# Boundary Conditions in a Step Potential

I'm trying to solve problem 2.35 in Griffith's Introduction to Quantum Mechanics (2nd edition), but it left me rather confused, so I hope you can help me to understand this a little bit better.

The aim of the problem is to find the probability that a particle with kinetic energy $E>0$ will reflect when it approaches a potential drop $V_0$ (a step potential).

I started with putting up the Schrödinger equations before and after the potential drop: $x<0: V(x)=0$ and $x>0: V(x)=-V_0$.

$\psi''+k^2\psi=0, x<0$

$\psi''+\mu^2\psi=0, x>0$

where $k=\sqrt{2mE}/\hbar$ and $\mu=\sqrt{2m(E+V_0)}/\hbar$

This would give me the general solutions

$\psi(x)=Ae^{ikx}+Be^{-ikx}, x<0$

$\psi(x)=Fe^{i\mu x}+Ge^{-i\mu x},x>0$

Now, I resonate that in order to have a physically admissable solution B=0 since the second term blows up when $x$ goes to $-\infty$ and F=0 since the first term in the second row blows up when $x$ goes to $\infty$. This would leave us with the solutions

$\psi(x)=Ae^{ikx}, x<0$

$\psi(x)=Ge^{-i\mu x},x>0$

which I then could use boundary conditions to solve. However, I realise that this is wrong since I need $B$ to calculate the refection probability. In the solution to this book they get the following general solutions (they don't say how the got them though).

$\psi(x)=Ae^{ikx}+Be^{-ikx}, x<0$

$\psi(x)=Fe^{i\mu x},x>0$

This is not very well explained in the book so I would really appriciate if someone could explain how to decide what parts of the general solutions that I should remove in order to get the correct general solution for a specific problem.

• as long as the momenta are real the exponentials will oscillate without blowing up, so you only have to be concerned with the cases where momenta are imaginary Commented Dec 27, 2014 at 23:40

$e^{-i k x}$ does not blow up as $x \rightarrow -\infty.$ You're thinking in terms of real exponentials, but this is a complex exponential. That is, as long as $k$ is real we have:

1. $\lim\limits_{x \rightarrow -\infty} e^{- k x} = \infty$
2. $\lim\limits_{x \rightarrow -\infty} e^{- i k x}$ does not exist (since $e^{- i k x} = \cos{kx} - i \sin{kx}$).

So that explains why the $B$ term is still there in the solution.

The reason that Griffiths discounts the $G$ term is because it represents a reflected wave traveling from the positive $x$ direction.

The problem at hand is of a particle coming from the $-x$ direction and encountering a sudden potential drop. The particle arriving gives rise to the term $A e^{i k x}$, as this is a traveling wave moving in the $+x$ direction.

When the particle encounters the barrier, it can either reflect (giving rise to the term $B e^{- i k x}$) or transmit (giving rise to the term $F e^{i \mu x}$).

Note that there is no circumstance under which a particle could be coming from the $+x$ direction in the region to the right of the potential drop, which is what the term $G e^{- i \mu x}$ would imply. A particle comes from the left, and then either moves forward to the right or reflects back to the left. A situation under which the particle travels towards the potential drop from the $+x$ direction is not physically admissible under this circumstance.

So we drop the term $G e^{- i \mu x}$, and we're left with

$\psi(x) = \begin{cases} A e^{i k x} + B e^{- i k x}, & x < 0 \\ F e^{i \mu x}, & x > 0 \end{cases}$

as described in the book.

• Also, to clarify: This wouldn't usually be called a finite square well, but rather a potential drop. A "finite square well" implies that there is another point further along the $x$ axis at which the potential jumps back up to the previous value. Commented Dec 27, 2014 at 23:47