__Rectangular box__

An electron is trapped in a two-dimensional, infinite
rectangular well of width L_{x} = 800 pm and L_{y} = 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 = E_{x}+ E_{y}. E = n_{x}^{2}π^{2}ħ^{2}/(2mL_{x}^{2}) + n_{y}^{2}π^{2}ħ^{2}/(2mL_{y}^{2}).

For the ground state n_{x}= n_{y}= 1, so E = π^{2}ħ^{2}/(2mL_{x}^{2}) + π^{2}ħ^{2}/(2mL_{y}^{2}).

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 .

Find the eigenvalues and eigenfunctions for a particle in a box with sides a, b,
c by solving the time independent Schroedinger equation (-ħ^{2}/(2m))∇^{2}Φ(**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

(-ħ^{2}/(2m))[∂^{2}Φ(x,y,z)/∂x^{2}+ ∂^{2}Φ(x,y,z)/∂y^{2}+ ∂^{2}Φ(x,y,z)/∂z^{2}] + 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

(-ħ^{2}/(2m))[Y(y)Z(z)∂^{2}X(x)/∂x^{2}+ X(x)Z(z) ∂^{2}Y(y)/∂y^{2}+ X(x)Y(y)∂^{2}Z(z)/∂z^{2}] = E X(x)Y(y)Z(z).

X(x)^{-1}∂^{2}X(x)/∂x^{2}+ Y(y)^{-1}∂^{2}Y(y)/∂y^{2}+ Z(z)^{-1}∂^{2}Z(z)/∂z^{2}+ (2m/ħ^{2})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/ħ^{2})E.

Write E = E_{x}+ E_{y}+ E_{z}.

∂^{2}X(x)/∂x^{2}+ (2m/ħ^{2})E_{x}X(x) = 0

∂^{2}Y(y)/∂y^{2}+ (2m/ħ^{2})E_{y}Y(y) = 0

∂^{2}Z(z)/∂z^{2}+ (2m/ħ^{2})E_{z}Z(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)^{½}sin(n_{x}πx/a), Y(y) = (2/b)^{½}sin(n_{y}πy/b), Z(z) = (2/c)^{½}sin(n_{z}πz/c),

n_{x}= 1, 2, 3, ... , n_{y}= 1, 2, 3, ... , n_{z}= 1, 2, 3, ... .

E_{nx}= n_{x}^{2}π^{2}ħ^{2}/(2ma^{2}), E_{ny}= n_{y}^{2}π^{2}ħ^{2}/(2mb^{2}), E_{nz}= n_{z}^{2}π^{2}ħ^{2}/(2mc^{2}).The eigenfunctions for the problem therefore are

Φ_{nx,ny,nz}(x,y,z) = (8/(abc))^{½}sin(n_{x}πx/a) sin(n_{y}πy/b) sin(n_{z}πz/c),

with the corresponding eigenvalues E_{nx,ny,nz}= [π^{2}ħ^{2}/(2m)][(n_{x}/a)^{2}+ (n_{y}/b)^{2}+ (n_{z}/c)^{2}].

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 (n_{x}^{2}+ n_{y}^{2}+ n_{z}^{2})E_{0}, with E_{0}= π^{2}ħ^{2}/(2mL^{2}). Many of the energy levels are degenerate. Only two spin ½ particles can occupy a state with a unique set of spatial quantum numbers n_{x}, n_{y}, n_{z}. - Details of the calculations:

(a) E_{ground state}= 3E_{0}. (n_{x}= n_{y}= n_{z}= 1.)

Considering only spatial degrees of freedom, E_{ground state}is non-degenerate.

(b) E_{1st excited state}= 6E_{0}. (One of the n_{i}equals 2, the other two equal 1.)

Considering only spatial degrees of freedom, E_{1st excited state}is 3 fold degenerate.

(c) We need to occupy the 10 lowest lying states with a unique set of spatial quantum numbers n_{x}, n_{y}, n_{z}.

Each will be occupied by 2 particles, one spin up and one spin down.

E_{lowest state}= 3E_{0}. (2 particles)

E_{1st excited state}= 6E_{0}. (6 particles) (One of the n_{i}equals 2, the other two equal 1.)

E_{2nd excited state}= 9E_{0}. (6 particles) (One of the n_{i}equals 1, the other two equal 2.)

