A particle of mass m is bound in a 2-dimensional isotropic oscillator
potential with a spring constant k.

(a) Write the Schroedinger equation for this system in
both Cartesian and polar coordinates.

(b) Separate the equation in polar
coordinates and solve the resulting equation in θ.
Demonstrate the connection between the θ solution and its classical analog.

(c) Write the resulting radial equation utilizing the θ
solution, but do not solve it.

Solution:

- Concepts:

The Schroedinger equation in polar coordinates, separation of variables - Reasoning:

We are asked to write the Schroedinger equation Hψ = Eψ for the system in polar coordinates and separate variables. -
Details of the calculation:

(a) The Schroedinger equation in Cartesian coordinates is

-[ħ^{2}/(2m)][∂^{2}/∂x^{2}+ ∂^{2}/∂y^{2}]ψ + ½k(x^{2}+ y^{2})ψ = Eψ.The Schroedinger equation in polar coordinates is

-[ħ^{2}/(2m)][∂^{2}/∂r^{2}+ (1/r) (∂/∂r) + (1/r^{2})(∂^{2}/∂θ^{2})]ψ + ½kr^{2}ψ = Eψ.(b) Let ψ = f(r)χ(θ). Then

-[ħ^{2}/(2m)][χ(θ)∂^{2}f(r)/∂r^{2}+ (1/r)χ(θ)∂f(r)/∂r + (1/r^{2})f(r)∂^{2}χ(θ)/∂θ^{2}] + ½kr^{2}f(r)χ(θ) = Ef(r)χ(θ).

Multiply by (-2mr^{2}/ħ^{2})/[f(r)χ(θ)].(r

^{2}/f(r))∂^{2}f(r)/∂r^{2}+ (r/f(r))∂f(r)/∂r + (2mr^{2}/ħ^{2})(E - ½kr^{2})

+ (1/χ(θ))∂^{2}χ(θ)/∂θ^{2}= 0.

The first term is a function of r only and the second term is a function of θ only. Both terms must be equal to a constant and the sum of these constants must be zero.

(1/χ(θ))∂^{2}χ(θ)/∂θ^{2}= -C. ∂^{2}χ(θ)/∂θ^{2}= -Cχ(θ). χ(θ) = exp(inθ), C = n^{2}.

n = integer since χ(θ) = χ(θ + 2π).

The θ solutions are eigenfunctions of L_{z}with eigenvalues nħ.

L_{z}= (ħ/i)∂/∂θ commutes with H, L_{z}is a constant of motion, quantum mechanically as well as classically.

(c) The resulting radial equation is

(r^{2}/f(r)))∂^{2}f(r)/∂r^{2}+ (r/f(r))∂f(r)/∂r + (2mr^{2}/ħ^{2})(E - ½kr^{2}) - n^{2}= 0.

∂^{2}f(r)/∂r^{2}+ (1/r)∂f(r)/∂r + (2m/ħ^{2})(E - ½kr^{2})f(r) - (n^{2}/r^{2})f(r) = 0.

(a) Write down the Hamiltonian of the three-dimensional isotropic harmonic
oscillator in spherical coordinates, and in Cartesian coordinates.

(b) Write down the ground-state wave function of the three-dimensional
isotropic harmonic oscillator in spherical coordinates, and in Cartesian
coordinates. What is its energy eigenvalue?

Solution:** **

- Concepts:

The 3-d isotropic harmonic oscillator - Reasoning:

We are asked to write down the Hamiltonian of the three-dimensional isotropic harmonic oscillator in spherical coordinates, and in Cartesian coordinates and to write down the ground-state wave function of the three-dimensional isotropic harmonic oscillator in spherical coordinates, and in Cartesian coordinates. - Details of the calculation:

(a) H = p^{2}/(2m) + mω^{2}r^{2}/2 is the Hamiltonian of the 3-d harmonic oscillator.

H = -(ħ^{2}/(2m)(∂^{2}/∂x^{2 }+ ∂^{2}/∂y^{2 }+ ∂^{2}/∂z^{2}) + mω^{2}(x^{2 }+ y^{2 }+ z^{2})/2) is the Hamiltonian in Cartesian coordinates.

H = -(ħ^{2}/(2m)((1/r)(∂^{2}/∂r^{2})r + L^{2}/(2mr^{2})) + mω^{2}r^{2}/2) withL

^{2}= -ħ^{2}[∂^{2}/∂θ^{2}+ (1/tanθ)∂/∂θ + (1/sin^{2}θ))∂^{2}/∂φ^{2}]

= -ħ^{2}[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin^{2}θ))∂^{2}/∂φ^{2}]is the Hamiltonian in spherical coordinates.

