### Square potentials (cube and sphere)

Rectangular box

#### Problem:

An electron is trapped in a two-dimensional, infinite rectangular well of width Lx = 800 pm and Ly = 1200 pm.  What is the electron's ground state energy?

Solution:

• Concepts:
The 1-D infinite square well
• Reasoning:
We are asked to find the ground-state energy of an electron in an infinite square well.
• Details of the calculation:
E = Ex + Ey.  E = nx2π2ħ2/(2mLx2)  + ny2π2ħ2/(2mLy2).
For the ground state nx = ny = 1, so E = π2ħ2/(2mLx2)  + π2ħ2/(2mLy2).
E = [(π*1.05*10-34)2/(2*9.1*10-31*(800*10-12)2) + (π*1.05*10-34)2/(2*9.1*10-31*(1200*10-12)2] J
= 1.22* 10-19 J = 0.76 eU .

#### Problem:

Find the eigenvalues and eigenfunctions for a particle in a box with sides a, b, c by solving the time independent Schroedinger equation (-ħ2/(2m))∇2Φ(r) + U(r)Φ(r)  = EΦ(r).  Do not just write down your answer but derive it.

Solution:

• Concepts:
The particle in a 3-dimensional box, separation of variables
• Reasoning:
We are asked to find the eigenfunctions and eigenvalues of the Hamiltonian of a particle in a 3D box.
• Details of the calculation:
U(x,y,z) = 0 if 0 < x < a AND 0 < y < b AND 0 < z < c,  U(x,y,z) = infinite otherwise.
In Cartesian coordinate the 3D time-independent Schroedinger equation is
(-ħ2/(2m))[∂2Φ(x,y,z)/∂x2 + ∂2Φ(x,y,z)/∂y2 + ∂2Φ(x,y,z)/∂z2] + U(x,y,z)Φ(x,y,z)  = EΦ(x,y,z).

We usually try to solve such equations by a technique called separation of variables.
For a particle trapped in a rectangular "infinite well" the potential is 0 inside the well and infinite outside the well.  In the region where the potential is zero we solve the Schroedinger equation by trying a solution of the form Φ(x,y,z) = X(x)Y(y)Z(z).
Then
(-ħ2/(2m))[Y(y)Z(z)∂2X(x)/∂x2 + X(x)Z(z) ∂2Y(y)/∂y2 + X(x)Y(y)∂2Z(z)/∂z2]  = E X(x)Y(y)Z(z).
X(x)-12 X(x)/∂x2 + Y(y)-12Y(y)/∂y2 + Z(z)-12 Z(z)/∂z2  + (2m/ħ2)E = 0
The first term is a function of x only, the second term is a function of y only, and the third term is a function of z only.  All three terms must be equal to constants and the sum of these constants must be equal to -(2m/ħ2)E.
Write E = Ex + Ey + Ez.
2X(x)/∂x2 + (2m/ħ2)ExX(x) = 0
2Y(y)/∂y2 + (2m/ħ2)EyY(y) = 0
2Z(z)/∂z2 + (2m/ħ2)EzZ(z) = 0
The boundary conditions are
X(x) = 0 at x = 0 and x = a.
Y(y) = 0 at y = 0 and y = b.
Z(z) = 0 at z = 0 and z = c.

The normalized solutions to the differential equations are
X(x) = (2/a)½sin(nxπx/a),  Y(y) = (2/b)½sin(nyπy/b),  Z(z) = (2/c)½sin(nzπz/c),
nx = 1, 2, 3, ...  ,  ny = 1, 2, 3, ...  ,  nz = 1, 2, 3, ...  .
Enx = nx2π2ħ2/(2ma2),  Eny = ny2π2ħ2/(2mb2),  Enz = nz2π2ħ2/(2mc2).

The eigenfunctions for the problem therefore are
Φnx,ny,nz(x,y,z) = (8/(abc))½sin(nxπx/a) sin(nyπy/b) sin(nzπz/c),
with the corresponding eigenvalues Enx,ny,nz = [π2ħ2/(2m)][(nx/a)2 + (ny/b)2 + (nz/c)2].

#### Problem:

A particle of mass m is in a cubical well of side L, corresponding to the potential energy function
U(r) = 0 if |x|, |y| and |z| < L/2,  U(r) = ∞ otherwise.
(a)  What is its ground-state energy?
(b)  What is the energy of the first excited state?  Is this energy level degenerate or non-degenerate?  Explain!
(c)  Suppose 20 identical, non-interacting particles of mass m and spin ½ are in this well.  What is the ground state energy of this system?  Is this ground state degenerate or non-degenerate?  Explain!

