Finite Potential Well: Wave Function and Energy
Table of contents
Imagine a quantum mechanical particle (e.g. an electron) in an infinitely high potential box \(W_{\text{pot}}(x)\):
0, & \frac{L}{2} \le x \le \frac{L}{2} \\
V_0, & x \gt \frac{L}{2} \end{cases} \end{align} $$
The potential box has length \( L \), with one end of the potential box at \( x = \frac{L}{2} \) and the other end at \( x = \frac{L}{2} \). Between these two points the potential shall be zero and outside it shall have a constant value \( V_0 \).
The energy \( W \) of the particle can now be either larger than the potential barrier \( W \gt V_0 \) (unbound state) or smaller \( W \lt V_0 \) (bound state).
Let us consider three regions of the potential box separately:
 Region #1 is the inside of the potential: \( \frac{L}{2} \le x \le \frac{L}{2} \).
 Region #2 is the right potential wall at \( x \gt \frac{L}{2} \).
 Region #3 is the left potential wall at \( x \lt \frac{L}{2} \).
Region #1: Inside the potential box
First we look at the region \( \frac{L}{2} \le x \le \frac{L}{2} \). This region is between the two potential walls.
To find the allowed energies \(W\) and the corresponding wavefunctions \(\psi_1(x)\) of the electron in this region, we need to solve the timeindependent Schrödinger equation. We need only consider the timeindependent Schrödinger equation because the potential \( W_{\text{pot}}(x) \) is timeindependent. According to the definition 1
of the potential box, in this region the potential is \( W_{\text{pot}}(x) = 0 \):
Let's first put the Schrödinger equation 2
into a more compact form. Bring \( \frac{\hbar^2}{2m} \) to the right side and set \( \alpha^2 := \frac{2m \, W}{\hbar^2} \). Then Eq. 2
will look like this:
Thereby the defined constant \( \alpha \) is always positive: \( \alpha^2 \gt 0 \), because all constants and especially the energy \( W \) are positive. Are you wondering why we used \( \alpha^2 \) and not \( \alpha \) for definition? Because now equation 3
is brought into a form corresponding to a differential equation whose solutions we know, namely harmonic functions:
Here we have to determine the constants \( A \) and \( B \) with the help of the normalization of the wave function.
Region #2: Right potential wall
Next, let's look at the right side of the potential, namely the right potential wall in the range: \( x \gt \frac{L}{2} \). In contrast to the interior of the potential well, in this region the potential is not zero, but has a constant value \(V_0\): \( W_{\text{pot}}(x) = V_0 \). For this reason, \( V_0 \) must of course be included in the Schrödinger equation:
Here we denote the wave function of the electron in the right potential wall by \( \psi_2(x) \). Let's transform Eq. 5
into a more compact form. Bring all constants to one side of the equation:
Define the constant \( \beta^2 := \frac{2m \, (V_0  W)}{\hbar^2} \). Thus Eq. 6
becomes nicely compact and we see directly what kind of differential equation we are dealing with:
Our goal is first to find out the allowed energies inside the potential box. That is, we assume that the energy \( W \) of the particle is smaller than \( V_0 \): \( W \lt V_0 \). Therefore the defined \( \beta^2 \) is always positive!
