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Show the transcendental equation as a function of zeta instead |

Show the transcendental equation as a function of energy instead |

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The finite square
well problem is defined by a potential energy function that is zero everywhere
except^{1}

*V*(*x*) = −|*V*_{0}|
−*a* < *x* < *a*.

Since the potential energy function is finite, quantum mechanically there
will be some *leakage* of the wave
function into the classically-forbidden regions specified by *x* > *a* and *x* < −*a*. We will have three regions in which we must solve the
time-independent Schrödinger equation. In Region I (*x* < −*a*)
and Region III (*x* > *a*), for bound-state solutions, *E* < 0, we
can write the time-independent Schrödinger equation as

[*d*^{2}/*dx*^{2 }− κ^{2}] ψ(*x*) = 0 ,
(11.1)

where^{2} κ^{2 }≡ 2*m*|*E*|/*ħ*^{2}.^{ } The equation above has the solutions

ψ(*x*) = *A*exp(+κ*x*) + *B*exp(−κ*x*).
(11.2)

Because the energy eigenfunction must be zero at ±∞ we have

ψ_{I}(*x*) = *A*exp(+κ*x*) and ψ_{III}(*x*) = *D*exp(−κ*x*)
(11.3)

for the solutions in Region I and III, respectively.

In Region II (−*a* < *x* < *a*),
we expect an oscillatory solution since the
energy is greater than the potential energy: *E* > *V*_{0} or |*V*_{0}| > |*E*|. In
this region we can write the time-independent Schrödinger as

[*d*^{2}/*dx*^{2 }+* k*^{2}] ψ(*x*) = 0 ,
(11.4)

where *k*^{2 }≡ 2*m*(|*V*_{0}| − |*E*|)/*ħ*^{2}.
The above equation has the solutions ψ_{II}(*x*) = *B*sin(*kx*) + *C*cos(*kx*)
which are valid solutions for Region II.

Next, we must match the solutions across the boundaries at *x* = −*a* and *x* = *a*. Matching the energy eigenfunctions across these boundaries means that the
wave functions *and* the first derivatives of the energy eigenfunctions must match
at each boundary so that we have a continuous and smooth energy eigenfunction (no jumps
or kinks).

Since the potential energy function is symmetric about the origin,
there are even *and* odd parity solutions to the
bound-state problem.^{3 }We begin by considering the even (parity)
solutions and therefore the ψ_{II}(x) = *C*cos(*kx*) solution in Region II.

Matching proceeds much like the scattering cases we considered in
Chapter 8. At *x* = −*a* we have the conditions

ψ_{I}(−*a*) = ψ_{II}(−*a*) → *A*exp(−κ*a*) = *C*cos(−*ka*),

and

ψ'_{I}(−*a*) = ψ'_{II}(−*a*) → *A*κ
exp(−κ*a*) = *Ck*sin(−*ka*).

From the symmetry in the problem, we need not consider the
boundary at *x* = *a* as it yields the exact same condition on energy eigenfunctions. We now divide the resulting two equations to give a condition for the existence
of even solutions: κ/*k* = tan(*ka*). This is actually a constraint on the allowed
energies, as both *k* and κ involve the energy.

We now consider the following substitutions in terms of dimensionless variables:

ζ ≡ *ka* = [2*m*(|*V*_{0}| − |*E*|)*a*^{2}/*ħ*^{2}]^{1/2}
(11.5)

and

ζ_{0} ≡ [2*m*|*V*_{0}|*a*^{2}/*ħ*^{2}]^{1/2}
(11.6)

where ζ_{0 } > ζ. Using these variables we have

[(ζ_{0}/ζ)^{2 }− 1]^{1/2} = tan(ζ) .
(11.7)

This equation is a transcendental equation for ζ which
itself is related via the preceding equations to the energy. In
addition, the above equation only has solutions for particular
values of ζ. We can solve this equation numerically or
graphically, and we choose graphically in the animation. The
right-hand side of the equation is shown in black and the
left-hand side of the transcendental equation is shown in red. ** In
the animation, ħ = 2m = 1. **You may also select the Show the transcendental equation as a function of energy instead link
to see the equations as a function of energy. You can change

The solution for the odd (parity) energy eigenfunctions proceeds like the even-parity case except that we use the sine solution in Region II:

ψ_{I}(−*a*) = ψ_{II}(−*a*)
→ *A*exp(−κ*a*) = *B*sin(−*ka*),

and

ψ'_{I}(−*a*) = ψ'_{II}(−*a*) → *A*κ
exp(−κ*a*) = *Bk* cos(−*ka*).

Again, we need not consider the equations for *x* = *a* because by symmetry, they
yield the same result. We again divide the two equations to give κ/*k* = −cot(*ka*),
and using the same substitutions as above yields: [(ζ_{0}/ζ)^{2 }− 1]^{1/2} = −cot(ζ).

This equation for the odd-parity solutions is shown in the
animation by checking the text box and moving the slider. Note
that as the potential energy well gets shallower and/or narrower,
ζ_{0} gets smaller, and it is possible for there to be no
intersections on the
graph which means that there will not be any odd-parity states. No matter how
shallow or narrow the symmetric finite potential energy well,
there will always be at least one bound state and it is an
even-parity state.
As ζ_{0} gets larger (meaning larger *a* and |*V*_{0}|), the
number of bound-state solutions increases. In addition, the
intersection of the curves on the graph approaches ζ = *n*π/2, with *n*
even. Again this means that the energy as measured relative to the bottom of the
well approaches that of the infinite square well of length 2*a*.

In order to find the energy eigenfunctions, we must solve for the
constants *A*, *B*, *C*, and *D*. This requires using the
matching equations and then normalizing the wave function. In
practice this is time consuming, instead you can view the
numerical solution by clicking the Show the energy eigenfunction and well instead link. To see other bound states, simply click-drag
in the energy level diagram and select a level. The selected level
will turn red.

___________________

^{1}Note the differences between
the potential energy functions describing the finite well and the
infinite well. The width of the finite well is 2*a* and its walls are at *V* = 0 while the well is at *V* =
−|*V*_{0}|. Bound states of the finite well, therefore, have *E* < 0. With the infinite square well, the width is
*L* and its walls are at *V* = ∞ while the well is at *V* = 0. Bound states of the infinite well,
therefore, have *E* > 0.

^{2}Since *E* < 0, we choose to write *E* = −|*E*| to avoid any
ambiguity in sign.

^{3}This is due to the fact that for
even potential energy functions, the Hamiltonian commutes with the
parity operator. As a consequence, there are even states in which ψ_{e}(−*x*) = ψ_{e}(*x*), and odd states in which
ψ_{o}(−*x*) = −ψ_{o}(*x*).