2.2. Finite-depth potential well

Here, we derive equations for the energies and wavefunction of the bound stationary states of a particle of mass \(m\) trapped in a potential well with finite depth \(V_0\) and width \(a\).

The potential-energy function is given by

(2.44)\[\begin{split}V(x) = \begin{cases} 0 &\text{ for } -a/2\le x\le+a/2\\ V_0 &\text{ otherwise} \end{cases}\end{split}\]

Since this function is piecewise constant, we can solve the Schrödinger equation for each region separately and then “glue” them together using wavefunction continuity requirements at the boundaries. Let us designate the three regions as L (left, \(x<-a/2\)), C (center, \(-a/2\le x\le +a/2\)), and R (right, \(x>a/2\)).

For the central region, the Schrödinger equation can be rearranged to

(2.45)\[\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_\mathrm{C}(x) = -\frac{2mE}{\hbar^2}\psi_\mathrm{C}(x) = - k^2\psi_\mathrm{C}(x)\]

with the abbreviation \(k = \sqrt{2mE}/\hbar\). One form of the general mathematical solution for this equation is \(\psi_\mathrm{C} = C\cos(kx) + D\sin(kx)\), with arbitrary constants \(C\) and \(D\).

For the right region R, the Schrödinger equation is

(2.46)\[\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_\mathrm{R}(x) = -\frac{2m(E-V_0)}{\hbar^2}\psi_\mathrm{R}(x)\]

Now, if \(E\ge V_0\), the particle has enough energy to escape from the well and become a free particle. These are unbound states. We are interested in solving for the bound states, where \(E<V_0\). In that case, the equation is best written as

(2.47)\[\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_\mathrm{R}(x) = \frac{2m(V_0-E)}{\hbar^2}\psi_\mathrm{R}(x) = K^2 \psi_\mathrm{R}(x)\]

with \(K = \sqrt{2m(V_0-E)}/\hbar\), which is real-valued for bound states. The general mathematical solutions for this differential equation can be written as \(\psi_\mathrm{R} = A\mathrm{e}^{-Kx}+A'\mathrm{e}^{Kx}\), with arbitrary constants \(A\) and \(A'\).

For the left region, L, \(\psi_\mathrm{L}\) can be obtained in an analogous way. Overall, we get the piecewise defined mathematical solutions of the Schrödinger equation:

(2.48)\[\begin{split}\psi(x) = \begin{cases} B\mathrm{e}^{Kx}+B'\mathrm{e}^{-Kx} &\text{for } x<-a/2\\ C\cos(kx) + D\sin(kx) &\text{for } -a/2\le x\le +a/2\\ A\mathrm{e}^{-Kx}+A'\mathrm{e}^{Kx} &\text{for } x>+a/2 \end{cases}\end{split}\]

Next, we need to identify the physically acceptable solutions among these functions, i.e. all those that are single-valued, normalizable, continuous, and smooth. The functions are all single-valued. However, they are in general not normalizable, since the \(A'\) term grows exponentially with increasing \(x\) and the \(B'\) terms grows exponentially with decreasing \(x\). Therefore, both these terms have to vanish for a physically acceptable wavefunction: \(A'=0\) and \(B'=0\).

In addition, \(\psi\) must be continuous and smooth everywhere. At the boundary at \(x = -a/2\), this gives

(2.49)\[\begin{split}\begin{align} \psi_\mathrm{L}(-a/2) &= \psi_\mathrm{C}(-a/2)\\ \left.\frac{\mathrm{d}\psi_\mathrm{L}}{\mathrm{d}x}\right|_{x=-a/2} &= \left.\frac{\mathrm{d}\psi_\mathrm{C}}{\mathrm{d}x}\right|_{x=-a/2} \end{align}\end{split}\]

which expands to

(2.50)\[\begin{split}\begin{align} B \mathrm{e}^{-Ka/2} &= C\cos(-ka/2) + D\sin(-ka/2) = C\cos(ka/2) - D\sin(ka/2) \\ K B \mathrm{e}^{-Ka/2} &= -k C \sin(-ka/2) + k D \cos(-ka/2) = k C \sin(ka/2) + k D \cos(ka/2) \end{align}\end{split}\]

