__Find eigenvalues and eigenfunctions__

Consider a 3-dimensional state space with orthonormal eigengbasis {|a>, |b>,
|c>}. The action of the Hamiltonian on the basis vectors is

H|a> = -iE|b> + iE|c>, H|b> = iE|a>, H|c> = -iE|a>, where E is a
real constant with units of energy.

(a) Construct the matrix of the Hamiltonian in the {|a>, |b>, |c>} basis.

(b) Find the energy eigenvalues and the corresponding normalized
eigenvectors.

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) <a|H|a> = <b|H|b> = <c|H|c> = <b|H|c> = <c|H|b> = 0.

<a|H|b> = <b|H|a>* = iE, <c|H|a> = <a|H|c>* = iE.

The matrix of the Hamiltonian in the given basis is

.

(b) Denote the eigenvalues by αE. To find α we solve

.

-α^{3}+ 2α = 0, α = 0, ±√2.

(1) eigenvalue 0: 0a_{1}+ ib_{1}- ic_{1}= 0, -ia_{1}= 0.

corresponding normalized eigenvector: (1/√2)(|b> + |c>).

(2) eigenvalue √2E: -√2a_{2}+ ib_{2}- ic_{2}= 0, -ia_{2}- √2b_{2}= 0.

corresponding eigenvector: (1/√2)|a> - (i/2)|b> + (i/2)|c>.

(2) eigenvalue -√2E: √2a_{3}+ ib_{3}- ic_{3}= 0, -ia_{3}+ √2b_{3}= 0.

corresponding eigenvector: (1/√2)|a> + (i/2)|b> - (i/2)|c>.

__Properties and consequences__

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.

Consider a quantum system for which the exact
Hamiltonian is H. Assume the quantum system is of bounded spatial
extend, so that it is known rigorously that the eigenstates of H, {|Ψ_{n}>},
are complete.

(a) Show that if |Ψ_{n}>
and |Ψ_{m}> are two eigenstates of H with eigenvalues E_{n} and E_{m}

with E_{n }≠ E_{m}, then <Ψ_{n}|Ψ_{m}> = 0.

(b) Suppose E_{n }= E_{m},
with n ≠ m. Can we still have <Ψ_{n}|Ψ_{m}> = 0
?

(c) The problem H|Ψ> = E|Ψ>
is very complicated but it is suggested that we use a trial function |Ψ_{trial}>
for |Ψ> and approximate E by E = <Ψ_{trial}|H|Ψ_{trial}>/<Ψ_{trial}|Ψ_{trial}>.

Show that E > E_{0}, where E_{0} is the lowest
eigenvalue of H.

Solution:

- Concepts:

Hermitian operators - Reasoning:

H is a Hermitian operator. The eigenvalues of a Hermitian operator are real. Every Hermitian operator has at least one basis of orthonormal eigenvectors. - Details of the calculation:

(a) H = H^{T}, H is a Hermitian operator

<Ψ_{n}|H|Ψ_{m}> = E_{m}<Ψ_{n}|Ψ_{m}> = <Ψ_{n}|H^{T}|Ψ_{m}> = E_{n}<Ψ_{n}|Ψ_{m}>.

E_{m}<Ψ_{n}|Ψ_{m}> - E_{n}<Ψ_{n}|Ψ_{m}> = 0. Since E_{n }≠ E_{m}, <Ψ_{n}|Ψ_{m}> must be zero.

(b) Yes, an orthonormal basis for the state space can be formed by the eigenfunctions of any Hermitian operator.

(c) Any wave function can be expanded in terms of the eigenfunctions of H.

|Ψ_{trial}> = ∑_{n}a_{n}|Ψ_{n}>, {|Ψ_{n}>} = orthonormal basis for the state space.

H|Ψ_{trial}> = ∑_{n}a_{n}E_{n}|Ψ_{n}>.

<Ψ_{trial}|H|Ψ_{trial}> = ∑_{n}a_{n}E_{n}<Ψ_{trial}|Ψ_{n}> = ∑_{m}∑_{n}a_{m}*a_{n}E_{n}<Ψ_{m}|Ψ_{n}>

= ∑_{m}∑_{n}a_{m}*a_{n}E_{n}δ_{mn}<Ψ_{n}|Ψ_{n}> = ∑_{n}|a_{n}|^{2}E_{n}<Ψ_{n}|Ψ_{n}> ≥ ∑_{n}|a_{n}|^{2}E_{0}<Ψ_{n}|Ψ_{n}>

since E_{0}is the lowest eigenvalue.

<Ψ_{trial}|H|Ψ_{trial}> ≥ E_{0}∑_{n}|a_{n}|^{2}<Ψ_{n}|Ψ_{n}> = E_{0}<Ψ_{trial}|Ψ_{trial}>

Consider two solutions to the one-dimensional
time-independent Schrödinger equation with same energy E:
ψ_{1}(x) and
ψ_{2}(x).