If we compare the differential equation 7
with the differential equation 3
, we see that they are not completely equal. In Eq. 3
there is a minus sign on the right side! So the solution of the Schrödinger equation 7
will not correspond to harmonic functions but to exponential functions:
For the solution 8
to be normalizable (and thus physical), \( \psi_2(x) \) must approach zero for large \( x \):
For this condition to be satisfied, the coefficient \(D\) must vanish: \( D = 0 \), because otherwise the exponential function \( \mathrm{e}^{\beta x} \) goes to infinity for large \( x \) and thus \(\psi_2(x)\) would also go to infinity. Thus, the normalizable solution of the Schrödinger equation 7
for the right potential wall is:
Region #3: Left potential wall
And finally we have to consider the left potential wall which is in the region \( x \lt \frac{L}{2} \). We denote the wave function of the electron in this region by \(\psi_3(x)\). The Schrödinger equation to be solved looks analogous to region #2:
The general solution of the differential equation 11
looks like for region #2. Just note that we have to use other constants \(F\) and \(G\) for this region:
Also, the wave function \( \psi_3(x) \) should approach zero for large, negative \(x\) values so that \( \psi_3(x) \) is normalizable:
For the condition 13
to be satisfied, the coefficient must be \( F = 0 \). Why? Because \( x \) is now negative and thus at \(F\, \mathrm{e}^{\beta x}\) both minus signs cancel each other out and the exponential function diverges. So the physically reasonable solution for region #3 is:
Let us combine our results 4
, 10
, and 14
:
C \, \mathrm{e}^{\beta x}, & x \gt \frac{L}{2} \\
G \, \mathrm{e}^{\beta x}, & x \lt \frac{L}{2} \end{cases} \end{align} $$
But we are not yet finished with the wave function! We have to connect these two separately determined parts of the wave function in such a way that the wave function \(\psi(x)\) is continuously differentiable and thus actually represents a solution of the Schrödinger equation (remember that the Schrödinger equation contains second derivative of the wave function and thus requires that the wave function is continuously differentiable).
Symmetric and antisymmetric wave function
The potential \( W_{\text{pot}}(x) \) is mirrored about \( x = 0 \). It can be shown that in the case of a mirror symmetric potential the wave functions of the electron are either symmetric or antisymmetric.
The symmetric part in Eq. 4
is the cosine term of \( \psi_1(x) \) and the antisymmetric part is the sine term of \( \psi_1 \). Let's denote the symmetric part with a "+" at the top and the antisymmetric part with "" at the top:
\psi^_1(x) ~&:=~ B \, \sin(\alpha x) \end{align} $$
The following connection conditions for symmetric (+) and antisymmetric () wavefunction tells that the wavefunction and its derivatives must coincide at the ends of the potential box. Damit erreichen wir, dass die symmetrische oder antisymmetrische Gesamtwellenfunktion \( \psi^{\pm}(x) \) stetig differenzierbar ist.

The wave function \(\psi^{\pm}_1\) must coincide with the wave function \(\psi_2\) at the right boundary wall, at the location \(x=\frac{L}{2}\):
$$ \begin{align} \psi^{\pm}_1\left(\frac{L}{2}\right) ~=~ \psi_2\left(\frac{L}{2}\right) \end{align} $$ 
The wave function \(\psi^{\pm}_1\) must coincide with the wave function \(\psi_3\) at the left boundary wall, at the location \(x=\frac{L}{2}\):
$$ \begin{align} \psi^{\pm}_1\left(\frac{L}{2}\right) ~=~ \psi_3\left(\frac{L}{2}\right) \end{align} $$ 
The derivative of the wave function \(\psi^{\pm}_1\) at the right potential wall must match the derivative of the wave function \(\psi_2\):
$$ \begin{align} \frac{\text{d} }{\text{d} x} \psi^{\pm}_1\left(\frac{L}{2}\right) ~=~ \frac{\text{d} }{\text{d} x}\psi_2\left(\frac{L}{2}\right) \end{align} $$ 
The derivative of the wave function \(\psi^{\pm}_1\) at the left potential wall must match the derivative of the wave function \(\psi_3\):
$$ \begin{align} \frac{\text{d} }{\text{d} x} \psi^{\pm}_1\left(\frac{L}{2}\right) ~=~ \frac{\text{d} }{\text{d} x}\psi_3\left(\frac{L}{2}\right) \end{align} $$
Specifically, let's use the wave function in 17
and in 18
:
A \cos\left(\alpha \frac{L}{2}\right) ~&=~ G \mathrm{e}^{\beta \frac{L}{2}} \end{align} $$
Here we have exploited the symmetry property of cosine \( \cos(\alpha \frac{L}{2}) = \cos(\alpha \frac{L}{2}) \). Set the two conditions equal and you get \(G=C\):
C \, \mathrm{e}^{\beta x}, & x \gt \frac{L}{2} \\
C \, \mathrm{e}^{\beta x}, & x \lt \frac{L}{2} \end{cases} \end{align} $$
Let us proceed analogously with the antisymmetric wave function. This time insert the antisymmetric wave function from 16
into the conditions 17
and 18
:
B \, \sin\left(\alpha \frac{L}{2}\right) ~&=~ G \, \mathrm{e}^{\beta \frac{L}{2}} \end{align} $$
Here we have exploited the antisymmetry of sine \( \sin(\beta \frac{L}{2}) = \sin(\beta \frac{L}{2}) \). By equating the two conditions, you get \(G ~=~ C\).
