### Mathematical foundations

#### Problem:

Let Ω be the operator defined bψ Ω = |Φ><ψ| where |Φ> and |ψ> are two vectors in a vector space V.
(a)  Under what conditions is Ω Hermitian?
(b)  Calculate Ω2.  Under what conditions is Ω a projector?

Solution:

• Concepts:
Mathematical foundations of quantum mechanics
• Reasoning:
An operator A is Hermitian if A = A. A Hermitian operator satisfies <ψ|A|Φ> = <Φ|A|ψ>*.
A projector is a Hermitian operator.  If Ω is a projector, then Ω2 = Ω.
• Details of the calculation:
(a)  Ω = |Φ><ψ|, Ω = |ψ><Φ|.
Ω = Ω  -->  Ω|ζ> = Ω|ζ>  for any |ζ> in V.
We need |Φ><ψ|ζ> = |ψ><Φ|ζ>.  This implies |Φ> = a|ψ>.
Then a|ψ><ψ|ζ> = |ψ><Φ|ζ> or a<ψ|ζ> =  <Φ|ζ>.
But we also have <Φ|ζ> = a*<ψ|ζ>.
Therefore a = a*.
We need |Φ> = a|ψ>, with a real.
(b) Ω2 = |Φ><ψ|Φ><ψ|.  We need <ψ|Φ> = 1 and |Φ> = a|ψ>.
This implies a<ψ|ψ> = 1, a = 1/<ψ|ψ>.

#### Problem:

Quantum mechanics is often conveniently formulated in terms of matrix operators.  Let {|i>} be an orthonormal basis for the state space.
(a)  Prove that for any operator A we have ∑ij|<i|A|j>|2 = Tr(AA), where Tr denotes the trace.
(b)  Derive the condition that must be satisfied for the product of two Hermitian operators to be itself a Hermitian operator.
(c)  Prove that the trace of a matrix operator is invariant under a change of representation, i.e. a change of basis.

Solution:

• Concepts:
Mathematical foundations of quantum mechanics
• Reasoning:
This is a linear algebra problem.
• Details of the calculation:
(a)  ∑ij|<i|A|j>|2 = ∑ij<i|A|j><j|A|i> = ∑i<i|AA|i> = Tr(AA), since ∑j|j><j| = I.
(b)  Let C = AB, where A and B are Hermitian operators.
For any |i>, |j> we have
<i|AB|j>* = <j|(AB)|i> = <j|BA|i> = <j|BA|i>.
If C is a Hermitian operator then <i|AB|j>* = <j|AB|i>, or
<j|BA|i> = <j|AB|i>. for any |i>, |j>.
C is a Hermitian operator implies that BA = AB, i.e. that A and B commute.
Let {|u>} be a second orthonormal basis for the state space.E.
Tr(A) = ∑i<i|A|i> = ∑iuu'<i|u><u|A|u'><u'|i> = ∑iuu'<u'|i><i|u><u|A|u'>
= ∑uu'<u'|u><u|A|u'> = ∑u'<u'|A|u'> = Tr(A),
since ∑j|j><j| = ∑u'|u'><u'| = I.

#### Problem:

Use the virial theorem for the eigenstates of atomic hydrogen, namely, <nlm|T|nlm> = -Enl, where Enl is the eigenenergy of the |nlm> bound state and T is the kinetic energy operator for an electron bound to a proton, to evaluate the sum ∑nlm|<nlm|p|n'l'm'>|2, where p is the momentum operator and the sum is over a complete set of bound and continuum states.  Take the proton mass to be infinite and give your answer in terms of En'l' and the mass me of the electron.

Solution:

• Concepts:
Mathematical foundations of quantum mechanics
• Reasoning:
The eigenvectors of every Hermitian operator form a basis for the vector space,
nlm | nlm ><nlm| = 1.
• Details of the calculation:
nlm |<nlm|p|n'l'm'>|2 = ∑nlm  <n'l'm'|p| nlm><nlm|p|n'l'm'>
= <n'l'm'|p2|n'l'm'> = 2me<n'l'm'|T|n'l'm'> = -2meEn'l'.
Note: p is a Hermitian operator, p = pT.

#### Problem:

Consider the matrix representation of the operator

(a)  Is T Hermitian?
(b)  Solve for the eigenvalues.  Are they real?
(c)  Determine the normalized eigenvectors.  Since eigenvectors are not unique to within a phase factor, arrange your eigenvectors so that the first component of each is positive and real.  Are they orthogonal?
(d)  Using the eigenvectors as columns, construct U, the inverse of the unitary matrix which diagonalizes T.  Use this to find this diagonalized version Td = UTU.
What is special about the diagonal elements?
(e)  Compare the determinant |T|, the trace Tr(T ), and eigenvalues of T to those of Td.

