Tunnel barrier potential within well

For a rectangular tunnel barrier of width $a$ and height $V$, that is $V(x)=V\theta(a/2-\vert x\vert)$, we have calculated the transfer matrix $M$ before (note that now the width is $a$ and not $2a$, $\sin(ix)=i\sinh(x)$, and $\cos(ix)=\cosh(x)$:

$$M_{11} = e^{ika}\left[\cosh (\kappa a) +i\frac{\varepsilon_-}{2}\sinh (\kappa a)\right ]$$

$$M_{12} = i\frac{\varepsilon_+}{2}\sinh (\kappa a)$$

$$\varepsilon_{\pm} := \frac{\kappa}{k}\pm \frac{k}{\kappa},\quad k = \sqrt{(2m/\hbar^2)E}, \quad \kappa = \sqrt{(2m/\hbar^2)(V-E)}.$$ (154)

From this and Eq. (2.76), we obtain

$$\pm 1 = -e^{ik(a-L)}\left[\cosh (\kappa a) +i\frac{\varepsilon_-}{2}\sinh (\kappa a)\right ] + i\frac{\varepsilon_+}{2}\sinh (\kappa a)$$ (155)

We multiply this equation by $e^{-ik(a-L)/2}$ and take the real part to obtain two equations for the even and the odd case. We can check that taking the imaginary part leads to the same result. Using

$$\coth (x/2) = \frac{\sinh x}{\cosh x -1} = \frac{\cosh x +1}{\sinh x},$$ (156)

we obtain

$$1 = \frac{\kappa}{k}\tan \left(k\frac{a-L}{2}\right)\tanh \left(\frac{\kappa a}{2}\right),$$

$$1 = \frac{\kappa}{k}\tan \left(k\frac{a-L}{2}\right)\coth \left(\frac{\kappa a}{2}\right), .$$ (157)