Solve Potential Barrier Homework: Find Transmission/Reflection Coeff.

[E92M3]

Homework Statement

Find the transmission and reflection coefficients for:
V(x)=\left\{\begin{matrix}<br /> V_0, &amp; -a\leq x\leq a\\ <br /> 0, &amp; elsewhere<br /> \end{matrix}\right.

Homework Equations

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

The Attempt at a Solution

I have successfully got the solution for:
E=V_0
and
E&gt;V_0
But I am having trouble with
E&lt;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}<br /> \frac{\partial \psi_{I}}{\partial x}<br /> \end{matrix}\right|_{x=-a}=\left.\begin{matrix}<br /> \frac{\partial \psi_{II}}{\partial x}<br /> \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}<br /> \frac{\partial \psi_{II}}{\partial x}<br /> \end{matrix}\right|_{x=a}=\left.\begin{matrix}<br /> \frac{\partial \psi_{III}}{\partial x}<br /> \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?

 

[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?