Consider the 1-dimensional, quantum mechanical system consisting of a single
structure-less particle.

How well do the classical equations of motion describe the
motion of this particle?

We have Ehrenfest’s theorem, d<X>/dt = <P>/m, d<P>/dt = -<dU(x)/dx>.

The classical equations of motion
work best when the potential varies slowly with distance.

Assume the potential is a slowly varying function of distance, U = U(x),
independent of time.

Then solutions of the form Ψ_{E}(x,t) = Φ_{E}(x) exp(-iEt/ħ) can be found.

Φ_{E}(x) is a
solution of

∂^{2}Φ(x)/∂x^{2} + k^{2}(x) Φ(x) = 0,
E > U,

or

∂^{2}Φ(x)/∂x^{2} - ρ^{2}(x)Φ(x) = 0,
E < U,

where k^{2}(x) = (2m(E - U(x))/ħ^{2}), ρ^{2}(x) =
(2m(U(x) - E)/ħ^{2}).

Let us try a solution of the form Φ(x) = A exp((i/ħ)S(x)).

[Any complex function of x can
be written as Φ(x) = A(x) exp(i S(x)/ħ).]

Substituting this solution
into the time-independent Schroedinger equation we obtain

iħd^{2}S/dx^{2} - (dS/dx)^{2} + ħ^{2}k^{2}(x)
= 0, or iħd^{2}S/dx^{2} - (dS/dx)^{2} - ħ^{2}ρ^{2}(x)
= 0.

(For a potential square well S(x) = ±ħkx inside the well and
±iħρx outside
the well.)

Assume that ħ can, in some
sense, be regarded as a small quantity and that S(x) can be expanded in powers of
ħ, S(x) = S_{0}(x) + ħS_{1}(x) + ... .

Then

iħd[dS_{0}/dx + ħdS_{1}/dx + ... ]/dx - [dS_{0}/dx +
ħdS_{1}/dx + ... ]^{2} + ħ^{2}k^{2}(x) = 0,
(E > U(x)).

We assume that |dS_{0}/dx| >> |ħdS_{1}/dx| and collect terms
with equal powers of ħ.

[Note: ħ^{2}k^{2}
is
zeroth order in ħ, since k(x) = (2m(E - U(x))/ħ^{2})^{1/2}.]

[dS_{0}/dx]^{2} + ħ^{2}k^{2}(x) = 0
--> S_{0} = ±∫^{x }ħk(x') dx'.

id^{2}S_{0}/dx^{2} - 2(dS_{0}/dx)(dS_{1}/dx)
= 0 --> S_{1} = ½i ln(k(x)).

We have used

id[S_{0}/dx]/dx = 2(dS_{0}/dx)(dS_{1}/dx),

idk/dx = 2k dS_{1}/dx, dS_{1} = (i/2)dk/dk.

Therefore Φ(x) = A k(x)^{-1/2 }exp(±i∫^{x }k(x') dx'.

In the classically allowed region S_{0} = ±∫^{x }ħk(x') dx'
counts the oscillations of the wave function. An increase of 2πħ
corresponds to an additional phase of 2π.

Similarly, in regions where E < U(x) we have

Φ(x) = A ρ(x)^{-1/2 }exp(±∫^{x }ρ(x') dx'.

This is the **WKB (Wentzel, Kramers, Brillouin) approximation**.

When is the WKB approximation valid?

For our first order expansion to be accurate we need that the magnitude of higher order
terms decreases rapidly.

We need |dS_{0}/dx| >> |ħdS_{1}/dx| or k >> |(2k)^{-1}dk/dx|.

The local deBroglie wavelength is λ = 2π/k.

Therefore we need |(λ /4π)(dλ/dx)| << λ, i.e. the change in
λ over a distance λ /4π is small compared to λ .

This holds when the potential varies
slowly and the momentum of the particle is nearly constant over several wavelength.

Near the classical turning points the WKB solutions become invalid, because k
goes to zero here. We have to find a way to connect an oscillating solution to an
exponential solution across a turning point if we want to solve barrier penetration
problems or find bound states.

For x < x_{1} the wave function is of the form

Φ_{1}(x) = A_{1} ρ^{-1/2 }exp(+∫^{x }ρ(x') dx'.

For x > x_{2} the wave function is of the form

Φ_{3}(x) = A_{3} ρ^{-1/2 }exp(-∫^{x }ρ(x') dx'.

In the region between x_{1} and x_{2} it is of the form

Φ_{2}(x) = A_{2} k^{-1/2 }exp(+i∫^{x }k(x') dx'
+ A_{2}' k^{-1/2 }exp(-i∫^{x }k(x') dx'.

At x = x_{1} and x = x_{2} the wave function Φ
and its derivative have to be continuous.

How do we apply these
boundary conditions?

Near x_{1} and x_{2} we expand the potential in a Taylor
series expansion in x and neglect all terms of order higher than 1.

Near x_{1} we have U(x) = E - K_{1}(x - x_{1}),
and near x_{2} we have U(x) = E - K_{2}(x + x_{2}), with
K_{1} and K_{2} positive constants.

In the neighborhood of x_{1} the time-independent Schroedinger equation
then becomes

∂^{2}Φ(x)/∂x^{2} + (2mK_{1}/ħ^{2})(x
- x_{1}) Φ(x) = 0,

and in the neighborhood of x_{2} the time-independent Schroedinger
equation becomes

∂^{2}Φ(x)/∂x^{2} - (2mK_{2}/ħ^{2})(x
- x_{2}) Φ(x) = 0.

Let us define z = -(2mK_{1}/ħ^{2})^{1/3}(x - x_{1}).

