}>}
basis are the same as the matrix elements of U^{†}ΩU
in the {|i>} basis. The matrix of U^{†}ΩU
in the {|i>} basis is diagonal, <i|U^{†}ΩU|j> = 0
if i ≠ j.

<i|U^{†}ΩU|j> = ∑_{kl}<i|U^{†}|k><k|Ω|l><l|U|j>
= ∑_{kl}U^{†}_{ik}Ω_{kl}U_{lj}.

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 ω_{i} and eigenvectors |ω_{i}>.

(c) Find the matrix of U in the {|i>} basis.

(d) Find the matrix of U^{†} in the {|i>} basis. (U
changes {|i>} into {|ω_{i}>}.)

(e) Verify that UU^{† }= U^{†}U = I.

(f) Verify that the matrix U^{†}ΩU in
the {|i>} basis is diagonal.

(a) Ω is Hermitian, Ω_{ij }= Ω_{ji}^{*}.

(b)> det(ωI - Ω) = 0, ω = 1, 0, -1.

ω_{1} = 1: -c_{1 }+ c_{3} = 0,
c_{2} = 0,

ω_{2} = 0: c_{1} = c_{3} = 0,
c_{2} = arbitrary,

ω_{3} = -1: c_{1}+c_{3} = 0,
c_{2} = 0.

are normalized eigenvectors.

(c) U_{ij} = <i|ω_{j}>

The matrix has the eigenvectors as its columns.

(d) U^{†}_{ij} = U_{ij}^{*
}(e) UU^{† }= U^{†}U = I.

(f) U^{†}Ω U is a diagonal
matrix with the eigenvalues along its diagonal.