*Michael Fowler, *

We have considered in some detail a particle trapped between infinitely high walls a distance $L$ apart, we found the wave function solutions of the time independent Schrödinger equation, and the corresponding energies. The essential point was that the wave function had to go to zero at the walls, because there is zero probability of finding the particle penetrating an infinitely high wall. This meant that the lowest energy state couldn't have zero energy (meaning the potential energy at the flat bottom of the well). Rather, the lowest energy state had to have the minimal amount of bending of the wave function necessary for it to be zero at the two walls but nonzero in between$\u2014$this corresponds to half a period of a sine or cosine (depending on the choice of origin), these functions being the solutions of Schrödinger's equation in the zero potential region between the walls. The sequence of wave functions (eigenstates) as the energy increases have 0, 1, 2, … zeros (nodes) in the well.

Let us now consider how this picture is changed if the potential at the
walls is *not* infinite. It will turn
out to be convenient to have the origin at the center of the well, so we take

$\begin{array}{l}V\left(x\right)={V}_{0},\text{\hspace{1em}}x\le -L/2,\\ V\left(x\right)=0,\text{\hspace{1em}}-L/2<x<L/2,\\ V\left(x\right)={V}_{0},\text{\hspace{1em}}L/2\le x.\end{array}$

Having the potential symmetric about the origin makes it easier to catalog the wave functions. For a symmetric potential, the wave functions can always be taken to be symmetric or antisymmetric. If a wave function $\psi \left(x\right)$ is a solution of Schrödinger's equation with energy $E,$ and the potential is symmetric, then $\psi \left(-x\right)$ is a solution with the same energy. This means that $\psi \left(x\right)+\psi \left(-x\right)$ and $\psi \left(x\right)-\psi \left(-x\right)$ are also solutions, since the equation is linear, and these are symmetric and antisymmetric respectively, and using them is completely equivalent to using the original $\psi \left(x\right)$ and its reflection $\psi \left(-x\right).$

How is the lowest energy state wave function affected by having finite
instead of infinite walls? Inside the
well, the solution to Schrödinger's equation is still of cosine form (it's a
state symmetric about the origin). However, since the walls are now finite, $\psi \left(x\right)$ cannot change slope discontinuously to a flat
line at the walls. It must instead connect smoothly with a function which is a
solution to Schrödinger's equation *inside the wall*.

The equation in the wall is

$$-\frac{{\hslash}^{2}}{2m}\frac{{d}^{2}\psi \left(x\right)}{d{x}^{2}}+{V}_{0}\psi \left(x\right)=E\psi \left(x\right)$$

and has two exponential solutions (say for $x\ge L/2$ ) one increasing to the right, the other decreasing,

${e}^{\alpha x}\text{and}{e}^{-\alpha x},\text{where}\alpha =\sqrt{2m\left({V}_{0}-E\right)/{\hslash}^{2}}.$

(We are assuming here that $E<{V}_{0},$ so the particle is bound to the well. We shall find this is always true for the lowest energy state.)

Let us try to construct the wave function for the energy $E$ corresponding to this lowest bound state. From the equation with ${V}_{0}=0,$ the wavefunction inside the well (let's assume it's symmetric for now) is proportional to $\mathrm{cos}kx,$ where $k=\sqrt{2mE/{\hslash}^{2}}.$

The wave function (and its derivative!) inside the well must match a sum of
exponential terms$\u2014$the wave
function in the *wall*$\u2014$at $x=L/2,$ so

$\begin{array}{c}\mathrm{cos}\left(kL/2\right)=A{e}^{\alpha L/2}+B{e}^{-\alpha L/2}\\ -k\mathrm{sin}\left(kL/2\right)=\alpha A{e}^{\alpha L/2}-\alpha B{e}^{-\alpha L/2}.\end{array}$

(By writing just a cosine term inside the well, we have left out the overall normalization constant. This can be put back in at the end.)

