### Other 3D systems

#### Problem:

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
-[ħ2/(2m)][∂2/∂x2 + ∂2/∂y2]ψ + ½k(x2 + y2)ψ = Eψ.

The Schroedinger equation in polar coordinates is
-[ħ2/(2m)][∂2/∂r2 + (1/r) (∂/∂r) + (1/r2)(∂2/∂θ2)]ψ + ½kr2ψ = Eψ.

(b)  Let ψ = f(r)χ(θ).  Then
-[ħ2/(2m)][χ(θ)∂2f(r)/∂r2 + (1/r)χ(θ)∂f(r)/∂r + (1/r2)f(r)∂2χ(θ)/∂θ2] + ½kr2f(r)χ(θ) = Ef(r)χ(θ).
Multiply by (-2mr22)/[f(r)χ(θ)].

(r2/f(r))∂2f(r)/∂r2 + (r/f(r))∂f(r)/∂r + (2mr22)(E - ½kr2)
+ (1/χ(θ))∂2χ(θ)/∂θ2 = 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))∂2χ(θ)/∂θ2 = -C.  ∂2χ(θ)/∂θ2 = -Cχ(θ).  χ(θ) = exp(inθ),  C = n2.
n = integer since χ(θ) = χ(θ + 2π).

The θ solutions are eigenfunctions of Lz with eigenvalues nħ.
Lz = (ħ/i)∂/∂θ commutes with H, Lz is a constant of motion, quantum mechanically as well as classically.
(c)  The resulting radial equation is
(r2/f(r)))∂2f(r)/∂r2 + (r/f(r))∂f(r)/∂r + (2mr22)(E - ½kr2) - n2 = 0.
2f(r)/∂r2 + (1/r)∂f(r)/∂r + (2m/ħ2)(E - ½kr2)f(r) - (n2/r2)f(r) = 0.

#### Problem:

(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 = p2/(2m) + mω2r2/2 is the Hamiltonian of the 3-d harmonic oscillator.
H = -(ħ2/(2m)(∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2) + mω2(x2 + y2 + z2)/2) is the Hamiltonian in Cartesian coordinates.
H = -(ħ2/(2m)((1/r)(∂2/∂r2)r + L2/(2mr2)) + mω2r2/2) with

L2 = -ħ2[∂2/∂θ2 + (1/tanθ)∂/∂θ + (1/sin2θ))∂2/∂φ2]
= -ħ2[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin2θ))∂2/∂φ2]

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(x2 + y2 + z2)/(2ħ)).

The ground-state wave function of the three-dimensional isotropic harmonic oscillator in spherical coordinates must be
ψ(r) = (mω/(πħ))¾ exp(-mωr2/(2ħ)).
Show:
The ground state has l = 0.
ψ0(r,θ,φ) = [u0(r)/r]Y00(θ,φ) is the ground-state wave function.
[-(ħ2/(2m))(∂2/∂r2) + ½kr2]u0(r) = Eu0(r) is the differential equation satisfied by u0(r),
with the boundary conditions that u0(0) = u0(∞) = 0, and the condition that u0(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.
u0(r) is proportional to r exp(-½mωr2/ħ),
ψ(r) = (mω/(πħ))¾ exp(-mωr2/(2ħ)) is the normalized ground state wave function.
The energy eigenvalue for the ground state is E0 = (3/2)ħω.

#### Problem:

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 R0, the potential can be approximated by
U(R) = ½k(RR0)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 Iz, and give your answer in terms of m, Iz, 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 - R0,  r2 = (R - R0)2,  U(R) = ½kr2 .
If the mutual interaction depends only on the distance r between the particles, then the eigenvalue equation for the relative motion becomes
-(ħ2/(2μ))∇2rΦ(r) + U(r)Φ(r) = Erf(r),
with μ being the reduced mass, μ = m1m2/(m1 + m2) = m/2.
Here -(ħ2/(2μ))∇2rΦ(r) + ½kr2Φ(r) = ErΦ(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,
[-(ħ2/(2μ))(∂2/∂r2) + ½kr2]u(r) = Eu(r).
This is the eigenvalue equation of a one-dimensional harmonic oscillator Hamiltonian.  The eigenvalues are
En = (n + ½)ħω, n = 0, 1, 2, ...,  ω2 = k/m,
These are the eigenenergies of the vibrational motion of the two nuclei .
[Typical vibrational frequencies are on the order of 1012 to 1014 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 = L2/(2μr2) = L2/(2Iz),  Iz = m1r12+ m2r22 = μr2
The eigenfunctions of H are the eigenfunctions of L2.
These are the spherical harmonics, Ylm(θ,φ).  The eigenvalues are El = l(l+1)ħ2/2Iz.
[Remarks:
It is customary to set B = ħ/(4πIz), and write El = Bhl(l+1).  The separation between adjacent levels is
El - El-1 = Bh(l(l+1) - (l-1)l] = 2Bhl.  It increases linear with l.  Each energy eigenvalue El 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)
Enl = (n + ½)ħω + l(l+1)ħ2/2Iz.

