How to plot the transmission coefficient as a function of E//V_0?

Mar 18, 2018

Here we follow what is readily found in McQuarrie & Simon (pp. 142-143). We found that:

$T = \left\{\begin{matrix}{\left[1 + \frac{{\sinh}^{2} \left[{v}_{0}^{1 / 2} {\left(1 - r\right)}^{1 / 2}\right]}{4 r \left(1 - r\right)}\right]}^{- 1} & 0 < r < 1 \\ {\left[1 + \frac{{\sin}^{2} \left[{v}_{0}^{1 / 2} {\left(r - 1\right)}^{1 / 2}\right]}{4 r \left(r - 1\right)}\right]}^{- 1} & r \ge 1\end{matrix}\right.$

where:

• $T$ is the transmission coefficient (i.e. the fraction of particles that tunnel).
• $r = \frac{E}{V} _ 0 = \frac{\epsilon}{v} _ 0$ is the ratio of the particle energy to the barrier height.
• v_0 = (2ma^2V_0)/ℏ^2 is a unitless variable for the height of the barrier.
• epsilon = (2ma^2E)/ℏ^2 is a unitless variable for the energy of the particle.

and because I thought it would be interesting, besides ${v}_{0} = 10$, I also graphed ${v}_{0} = 10 , 20 , 40$:

It is rather interesting that the higher the barrier gets, the more values of $E$ there are that allow for 100% transmission. Here we have:

$\underline{{V}_{0} \text{ "" "" "" "" "E//V_0" "" "" "%T" }}$
10ℏ^2//2ma^2" "color(white)(.)1.99" "" "" "" "100%
$\text{ "" "" "" "" "" } 4.95$

20ℏ^2//2ma^2" "color(white)(.)1.49" "" "" "" "
$\text{ "" "" "" "" "" } 2.97$

40ℏ^2//2ma^2" "color(white)(.)1.25" "" "" "" "
$\text{ "" "" "" "" "" } 1.99$
$\text{ "" "" "" "" "" } 3.22$
$\text{ "" "" "" "" "" } 4.95$

It just so happens that at the classical limit, we have

lim_(V_0 -> oo) T = 100% for $E > {V}_{0}$,

with a rigid cutoff at $E = {V}_{0}$. That's just what we expect. Classically, particles can't tunnel unless they have enough energy. Here is the same graph at ${v}_{0} = {10}^{6}$:

DISCLAIMER: LONG AND RIGOROUS DERIVATION!

The potential is set up as:

$V \left(x\right) = \left\{\begin{matrix}0 & x < 0 \\ {V}_{0} & 0 < x < a \\ 0 & x > a\end{matrix}\right.$

We of course set $E < {V}_{0}$ to allow tunnelling, and the eigenfunctions are:

${\psi}_{I} \left(x\right) = A {e}^{i k x} + B {e}^{- i k x} , \text{ } x < 0$
${\psi}_{I I} \left(x\right) = C {e}^{k ' x} + D {e}^{- k ' x} , \text{ } 0 < x < a$
${\psi}_{I I I} \left(x\right) = F {e}^{i k x} + G {e}^{- i k x} , \text{ } x > a$

where k = sqrt((2mE)/ℏ^2) and k' = sqrt((2m(V_0 - E))/ℏ^2).

If we restrict the particle to only come in from the left, $G = 0$. Then, the transmission coefficient is given by

$T = \frac{| F {|}^{2}}{| A {|}^{2}}$

The continuity conditions are:

$\underline{{\psi}_{I} \left(0\right) = {\psi}_{I I} \left(0\right)}$:
$A + B = C + D$ $\text{ "" "" "" "" "" } \boldsymbol{\left(1\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I} \left(0\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I} \left(0\right)}{\mathrm{dx}}}$:
$i k \left(A - B\right) = k ' \left(C - D\right)$ $\text{ "" "" } \boldsymbol{\left(2\right)}$

$\underline{{\psi}_{I I} \left(a\right) = {\psi}_{I I I} \left(a\right)}$:
$C {e}^{k ' a} + D {e}^{- k ' a} = F {e}^{i k a}$ $\text{ "" "" } \boldsymbol{\left(3\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I I} \left(a\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I I} \left(a\right)}{\mathrm{dx}}}$:
$k ' C {e}^{k ' a} - k ' D {e}^{- k ' a} = i k F {e}^{i k a}$$\text{ } \boldsymbol{\left(4\right)}$