Solving these equations for the coefficients $A,B$ in the usual way, we find that in general the cosine solution inside the well goes smoothly into a linear combination of exponentially increasing and decreasing terms in the wall. (By the symmetry of the problem, the same thing must happen for $x<-L/2.$ )

*However, this cannot in general represent a bound state in the well*. The increasing solution increases *without
limit* as $x$ goes to infinity, so since the square of the
wave function is proportional to the probability of finding the particle at any
point, the particle is infinitely more likely to be found at infinity than
anywhere else. It got away! This clearly makes no sense$\u2014$we're
trying to find wave functions for particles that stay in, or at least close to,
the well. We are forced to conclude that
the *only* exponential wave function that makes sense is the one for which
$A$ *is exactly zero*, so that there is only
a *decreasing* wave in the wall.

Requiring the decreas ing wave function, $A=0,$ means that only a discrete set of values of $k,$ or $E,$ satisfy the boundary condition equations above. They are most simply found by taking $A=0$ and dividing one equation by the other to give:

$\mathrm{tan}\left(kL/2\right)=\alpha /k.$

This cannot be solved analytically, but is easy to solve graphically by plotting the two sides as functions of $k$ (recall $\alpha =\sqrt{2m\left({V}_{0}-E\right)/{\hslash}^{2}},$ and $k=\sqrt{2mE/{\hslash}^{2}}$ ) and finding where the curves intersect.

A more straightforward approach is to solve the Schrödinger equation numerically, using a spreadsheet, and explore how the wave function varies as $E$ is changed. This method has the great advantage that the spreadsheet can be readily adapted to any reasonable potential. We present first the simplest, least sophisticated method, which gives an accuracy of around one percent for energy values using a few hundred rows of the spreadsheet.

*To simplify adapting
the spreadsheet to the equation, we choose units so that* ${\hslash}^{2}/2m=1.$

From now on, in this section we
shall denote the wave function by *f*(*x*) or just* f *instead
of *ψ*(*x*) , to lessen headaches in setting up the spreadsheet.

Schrödinger's Equation is then:

$$\frac{{d}^{2}f\left(x\right)}{d{x}^{2}}=\left(V\left(x\right)-E\right)f\left(x\right).$$

To solve this using
the spreadsheet, it is necessary to discretize the derivative. We replace the
continuous variable *x* by a sequence of equally spaced points *dx*
apart, such as 0, *dx*, 2*dx*, 3*dx*, … . Here *dx* is of
course finite, we should really call it Δ*x*.

We approximate the second derivative by:

$${\left(\frac{{d}^{2}f\left(x\right)}{d{x}^{2}}\right)}_{x={x}_{j}}=\frac{f\left({x}_{j+1}\right)-2f\left({x}_{j}\right)+f\left({x}_{j-1}\right)}{{\left(dx\right)}^{2}}$$

so the differential equation is replaced by the difference equation:

$$\frac{f\left({x}_{j+1}\right)-2f\left({x}_{j}\right)+f\left({x}_{j-1}\right)}{{\left(dx\right)}^{2}}=\left(V\left({x}_{j}\right)-E\right)f\left({x}_{j}\right).$$

Notice that from this equation, if we know *f*(*x** _{j}*),

A second-order differential equation
like this needs *two* boundary conditions for the solution to be defined. We can specify the initial value of *f*
and its derivative *df*/*dx*. But this is just equivalent to giving
the first two members of the discrete series, *f*(*x*_{0})
and *f*(*x*_{1}), say. We can then use the difference equation to
find *f*(*x*_{2}), then use it again to find *f*(*x*_{3})
and so on. This is what the spreadsheet
does for us: we find *f*(*x*_{j}_{+1})
row by row using the difference equation above written:

$$f\left({x}_{j+1}\right)=\left(2+{\left(dx\right)}^{2}\left(V\left({x}_{j}\right)-E\right)\right)f\left({x}_{j}\right)-f\left({x}_{j-1}\right).$$

Recall that by taking the origin in the center of the square well, we argued that we need only look at wave functions that were symmetric or antisymmetric about the origin. If you try to sketch such wave functions, you will find that symmetric wave functions must have zero slope at the origin, and antisymmetric wave functions must be zero at the origin.

Furthermore, if we know the wave function in the right-hand
half, that is, for *x* > 0, we know it for all *x*, from the
symmetry. Hence, we need only solve Schrödinger's equation going from the
origin to the right - we can take *x*_{0} = 0, *x*_{1}
= *dx*, etc.

