A more complicated case

We consider a particle of mass $m$ confined in a one-dimensional potential well with infinitely high walls at $x=\pm L/2$. Within the well, i.e. within the interval $[-a/2,a/2]$, $a<L$, there is a symmetric potential $V(x)=V(-x)>0$, see Fig. 2.3.

Figure: Left: Symmetric potential $V(x)$ within a one-dimensional potential well. Right: special case of a rectangular potential.

We wish to determine the stationary bound states with energy $E$ and the possible energy eigenvalues $E$ for this potential. Because the potential is symmetric around the origin, the eigenstate wave functions must have even or odd parity $\psi_{\pm}(x)=\pm\psi(-x)$. For $\vert x\vert>a/2$, the wave function must be a superposition of plane waves that has to vanish at the boundaries $\pm L/2$. Therefore, we can set

$$\psi_{\pm}(x)=\left \{ \begin{array}{cc} A \sin k(x+L/2)= a_1... ...L/2-x)= a_3e^{ikx}+b_3e^{-ikx} & a/2 <x<L/2 \end{array}\right.,$$ (147)

where $A$ is a complex constant, $k=\sqrt{(2m/\hbar^2)E}$ and $\phi_{\pm}(x)$ the wave function within the potential region $\vert x\vert<a/2$. It is convenient to write the even ($+$) and odd ($-$) wave functions within one line, using the definitions

$$\psi_{\pm}(x) = \psi_{L}(x)+\phi_{\pm}(x) \pm \psi_{R}(x),$$

$$\psi_{L}(x) := \left \{ \begin{array}{cc} A \sin k(x+L/2)= & -L/2<x<-a/2 \\ 0 & else \end{array}\right.$$

$$\psi_{R}(x) := \left \{ \begin{array}{cc} A \sin k(L/2-x)= & a/2<x<L/2 \\ 0 & else \end{array}\right.$$ (148)

Here, $\psi_L(x)$ is localized only in the left part $x<-a/2$ and $\psi_R(x)$ is localized only in the right part $x>a/2$ of the well.

We now use our transfer matrix formalism to obtain the equation that determines the possible energy values $E$: The solution on the left of the potential $V(x)$ is connected to the solution on the right, cf. eq. (2.54),

$$\left( \begin{array}{c} a_1 \\ b_1 \end{array}\right)= \left... ...{array}{c} \mp e^{-ikL/2} \\ \pm e^{ikL/2} \end{array}\right).$$

Using

$$\sin k(L/2+x) = \frac{1}{2i}\left(e^{ik(L/2+x)} - e^{-ik(L/2+x)}\right)$$

$$\sin k(L/2-x) = \frac{1}{2i}\left(e^{ik(L/2-x)} - e^{-ik(L/2-x)}\right),$$ (149)

we identify

$$a_1 = \frac{A}{2i} e^{ikL/2},\quad b_1 = -\frac{A}{2i} e^{-ikL/2}$$

$$a_3 = \mp \frac{A}{2i} e^{-ikL/2},\quad b_3 = \pm \frac{A}{2i} e^{ikL/2},$$ (150)

which yields two linear equations

$$e^{ikL/2} = \mp M_{11}e^{-ikL/2} \pm M_{12}e^{ikL/2}$$

$$- e^{-ikL/2}= \mp M_{12}^* e^{-ikL/2} \pm M_{22} e^{ikL/2}.$$ (151)

Here, the upper sign always holds for the even solution $\psi_+(x)$ while the lower sign holds for the odd solution $\psi_-(x)$. In fact, for a symmetric potential $V(x)=V(-x)$, $M_{22}^*=M_{11}$ and $M_{21}=M_{12}$ such that the second of the above equations is just the conjugate complex of the first. The condition that determines the possible wave vectors $k$ and therewith the energies $E=\hbar^2k^2/2m$ is

$$\pm 1 = -M_{11}(k)e^{-ikL} + M_{12}(k),$$ (152)

where we explicitly indicated the $k$-dependence of the transfer matrix elements.