# Transmission amplitude for asymmetrical potential barrier

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

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, $$-\infty , and similarly let $$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)$$.