As we discussed in the preceding section, on integrating out from the origin
with energy *E* set less than *V*_{0}, once we cross over
into the wall the solution is some combination of an exponentially increasing
function and an exponentially decreasing function,

$f\left(x\right)=A\left(E\right){e}^{\alpha x}+B\left(E\right){e}^{-\alpha x}$

where the exponential coefficient *α*
is positive, and depends on *E*$\u2014$it was
defined in the preceding section.

Let us first look at the *symmetric* solutions for very low energies,
so take *f*(0) = 1, *f'*(0)
=0. (Note that we cannot find the correct overall *normalization constant*
until we find the solution, then integrate its square over all space$\u2014$this can
always be done later, and is unnecessary for analyzing the properties of the
state).

Let us begin with the trivial case *E* = 0. For zero energy, inside the
well *E* - *V* will of course be identically zero, so from
Schrödinger's equation the *slope* of* f*(*x*) can never change,
consequently *f*(*x*) = 1 for all *x* < *L*. On reaching the wall, this wave function and
its derivative connect smoothly from inside to outside if *A* = *B* =
$\mathrm{\xbd}$. It is
clear that as we keep going to the right, the *A* term in the above
equation dominates and *f*(*x*) diverges, signaling that there is no
state localized in the well at *E* = 0.

At this point, you should create an Excel
spreadsheet, with a graph showing *f*(*x*) and *V*(*x*)-*E*
as functions of *x*. Details
of how this might be done are given in the accompanying homework assignment. *If
you have a working spreadsheet, it will be much easier to understand the
following discussion!*

As we now increase *E* from zero, the symmetric wave function, having zero
slope at the center of the well, will begin to curve downwards on moving away
from the center, and as the energy increases so does the downward curvature. This naturally changes the mix of increasing
and decreasing exponentials needed to connect smoothly at the wall, and in fact
as a function of *E*, *A*(*E*) changes sign at a certain value
we will call *E*_{0}. For
energies just below *E*_{0}, *f*(*x*) diverges to plus
infinity for large *x*. For
energies above *E*_{0}, it diverges to minus infinity for large *x*.*
Exactly at E _{0}, f(x) goes to zero for large x*. This is the wave function we are looking for: it
corresponds to a particle localized close to the well, and in fact is the
lowest possible energy$\u2014$the

The first problem is to find this value *E*_{0}.
If we just guess a value of *E*,
the wave function will almost certainly diverge for large *x*. The way to find *E*_{0} is to
notice that for *E* below *E*_{0}, *f*(*x*) goes to
large positive values on the far right, for *E* just above *E*_{0},
*f*(*x*) goes to large negative values on the far right. So we take an increasing set of *E*
values starting near zero, and watch for the tail to wag! When this happens, we back up half way, then
back or forward as necessary, choosing a set of *E* values that bracket *E*_{0}.

Actually, this rather tedious process can be automated in Excel, but it's
worth going through it manually a time or two to get a feeling for how the wave
functions behave. A general feature to
check is that for *E* > *V*(*x*) they are oscillatory, sine
or cosine like, although with changing wavelengths in general; whereas for *E*
< *V*(*x*) they have exponential character. You can see why this is by looking at the
equation, but it’s worth bearing in mind as you examine the curves.

Now to a bit of automation. The strategy we are following is to adjust *E*
until the function *f*(*x*) goes to zero for large *x*, that is,
far down at the bottom of our column having several hundred entries. Since we would like to know how well we are
doing, it’s convenient to have the spreadsheet copy that value to some cell
near the top. So, if the end of our
column of *f*(*x*_{i})'s is F600, say, then we enter =$F$600
in some convenient square near the top, say B22. Now we can keep an eye on the distant value
as we adjust *E*. We can also get
Excel to do this for us as follows (I assume the value of *E* is entered
into square B15):

Click Tools, Goal Seek. Then fill out the dialog box: Set Cell B22 to value 0 by changing cell B15. (You can fill in the cell addresses by just clicking on the appropriate cells.) Click OK. It will then calculate and come up with a value. Click OK again.