Solution:

• Concepts:
Matrix representation of operators.  (See representations.)
• Reasoning:
We are given the matrix of an operator in a particular basis. We are asked to find the eigenvalues of the operator and the eigenvectors of the operator in the given basis. We are then asked to make a unitary transformation to a basis in which the matrix of the operator is diagonal.
• Details of the calculation:
(a)  Tij = Tji*,  T is Hermitian.
(b)  eigenvalues α:  (1 - α )(-α) - 2 = 0.  α+ = 2,  α- = -1.  The eigenvalues are real.
(c)  eigenvector for α+:

eigenvector for α-:

The eigenvectors are orthogonal.  <+|-> = <-|+> = 0.

(d)  (See representations.)

UU = UU = I.

The diagonal values now are equal to the eigenvalues.

(e)  det(T) = det(Td) = -2,  Tr(T) = Tr(Td) = 1.  Both matrices have the same eigenvalues.

#### Problem:

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:

• Concepts:
The eigenvalues and eigenvectors of a Hermitian operator.
• Reasoning:
We are given enough information to construct the matrix of the Hermitian operator H in some basis.  To find the eigenvalues E we set the determinant of the matrix (H - EI) equal to zero and solve for E.  To find the corresponding eigenvectors {|Ψ>}, we substitute each eigenvalue E back into the equation (H  -E*I)|Ψ> = 0 and solve for the expansion coefficients of |Ψ> in the given basis.
• Details of the calculation:
(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.
(See representations.)
.

#### Problem:

Consider a three-state system.  Let {|ω1>, |ω2>, |ω3>} be the non-degenerate orthonormal eigenbasis of the Hermitian operator Ω with |ω1> having the smallest and |ω3> the largest eigenvalue.  Let U be the unitary transformation from the {|ω1>, |ω2>, |ω3>} basis to another basis {|t1>, |t2>, |t3>} of the state space.
In the {|t1>, |t2>, |t3>} basis the matrix of the operator Ω is

Ω =

 0 2 0 2 0 0 0 0 1

.

(a)  Find the eigenvectors Ω in the {|t1>, |t2>, |t3>}.
(b)  Find the matrix U.
(c)  The matrix of another operator M in the {|ω1>, |ω2>, |ω3>} basis is

M =

 0 0 i 0 i 0 -i 0 0

.

What is the matrix of M in the {|t1>, |t2>, |t3>} basis?

Solution:

• Concepts:
Change of representation
• Reasoning:
|ti> = ∑jj><ωj|ti> = ∑jUjii>.
The matrix elements of U and U are Uij = <ωi|tj>,   Uij = <tij>, respectively.
The eigenvectors of Ω in the {|t1>, |t2>, |t3>} basis are the columns of the U matrix.
• Details of the calculation:
(a)  The eigenvalues of Ω are λ, where

 -λ 2 0 2 -λ 0 0 0 1-λ

= 0.

λ2(1 - λ) - 4(1 - λ) = 0.  The solutions to this equation are λ1 = -2, λ2 = 1, λ3 = 2.
The corresponding normalized eigenvectors of Ω in the {|t1>, |t2>, |t3>} basis are

|1> = 2

 -1 1 0

,   |2> =

 0 0 1

,   |3> = 2

 1 1 0

.

(b)  (See representations.)

The matrix of U is

U  =

 -2-½ 0 2-½ 2-½ 0 2-½ 0 1 0

.

The matrix of U is

U  =

 -2-½ 2-½ 0 0 0 1 2-½ 2-½ 0

.

(c)  An operator is defined by what is does to the basis vectors.
Let M|A> = |B>.  Let the coordinates of the vector A in the {|ω1>, |ω2>, |ω3>} basis be (a1, a2, a3) and in the {|t1>, |t2>, |t3>} basis be  (a'1, a'2, a'3).
We have a'i = <ti|A> =  ∑j<ti|uj><uj|A>  = ∑jUijaj.

Let the coordinates of the vector B in the {|ω1>, |ω2>, |ω3>} basis be (b1, b2, b3).
Its coordinates in the {|t1>, |t2>, |t3>} basis are b'i =  ∑jUijbj.
M|A> = |B> --> bj = ∑kMjk|ak> or b'j = ∑kM'jk|a'k>.
b'i =  ∑jUijbj = ∑jkUijMjl|al> = ∑jklUijMjlUlk|a'k>.
Therefore M'jl =  ∑jlUijMjlUlk,  or  M' =  UMU,
We therefore have M' = UMU.

M' =

 0 -i 0 i 0 0 0 0 i

.