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 Ω^{2}. 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 Ω^{2}= 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) Ω^{2}= |Φ><ψ|Φ><ψ|. We need <ψ|Φ> = 1 and |Φ> = a|ψ>.

This implies a<ψ|ψ> = 1, a = 1/<ψ|ψ>.

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|B^{†}A^{†}|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.

Use the virial theorem for the eigenstates of atomic
hydrogen, namely, <nlm|T|nlm> = -E_{nl}, where E_{nl} 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 E_{n'l'}
and the mass m_{e} 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'|p^{2}|n'l'm'> = 2m_{e}<n'l'm'|T|n'l'm'> = -2m_{e}E_{n'l'}.

Note:**p**is a Hermitian operator,**p**=**p**^{T}.

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 T^{d} = UTU^{†}.

What is special about the diagonal elements?

(e) Compare the determinant |T|, the trace Tr(T ), and eigenvalues of T to
those of T^{d}.

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) T_{ij}= T_{ji}*, 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.)

U^{†}U = UU^{†}= I.

The diagonal values now are equal to the eigenvalues.

(e) det(T) = det(T^{d}) = -2, Tr(T) = Tr(T^{d}) = 1. Both matrices have the same eigenvalues.

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.

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.)

.

Consider a three-state system. Let {|ω_{1}>, |ω_{2}>,
|ω_{3}>} be the non-degenerate orthonormal eigenbasis of the
Hermitian operator Ω with |ω_{1}> having the smallest and |ω_{3}>
the largest eigenvalue. Let U be the unitary transformation from the {|ω_{1}>,
|ω_{2}>, |ω_{3}>} basis to another basis {|t_{1}>, |t_{2}>,
|t_{3}>} of the state space.

In the {|t_{1}>, |t_{2}>, |t_{3}>} 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 {|t_{1}>, |t_{2}>, |t_{3}>}.

(b) Find the matrix U.

(c) The matrix of another operator M in the {|ω_{1}>,
|ω_{2}>, |ω_{3}>} basis is

M =

0 | 0 | i | ||

0 | i | 0 | ||

-i | 0 | 0 |

.

What is the matrix of M in the {|t_{1}>, |t_{2}>, |t_{3}>}
basis?

Solution:

- Concepts:

Change of representation - Reasoning:

|t_{i}> = ∑_{j}|ω_{j}><ω_{j}|t_{i}> = ∑_{j}U_{ji}|ω_{i}>.

The matrix elements of U and U^{†}are U_{ij}= <ω_{i}|t_{j}>, U_{ij}^{†}= <t_{i}|ω_{j}>, respectively.

The eigenvectors of Ω in the {|t_{1}>, |t_{2}>, |t_{3}>} 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.

λ^{2}(1 - λ) - 4(1 - λ) = 0. The solutions to this equation are λ_{1}= -2, λ_{2}= 1, λ_{3}= 2.

The corresponding normalized eigenvectors of Ω in the {|t_{1}>, |t_{2}>, |t_{3}>} basis are|1> = 2

^{-½}-1 1 0 , |2> =

0 0 1 , |3> = 2

^{-½}1 1 0 .

(b) (See representations.)

The matrix of U

^{†}isU

^{†}=-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 {|ω_{1}>, |ω_{2}>, |ω_{3}>} basis be (a_{1}, a_{2}, a_{3}) and in the {|t_{1}>, |t_{2}>, |t_{3}>} basis be (a'_{1}, a'_{2}, a'_{3}).

We have a'_{i}= <t_{i}|A> = ∑_{j}<t_{i}|u_{j}><u_{j}|A> = ∑_{j}U^{†}_{ij}a_{j}.Let the coordinates of the vector B in the {|ω

_{1}>, |ω_{2}>, |ω_{3}>} basis be (b_{1}, b_{2}, b_{3}).

Its coordinates in the {|t_{1}>, |t_{2}>, |t_{3}>} basis are b'_{i}= ∑_{j}U^{†}_{ij}b_{j}.

M|A> = |B> --> b_{j}= ∑_{k}M_{jk}|a_{k}> or b'_{j}= ∑_{k}M'_{jk}|a'_{k}>.

b'_{i}= ∑_{j}U^{†}_{ij}b_{j}= ∑_{jk}U^{†}_{ij}M_{jl}|a_{l}> = ∑_{jkl}U^{†}_{ij}M_{jl}U_{lk}|a'_{k}>.

Therefore M'_{jl}= ∑_{jl}U^{†}_{ij}M_{jl}U_{lk}, or M' = U^{†}MU,

We therefore have M' = U^{†}MU.M' =

0 -i 0 i 0 0 0 0 i .