Analyze the odd bound state wave functions for the finite square well. Derive the transcendental equation for the allowed energies and solve it graphically. Examine the two limiting cases. Is there always an odd bound state?

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

First, let's write the potential in the following form:

$V\left(x\right)=\left\{\begin{array}{c}0,x\le -a\\ -V_0,-a<x<a\\ 0,x\ge a\end{array}\right.$

Now, the stationary Schrödinger equation reads:

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

Next, we shall split the $x$ axis into three regions, following the same logic as the piecewise form of the potential (region I correspond to the area left of the well, region II corresponds to the well, and region IIl to the area right of the well).

Since the potential is symmetric, the wave functions consist of a set of even and odd wave functions. We shall observe the odd wave functions. Since we are observing the case where $E<0$ the stationary Schrödinger equation in the region, I read:

$-\frac{{ħ}^2}{2m}\frac{d}{d{x}^2}\psi \left(x\right)=\left|E\right|\psi \left(x\right)$

Now, using the abbreviation,$k^2=2m\left|E\right|l{ħ}^2$the previous equation becomes:

$\frac{d}{d{x}^2}\psi \left(x\right)-k^2\psi \left(x\right)=0$

This is a simple linear differential equation, for which the solution in the most general form is:

${\psi }_l\left(x\right)=A_1{e}^{kx}+B_1{e}^{-kx}$

Now, in region II, the stationary Schrödinger equation in the region I read:

$-\frac{{ħ}^2}{2m}\frac{d}{dx}\psi \left(x\right)+\left(\left|E\right|-V_0\right)\psi \left(x\right)=0$

Proceeding in the same manner, we can define the following abbreviation:

$k^2=\frac{2m\left(V_0-\left|E\right|\right)}{{ħ}^2}$

And using it, we obtain the following form of the Schrödinger equation:

$\frac{d}{d{x}^2}\psi \left(x\right)+k^2\psi \left(x\right)=0$

Which is, again, a simple linear differential equation having the solutions of the form:

${\psi }_{ll}\left(x\right)=A_2\cos \left(kx\right)+B_2\sin \left(kx\right)$

We can very easily obtain the solutions to the problem in region III due to the fact that the problem is the same as in region I; therefore, the total solution is given by:

$$\begin{gathered} {\psi }_l\left(x\right)=A_1{e}^{kx}+B_1{e}^{-kx} \\ {\psi }_{ll}\left(x\right)=A_2\cos \left(kX\right)+B_2\sin \left(kX\right) \\ {\psi }_{lll}\left(x\right)=A_3{e}^{kx}+B_3{e}^{-kx} \end{gathered}$$

We now have the total solution. We need to observe the boundary conditions.

First of all, since we are assuming a bound state, we require our wave function to vanish as$x\to \underline{+}\infty$.

Therefore, we conclude that $B_1=A_3=0$, since these terms diverge at infinities. Secondly, we are observing odd states, so$\psi \left(-x\right)=\psi \left(x\right)$, therefore, we conclude that $A_1=-B_3=c$and $A_2=0$, since the cosine is an even function.

For simplicity, we shall define$B_2=b$. After these conclusions, our solution has the following form:

$$\begin{gathered} {\psi }_l\left(x\right)=c{e}^{kx} \\ {\psi }_{ll}\left(x\right)=b\sin \left(kx\right) \\ {\psi }_{lll}\left(x\right)=-c.e^{-kx} \end{gathered}$$

Next, we need to connect the wave functions at the meeting points of different regions, as well as their derivatives, since both need to be continuous (the only case where the derivative is discontinuous is when the potential involves a delta function). It suffices to connect them at either $x=-a$or $x=a$, since both give the same set of linear equations.

By requiring the continuity of the wave function and the continuity of the derivatives at$x=a$,we obtain the following set of linear equations:

$$\begin{gathered} -c.e^{-ka}=b.\sin \left(ka\right) \\ kc.e^{-ka}=kb.\cos \left(ka\right) \end{gathered}$$

We can put this homogeneous set of equations into a matrix form, and the condition for it to have a non-trivial solution is given by the requirement for the determinant of the system to vanish:

localid="1658232743867" $$\begin{gathered} \left|\begin{array}{cc}-e^{-ka}& -\sin \left(ka\right)\\ k{e}^{-ka}& -k\cos \left(ka\right)\end{array}\right|=0 \\ -e^{-ka}x-k\cos \left(ka\right)+k{e}^{-ka}\times -\sin \left(ka\right)=0 \\ \tan \left(ka\right)=-\frac{k}{k} \end{gathered}$$

This is a transcendental equation, and before we proceed, we need to observe the relation between $k^2$and $k^2$, namely:

$k^2+k^2=\frac{2m{V}_0}{{ħ}^2}={\Omega }^2$

After squaring the transcendental equation and inserting the relation between$k^2$and$k^2$, we obtain the following equation:

${\tan }^2\left(ka\right)=\frac{k^2}{{\Omega }^2-k^2}$

It is important to notice that the variable here is .

The following graph represents the latter equation (for some fixed values $a$of and $\Omega$), the red graph represents the tangent function, and the green graph represents the right side.

The intersections of the two give the appropriate values of $\kappa$.

Again, it is possible for the two not to intersect if the depth of the well $V_0$is not large enough.

We have squared the transcendental equation to avoid the confusion when choosing the sign of$k$after taking the square root of the equation (1).

However, we could've not squared it, but then we would be required to observe different cases for different signs of$k$.

It is possible not to have solutions to the equation, depending on the depth and strength of the well.

The graphical representation of such a case is shown below.

The requirement for such a case would be if the tangent graph never intersects with the rational function, meaning that the vertical asymptote would be lower than the one of the tangent squared graphs. In other words:

$$\begin{gathered} \Omega <\frac{\mathrm{\pi }}{2a} \\ {V}_0<\frac{{ħ}^2{\mathrm{\pi }}^2}{8m{a}^2} \end{gathered}$$

If the value of$V_0$is lower than the latter, the bound state will not exist. Next, we can observe the other limiting case, where$V_0\to \infty$.

Then, the transcendental equation reduces to:

$$\begin{gathered} {\tan }^2\left(ka\right)=0 \\ \sin \left(ka\right)=0 \\ k=\frac{n\mathrm{\pi }}{a},\mathrm{n}\mathrm{being}\mathrm{an}\mathrm{integer}. \end{gathered}$$

Which corresponds to the energy spectrum of the odd states of the infinite potential well.