(b) The ground-state wave function of the three-dimensional isotropic harmonic oscillator in Cartesian coordinates is

ψ(**r**) = (mω/(πħ))^{¾}exp(-mω(x^{2}+ y^{2}+ z^{2})/(2ħ)).

The ground-state wave function of the three-dimensional isotropic harmonic oscillator in spherical coordinates must be

ψ(**r**) = (mω/(πħ))^{¾}exp(-mωr^{2}/(2ħ)).

Show:

The ground state has l = 0.

ψ_{0}(r,θ,φ) = [u_{0}(r)/r]Y_{00}(θ,φ) is the ground-state wave function.

[-(ħ^{2}/(2m))(∂^{2}/∂r^{2}) + ½kr^{2}]u_{0}(r) = Eu_{0}(r) is the differential equation satisfied by u_{0}(r),

with the boundary conditions that u_{0}(0) = u_{0}(∞) = 0, and the condition that u_{0}(r) has no nodes in the region 0 < r < ∞.

The wave function corresponding to the first excited state of the 1D harmonic oscillator is a solution which satisfies these conditions.

u_{0}(r) is proportional to r exp(-½mωr^{2}/ħ),

ψ(**r**) = (mω/(πħ))^{¾}exp(-mωr^{2}/(2ħ)) is the normalized ground state wave function.

The energy eigenvalue for the ground state is E_{0}= (3/2)ħω.

Consider a diatomic molecule
consisting of two identical atoms with mass m interacting via a potential, U(**R**),
where **R** is the separation between the atoms.

Near its minimum at **R**_{0},
the potential can be approximated by

U(**R**) =
½k(**R** − **R**_{0})^{2}.

Write down the energies of the
low lying states of the diatomic molecule, assuming that these are rotational
and/or vibrational excitations. Define the moment of inertia of the diatomic
molecule about the symmetry axis to be I_{z}, and give your answer in
terms of m, I_{z}, and the parameters appearing in the potential. Make
sure you specify allowed values of all quantum numbers appearing in your answer.

Solution:

- Concepts:

The time-independent Schroedinger equation for two interacting particles - Reasoning:

Let**r**=**R**–**R**_{0}, r^{2}= (**R**–**R**_{0})^{2}, U(**R**) = ½kr^{2}.

If the mutual interaction depends only on the distance r between the particles, then the eigenvalue equation for the relative motion becomes

-(ħ^{2}/(2μ))∇^{2}_{r}Φ(**r**) + U(r)Φ(**r**) = E_{r}f(**r**),

with μ being the reduced mass, μ = m_{1}m_{2}/(m_{1}+ m_{2}) = m/2.

Here -(ħ^{2}/(2μ))∇^{2}_{r}Φ(**r**) + ½kr^{2}Φ(**r**) = E_{r}Φ(**r**). - Details of the calculation:

Let us first consider only**small amplitude vibrations**.

If the angular momentum of the nuclei is zero, then the eigenvalue equation reduces to the radial equation with l = 0,

[-(ħ^{2}/(2μ))(∂^{2}/∂r^{2}) + ½kr^{2}]u(r) = Eu(r).

This is the eigenvalue equation of a one-dimensional harmonic oscillator Hamiltonian. The eigenvalues are

E_{n}= (n + ½)ħω, n = 0, 1, 2, ..., ω^{2}= k/m,

These are the eigenenergies of the vibrational motion of the two nuclei .

[Typical vibrational frequencies are on the order of 10^{12}to 10^{14}Hz and fall in the infrared.]Let us now study the

**rotational motion**of molecules. We neglect vibrations and assume that the distance r between the nuclei remains constant. We then have a**rigid rotator**. We have

H = L^{2}/(2μr^{2}) = L^{2}/(2I_{z}), I_{z}= m_{1}r_{1}^{2}+ m_{2}r_{2}^{2}= μr^{2}.

The eigenfunctions of H are the eigenfunctions of L^{2}.

These are the spherical harmonics, Y_{lm}(θ,φ). The eigenvalues are E_{l }= l(l+1)ħ^{2}/2I_{z}.