The curve below was generated by this procedure. It is worth noting that there is a definite probability of finding the particle inside the wall, that is to say, the wave function is nonzero there (and, it turns out, for reasonable, physical values of the parameters):

All square well potentials in one dimension, however shallow, have a
localized ground state with this general shape. Whether or not there are other eigenstates
with other eigenvalues depends on the depth of the potential. For a sufficiently shallow potential, there is
only one state. An infinitely deep well,
as we discussed earlier, has an infinite number of bound states. As the well depth increases from zero, states
are bound sequentially. These higher
eigenstates are called* excited states*. A particle in one of them will usually decay
to a lower state, emitting a photon, just as in the Bohr atom.

Let us now see how the wave function develops as *E*
increases beyond *E*_{0} for a sufficiently deep well. As the energy is increased, the cosine term
inside the well has tighter curvature, and the exponentially increasing term
for large *x* is large and negative. However, continuing to increase the energy,
eventually the cosine term bottoms out inside the well and begins to turn up
again before reaching the wall. At a
certain energy, it again becomes possible to match it to a decreasing function:

To find this new eigenvalue, we can get close to it then use: Tools, Goal Seek just as we did for the ground state$\u2014$see above.

To really automate this procedure, it's worth creating a macro. So, go to Tools, Record Macro, Record New Macro. Call the Macro "Eigenvalue", and put in the shortcut Ctrl+e. In the Record Macro box, click OK to begin recording. Then Click Tools, Goal Seek. Then fill out the dialog box just as we did before, that is, Set Cell B22 to value 0 by changing cell B15. Click OK. It will then calculate and come up with a value. Click OK again. That is all we want to record, so click Tools, Macro, Stop Recording.

*You can check to see if your macro is working by changing the value of E
in B15, then hit Ctrl+e. (If you forgot to assign the shortcut to your macro,
you can do it now by clicking Tools, Macro, Macros, click on your macro and
click Options. You can fill in the shortcut key in the Options box.)*

*To get really fancy, it’s nice to run this macro by clicking on
something. Insert a textbox somewhere in
your spreadsheet, and write “Find Eigenvalue” on it. With the textbox selected, right click on one
of the sizing handles and click “Assign Macro” . Choose your “Eigenvalue” macro. Now it will run if you click on the textbox.*

A point to notice about the wave function pictured above is that has a node
at about *x* = 2. (Actually, of course, this is just *half *the wave
function for the complete well-there will be another node at *x* = -2.) As
we continue to increase *E*, the high-*x* tail of the wave function
wags up and down. Each time it crosses the axis there is an allowable wave
function. Furthermore, each time it crosses the axis the wave function collects
another node for *x* > 0. Thus, the complete wave functions generated
by this method have 0, 2, 4, 6, … nodes. From the symmetry of the problem, the
allowable wave functions with an *odd* number of nodes must have one node
at the origin. They can be generated by taking as initial conditions that *f*
is zero at the origin, and has finite slope.

Schrödinger's equation in the form

$\frac{{d}^{2}f\left(x\right)}{d{x}^{2}}=\left(V\left(x\right)-E\right)f\left(x\right)$

can be interpreted by saying that the left-hand side, the rate of change of
slope, is the *curvature*$\u2014$so the
curvature of the function is equal to (*V*(*x*) - *E*)*f*(*x*).
This means that if *E* > *V*(*x*),
for *f*(*x*) positive *f*(*x*) is curving negatively, for *f*(*x*)
negative *f*(*x*) is curving positively. *In both cases, f(x) is always curving
towards the axis*. This means that for *E* > *V*(*x*), *f*(*x*) has a kind of stability: its
curvature is always bringing it back towards the axis, so it has oscillatory
character.

On the other hand, for *V*(*x*) > *E*, the curvature is
always *away* from the axis. This means that *f*(*x*) tends to
diverge to infinity. Only under exactly
the right conditions will this curvature be just enough to bring the wave
function to zero as *x* goes to infinity. (As *f*(*x*) tends to zero, the
curvature tends to zero, too.)

It is worth examining the wave functions generated by the spreadsheet to see
just how the curvature changes as *V*(*x*) - *E*, or for that
matter *f*(*x*), changes sign.