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
Best Answer
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.