Solution:

• Concepts:
The energy eigenvalues of the 3D infinite square well.
• Reasoning?
The energy eigenvalues of the 3D infinite square well are (nx2 + ny2 + nz2)E0, with E0 = π2ħ2/(2mL2).  Many of the energy levels are degenerate.  Only two spin ½ particles can occupy a state with a unique set of spatial quantum numbers nx, ny, nz.
• Details of the calculations:
(a)  Eground state = 3E0.  (nx = ny = nz = 1.)
Considering only spatial degrees of freedom,  Eground state is non-degenerate.
(b) E1st excited state = 6E0.  (One of the ni equals 2, the other two equal 1.)
Considering only spatial degrees of freedom,  E1st excited state is 3 fold degenerate.
(c)  We need to occupy the 10 lowest lying states with a unique set of spatial quantum numbers nx, ny, nz.
Each will be occupied by 2 particles, one spin up and one spin down.
Elowest state = 3E0.  (2 particles)
E1st excited state = 6E0.  (6 particles)  (One of the ni equals 2, the other two equal 1.)
E2nd excited state = 9E0.  (6 particles)  (One of the ni equals 1, the other two equal 2.)
E3rd excited state = 11E0.  (6 particles)  (One of the ni equals 3, the other two equal 1.)
For the 20 fermions Eground state = 2*3E0 + 6*6E0 + 6*9E0 + 6*11E0 = 162E0.
There is only one possible state with this energy, Eground state of the 20 fermions is non-degenerate.

Spherical box

#### Problem:

(a)  Determine the energy levels and normalized wave functions ψ(r) of a particle with zero angular momentum in a spherical "potential well" U(r) = 0, (r < a), U(r) = ∞, (r > a).
(b)  Determine the average value of r and r2 for each of the energy eigenstates with l = 0.

Solution:

• Concepts:
Three-dimensional "square" potentials
• Reasoning:
We are asked to determine the energy levels and normalized wave functions ψ(r) of a particle in a three dimensional square potential.
• Details of the calculation:
(a)  The Hamiltonian is H = -(h2/(2m)(1/r)(∂2/∂r2)r + L2/(2mr2) + U(r).
The wave function ψklm(r,θ,φ) = Rkl(r)Ylm(θ,φ) = [ukl(r)/r]Ylm(θ,φ) is a product of a radial function Rkl(r) and the spherical harmonic Ylm(θ,φ).
The differential equation for ukl(r) is
[-(ħ2/(2m)(∂2/∂r2) + ħ2l(l+1)/(2mr2) + U(r)]ukl(r) = Eklukl(r).
For l = 0 we have [-(ħ2/(2m)(∂2/∂r2) + U(r)]uk0(r) = Ek0uk0(r).
With U(r) = 0, (r < a), U(r) = ∞, (r > a) the solutions are
uk0(r) = A sinkr, ka = nπ, k = nπ/a.
We therefore label uk0(r) with the quantum number n as un0(r).
En0 =  n2π2ħ2/(2ma2) are the energy levels when l = 0.
∫|ψn00(r,θ,φ)|2d3r = 1, A20a sin2(nπr/a)dr = 1,  A = (2/a)½.
ψn00(r,θ,φ) = (1/(2πa)½)sin(nπr/a)/r.
(b)  <r> = (2/a)∫0a rsin2(nπr/a)dr = a/2.
<r2> = (2/a)∫0a r2sin2(nπr/a)dr = a2/3 - a2/(2n2π2).

#### Problem:

Assume that the potential energy of the deuteron is given by
U(r) = -U0, r < r0; U(r) = 0, r ≥ r0.
(a)  Show that the ground state of the deuteron possesses zero orbital angular momentum (l = 0).  Since this is true for any central potential, you may not need the detailed nature of the square well potential.
(b)  Assume that l = 0 and estimate the value of U0 under the additional condition that the value of the binding energy is much smaller than U0.

Solution:

• Concepts:
Three-dimensional square potentials
• Reasoning:
We have to investigate the properties of the ground state in a three dimensional square potential.
• Details of the calculation:
(a)  The Hamiltonian of the relative motion of the two particles is
H = pr2/(2μ) + L2/(2μr2) + U(r).
H = H0 + L2/(2μr2),  H0 = pr2/(2μ) + U(r).
Let ψl be the lowest energy eigenfunction of H with orbital angular momentum quantum  number l.
l =  [H0 +  l(l + 1)ħ2/(2μr2)]ψl
Elmin = ∫ψl*[H0 +  l(l + 1)ħ2/(2μr2)]ψl d3r.
Let ψl+1 be the lowest energy eigenfunction of H with orbital angular momentum quantum number l + 1.
l+1 =  [H0 +  (l + 1)(l + 2)ħ2/(2μr2)]ψl+1
El+1min = ∫ψl+1*[H0 +  (l + 1)(l + 2)ħ2/(2μr2)]ψl+1 d3r.
El+1min = ∫ψl+1*[H0 +  (l + 1) l ħ2/(2μr2)]ψl+1 d3r + ∫ψl+1*[2(l + 1) ħ2/(2μr2)]ψl+1 d3r.
∫ψl*[H0 +  l(l + 1)ħ2/(2μr2)]ψl d3r ≤ ∫ψl+1*[H0 + l (l + 1)ħ2/(2μr2)]ψl+1 d3r
since ψl is assumed to be the lowest energy eigenfunction of the Hamiltonian
H0 + l(l+1)ħ2/(2μr2).
[The operator H0 + l(l+1)ħ2/(2μr2) = pr2/(2μ) + U'(r) does not operate on the angular coordinates.  We can write its eigenfunctions as [u(r)/r]f(θ,φ), where f(θ,φ) is arbitrary.  The eigenvalues are determined by
(-ħ2/(2μ))(d2/dr2)u(r) + U'(r)u(r) = Eu(r).  By assumption, the lowest energy eigenfunction u(r) is ul(r).]
Therefore   Elmin ≤ El+1min and the ground state possesses zero orbital angular momentum.
(b)  The relative motion of the two particles is described in the same way as the motion of a fictitious particle of reduced mass m in a central potential.  Therefore we have
Φ100(r) = R10(r)Y00θ,φ) = [u10(r)/r]Y00(θ,φ).
[(-ħ2/(2μ))(∂2/∂r2)+ U(r)]u10(r) = E10u10(r)
The problem is reduced to a one dimensional "square well problem" with E10 = E < 0.