#### Problem:

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, Lz, and show that its eigenvalues are good quantum numbers for all of the non-degenerate energy eigenstates.  What are the possible eigenvalues of Lz?
(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 L2 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 = p2/(2m) + U(r),  U(r) = -Ze2/|r - ak| - Ze2/|r + ak|.
The Schroedinger equation is
(iħ∂/∂t)|Ψ> = H(t)|Ψ>,  (iħ∂/∂t)|Ψ> = [-ħ2/(2m))∇2 + U(r)]|Ψ>,
In spherical coordinates we have
H = -(ħ2/(2m))(1/r)(∂2/∂r2)r + L2/(2mr2) + U(r),
L2 = -ħ2[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin2θ))∂2/∂φ2],  Lz = (ħ/i)∂/∂φ.
U(r) = -Ze2/(r2 + a2 - 2ra cosθ)½ -  Ze2/(r2 + a2 + 2ra cosθ)½.

(b)  We know [ L2,Lz] = 0, therefore [ H,Lz] = 0, since the L2 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 Lz.  The possible eigenvalues of Lz 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,L2] = 0, since changing θ to π - θ and φ to π + φ does not change L2.
[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
L2 = -ħ2[∂2/∂θ2 + (1/tanθ)∂/∂θ + (1/sin2θ))∂2/∂φ2]
= -ħ2[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin2θ))∂2/∂φ2].
L2U(r)ψ(r) ≠ LU(r)2ψ(r), [L2,H] ≠ 0, since [L2,U(r)] ≠ 0.
No common eigenbasis of L2 and H exists.

#### Problem:

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 Lz and Pz.  Show from this that the wave function associated with the stationary states of the particle can be written as
ψnmk(ρ, φ, z) = fnm(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 fnm(ρ).

Solution:

• Concepts:
The Hamiltonian and Schroedinger equation in polar coordinates, the  Lz and Pz operators, commuting variables
• Reasoning:
We are asked to write the Hamiltonian H in cylindrical coordinates and show that H commutes with Lz and Pz.
• Details of the calculation:
(a)  H = -[ħ2/(2μ)]∇2 + U(ρ).  Here μ is the mass of the particle.
In cylindrical coordinates
H = -[ħ2/(2μ)][∂2/∂ρ2 + (1/ρ) (∂/∂ρ) + (1/ρ2)(∂2/∂φ2) + ∂2/∂z2] + U(ρ).
Using
Lz = [ħ/i]∂/∂φ,  Pz = [ħ/i]∂/∂z,  Lz2 = -ħ22/∂φ2,  Pz2 = -ħ22/∂z2,
we can write
H = Pρ2/(2μ) + Lz2/(2μρ2) + Pz2/(2μ) + U(ρ).
Here Pρ2 = -ħ2[∂2/∂ρ2 + (1/ρ) (∂/∂ρ)].
Lz depends only on φ, Pz depends only on z, and Pρ2 depends only on ρ.
Therefore [Lz,Pρ2] = [Lz,Pz ], = [Pρ2,Pz ] = 0 and [H,Lz] = [H,Pz ] = 0.
We can find a common eigenbasis of H, Lz and Pz.
This means separation of variables in cylindrical coordinates is possible.
{exp(imφ)} = eigenbasis of Lz, m = 0, ±1, ±2, ... .
{exp(ikz)} = eigenbasis of Pz, 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) = ∑mexp(imφ)∫dkfkm(ρ)exp(ikz), since k is a continuous index.
The functions fkm(ρ) exp(imφ) exp(ikz) is an eigenfunction of H if
H fkm(ρ) exp(imφ) exp(ikz) = E fkm(ρ) exp(imφ) exp(ikz),
[Pρ2/(2μ) + Lz2/(2μρ2) + Pz2/(2μ) + U(ρ)]fkm(ρ) exp(imφ) exp(ikz)
= E fkm(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 ħ2k2/(2μ).
Define E' = E - ħ2k2/(2μ).  E' is the energy associated with the motion perpendicular to the z-axis.
[Pρ2/(2μ) + ħ2m2/(2μρ2) + U(ρ)]f(ρ) = E' f(ρ).
The eigenfunction fkm(ρ) does not depend on k, only on m.  We label different possible solution for the same m with the index n.
[Pρ2/(2μ) + ħ2m2/(2μρ2) + U(ρ)]fnm(ρ) = E'nmfnm(ρ).

(b)  [-(ħ2/(2μ))[∂2/∂ρ2 + (1/ρ) (∂/∂ρ)] + ħ2m2/(2μρ2) + U(ρ)]fnm(ρ) = E'nm nm(ρ).
[[∂2/∂ρ2 + (1/ρ) (∂/∂ρ)] - m22 - V(ρ)]fnm(ρ) = λnmfnm(ρ).
Here E'nm = ħ2λnm2/(2μ) and U(ρ) = ħ2V(ρ)/(2μ).