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.

  • $\begingroup$ 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 $\endgroup$
    – Phoenix87
    Commented Dec 27, 2014 at 23:40

1 Answer 1


$e^{-i k x}$ 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.

Think about it this way:

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.

  • $\begingroup$ 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. $\endgroup$
    – user_35
    Commented Dec 27, 2014 at 23:47

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