E_{3rd excited state}= 11E_{0}. (6 particles) (One of the n_{i}equals 3, the other two equal 1.)

For the 20 fermions E_{ground state}= 2*3E_{0}+ 6*6E_{0}+ 6*9E_{0}+ 6*11E_{0}= 162E_{0}.

There is only one possible state with this energy, E_{ground state}of the 20 fermions is non-degenerate.

__Spherical box__

(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 r^{2} 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 = -(h^{2}/(2m)(1/r)(∂^{2}/∂r^{2})r + L^{2}/(2mr^{2}) + U(r).

The wave function ψ_{klm}(r,θ,φ) = R_{kl}(r)Y_{lm}(θ,φ) = [u_{kl}(r)/r]Y_{lm}(θ,φ) is a product of a radial function R_{kl}(r) and the spherical harmonic Y_{lm}(θ,φ).

The differential equation for u_{kl}(r) is

[-(ħ^{2}/(2m)(∂^{2}/∂r^{2}) + ħ^{2}l(l+1)/(2mr^{2}) + U(r)]u_{kl}(r) = E_{kl}u_{kl}(r).

For l = 0 we have [-(ħ^{2}/(2m)(∂^{2}/∂r^{2}) + U(r)]u_{k0}(r) = E_{k0}u_{k0}(r).

With U(r) = 0, (r < a), U(r) = ∞, (r > a) the solutions are

u_{k0}(r) = A sinkr, ka = nπ, k = nπ/a.

We therefore label u_{k0}(r) with the quantum number n as u_{n0}(r).

E_{n0}= n^{2}π^{2}ħ^{2}/(2ma^{2}) are the energy levels when l = 0.

∫|ψ_{n00}(r,θ,φ)|^{2}d^{3}r = 1, A^{2}∫_{0}^{a}sin^{2}(nπr/a)dr = 1, A = (2/a)^{½}.

ψ_{n00}(r,θ,φ) = (1/(2πa)^{½})sin(nπr/a)/r.

(b) <r> = (2/a)∫_{0}^{a}rsin^{2}(nπr/a)dr = a/2.

<r^{2}> = (2/a)∫_{0}^{a}r^{2}sin^{2}(nπr/a)dr = a^{2}/3 - a^{2}/(2n^{2}π^{2}).

Assume that the potential energy of the deuteron is given by

U(r) = -U_{0}, r < r_{0};
U(r) = 0, r ≥ r_{0}.

(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 U_{0}
under the additional condition that the value of the binding energy is much
smaller than U_{0}.

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 = p_{r}^{2}/(2μ) + L^{2}/(2μr^{2}) + U(r).

H = H_{0}+ L^{2}/(2μr^{2}), H_{0}= p_{r}^{2}/(2μ) + U(r).

Let ψ_{l}be the lowest energy eigenfunction of H with orbital angular momentum quantum number l.

Hψ_{l}= [H_{0}+ l(l + 1)ħ^{2}/(2μr^{2})]ψ_{l},

E_{l}^{min}= ∫ψ_{l}^{*}[H_{0}+ l(l + 1)ħ^{2}/(2μr^{2})]ψ_{l}d^{3}r.

Let ψ_{l+1}be the lowest energy eigenfunction of H with orbital angular momentum quantum number l + 1.

Hψ_{l+1}= [H_{0}+ (l + 1)(l + 2)ħ^{2}/(2μr^{2})]ψ_{l+1},

E_{l+1}^{min}= ∫ψ_{l+1}^{*}[H_{0}+ (l + 1)(l + 2)ħ^{2}/(2μr^{2})]ψ_{l+1}d^{3}r.

E_{l+1}^{min}= ∫ψ_{l+1}^{*}[H_{0}+ (l + 1) l ħ^{2}/(2μr^{2})]ψ_{l+1}d^{3}r + ∫ψ_{l+1}^{*}[2(l + 1) ħ^{2}/(2μr^{2})]ψ_{l+1}d^{3}r.

∫ψ_{l}^{*}[H_{0}+ l(l + 1)ħ^{2}/(2μr^{2})]ψ_{l}d^{3}r ≤ ∫ψ_{l+1}^{*}[H_{0}+ l (l + 1)ħ^{2}/(2μr^{2})]ψ_{l+1}d^{3}r

since ψ_{l}is assumed to be the lowest energy eigenfunction of the Hamiltonian

H_{0}+ l(l+1)ħ^{2}/(2μr^{2}).