Now, we take $i k \left(1\right) + \left(2\right)$ to get:

$\textcolor{g r e e n}{2 i k A = \left(i k + k '\right) C + \left(i k - k '\right) D}$

Then, take $k ' \left(3\right) + \left(4\right)$ to get $C$ in terms of $F$:

$2 k ' C {e}^{k ' a} = \left(k ' + i k\right) F {e}^{i k a}$

$\textcolor{g r e e n}{C = \left(\frac{k ' + i k}{2 k '}\right) F {e}^{i k a} {e}^{- k ' a}}$

and take $k ' \left(3\right) - \left(4\right)$ to get $D$ in terms of $F$:

$2 k ' D {e}^{- k ' a} = \left(k ' - i k\right) F {e}^{i k a}$

$\textcolor{g r e e n}{D = \left(\frac{k ' - i k}{2 k '}\right) F {e}^{i k a} {e}^{k ' a}}$

Plug it back into the first result to get:

$2 i k A = \left(i k + k '\right) \left(\frac{k ' + i k}{2 k '}\right) F {e}^{i k a} {e}^{- k ' a} + \left(i k - k '\right) \left(\frac{k ' - i k}{2 k '}\right) F {e}^{i k a} {e}^{k ' a}$

$= \left[\left(i k - k '\right) \left(k ' - i k\right) {e}^{k ' a} + \left(i k + k '\right) \left(k ' + i k\right) {e}^{- k ' a}\right] \frac{F {e}^{i k a}}{2 k '}$

$= \left[\left({k}^{2} - k {'}^{2} + 2 i k k '\right) {e}^{k ' a} + \left(k {'}^{2} - {k}^{2} + 2 i k k '\right) {e}^{- k ' a}\right] \frac{F {e}^{i k a}}{2 k '}$

$= \left[\left({k}^{2} - k {'}^{2}\right) \left({e}^{k ' a} - {e}^{- k ' a}\right) + 2 i k k ' \left({e}^{k ' a} + {e}^{- k ' a}\right)\right] \frac{F {e}^{i k a}}{2 k '}$

Now, we use the identities $2 \sinh \left(x\right) = {e}^{x} - {e}^{- x}$ and $2 \cosh \left(x\right) = {e}^{x} + {e}^{- x}$ to get:

$2 i k A = \left[2 \left({k}^{2} - k {'}^{2}\right) \sinh \left(k ' a\right) + 4 i k k ' \cosh \left(k ' a\right)\right] \frac{F {e}^{i k a}}{2 k '}$

Solve for $\frac{F}{A}$ to get:

$\frac{F}{A} = \frac{2 i k}{\left[2 \left({k}^{2} - k {'}^{2}\right) \sinh \left(k ' a\right) + 4 i k k ' \cosh \left(k ' a\right)\right] {e}^{i k a} / \left(2 k '\right)}$

$= \frac{4 i k k ' {e}^{- i k a}}{2 \left({k}^{2} - k {'}^{2}\right) \sinh \left(k ' a\right) + 4 i k k ' \cosh \left(k ' a\right)}$

Using the definition of $T$,

T -= (F^"*"F)/(A^"*"A) = [(4ikk'e^(-ika))/[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]]^"*" [(4ikk'e^(-ika))/[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]]

$= \frac{4 {k}^{2} k {'}^{2}}{{\left({k}^{2} - k {'}^{2}\right)}^{2} {\sinh}^{2} \left(k ' a\right) + 4 {k}^{2} k {'}^{2} {\cosh}^{2} \left(k ' a\right)}$

Next, use the identity that ${\cosh}^{2} \left(x\right) = 1 + {\sinh}^{2} \left(x\right)$ to get:

$T = \frac{4 {k}^{2} k {'}^{2}}{{\left({k}^{2} - k {'}^{2}\right)}^{2} {\sinh}^{2} \left(k ' a\right) + 4 {k}^{2} k {'}^{2} + 4 {k}^{2} k {'}^{2} {\sinh}^{2} \left(k ' a\right)}$

