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> = |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,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]  =iħnP^n-1, then [Q,Pⁿ⁺¹] = PiħnP^n-1 + iħPⁿ = iħ(n+1)Pⁿ.
  [Q,Pⁿ] = iħnP^n-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] = -iħnQ^n-1, then [P,Qⁿ⁺¹] = -QiħnQ^n-1 - iħQⁿ =- iħ(n+1)Qⁿ.
  [P,Qⁿ] = -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> = ∑_jc_j|j> we have
<ψ|exp(i f(A))|ψ> = ∑_j|c_j|²exp(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 {|ω₁>, |ω₂>, |ω₃>, ... , |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> =  ∑_klU^†_ikΩ_klU_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.
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)  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.