Eigenvalues and eigenfunctions

Find eigenvalues and eigenfunctions

Problem:

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

.
3 + 2α = 0,  α = 0, ±√2.
(1)  eigenvalue 0:  0a1 + ib1 - ic1 = 0, -ia1 = 0.
corresponding normalized eigenvector:  (1/√2)(|b> + |c>).
(2)  eigenvalue √2E:  -√2a2 + ib2 - ic2 = 0, -ia2 - √2b2 = 0.
corresponding eigenvector:  (1/√2)|a> - (i/2)|b> + (i/2)|c>.
(2)  eigenvalue -√2E:  √2a3 + ib3 - ic3 = 0, -ia3 + √2b3 = 0.
corresponding eigenvector:  (1/√2)|a> + (i/2)|b> - (i/2)|c>.

Properties and consequences

Problem:

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> = ai|i>,  A|j> = aj|j>.
A(|i> + |j>) = ai|i> + aj|j> ≠ aij(|i> + |j>) unless ai = aj, i.e.  unless |i> and |j> have the same eigenvalue.

Problem:

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, {|Ψn>}, are complete.
(a) Show that if |Ψn> and |Ψm> are two eigenstates of H with eigenvalues En and Em
with En ≠ Em, then <Ψnm> = 0.
(b) Suppose En = Em, with n ≠ m.  Can we still have <Ψnm> = 0 ?
(c) The problem H|Ψ> = E|Ψ> is very complicated but it is suggested that we use a trial function |Ψtrial> for |Ψ> and approximate E by E = <Ψtrial|H|Ψtrial>/<Ψtrialtrial>.
Show that E > E0, where E0 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 = HT, H is a Hermitian operator
n|H|Ψm> = Emnm> =  <Ψn|HTm> = Ennm>.
Emnm> - Ennm> = 0.  Since En ≠ Em, <Ψnm> 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.
trial> = ∑nann>,  {|Ψn>} = orthonormal basis for the state space.
H|Ψtrial> = ∑nanEnn>.
trial|H|Ψtrial> = ∑nanEntrialn> = ∑mnam*anEnmn>
= ∑mnam*anEnδmnnn> = ∑n|an|2Ennn> ≥ ∑n|an|2E0nn>
since E0 is the lowest eigenvalue.
trial|H|Ψtrial> ≥ E0n|an|2nn> = E0trialtrial>

Problem:

Consider two solutions to the one-dimensional time-independent Schrödinger equation with same energy E: ψ1(x) and ψ2(x).
(a)  Prove that regardless of the potential energy function U(x),

ψ2(x) ∂ψ1(x)/∂x - ψ1(x) ∂ψ2(x)/∂x = C,

where C is a constant.
(b)  By considering the boundary conditions, show that if ψ1(x) and ψ2(x) are bound state solutions, then C = 0.  From this, show that ψ2(x)  = γ ψ1(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:
2ψ1(x)/∂x2 - (2m/ħ2)(E – U(x))ψ1(x) = 0.
2ψ2(x)/∂x2 - (2m/ħ2)(E – U(x))ψ2(x) = 0.
Multiplying the first equation by ψ2(x) and the second equation by ψ1(x) and subtracting we obtain ψ2(x) ∂2ψ1(x)/∂x2 - ψ1(x)∂2ψ2(x)/∂x2 = 0.
∂/∂x(ψ2(x) ∂ψ1(x)/∂x - ψ1(x) ∂ψ2(x)/∂x) = ψ2(x) ∂2ψ1(x)/∂x2 - ψ1(x)∂2ψ2(x)/∂x2 = 0.
Therefore ψ2(x) ∂ψ1(x)/∂x - ψ1(x) ∂ψ2(x)/∂x = C. (Integration)
(b)  For bound states the wave function must vanish as x --> ± ∞.  Therefore C = 0.
ψ2(x) ∂ψ1(x)/∂x - ψ1(x) ∂ψ2(x)/∂x = 0.
(1/ψ1(x)) ∂ψ1(x)/∂x = (1/ψ2(x)) ∂ψ2(x)/∂x.
∂/∂x(lnψ1(x)) = ∂/∂x(lnψ2(x)).
This integrates to lnψ1(x) = lnψ2(x) + c,  ψ2(x)  = γ ψ1(x), where γ = ec.
ψ2(x) can differ from ψ1(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.
ψ1(x) = exp(i(2mE/ħ2)½x) and ψ2(x) = exp(-i(2mE/ħ2)½x) are two degenerate continuum eigenstates.

Problem:

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:
(-ħ2/2m)(∂2/∂x2)Ψ(x) + U(x)Ψ(x) = EΨ(x).
(∂/∂x)(1/cosh(λx)) = -(sinh(λx)/cosh2(λx))λ.
(∂2/∂x2)(1/cosh(λx)) = -(λ2/cosh(λx))[1 - 2sinh2(λx)/cosh2(λx)]
2/2m)Aλ2/cosh(λx))[1 - 2sinh2(λx)/cosh2(λx)] + U(x)A/cosh(λx) = EA/cosh(λx).
2λ2/2m)[1 - 2sinh2(λx)/cosh2(λx)] = E - U(x).
As x --> ∞,  U(x) --> 0, and sinh(λx)/cosh(λx) --> 1.  Therefore E = -(ħ2λ2/2m).
U(x) = -(ħ2λ2/m) + (ħ2λ2/m) sinh2(λx)/cosh2(λx)] = -(ħ2λ2/m)(1/cosh2(λx)])