(a) Prove that regardless of the potential energy function
U(x),

ψ_{2}(x) ∂ψ_{1}(x)/∂x
- ψ_{1}(x) ∂ψ_{2}(x)/∂x
= C,

where C is a constant.

(b) By considering the boundary conditions, show that if
ψ_{1}(x) and
ψ_{2}(x) are bound state
solutions, then C = 0. From this, show that ψ_{2}(x) = γ
ψ_{1}(x)
for some constant γ, thus proving that there are no degenerate bound state
solutions to the one-dimensional time-independent Schrödinger equation.

(c) Why does this theorem
fail for continuum eigenstates (i.e. unbound states)? Give a specific example
of degenerate continuum eigenstates in one dimension.

Solution:

- Concepts:

The time-independent Schroedinger equation in one dimension - Reasoning:

We are supposed to show that in 1 dimension bound state solutions of the Schroedinger equation are non-degenerate. - Details of the calculation:

(a) Given:

∂^{2}ψ_{1}(x)/∂x^{2}- (2m/ħ^{2})(E – U(x))ψ_{1}(x) = 0.

∂^{2}ψ_{2}(x)/∂x^{2}- (2m/ħ^{2})(E – U(x))ψ_{2}(x) = 0.

Multiplying the first equation by ψ_{2}(x) and the second equation by ψ_{1}(x) and subtracting we obtain ψ_{2}(x) ∂^{2}ψ_{1}(x)/∂x^{2}- ψ_{1}(x)∂^{2}ψ_{2}(x)/∂x^{2}= 0.

∂/∂x(ψ_{2}(x) ∂ψ_{1}(x)/∂x - ψ_{1}(x) ∂ψ_{2}(x)/∂x) = ψ_{2}(x) ∂^{2}ψ_{1}(x)/∂x^{2}- ψ_{1}(x)∂^{2}ψ_{2}(x)/∂x^{2}= 0.

Therefore ψ_{2}(x) ∂ψ_{1}(x)/∂x - ψ_{1}(x) ∂ψ_{2}(x)/∂x = C. (Integration)

(b) For bound states the wave function must vanish as x --> ± ∞. Therefore C = 0.

ψ_{2}(x) ∂ψ_{1}(x)/∂x - ψ_{1}(x) ∂ψ_{2}(x)/∂x = 0.

(1/ψ_{1}(x)) ∂ψ_{1}(x)/∂x = (1/ψ_{2}(x)) ∂ψ_{2}(x)/∂x.

∂/∂x(lnψ_{1}(x)) = ∂/∂x(lnψ_{2}(x)).

This integrates to lnψ_{1}(x) = lnψ_{2}(x) + c, ψ_{2}(x) = γ ψ_{1}(x), where γ = e^{c}.

ψ_{2}(x) can differ from ψ_{1}(x) only by a multiplicative constant.

(c) Continuum eigenstates do not necessarily vanish as x --> ± ∞, so C is not necessarily zero.

Example: Let U(x) = 0, we have a free particle.

ψ_{1}(x) = exp(i(2mE/ħ^{2})^{½}x) and ψ_{2}(x) = exp(-i(2mE/ħ^{2})^{½}x) are two degenerate continuum eigenstates.

A particle of mass m moves in one dimension. It is remarked that the exact eigenfunction for the ground state is Ψ(x) = A/cosh(λx), where λ is a constant, and A is the normalization constant. Assuming that the potential U(x) vanishes at infinity, derive the ground state energy and also U(x).

Solution:

- Concepts:

The Schroedinger equation - Reasoning:

The given wave function is a solution of the 1-dimensional, time-independent Schroedinger equation. - Details of the calculation:

(-ħ^{2}/2m)(∂^{2}/∂x^{2})Ψ(x) + U(x)Ψ(x) = EΨ(x).

(∂/∂x)(1/cosh(λx)) = -(sinh(λx)/cosh^{2}(λx))λ.

(∂^{2}/∂x^{2})(1/cosh(λx)) = -(λ^{2}/cosh(λx))[1 - 2sinh^{2}(λx)/cosh^{2}(λx)]

(ħ^{2}/2m)Aλ^{2}/cosh(λx))[1 - 2sinh^{2}(λx)/cosh^{2}(λx)] + U(x)A/cosh(λx) = EA/cosh(λx).

(ħ^{2}λ^{2}/2m)[1 - 2sinh^{2}(λx)/cosh^{2}(λx)] = E - U(x).

As x --> ∞, U(x) --> 0, and sinh(λx)/cosh(λx) --> 1. Therefore E = -(ħ^{2}λ^{2}/2m).

U(x) = -(ħ^{2}λ^{2}/m) + (ħ^{2}λ^{2}/m) sinh^{2}(λx)/cosh^{2}(λx)] = -(ħ^{2}λ^{2}/m)(1/cosh^{2}(λx)])