[The operator H_{0}+ l(l+1)ħ^{2}/(2μr^{2}) = p_{r}^{2}/(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

(-ħ^{2}/(2μ))(d^{2}/dr^{2})u(r) + U'(r)u(r) = Eu(r). By assumption, the lowest energy eigenfunction u(r) is u_{l}(r).]

Therefore E_{l}^{min}≤ E_{l+1}^{min}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

Φ_{100}(**r**) = R_{10}(r)Y00θ,φ) = [u_{10}(r)/r]Y_{00}(θ,φ).

[(-ħ^{2}/(2μ))(∂^{2}/∂r^{2})+ U(r)]u_{10}(r) = E_{10}u_{10}(r)

The problem is reduced to a one dimensional "square well problem" with E_{10}= E < 0.

Define k^{2}= (2μ/ħ^{2})(E + U_{0}), ρ^{2}= (2μ/ħ^{2})(-E), and k_{0}^{2}= (2μ/ħ^{2})U_{0}.

Let r < r_{0}define region 1 and r > r_{0}define region 2. The coordinate r is never negative. Therefore

Φ_{1}(r) = A sin(kr), Φ_{1}(0) = 0.

Φ_{2}(r) = Bexp(-ρr), Φ_{2}(∞) = 0.

At r = r_{0}we need hat Φ(r_{0}) and (∂/∂r)Φ(r)|_{r0}are continuous.

A sin(kr_{0}) = Bexp(-ρr_{0}).

kA cos(kr_{0}) = -ρBexp(-ρr_{0}).

Therefore cot(kr_{0}) = -ρ/k.

1/sin^{2}(kr_{0}) = 1 + cot^{2}(kr_{0}) = (k^{2}+ ρ^{2})/k^{2}= k_{0}^{2}/k^{2}.

We can find a graphical solution by plotting |sin(kr_{0})| and k/k_{0}versus k. The intersections of the two plots in regions where cot(kr_{0}) < 0 gives the values of k for which a solution exist.

We need nπ/2 < kr_{0}< (n + 1)π/2, n = odd.

For only one solution to exist we need the following condition for the slope 1/k_{0}.

2r_{0}/π > 1/k_{0}> 2r_{0}/(3π).

If |E|<<|U_{0}| then k/k_{0}~ 1. Then |sinkr_{0}| ~ |sink_{0}r_{0}| = 1, k_{0}r_{0}~ π/2.

k_{0}^{2}= (π/2r_{0})^{2}, U_{0}= (π/2r_{0})^{2}ħ^{2}/(2μ) ~(π/2r_{0})^{2}ħ^{2}/m_{p}~ 100 MeV with r_{0}= 10^{-15}m.

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) = -|U_{0}| for a < r < b,

U(r) = 0 for r > b.

(a) Find the condition satisfied by a, b, and U_{0}
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

[(-ħ^{2}/(2μ))(∂^{2}/∂r^{2})+ U(r)]u_{k0}(r) = E_{k0}u_{k0}(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 k^{2}= (2μ/ħ^{2})(E + U_{0}), ρ^{2}= (2μ/ħ^{2})(-E), and k_{0}^{2}= (2μ/ħ^{2})U_{0}.

Let us define x = r - a and c = b - a.

In region 2 we have Φ_{2}(x) = A sin(kx), since Φ_{2}(0) = 0.

In region 3 we have Φ_{3}(x) = Bexp(-ρx), since Φ_{2}(∞) = 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/sin^{2}(kc) = 1 + cot^{2}(kc) = (k^{2}+ ρ^{2})/k^{2}= k_{0}^{2}/k^{2}.

We can find a graphical solution by plotting |sin(kc)| and k/k_{0}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/k_{0}> 2c/π, then no solution exists.

For a solution to exist we need k_{0}> π/(2c), or (2μU_{0})^{½}> πħ/(2c).

We need (2μU_{0})^{½}> πħ/(2b - 2a).

(b) Since He does not solidify no bound state exists and we can assume that the condition (2mU_{0})^{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.