The Tunnel Barrier: Transmission Coefficient

Next, we consider a potential that has the form of a rectangular barrier. In (2.42), we set $N=2$, $x_2=-x_1=a$, $V_1=V_3=0$, and $V_2=V>0$. Let us recall the matrices $T_1$ and $T_2$ for `step-up' and `step-down',

$$T_1 = \frac{1}{2}\left( \begin{array}{cc} (1+r)e^{-i\delta_-a} & (1... ... (1-r)e^{-i\delta_+a} & (1+r)e^{i\delta_-a} \end{array}\right)$$ (140)

$$T_2 = \frac{1}{2}\left( \begin{array}{cc} (1+1/r)e^{-i\delta_-a} & ... ...d \delta_{\pm} \equiv k_2\pm k_1,\quad r\equiv \frac{k_2}{k_1}.$$

We multiply the matrices $T_1$ and $T_2$ in order to obtain $M=T_1T_2$,

$$M = \left( \begin{array}{cc} M_{11} & M_{12} \\ M_{21} & M_{22} \end{array}\right)$$

$$M_{11} = \frac{1}{4}\left[ \frac{(1+r)^2}{r}e^{-2i\delta_-a} - \frac{(1-r)... ...2+k_2^2)^2}{k_1k_2}e^{-2ik_2a} -\frac{(k_1^2-k_2^2)^2}{k_1k_2}e^{2ik_2a}\right]$$

$$= e^{2ik_1a}\left[\frac{k_1^2+k_2^2}{2k_1k_2}i\sin(-2k_2a)+\cos(2k_2a)\right]=M_{22}^*$$

$$M_{12} = \frac{k_1^2-k_2^2}{4k_1k_2}2i\sin (2k_2a)=-M_{21}.$$ (141)

Use $(k_1^2+k_2^2)^2 =(k_1^2-k_2^2)^2 +4k_1^2k_2^2$ to find

$$\vert M_{11}\vert^2 = 1+\frac{(k_1^2-k_2^2)^2}{4k_1^2k_2^2}\sin^2(2k_2a).$$ (142)

CASE 1: $E>V$.

In this case, both $k_1$ and $k_2$ are real, and we find

$$T= \frac{1}{\frac{(k_1^2-k_2^2)^2}{4k_1^2k_2^2}\sin^2(2k_2a)+1}= ... ...lpha\sqrt{E/V-1})}{4(E/V)(E/V-1)}},\quad \alpha= \sqrt{\frac{2mVa^2}{\hbar^2}}.$$ (143)

CASE 2: $E<V$.

In this case, $k_2=i\kappa_2:=i\sqrt{(2m/\hbar^2)(V-E)}$ is complex, and we find

$$T= \frac{1}{\frac{(k_1^2+\kappa_2^2)^2}{4k_1^2\kappa_2^2} \sinh^2... ...lpha\sqrt{1-E/V})}{4(E/V)(1-E/V)}},\quad \alpha= \sqrt{\frac{2mVa^2}{\hbar^2}},$$ (144)

where we used

$$\sin ix = \frac{1}{2i}\left(e^{iix} -e^{-iix}\right) = i \sinh x.$$ (145)