Define k2 = (2μ/ħ2)(E + U0), ρ2 = (2μ/ħ2)(-E), and k02 = (2μ/ħ2)U0.
Let r < r0 define region 1 and r > r0 define region 2.  The coordinate r is never negative.  Therefore
Φ1(r) = A sin(kr),  Φ1(0) = 0.
Φ2(r) = Bexp(-ρr),  Φ2(∞) = 0.
At r = r0 we need hat Φ(r0) and (∂/∂r)Φ(r)|r0 are continuous.
A sin(kr0) = Bexp(-ρr0).
kA cos(kr0) = -ρBexp(-ρr0).
Therefore cot(kr0) = -ρ/k.
1/sin2(kr0) = 1 + cot2(kr0) = (k2 + ρ2)/k2  = k02/k2.
We can find a graphical solution by plotting |sin(kr0)| and k/k0 versus k.  The intersections of the two plots in regions where cot(kr0) < 0 gives the values of k for which a solution exist.
We need  nπ/2 < kr0 < (n + 1)π/2, n = odd.
For only one solution to exist we need the following condition for the slope 1/k0.
2r0/π > 1/k0 > 2r0/(3π).

If |E|<<|U0|   then  k/k0 ~ 1.  Then  |sinkr0| ~ |sink0r0| = 1,  k0r0 ~ π/2.
k02 = (π/2r0)2, U0 = (π/2r0)2ħ2/(2μ) ~(π/2r0)2ħ2/mp ~ 100 MeV with r0 = 10-15 m.

#### Problem:

Let us represent the interaction between two He atoms by the sum of a short-range repulsive part and a long-range attractive part.
U(r) = + ∞   for r < a,
U(r) = -|U0|   for a < r < b,
U(r) = 0   for r > b.
(a)  Find the condition satisfied by a, b, and U0 for a bound state to exist in the relative motion of two He atoms.
(b)  Experimental evidence indicates that He does not solidify at atmospheric pressure, even at ~0 K.  What do you think this implies about the effective helium-helium interaction?

Solution:

• Concepts:
Three-dimensional square potentials
• Reasoning:
The radial equation for a particle in a central potential with l = 0 is
[(-ħ2/(2μ))(∂2/∂r2)+ U(r)]uk0(r) = Ek0uk0(r)
Here U(r) is a square potential, and μ is the reduced mass.
• Details of the calculation:
(a)  We have to solve a square potential problem in the region 0 < r < ∞.
Let region 1 extend from r = 0 to r = a, region 2 from r = a to r = b, and region 3 from r = b to infinity.
In region 1 u = 0.  We need u(a) = u(∞) = 0.
For bound states we have E < 0.
Define k2 = (2μ/ħ2)(E + U0), ρ2 = (2μ/ħ2)(-E), and k02 = (2μ/ħ2)U0.
Let us define x = r - a and c = b - a.
In region 2 we have Φ2(x) = A sin(kx), since  Φ2(0) = 0.
In region 3 we have Φ3(x) = Bexp(-ρx), since Φ2(∞) = 0.
At x = c we need hat Φ(c) and (∂/∂x)Φ(x)|c are continuous.
A sin(kc) = Bexp(-ρc)
kA cos(kc) = -ρBexp(-ρc)
Therefore cot(kc) = -ρ/k.
1/sin2(kc) = 1 + cot2(kc) = (k2 + ρ2)/k2  = k02/k2.
We can find a graphical solution by plotting |sin(kc)| and k/k0 versus k.  The intersections of the two plots in regions where cot(kc) <  0 gives the values of k for which a solution exist.

If the slope 1/k0 > 2c/π, then no solution exists.
For a solution to exist we need k0 > π/(2c), or (2μU0)½ > πħ/(2c).
We need (2μU0)½ > πħ/(2b - 2a).
(b)  Since He does not solidify no bound state exists and we can assume that the condition (2mU0)½ > πħ/(2b - 2a) is not satisfied.  The effective helium-helium interaction is either not strong enough or too short range to support a bound state.