$= \frac{4}{4 + {\left({k}^{2} - k {'}^{2}\right)}^{2} / \left({k}^{2} k {'}^{2}\right) {\sinh}^{2} \left(k ' a\right) + 4 {\sinh}^{2} \left(k ' a\right)}$

$= \frac{4}{4 + \frac{{k}^{4} - 2 {k}^{2} k {'}^{2} + k {'}^{4}}{{k}^{2} k {'}^{2}} {\sinh}^{2} \left(k ' a\right) + \frac{4 {k}^{2} k {'}^{2}}{{k}^{2} k {'}^{2}} {\sinh}^{2} \left(k ' a\right)}$

$= \frac{4}{4 + \frac{{k}^{4} + 2 {k}^{2} k {'}^{2} + k {'}^{4}}{{k}^{2} k {'}^{2}} {\sinh}^{2} \left(k ' a\right)}$

$= \frac{4}{4 + {\left({k}^{2} + k {'}^{2}\right)}^{2} / \left({k}^{2} k {'}^{2}\right) {\sinh}^{2} \left(k ' a\right)}$

Then, insert $k$ and $k '$ in terms of $E$ and ${V}_{0}$ to get:

T = (4)/(4 + ((2mV_0)/ℏ^2)^2/(((2m)/ℏ^2)^2 E(V_0 - E))sinh^2(sqrt((2ma^2(V_0 - E))/ℏ^2)))

= 4/(4 + (V_0^2)/(E(V_0 - E))sinh^2(sqrt((2ma^2(V_0 - E))/ℏ^2)))

Now, let the unitless quantities v_0 = (2ma^2V_0)/ℏ^2, and epsilon = (2ma^2E)/ℏ^2. This means that:

V_0 = (ℏ^2v_0)/(2ma^2), " "E = (ℏ^2epsilon)/(2ma^2)

and we then rewrite $T$ as:

T = 4/(4 + (((ℏ^2)/(2ma^2))^2v_0^2)/((ℏ^2epsilon)/(2ma^2)((ℏ^2v_0)/(2ma^2) - (ℏ^2epsilon)/(2ma^2)))sinh^2(sqrt(v_0 - epsilon)))

= 4/(4 + (cancel(((ℏ^2)/(2ma^2))^2)v_0^2)/(cancel(((ℏ^2)/(2ma^2))^2) epsilon(v_0 - epsilon))sinh^2(sqrt(v_0 - epsilon)))

$= \frac{1}{1 + \frac{{v}_{0}^{2}}{4 \epsilon \left({v}_{0} - \epsilon\right)} {\sinh}^{2} \left(\sqrt{{v}_{0} - \epsilon}\right)}$

Finally, let $r = \frac{E}{V} _ 0 = \frac{\epsilon}{v} _ 0$ so that we can graph $T$ vs. $E / {V}_{0} \equiv r$:

$T = \frac{1}{1 + \frac{\cancel{{v}_{0}^{2}}}{\frac{4 \epsilon}{v} _ 0 \cancel{{v}_{0}^{2}} \left(1 - r\right)} {\sinh}^{2} \left(\sqrt{{v}_{0} \left(1 - r\right)}\right)}$

$= \overline{\underline{|}} \text{ "1/(1 + [sinh^2[v_0^(1//2)(1 - r)^(1//2)]]/(4r(1 - r)))" } |$

Now, when we plot this, if we want $r > 1$, i.e. $E > {V}_{0}$, then we would need to use the identity $\sinh \left(i x\right) = i \sin x$. So, we separate our result into two segments:

$\textcolor{b l u e}{T = \left\{\begin{matrix}{\left[1 + \frac{{\sinh}^{2} \left[{v}_{0}^{1 / 2} {\left(1 - r\right)}^{1 / 2}\right]}{4 r \left(1 - r\right)}\right]}^{- 1} & 0 < r < 1 \\ {\left[1 + \frac{{\sin}^{2} \left[{v}_{0}^{1 / 2} {\left(r - 1\right)}^{1 / 2}\right]}{4 r \left(r - 1\right)}\right]}^{- 1} & r \ge 1\end{matrix}\right.}$

Let's choose ${v}_{0} = 10$. Then this produces the following graph: