0
$\begingroup$

The problem is in 1D. So if I have a potential barrier $V(x)$ from $[-a,a]$ where $V(x)$ can be any function (also an asymmetrical function). Is the transmission amplitude for a particle travelling through that barrier from the right (from + to -) different from the transmission amplitude for a particle coming from the left (from - to +)?

Intuitively I'd say that it doesnt make a difference but I wanted to proof it.

My approach was to calculate the transmission amplitudes for both cases and compare them. For the particle comming from the left I got.
$$e^{-ika}+R_le^{ika}=\Psi_l(-a)$$ $$ik(e^{-ika}-R_le^{ika})=\Psi_l'(-a)$$ $$T_le^{ika}=\Psi_l(a)$$ $$ikT_le^{ika}=\Psi_l'(a)$$ where $\Psi_l$ fulfills the Schrödinger equation $\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi_l+V(x)\Psi_l=E\Psi_l$ for $x\in[-a,a]$ for the particle coming from the right I got $$e^{-ika}+R_re^{ika}=\Psi_r(-a)$$ $$ik(e^{-ika}-R_re^{ika})=\Psi_r'(-a)$$ $$T_re^{ika}=\Psi_r(a)$$ $$ikT_re^{ika}=\Psi_r'(a)$$ where $\Psi_r$ fulfills the Schrödinger equation $\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi_r+V(-x)\Psi_r=E\Psi_r$ for $x\in[-a,a]$
if one looks now at the third and fourth equation for the left and for the right particle and compare them one can see that $T_l=T_r$ only if $\Psi_l(a)=\Psi_r(a)$ and $\Psi_l'(a)=\Psi_r'(a)$. So it seems like the transmission amplitudes are generally actually not the same. Only if $V(x)=V(-x)$ so if its symmetrical around 0. But I cant make up an example for $V(x)$ that is asymmetrical and is calculateable. The Schrödinger euqation becomes just too hard to solve for these $V(x)$. Does someone have a good example ? Or was my calculation maybe wrong and the amplitudes are indeed always the same. But my calculation was right, is there a good explanation for why the differ in gerneral? So many questions. I hope some can be answered

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Consider your one-dimensional Schroedinger equation $$ -\frac{d^2\psi}{dx^2} +V(x)\psi=E\psi $$ where $V(x)$ is zero except in a finite interval $[-a,a]$ near the origin. $V$ doe not have to be left-right symmetric.

Let $L$ denote the left asymptotic region, $-\infty <x<-a$, and similarly let $R$ denote $\infty>x>a$. For $E=k^2$ and $k>0$ there will be scattering solutions of the form $$ \psi_{L,k}(x)= \cases{e^{ikx} +r_L(k)e^{-ikx},&$ x\in L,$\cr t_L(k)e^{ikx},&$ x\in R$,} $$ describing waves incident on the potential $V(x)$ from the left. For the same $k>0$ there will be solutions with waves incident from the right
$$ \psi_{R,k}(x)=\cases{ t_R(k)e^{-ikx},&$ x \in L,$\cr e^{-ikx}+r_R(k)e^{ikx},&$ x\in R.$} $$
The wavefunctions in $[-a,a]$ will naturally be more complicated. Observe that $[\psi_k(x)]^*$ is also a solution of the Schr{"o}dinger equation.

Now recall that for any solutions $u_1$, $u_2$ of the same $E$ Schroedinger equation the Wronskian $$ W(x)=u_1u_2'-u_1'u_2 $$ is independent of $x$. By choosing $u_1$, $u_2$ to be $\psi_R(x)$ and $\psi_L(x)$ you should be able to evaluate the Wronkian in the left and right regions and show that $t_L(k)=t_R(k)$.

The reflection coefficients need not be the same. They can differ by a phase.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.