[Remarks:

It is customary to set B = ħ/(4πI_{z}), and write E_{l }= Bhl(l+1). The separation between adjacent levels is

E_{l}- E_{l-1}= Bh(l(l+1) - (l-1)l] = 2Bhl. It increases linear with l. Each energy eigenvalue E_{l }is (2l+1) fold degenerate.

Pure rotational spectra fall in the very far infrared or the microwave region.]

For the vibrational and rotational motion we have for small amplitude oscillations (so that we can neglect the change in the moment of inertia)

E_{nl}= (n + ½)ħω + l(l+1)ħ^{2}/2I_{z}.

A particle of charge -e and mass m is under the influence of two
stationary heavy nuclei, each with charge Ze positioned at z = ±a. We shall further assume that the
particle is spinless and non relativistic.

(a) What is the Hamiltonian and the Schroedinger equation of the system.

(b) Define the angular momentum operator along the z - direction, L_{z}, and show that its eigenvalues are good quantum
numbers for
all of the non-degenerate energy eigenstates. What are the possible eigenvalues
of L_{z}?

(c) Define the parity operator P and show that parity is a good
quantum number for all the non-degenerate energy eigenstates. What are the
possible eigenvalues of P?

(d) Define the total angular momentum operator **L** and show that
the eigenvalues of L^{2} are not good quantum numbers for the
energy eigenstates.

Solution:

- Concepts:

The Schroedinger equation in three dimensions, the orbital angular momentum operator, the parity operator - Reasoning:

We are asked to write down the Schroedinger equation for the system and to define the orbital angular momentum operator and the parity operator. We check if the eigenvalues of an operator are good quantum numbers by checking if the operator commutes with H. -
Details of the calculation:

(a) The Hamiltonian of the electron is

H = p^{2}/(2m) + U(**r**), U(**r**) = -Ze^{2}/|**r**- a**k**| - Ze^{2}/|**r**+ a**k**|.

The Schroedinger equation is

(iħ∂/∂t)|Ψ> = H(t)|Ψ>, (iħ∂/∂t)|Ψ> = [-ħ^{2}/(2m))∇^{2}+ U(**r**)]|Ψ>,

In spherical coordinates we have

H = -(ħ^{2}/(2m))(1/r)(∂^{2}/∂r^{2})r + L^{2}/(2mr^{2}) + U(r),

L^{2}= -ħ^{2}[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin^{2}θ))∂^{2}/∂φ^{2}], L_{z}= (ħ/i)∂/∂φ.

U(**r**) = -Ze^{2}/(r^{2}+ a^{2}- 2ra cosθ)^{½}- Ze^{2}/(r^{2}+ a^{2}+ 2ra cosθ)^{½}.

(b) We know [ L^{2},L_{z}] = 0, therefore [ H,L_{z}] = 0, since the L^{2}term is the only one with a φ dependence. If two operators commute, then a common eigenbasis can be found. An eigenstate with a non-degenerate eigenvalue of H is also an eigenstate of L_{z}. The possible eigenvalues of L_{z}are nħ where n is an integer.

(c) We define the parity operator through its action on any ψ(**r**).

Pψ(**r**) = ψ(-**r**).**r**= (r,θ,φ), -**r**= (r,π - θ,π + φ).

[P,L^{2}] = 0, since changing θ to π - θ and φ to π + φ does not change L^{2}.

[P,U(**r**)] = 0. since changing θ to π - θ does not change U(**r**).

Therefore [P,H] = 0. As above, an eigenstate with a non-degenerate eigenvalue of H is also an eigenstate of P. The eigenvalues of P are ±1.(d) In spherical coordinates

L^{2}= -ħ^{2}[∂^{2}/∂θ^{2}+ (1/tanθ)∂/∂θ + (1/sin^{2}θ))∂^{2}/∂φ^{2}]

= -ħ^{2}[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin^{2}θ))∂^{2}/∂φ^{2}].

L^{2}U(**r**)ψ(**r**) ≠ LU(**r**)^{2}ψ(**r**), [L^{2},H] ≠ 0, since [L^{2},U(**r**)] ≠ 0.

No common eigenbasis of L^{2}and H exists.

Let ρ, φ, z be the
cylindrical coordinates of a spinless particle. Assume the potential
energy of this particle depends only on ρ and not on φ
and z.

(a) Write in cylindrical coordinates the differential operator associated
with the Hamiltonian. Show that H commutes with L_{z} and P_{z}.
Show from this that the wave function associated with the stationary states of
the particle can be written as

ψ_{nmk}(ρ, φ,
z) = f_{nm}(ρ) exp(imφ) exp(ikz),

where the values that can be taken on by the indices m and k are to be
specified.