Then we obtain
∂^{2}Φ/∂z^{2} - zΦ = 0 near x_{1}.

The solutions of this equation which vanish asymptotically as z --> ∞ or x
--> ∞ are the **Airy functions**. They are defined through

Ai(z) = π^{-1}∫_{0}^{∞} cos(s^{3}/3 + sz)ds

which for large
|z| has the asymptotic form

Ai(z) ≈ (2 √π z^{1/4})^{-1} exp(-2z^{3/2}/3),
(z > 0),

and

Ai(z) = (√π (-z)^{1/4})^{-1} sin(2(-z)^{3/2}/3
+ π/4), (z < 0).

If the energy is high enough the linear approximation to the potential remains valid
over many wavelength.

The Airy functions can therefore be the connecting wave
functions
through the turning point at x_{1}.

If we define z = (2mK_{2}/ħ^{2})^{1/3}(x - x_{2}) then we find
∂^{2}Φ/∂z^{2} - zΦ = 0 near x = x_{2}
and the Airy functions can also be the connecting wave functions through the turning point
at x_{2}.

In the neighborhood of x_{1} we have
k^{2} = -ρ^{2} = (2mK_{1}/ħ^{2})^{1/3}(x
- x_{1}) = -(2mK_{1}/ħ^{2})^{2/3}z

Therefore ∫_{x1}^{x }ρ dx' = (2mK_{1}/ħ^{2})^{1/3}∫_{x1}^{x}
√z dx' = -∫_{0}^{z} √z' dz' = -2z^{3/2}/3.

Similarly ∫_{x1}^{x }k dx' = (2mK_{1}/ħ^{2})^{1/3}∫_{x1}^{x}
√(-z) dx' = -∫_{0}^{z} √(-z') dz' = -2(-z)^{3/2}/3.

By comparing this with the asymptotic forms of the Airy functions we note that

Φ_{1}(x) = A_{1} ρ^{-1/2 }exp(+∫_{x1}^{x }ρ(x') dx' (x < x_{1})

must continue on the right
side as

Φ_{2}(x) = 2A_{1} k^{-1/2 }
sin(∫_{x1}^{x }kdx' + π/4) (x > x_{1}).

In the neighborhood of x_{2 }we similarly find that

Φ_{3}(x) = A_{3} ρ^{-1/2 }exp(-∫_{x2}^{x }ρ(x')
dx' (x > x_{2})

must continue in region 2 as

Φ_{2}(x) = 2A_{3} k^{-1/2 }
sin(∫_{x}^{x2 }(kdx' + π/4) (x < x_{2}).

Both expressions for Φ_{2}(x) are
approximations to the same eigenfunction. We therefore need

2A_{1} k^{-1/2 }sin(∫_{x1}^{x }(kdx '
+ π/4) = 2A_{3} k^{-1/2 }sin(∫_{x}^{x2 }(kdx' + π/4).

Writing ∫_{x1}^{x = }∫_{x1}^{x2}
- ∫_{x}^{x2}, we require

sin[∫_{x1}^{x2 }k dx' - ∫_{x}^{x2}
kdx' + π/4)] = (A_{3}/A_{1})sin[∫_{x}^{x2 }
k dx' + π/4].

This condition is only satisfied if ∫_{x1}^{x2 }k dx
= (n + ½)π, n = integer.

This can be rewritten as ∫_{x1}^{x2 }p dx =
(n + ½)h/2, or ∫_{cylce}^{ }p dx = (n + ½)h.

Here ∫_{cylce }denote an integral over one
complete cycle of the classical motion.

The WKB method for bound states therefore leads to
the **Wilson- Sommerfeld quantization rule** except that
n is replaced by (n + ½).

It leads to a **quantization of the classical action**
J = ∫_{cylce}^{ }p dq.

If we let n = 0, 1, 2, ..., then n counts the number of zero of the wave
function in the well. We may also let n = 1, 2, 3, ... , and write ∫_{x1}^{x2 }k dx
= (n - ½)π. Then n counts the number of anti-nodes of the wave
function in the well. If our potential well has one or two vertical walls, the results of the WKB
approximation differ only in the number that is subtracted from n = 1, 2, 3, ...
, respectively.

For one vertical wall we have ∫_{x1}^{x2 }k dx = (n - ¼)π
and for two vertical walls we have ∫_{x1}^{x2 }k dx = nπ.

Since the WKB approximation works best for large n in the
semi-classical regime, this distinction is more in appearance than in substance.

**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 ∫_{xmin}^{xmax}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 k^{2}= (2m/ħ^{2})(E - U(x)). - Details of the calculation:

For stationary bound states we want ∫_{xmin}^{xmax}pdx = (n + ½)πħ, n = 0, 1, 2, ... .

In our problem k^{2}(x) = 2m(E - F|x|)/ħ^{2}.

We need (2m)^{1/2}∫_{xmin}^{xmax}dx (E - F|x|)^{1/2}= 2(2m)^{1/2}∫_{x0}^{xmax}dx (E - Fx)^{1/2}= (n + ½)πħ.

Here Fx_{max}= E, x_{max }= E/F.

Therefore

-2(2m)^{1/2}(2/(3F)) (E - F|x|)^{3/2}|_{x0}^{xmax}= 4(2m)^{1/2}/(3F))E^{3/2}= (n + ½)πħ.

E = [3π(n + ½)/(4√2)]^{2/3}(Fħ)^{2/3}/m^{1/3}.

For n = 1 (the first excited state), for example, we have

E = 1.84 (Fħ)^{2/3}/m^{1/3}.

T ≈ exp[-2∫

To the same approximation R = 1 - T.