Dividing the second by the first equation and by \(k\) gives

(2.51)\[\frac{K}{k} = \frac{C \sin(ka/2) + D \cos(ka/2)}{C\cos(ka/2) - D\sin(ka/2)}\]

For the boundary at \(x=+a/2\), the continuity and smoothness conditions are

(2.52)\[\begin{split}\begin{align} \psi_\mathrm{C}(a/2) &= \psi_\mathrm{R}(a/2)\\ \left.\frac{\mathrm{d}\psi_\mathrm{C}}{\mathrm{d}x}\right|_{x=a/2} &= \left.\frac{\mathrm{d}\psi_\mathrm{R}}{\mathrm{d}x}\right|_{x=a/2} \end{align}\end{split}\]

and expand to

(2.53)\[\begin{split}\begin{align} A \mathrm{e}^{-Ka/2} &= C\cos(ka/2) + D\sin(ka/2)\\ -K A \mathrm{e}^{-Ka/2} &= -k C \sin(ka/2) + k D \cos(ka/2) \end{align}\end{split}\]

Dividing the second by the first equation and by \(-k\) gives

(2.54)\[\frac{K}{k} = \frac{C \sin(ka/2) - D \cos(ka/2)} {C\cos(ka/2) + D\sin(ka/2)}\]

The two equations for \(K/k\) have right-hand sides that differ only in the signs. The two right-hand sides must be equal, and the only way to achieve this is that either \(C\) or \(D\) is zero. This results in two sets of physically acceptable wavefunctions:

(2.55)\[\begin{split}\begin{align} C\neq0:\qquad&\psi_\mathrm{C} = C\cos(ka/2)\\ D\neq0:\qquad&\psi_\mathrm{C} = D\sin(ka/2) \end{align}\end{split}\]

The first set of functions is symmetric with respect to \(x=0\), and the second set is antisymmetric.

Using the continuity equations, we can derive the relations between \(C\) (and \(D\)) and \(A\) and \(B\):

(2.56)\[\begin{split}\begin{align} \text{symmetric functions:}\qquad & A = B = C \cos(ka/2)/\mathrm{e}^{-Ka/2}\\ \text{antisymmetric functions:}\qquad & A = -B = D \sin(ka/2)/\mathrm{e}^{-Ka/2} \end{align}\end{split}\]

The last remaining constants we have to determine are \(C\) (for the symmetric functions) and \(D\) (for the antisymmetric functions). They can be obtained by requiring the wavefunctions to be normalized.

The symmetric wavefunctions have energies determined by

(2.57)\[\frac{K}{k} = \sqrt{\frac{V_0-E}{E}} = \tan\frac{a\sqrt{2mE}}{2\hbar}\]

and the antisymmetric wavefunctions have energies determined by

(2.58)\[\frac{K}{k} = \sqrt{\frac{V_0-E}{E}} = -\cot\frac{a\sqrt{2mE}}{2\hbar}\]

Unfortunately, neither of these equations can be solved for \(E\) analytically. Instead, numerical methods must be used.

To summarize, we started out with the Schrödinger equation and obtained mathematical solutions that contained seven arbitrary constants (\(A\), \(A'\), \(B\), \(B'\), \(C\), \(D\), and \(E\)). Continuity at \(\pm a/2\) imposed two conditions, smoothness at \(\pm a/2\) imposed two conditions, square integrability (i.e. non-divergence at \(x=\pm\infty\)) introduced two more. These six conditions allowed us to reduce the arbitrary constants in the wavefunctions to one, an overall scaling factor, which is eliminated by the normalization requirement.The six conditions alss imposed requirements on \(E\): only a finite number of discrete values of \(E\) are possible.