# Potential step and exponential decay?

Let us say we have a wave going from a region ($x<0$) where the potential is $U_1$ to a region ($x>0$) where the potential is $U_2$. The wave function in the second region takes the form: $$\phi=e^{i(\omega t-k_2x)}$$ That is a wave travelling to the right. This was assuming that $k_2$ is real. If we know take $k_2$ to be imaginary so $k_2=i\kappa$ where $\kappa$ is real and positive then: $$\phi =e^{i(\omega t)+\kappa x}$$ But this increases with increasing $x$, exactly the opposite that you would expect. Why is this wrong?

• why should $\kappa$ be positive? – danimal May 20 '15 at 9:45
• @danimal since $k_2=\sqrt{\frac{2m}{\hbar^2}(E-U_2)}$ and thus $\kappa = \sqrt{\frac{2m}{\hbar^2}(U_2-E)}$ which is positive – Quantum spaghettification May 20 '15 at 10:04
• Hi. Maybe you should post, when you have some time the rest of the calculations you've made. – Constantine Black May 20 '15 at 10:04
• Is E greater than the potential 1 and 2? Thanks – Constantine Black May 20 '15 at 10:13
• @ConstantineBlack when we say that $k_2$ is imaginary then $U_1 \le E \le U_2$ – Quantum spaghettification May 20 '15 at 10:27

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and this is not a problem if there is a valid and sufficient (w.r.t. the experiment) alternative solution. The reasoning behind this is that if that solution ($\psi_\mathrm{infinite}$ was a part of the total solution,

$\Psi = A \psi_\mathrm{finite} + B \psi_\mathrm{infinite}$,

any real solution $\Psi$ (with finite energy/density/...) must have $B=0$.

In this case, the other solution, as already said in the comments, comes from the negative square root, which gives $\kappa$ a negative value. Note that even for the non-imaginary case of $k_2$, you also have two solutions: a left- and right-moving wave $\propto \mathrm{e}^{\pm \mathrm{i} k_2 x}$. So it is only logical to consider both signs.

To find the solution to your problem:

Area 1: U1,E>U1

The Schrodinger equation and it's solution is: $$ψ'' + {2m \over \hbar ^2}(E-U1)ψ=0 ---> ψ_A=e^{ik_1 x} + c_1e^{-i k_1 x}$$ where $k_1 ^2 = {2m \over \hbar ^2}(E - U1)$.

Area 2: U2, E

As in the above: $$ψ'' - {2m \over \hbar ^2}(U2-E)ψ=0--> Ψ_Β = c_2 e^{k_2 x} + c_3 e^{-k_2 x}$$ where $k_2 ^2 = {2m \over \hbar ^2}(U2-E)$. Note that $U2-E>0$

Note that in area 2 when x goes to infinity, the solution of the wavefunction must be zero. Thus $c_2 =0$. Also from the boundaries conditions on zero$$ψ_Α(0)=ψ_Β(0)$$ and $$ψ' _Α (0) =ψ' _Β(0)$$ So, we have: $$1+c_1 = c_3$$ and $${ik_1(1-c_1) \over k_2}=-c_3$$. Thus, we know the factors $c_1 and c_3$ and the final solution is: $$ψ_Α= e^{ik_1 x } +c_1 e^{-i k_1 x}~,~~~~~~ψ_B=c_3 e^{-k_2 x}$$

Also one can calculate from here the reflection and transition factors are: $$R=|c_1|^2 =1--->T=0$$.

I hope this helps.