Find the average kinetic energy and the average potential energy of a
particle in the ground state of a simple harmonic oscillator with frequency ω_{0}.

Solution:

- Concepts:

Virial theorem - Reasoning:

<T> = <U> for the harmonic oscillator. - Details of the calculation:

<E> = <T> + <U> = ½ħω_{0}for the ground state.

Therefore <T> = <U> = ¼ħω_{0}.

Consider
the one-dimensional problem in which a particle of mass m and charge -q is placed
in a harmonic oscillator potential U(x) = ½mω^{2}x^{2} in the presence of an electric field E = E_{0}.

(a) Write down the Hamiltonian H.

(b) Find the eigenvalues of H.

(c) Find <x> for all eigenstates
of H.

Solution:

- Concepts:

The eigenfunctions of the harmonic oscillator - Reasoning:

By changing the variable x to x' = x + qE_{0}/(mω^{2}) we can make H look like the Hamiltonian of a simple harmonic oscillator. - Details of the calculation:

(a) H = ½P^{2}/m + ½mω^{2}X^{2}+ qE_{0}X.

(b) H|Φ> = E|Φ> ∂^{2}Φ(x)/∂x^{2}+ (2m(E - ½mω^{2}x^{2}- qE_{0}x)/ħ^{2})Φ(x) = 0.

Let us try to complete the square.

∂^{2}Φ(x)/∂x^{2}+ (2m(E - ½mω^{2}(x + qE_{0}/(mω^{2}))^{2}- q^{2}E_{0}^{2}/(2mω^{2}))/ħ^{2})Φ(x) = 0.

Let x' = x + qE_{0}/(mω^{2}), E' = E - q^{2}E_{0}^{2}/(2mω^{2}). Then

∂^{2}Φ(x')/∂x^{'2}+ (2m(E' - ½mω^{2}x'^{2})/ħ^{2})Φ(x') = 0.

This is the equation for a harmonic oscillator in the absence of an electric field.

E_{n}' = (n + ½)ħω, E_{n}= (n + ½)ħω - q^{2}E_{0}^{2}/(2mω^{2}).

(c) <x’> = 0 for all eigenstates, <x> = -qE_{0}/(mω^{2}) for all eigenstates.

(a) Give and
sketch the probability distribution for the second lowest energy solution of the
simple quantum mechanical harmonic oscillator,

-(ħ^{2}/2m)(d^{2}/dx^{2})Φ(x)
+ ½kx^{2}Φ(x) = EΦ(x),

including the classical oscillator limits for the amplitude of oscillation.

(b) Assume
ψ(x,t = 0) = C_{0}Φ_{0}(x)
+ C_{1}Φ_{1}(x), and
show that at later time

ψ(x,t) = C_{0}Φ_{0}(x)exp(-iE_{0}t/ħ)
+ C_{1}Φ_{1}(x)exp(-iE_{1}t/ħ).

(c) Compute and sketch |ψ(x,t)|^{2}
for different times.

Solution:

- Concepts:

The eigenstates of the harmonic oscillator, evolution operator - Reasoning:

We are asked to sketch an eigenfunction of the harmonic oscillator and to find ψ(x,t) given ψ(x,0) - Details of the calculation:

(a) The three lowest eigenfunctions of the harmonic oscillator are (with k = mω^{2})

Φ_{0}(x) = (mω/(πħ))^{¼}exp(-½mωx^{2}/ħ),

Φ_{1}(x) = ((4/π)(mω/ħ)^{3})^{¼}x exp(-½mωx^{2}/ħ),

Φ_{2}(x) = (mω/(4πħ))^{¼}[2mωx^{2}/ħ - 1] exp(-½mωx^{2}/ħ).

The probability densities are given by |Φ_{i}(x)|^{2}.

Below is a sketch of these functions with mω/h = 1.

The functions Φ_{1}(x), Φ_{2}(x) and Φ_{3}(x)

The functions |Φ_{1}(x)|^{2}, |Φ_{2}(x)|^{2}and |Φ_{3}(x)|^{2}

E_{n}= (n+½)ħω.

For Φ_{n}(x) the classical oscillator limits for the amplitude of oscillation is found from

½kA^{2}= (n + ½)ħω, ½mω^{2}A^{2}= (n+½)ħω, A^{2}= (2n + 1)ħ/(mω), A = ((2n + 1)ħ/(mω))^{½}.

