### Harmonic potentials, eigenvalues and eigenfunctions

#### Problem:

Find the average kinetic energy and the average potential energy of a particle in the ground state of a simple harmonic oscillator with frequency ω0.

Solution:

• Concepts:
Virial theorem
• Reasoning:
<T> = <U> for the harmonic oscillator.
• Details of the calculation:
<E> = <T> + <U> = ½ħω0 for the ground state.
Therefore <T> = <U> = ¼ħω0.

#### Problem:

Consider the one-dimensional problem in which a particle of mass m and charge -q is placed in a harmonic oscillator potential U(x) = ½mω2x2 in the presence of an electric field E = E0.
(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 + qE0/(mω2) we can make H look like the Hamiltonian of a simple harmonic oscillator.
• Details of the calculation:
(a)  H = ½P2/m + ½mω2X2 + qE0X.
(b)  H|Φ> = E|Φ>  ∂2Φ(x)/∂x2 + (2m(E - ½mω2x2 - qE0x)/ħ2)Φ(x) = 0.
Let us try to complete the square.
2Φ(x)/∂x2 + (2m(E - ½mω2(x + qE0/(mω2))2 - q2E02/(2mω2))/ħ2)Φ(x) = 0.
Let x' = x + qE0/(mω2),  E' = E - q2E02/(2mω2).  Then
2Φ(x')/∂x'2 + (2m(E' - ½mω2x'2)/ħ2)Φ(x') = 0.
This is the equation for a harmonic oscillator in the absence of an electric field.
En' = (n + ½)ħω,  En = (n + ½)ħω - q2E02/(2mω2).
(c)  <x'> = 0  for all eigenstates,  <x> = -qE0/(mω2) for all eigenstates.

#### Problem:

(a)   Give and sketch the probability distribution for the second lowest energy solution of the simple quantum mechanical harmonic oscillator,
-(ħ2/2m)(d2/dx2)Φ(x) + ½kx2Φ(x) = EΦ(x),
including the classical oscillator limits for the amplitude of oscillation.
(b)   Assume ψ(x,t = 0) = C0Φ0(x) + C1Φ1(x), and show that at later time
ψ(x,t) = C0Φ0(x)exp(-iE0t/ħ) + C1Φ1(x)exp(-iE1t/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 = mω2)
Φ0(x) = (mω/(πħ))¼exp(-½mωx2/ħ),
Φ1(x) = ((4/π)(mω/ħ)3)¼ x exp(-½mωx2/ħ),
Φ2(x) = (mω/(4πħ))¼ [2mωx2/ħ - 1] exp(-½mωx2/ħ).
The probability densities are given by |Φi(x)|2.
Below is a sketch of these functions with mω/h = 1.

The functions Φ1(x), Φ2(x) and Φ3(x)

The functions |Φ1(x)|2, |Φ2(x)|2 and |Φ3(x)|2
En = (n+½)ħω.
For Φn(x) the classical oscillator limits for the amplitude of oscillation is found from
½kA2 = (n + ½)ħω,  ½mω2A2 = (n+½)ħω,  A2 =  (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,t0)|ψ(t0)>,  U(t,t0) = exp(-(i/ħ)H(t-t0)).
ψ(x,t) = exp(-(i/ħ)H(t))[C0Φ0(x) + C1Φ1(x)] =
C0Φ0(x)exp(-iE0t/ħ) + C1Φ1(x)exp(-iE1t/ħ).
ψ(x,t) = exp(-mωx2/(2ħ))exp(-iωt)[C'0 exp(iωt/2) + C1'x exp(-iωt/2)].
|ψ(x,t)|2 = exp(-mωx2/ħ) [|C'0|2 + |C1'|2x2 + 2Re(C'0C1'x)cos(ωt)].
Choose mω/ħ = 1, C'1 = C'2 = 1.
Then |ψ(x,t)|2 = exp(-x2) [1 + x2 + 2xcos(ωt)].

ωt = 0

ωt = π/2

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

#### Problem:

A non-relativistic quantum mechanical particle of mass m is in a 1-dimensional potential,
U(x) = ax2 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) = ½mω2x2 are
Φn(x) = (n! 2n)(β/√π)½Hn(η) exp(-½η2),  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
En = (n + ½ )ħω = (n + ½ )ħ(2a/m)½, n = odd.
(b)  The ground state wave function is Φ1(x) = N1 x exp(-(2ma)½x2/(2ħ)), x > 0,
with N1 = (2/π¼)((2ma)½/ħ)¾
The most probable position is found from d|Φ1(x)|2/dx = 0.
x - (2ma)½x3/ħ = 0, x = ħ½/ (2ma)¼.
(c)  The the expectation value of x is
<x> = ∫0 x |Φ1( x )|2d x = N122/(2ma))∫0 y3 exp(-y 2) dy
= (4/√π)[(2ma)¾3/2](½ħ2/(2ma)) = 1.13√ħ/(2ma)¼.
<x> = 1.13 ħ½/(2ma)¼.

#### Problem:

Write down the energy eigenvalues for the one-dimensional Schroedinger equation with
U(x) = ½ m ω2x2 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 ω2x2 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, ...