Even and odd solutions for the potential well

Applied to our potential well, we can classify the solutions into even and odd,

$$\psi_e(x)=\left\{ \begin{array}{cc} a_1 e^{\kappa x}, & \\ a_2 \c... ...n({kx}) , & -a <x\le a \\ A_1 e^{-\kappa x}, & a <x< \infty \end{array} \right.$$ (108)

The wave function $\psi(x)$ and its derivative $\psi'(x)$ have to be continuous at $x=\pm a$. Therefore, also the logarithmic derivative

$$\frac{\psi'(x)}{\psi(x)}=\frac{d}{dx}\log \psi(x)$$ (109)

has to be continuous. This is a convenient way to obtain an equation that relates $k$ and $\kappa$ and determines the possible energy values: we calculate the logarithmic derivative for $x=a\pm \varepsilon$, $\varepsilon \to 0$ which yields

$$-\kappa = -k \tan (ka),$$

$$-\kappa = k \cot (ka),$$

$$k = \sqrt{({2m}/{\hbar^2})\left(-\vert E\vert+V\right)},\quad \kappa = \sqrt{({2m}/{\hbar^2}) \vert E\vert}.$$ (110)

These are transcendent equations for the energy $E$: we introduce auxiliary dimensionless variables

$$x\equiv ka,\quad y = \kappa a \leadsto x^2+y^2 = r^2\equiv \frac{2ma^2V}{\hbar^2},\quad V>0.$$ (111)

The equations

$$y = x\tan x,\quad x^2+y^2 = r^2$$

$$y = -x\cot x,\quad x^2+y^2 = r^2$$ (112)

describe two curves in the $x$-$y$-plane, i.e. the circle $x^2+y^2=r^2$, with

$$r\equiv \sqrt{\frac{2m}{\hbar^2}a^2 V},$$ (113)

and the curve $y=x\tan(x)$ ( $y=-x\cot(x)$ for the odd solution), whose intersections determine a fixed number of points $(x_n,y_n)$ in the quadrant of positive $x$ and $y$. These determine the energy eigenvalues $E_n$ via the definition of $k$ and $\kappa$. Of course, the $E_n$ depend on the value of the parameter $r$ which in turn is determined by the depth of the potential well $V$, its width $a$ and the particle mass $m$.

Figure: Graphical solution of (2.35) for $r= 10$ (left) and $r=2$ (right).

To obtain precise values for the possible energy eigenvalue $E_n$, one has to numerically solve (2.35). A convenient method to obtain a qualitative picture, however, is the graphical solution of the transcendent equations as shown in Fig. (2.1). The intersections $y_n$, $n=1,2,...$ of the $x \tan(x)$- or $-x \cot(x)$-curves with the circle of radius $r$ determine $E_n$ via $E_n= - (\hbar^2/2ma^2)y_n^2$ (remember that we have required $E_n<0$)

1. There are only a finite number $N$ of solutions for the energies $E_1<E_2<...<E_N$ depending on the value of the parameter $r$.

2. The wave function corresponding to the lowest eigenvalue $E_1$ is even. Even and odd solutions alternate when `climbing up' the ladder of possible eigenvalues $E_n$.