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?
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.
We usually try to solve such equations by a technique called separation of
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).
(-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)-1∂2 X(x)/∂x2 + Y(y)-1∂2Y(y)/∂y2 + Z(z)-1∂2 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].
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!
(a) Determine the energy levels and normalized wave functions ψ(r)
of a particle with zero angular momentum in a spherical "potential well"
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.
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.
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.
Hψl = [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.
Hψl+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
u1(r) = A sin(kr), u1(0) = 0.
u2(r) = Bexp(-ρr), u2(∞) = 0.
At r = r0 we need hat u(r0) and (∂/∂r)u(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.
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?