C \, \mathrm{e}^{\beta x}, & x \gt \frac{L}{2} \\
C \, \mathrm{e}^{\beta x}, & x \lt \frac{L}{2} \end{cases} \end{align} $$
Energies of the electron within the potential box
Next, we want to find the associated energies \(W^{\pm}\) of the electron when described by the symmetric wavefunction 22
or antisymmetric wavefunction 24
.
Let's consider the symmetric case first. Write out the conditions 17
and 19
(in 19
you have to differentiate once with respect to \(x\)). Then you get:
\alpha \, A \, \sin\left(\alpha \frac{L}{2}\right) ~&=~  \beta \, C \, \mathrm{e}^{\beta \frac{L}{2}} \end{align} $$
Divide the lower equation by the upper equation so that the constants \(A\), \(C\) and the exponential function cancel out. The ratio of sine to cosine corresponds to the tangent function:
Set \( \beta \) and \( \alpha \) and cancel out what can be canceled out. Then you get an equation which does not contain unknown constants. With this equation you can calculate the energy \(W^+\) of the electron when it is inside the potential well, i.e., in the bound state:
Note that this is a transcendental equation, which means: you can't solve it for \(W^+\). But it doesn't matter as long as we can graph it. To do this, we plot the function on the left side of the equation and the function on the right side of the equation. The intersections of the two functions give you the allowed energies \(W^+\) of the electron when described by the symmetric wave function.
Analogously you get the energy \(W^\) of the antisymmetric wave function 24
. Use conditions 18
and 20
for this (again, you must differentiate with respect to \(x\) for Eq. 20
):
\alpha \, B \, \cos\left(\alpha \frac{L}{2}\right) ~&=~  \beta \, C \, \mathrm{e}^{\beta \frac{L}{2}} \end{align} $$
Divide in 28
the lower equation by the upper one, so that we get the same coefficient \( \beta / \alpha \) on the righthand side as in 26
. The ratio of cosine by sine is exactly equal to cotangent function:
Insert \( \alpha \) and \( \beta \):
Limit case: Infinitely high potential well
To derive the energy of the electron in an infinite potential well from the more general case 27
and 30
, you must of course let the potential walls go to infinity: \( V_0 \rightarrow \infty \). In this case, the right sides of 27
and 30
become infinite:
\cot\left( \frac{L}{2}\sqrt{\frac{2m}{\hbar^2} \, W^} \right) ~&=~ \infty
\end{align} $$
So you have a horizontal straight line which runs to infinity and thus intersects all asymptotes of the tangent and cotangent functions. So there are infinitely many energy levels, which the electron can occupy in the infinite potential well.
To determine the intersection points of the tangent and cotangent (and thus also energy levels) with the straight line at infinity, you must first ask yourself at which xvalues the asymptotes of the tangent and cotangent lie! For cotangents, the asymptotes are at multiples of \( \pi \): \( n \, \pi \) with \( n \in \{ 1,2,3... \} \). So set \( n \, \pi \) equal to the argument of cotangent in 31
:
The asymptotes of tangent on the other hand, are at \( \{ \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}... \} \), so at \( (n\frac{1}{2})\pi \). Set the argument of tangent in 31
equal to \( (n\frac{1}{2})\pi \):
Rearrange 32
and 33
for the energy \(W^{\pm}\):
W^_n ~&=~ \frac{h^2}{2m \, L^2} \, n^2 \end{align} $$
If you transform the equations a bit, you will be able to see that for the energies of the antisymmetric solution all even natural numbers are traversed and for the energies of the symmetric solution all odd natural numbers are traversed:
W^_n ~&=~ \frac{h^2}{8m \, L^2} \, (2n)^2 \end{align} $$
You can combine the allowed energy levels 35
of an electron in an infinite potential box into one equation if you take the union of the allowed energies \( W^+_n \cup W^_n \):