(At the classical oscillator limits for the amplitude of oscillation the kinetic energy is zero and the potential energy equals the total energy.)

(b) |ψ(t)> = U(t,t_{0})|ψ(t_{0})>, U(t,t_{0}) = exp(-(i/ħ)H(t-t_{0})).

ψ(x,t) = exp(-(i/ħ)H(t))[C_{0}Φ_{0}(x) + C_{1}Φ_{1}(x)] =

C_{0}Φ_{0}(x)exp(-iE_{0}t/ħ) + C_{1}Φ_{1}(x)exp(-iE_{1}t/ħ).

ψ(x,t) = exp(-mωx^{2}/(2ħ))exp(-iωt)[C'_{0 }exp(iωt/2) + C_{1}'x exp(-iωt/2)].

|ψ(x,t)|^{2}= exp(-mωx^{2}/ħ) [|C'_{0}|^{2}+ |C_{1}'|^{2}x^{2}+ 2Re(C'_{0}C_{1}'x)cos(ωt)].

Choose mω/ħ = 1, C'_{1}= C'_{2}= 1.

Then |ψ(x,t)|^{2}= exp(-x^{2}) [1 + x^{2}+ 2xcos(ωt)].

ωt = 0

ωt = π/2

ωt = π

<x>(t) oscillates with frequency ω like the classical coordinate x.

A non-relativistic quantum mechanical particle of mass m is
in a 1-dimensional potential,

U(x) = ax^{2} for x > 0 and U(x) = ∞ for x < 0.

(a) Find the possible energy eigenvalues and normalized eigenfunctions.

(b) The particle is in the ground state. At what position x is it most likely
to be found?

(c) What is the expectation value of the particle position?

Solution:

- Concepts:

The harmonic oscillator - Reasoning:

For x > 0, the given potential is identical to the harmonic oscillator potential.

Odd harmonic oscillator energy eigenfunctions are zero at x = 0 and, satisfy all boundary conditions for x > 0. They are eigenfuctions of H for the given potential for x > 0. For x < 0, the eigenfunctions of the given H are zero. - Details of the
calculation:

(a) The harmonic oscillator eigenfunctions with V(x) = ½mω^{2}x^{2}are

Φ_{n}(x) = (n! 2^{n})^{-½ }(β/√π)^{½}H_{n}(η)^{ }exp(-½η^{2}), where η = (mω/ħ)^{½}x = βx.If we replace ω by (2a/m)

^{½}and multiply the functions with n = odd by √2, we obtain the normalized eigenfunctions for the given H. The eigenvalues are

E_{n}= (n + ½ )ħω = (n + ½ )ħ(2a/m)^{½}, n = odd.

(b) The ground state wave function is Φ_{1}(x) = N_{1 }x exp(-(2ma)^{½}x^{2}/(2ħ)), x > 0,

with N_{1}= (2/π^{¼})((2ma)^{½}/ħ)^{¾}

The most probable position is found from d|Φ_{1}(x)|^{2}/dx = 0.

x – (2ma)^{½}x^{3}/ħ = 0, x = ħ^{½}/ (2ma)^{¼}.

(c) The the expectation value of x is

<x> = ∫_{0}^{∞}x |Φ_{1}( x )|^{2}d x = N_{1}^{2}(ħ^{2}/(2ma))∫_{0}^{∞}y^{3}exp(-y^{ 2}) dy

= (4/√π)[(2ma)^{¾}/ħ^{3/2}](½ħ^{2}/(2ma)) = 1.13√ħ/(2ma)^{¼}.

<x> = 1.13 ħ^{½}/(2ma)^{¼}.

Write down the energy eigenvalues for the one-dimensional Schroedinger
equation with

U(x) = ½ m ω^{2}x^{2} for x > 0 and U(x) = ∞ for x < 0.

Solution:

- Concepts:

The eigenfunctions of the 1D harmonic oscillator - Reasoning:

The odd eigenfunctions of the harmonic oscillator with U(x) = ½ m ω^{2}x^{2}for all x are eigenfunctions of H for this potential. - Details of the calculation:

The eigenvalues therefore are E = (n + ½)ħω, n = odd

or rewriting, E = (2n + 3/2)ħω, n = 0, 1, 2, …