# Homework Help: Potential Barrier

1. Oct 3, 2009

### E92M3

1. The problem statement, all variables and given/known data
Find the transmission and reflection coefficients for:
$$V(x)=\left\{\begin{matrix} V_0, & -a\leq x\leq a\\ 0, & elsewhere \end{matrix}\right.$$

2. Relevant equations
$$\frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}\psi+V\psi=E\psi$$

3. The attempt at a solution
I have successfully got the solution for:
$$E=V_0$$
and
$$E>V_0$$
But I am having trouble with
$$E<V_0$$
Here's my attempt:
First, I denoted the left of the barrier as region I, within the barrier as region II, and to the right of the barrier as region III.
Rearranging the schrodinger Equation I get:
$$\frac{\partial^2 }{\partial x^2}\psi_{I}=\frac{-2m}{\hbar^2}(E)\psi_{I}$$
$$\frac{\partial^2 }{\partial x^2}\psi_{II}=\frac{2m}{\hbar^2}(V_0-E)\psi_{II}$$
$$\frac{\partial^2 }{\partial x^2}\psi_{III}=\frac{-2m}{\hbar^2}(E)\psi_{III}$$
The solutions are:
$$\psi_{I}=Ae^{ikx}+Be^{-ikx}$$
$$\psi_{II}=Ce^{-lx}+De^{lx}$$
$$\psi_{I}=Fe^{ikx}$$
where:
$$k=\frac{\sqrt{2mE}}{\hbar}$$
$$l=\frac{\sqrt{2m(V_0-E)}}{\hbar}$$
The -ikx term omitted since there's assumed to be no incoming wave from +x-direction.
Applying the boundary conditions:
$$\psi_{I}(-a)=\psi_{II}(-a)\Rightarrow Ae^{-ika}+Be^{ika}=Ce^{la}+De^{-la} \Rightarrow (1)$$

$$\left.\begin{matrix} \frac{\partial \psi_{I}}{\partial x} \end{matrix}\right|_{x=-a}=\left.\begin{matrix} \frac{\partial \psi_{II}}{\partial x} \end{matrix}\right|_{x=-a}\Rightarrow ik(Ae^{-ika}-Be^{ika})=l(-Ce^{la}+De^{-la})\Rightarrow (2)$$

$$\psi_{II}(a)=\psi_{III}(a)\Rightarrow Ce^{-la}+De^{la}=Fe^{ika}\Rightarrow (3)$$

$$\left.\begin{matrix} \frac{\partial \psi_{II}}{\partial x} \end{matrix}\right|_{x=a}=\left.\begin{matrix} \frac{\partial \psi_{III}}{\partial x} \end{matrix}\right|_{x=a}\Rightarrow l(-Ce^{-la}+De^{la})=ikFe^{ika}\Rightarrow (4)$$

From (3) we have:

$$C=Fe^{ika}e^la-De{2la}$$

Putting that into (4) we get:

$$D=\frac{1}{2}(1+\frac{ik}{l})Fe^{ika}e^{-la}$$

From (3) we also have:

$$D=Fe^{ika}e^-{la}-Ce^{-2la}$$

Putting that into (4) we now get:

$$C=\frac{1}{2}(1-\frac{ik}{l})Fe^{ika}e^{la}$$

Rearranging (2) we get:

$$Ae^{-ika}-Be^{ika}=\frac{il}{k}(De^{-la}-Ce^{la})$$

Summing the equation above with (1) we get::

$$2Ae^{-ika}=Ce^{la}+De^{-la}-\frac{il}{k}(De^{-la}-Ce^{la})=Ce^{la}(1+\frac{il}{k})+De^{-la}(1-\frac{il}{k})$$

Now we put in the the C and D that we got from playing with (3) and (4) before and get:

$$2Ae^-{ika}=\frac{1}{2}(1-\frac{ik}{l})Fe^{ika}e^{la}e^{la}(1+\frac{il}{k})+\frac{1}{2}(1+\frac{ik}{l})Fe^{ika}e^{-la}e^{-la}(1-\frac{il}{k})$$

$$2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (1-\frac{ik}{l})(1+\frac{il}{k}) e^{2la} +\frac{1}{2} (1+\frac{ik}{l})(1-\frac{il}{k}) e^{-2la} \right ]$$

$$2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (1+\frac{il}{k}-\frac{ik}{l}+1)e^{2la}+\frac{1}{2} (1-\frac{il}{k}+\frac{ik}{l}+1)e^{-2la} \right ]$$

$$2Ae^{-ika}=Fe^{ika}\left [\frac{1}{2} (2+ i\frac{l^2-k^2}{kl})e^{2la}+\frac{1}{2} (2-i\frac{l^2-k^2}{kl} )e^{-2la} \right ]$$

$$2Ae^{-ika}=Fe^{ika}\left [e^{2la}+ i\frac{l^2-k^2}{kl}\frac{1}{2} e^{2la}+e^{-2la} -\frac{1}{2}i\frac{l^2-k^2}{kl} e^{-2la} \right ]$$

$$2Ae^{-ika}=Fe^{ika}\left [(e^{2la}+e^{-2la})+( i\frac{l^2-k^2}{kl})(\frac{1}{2} e^{2la} -\frac{1}{2} e^{-2la} )\right ]$$

$$2Ae^{-ika}=Fe^{ika}\left [2cosh(2la)+( i\frac{l^2-k^2}{kl})sinh(2la)\right ]$$

$$F=\frac{Ae^{-2ika}}{cosh(2la)+( i\frac{l^2-k^2}{2kl})sinh(2la)}$$

$$T=\frac{\left | F \right |^2}{\left | A \right |^2}=\frac{e^{-2ika}}{cosh(2la)+( i\frac{l^2-k^2}{2kl})sinh(2la)}\frac{e^{2ika}}{cosh(2la)-( i\frac{l^2-k^2}{2kl})sinh(2la)}$$

$$=\frac{1}{cosh^2(2la)+( \frac{l^2-k^2}{2kl})^2sinh^2(2la)}$$

$$=\frac{1}{1+sinh^2(2la)+\frac{1}{4}( \frac{l^2}{k^2}+\frac{k^2}{l^2}-2)sinh^2(2la)}$$

I am now stuck... I can't get the sinh(2la) that I wanted. Did I do snmething wrong?

Last edited: Oct 3, 2009
2. Oct 3, 2009

### gabbagabbahey

Everything look fine to me (although you could simplify $\frac{l^2}{k^2}+\frac{k^2}{l^2}$ significantly)....do you know what your final answer is supposed to look like?

Last edited: Oct 3, 2009
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