### WKB approximation

#### Problem:

Use the WKB approximation to derive the energy levels of a particle of mass m confined to the one-dimensional potential U(x) = F|x|.

Solution:

• Concepts:
The WKB approximation
• Reasoning:
We are instructed to use the WKB approximation.  The WKB approximation requires that  ∫xminxmax pdx = (n + ½)(h/2) or ∫cylce p dx  = ∫cylce ħk dx  = (n + ½)h, n = 0, 1, 2, ... .   Here ∫cylce denotes an integral over one complete cycle of the classical motion and  k2 = (2m/ħ2)(E - U(x)).
• Details of the calculation:
For stationary bound states we want ∫xminxmax pdx = (n + ½)πħ, n = 0, 1, 2, ... .
In our problem k2(x) = 2m(E - F|x|)/ħ2.
We need  (2m)½xminxmax dx (E - F|x|)½ = 2(2m)½x0xmax dx (E - Fx)½ = (n + ½)πħ.
Here Fxmax = E,  xmax = E/F.
Therefore
-2(2m)½ (2/(3F)) (E - F|x|)3/2|x0xmax =  4(2m)½/(3F))E3/2  = (n + ½)πħ.
E = [3π(n + ½)/(4√2)](Fħ)/m.
For n = 1 (the first excited state), for example, we have
E = 1.84 (Fħ)/m.

#### Problem:

A particle of mass m moves in a one-dimensional potential energy function U(x) given by
U(x) = ∞         for x < 0,
U(x) = ax        for x > 0.
The energy E = p2/(2m) + ax must be positive, so the particle moves between x = 0 and x = a-1E, the classical turning points.
Use the Bohr-Sommerfeld quantization rule,
∮pdx =  (n + γ)h,
where n = 0, 1. 2, … and 0 < γ < 1 is a constant, to determine the energy levels En.

Solution

• Concepts:
The Bohr-Sommerfeld quantization rule
• Reasoning:
We are asked to use he Bohr-Sommerfeld quantization rule.  With the appropriate choice of γ, this becomes the WKB approximation.
• Details of the calculation:
For stationary bound states we want ∫0xmax pdx = (n + γ)h/2, n = 0, 1, 2, ... .
In our problem k2(x) = 2m(E - a|x|)/ħ2.
We need  (2m)½0xmax dx (E - a|x|)½ = (2m)½x0xmax dx (E - ax)½ = (n + γ)h/2.
Here axmax = E,  xmax = E/a.
Therefore
-(2m)½ (2/(3a)) (E - a|x|)3/2|x0xmax =  2(2m)½/(3a))E3/2  = (n + γ)h/2.
E = (n + γ)[¾(h2a2/(2m))½].

#### Problem:

A particle of a given energy E > 0 is confined by a potential energy function which is given by U(x) = c|x|.
(a)  Describe in detail, quantitatively and qualitatively as best as you can, what an excited state wave function looks like for region E > U(x) and region E < U(x).
(b)  Introduce the parity operator P and prove that each eigenstate with an eigenenergy E is also an eigenstate of the parity operator P.
(c)  With the energy diagram ordered accordingly to the magnitude of the eigenenergy E, show that the ground state (lowest eigenenergy state) is an even parity state, followed by the next excited state being an odd parity state.  All subsequent excited states possess even or odd parity alternatively.

Solution:

• Concepts:
The eigenvalue equation for the Hamiltonian H, the WKB approximation, the parity operator
• Reasoning:
We are asked to examine the properties of excited, bound-state solutions of HΦ(x) = EΦ(x).  We can use the WKB approximation as a guide.
• Details of the calculation:
(a)  The time-independent Schrödinger equation is
2Φ(x)/∂x2 + k2(x) Φ(x) = 0,
with
k2(x) = (2m(E - U(x))/ħ2).
Let us concentrate in the region E > U(x).  For excited states the WKB approximation can be used.
Φ(x)  = A k(x)exp(±i∫x k(x') dx'.
except near the turning points.
x k(x') dx' counts the oscillations of the wave function.
Near x = 0 the wave function oscillates rapidly since k is large.  Near the turning points it oscillates less rapidly since k is much smaller.  The amplitude increases as we approach the turning points.
Now concentrate on the region where E < U(x).  In this region the wave function decreases rapidly as we move towards larger |x| away from the turning points.

(b)  The parity operator is defined through PΦ(x) = Φ(-x).
[P,H] = PH - HP.  If U(x) = U(-x) then [P,H] = 0 since
PHΦ(x) = P((-ħ2/(2m))∂/∂x2 + U(x))Φ(x)
= (-ħ2/(2m))∂2/∂x2 + U(-x))Φ(-x)
= (-ħ2/(2m))∂2/∂x2 + U(x))Φ(-x)
HPΦ(x) .
Since P and H commute, a basis of common eigenfunctions can be found.  The eigenstates of H are not degenerate.  They are found from matching the solutions for E < U(x) to the solutions for E > U(x) at the turning points.  Each successive match requires an additional half oscillation and therefore a different k(x) (or E).  Since the eigenvalues of H are not degenerate a unique common eigenbasis exists, all eigenfunctions of H are eigenfunctions of P.

(c)  The ground state is the state with the smallest number of oscillations which satisfies the matching condition at the turning points.  We need only ½ oscillation, the state has even parity.  The state with the next lowest energy must have an additional ½ oscillation to satisfy the boundary conditions, it has odd parity.  (Proceed in the same fashion up the energy ladder.)

#### Problem:

Let U(x) = ∞ for x < 0, U(x) = ½mω2x2 for x > 0.  Use the WKB approximation to find the energy levels of a particle of mass m in this potential.  Compare the WKB energies with the exact energies for this potential.

Solution:

• Concepts:
The WKB approximation
• Reasoning:
We are instructed to use the WKB approximation.
For stationary bound states in a potential well with one vertical wall we want ∫0xmax pdx = (n - ¼)πħ, n = 1, 2, 3, ... .
• Details of the calculation:
p = (2m(E - U(x))½.
For the given potential energy function
0xmax (2m(E - ½mω2x2)½dx = (n - ¼)πħ.
xmax = (2E/(mω2))½.
0xmax (2m(E - ½mω2x2)½dx = mω∫0xmax (2E/(mω2) - x2)½dx
= ½mω(2E/(mω2))(sin-11 - sin-10) = (E/ω)(π/2).
(E/ω)(π/2) = (n - ¼)πħ,  E = (2n - ½)ħω.
These are the exact energy levels for n = 1, 2, 3, ... .