(b) Write in cylindrical coordinates the eigenvalue equation of the
Hamiltonian of the particle. Derive from it the differential equation
which yields f_{nm}(ρ).

Solution:

- Concepts:

The Hamiltonian and Schroedinger equation in polar coordinates, the L_{z}and P_{z}operators, commuting variables - Reasoning:

We are asked to write the Hamiltonian H in cylindrical coordinates and show that H commutes with L_{z}and P_{z}. - Details of the calculation:

(a) H = -[ħ^{2}/(2μ)]∇^{2}+ U(ρ). Here μ is the mass of the particle.

In cylindrical coordinates

H = -[ħ^{2}/(2μ)][∂^{2}/∂ρ^{2}+ (1/ρ) (∂/∂ρ) + (1/ρ^{2})(∂^{2}/∂φ^{2}) + ∂^{2}/∂z^{2}] + U(ρ).

Using

L_{z}= [ħ/i]∂/∂φ, P_{z}= [ħ/i]∂/∂z, L_{z}^{2}= -ħ^{2}∂^{2}/∂φ^{2}, P_{z}^{2}= -ħ^{2}∂^{2}/∂z^{2},

we can write

H = P_{ρ}^{2}/(2μ) + L_{z}^{2}/(2μρ^{2}) + P_{z}^{2}/(2μ) + U(ρ).

Here P_{ρ}^{2 }= -ħ^{2}[∂^{2}/∂ρ^{2}+ (1/ρ) (∂/∂ρ)].

L_{z}depends only on φ, P_{z}depends only on z, and P_{ρ}^{2}depends only on ρ.

Therefore [L_{z},P_{ρ}^{2}] = [L_{z},P_{z}], = [P_{ρ}^{2},P_{z}] = 0 and [H,L_{z}] = [H,P_{z}] = 0.

We can find a common eigenbasis of H, L_{z}and P_{z}.

This means separation of variables in cylindrical coordinates is possible.

{exp(imφ)} = eigenbasis of L_{z}, m = 0, ±1, ±2, ... .

{exp(ikz)} = eigenbasis of P_{z}, with k = -∞ to +∞ a continuous index.

Any function ψ(ρ, φ, z) can be written as a linear combination of functions f(ρ) exp(imφ) exp(ikz),

ψ(ρ, φ, z) = ∑_{m}exp(imφ)∫dkf_{km}(ρ)exp(ikz), since k is a continuous index.

The functions f_{km}(ρ) exp(imφ) exp(ikz) is an eigenfunction of H if

H f_{km}(ρ) exp(imφ) exp(ikz) = E f_{km}(ρ) exp(imφ) exp(ikz),

[P_{ρ}^{2}/(2μ) + L_{z}^{2}/(2μρ^{2}) + P_{z}^{2}/(2μ) + U(ρ)]f_{km}(ρ) exp(imφ) exp(ikz)

= E f_{km}(ρ) exp(imφ) exp(ikz).

The plane wave exp(ikz) represents a stream of particles moving with uniform speed along the z-axis. The energy associated with each particle is ħ^{2}k^{2}/(2μ).

Define E' = E - ħ^{2}k^{2}/(2μ). E' is the energy associated with the motion perpendicular to the z-axis.

[P_{ρ}^{2}/(2μ) + ħ^{2}m^{2}/(2μρ^{2}) + U(ρ)]f(ρ) = E' f(ρ).

The eigenfunction f_{km}(ρ) does not depend on k, only on m. We label different possible solution for the same m with the index n.

[P_{ρ}^{2}/(2μ) + ħ^{2}m^{2}/(2μρ^{2}) + U(ρ)]f_{nm}(ρ) = E'_{nm}f_{nm}(ρ).

(b) [-(ħ^{2}/(2μ))[∂^{2}/∂ρ^{2}+ (1/ρ) (∂/∂ρ)] + ħ^{2}m^{2}/(2μρ^{2}) + U(ρ)]f_{nm}(ρ) = E'_{nm}_{nm}(ρ).

[[∂^{2}/∂ρ^{2}+ (1/ρ) (∂/∂ρ)] - m^{2}/ρ^{2}- V(ρ)]f_{nm}(ρ) = λ_{nm}f_{nm}(ρ).

Here E'_{nm}= ħ^{2}λ_{nm}^{2}/(2μ) and U(ρ) = ħ^{2}V(ρ)/(2μ).