Say we are dealing with a 1-d non-relativistic quantum mechanics problem with a time independent potential $V\left(x\right)$. This satisfies the Schrodinger equation:

$$ i \hbar \frac{\partial}{\partial t} \psi\left(x,t\right) = \left[\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V\left(x\right)\right] \psi\left(x,t\right) $$

We can solve the TISE, and find the corresponding energy eigenstates of the Hamiltonian: $\psi_{k}$. We can write any wave packet in the following form:

$$\psi\left(x,t\right) = \sum_{n} a_{n} \psi_{n}e^{-iE_{n}t} + \int^{\infty}_{-\infty} \psi\left(k\right) a\left(k\right) e^{-i E\left(k\right) t}dk $$

Now we further assume that $V\left(x\right)$ monotonically increases from 0 at $x= -\infty$ to $+ \infty$ at $x=+\infty$. Imagine we scatter an arbitrary wave packet coming from the left so that $\psi\left(x,0\right)$ is moving toward higher $V\left(x\right)$.

Question: Can we say anything about what part of the wave packet will be transmitted and reflected just with this information?

My intuition is that because $V\left(+\infty\right) = \infty $, all of the wave packet should be reflected independently of its original shape. However, I am not sure because maybe the bounded energy eigenstates might not be 0 there...?

What if we imagine that the original wave packet is similar to a gaussian wave packet that is centered at $a \ll 0 $ at $ t= 0 $? so that the wave packet is essentially in $V\left(x\right) = 0$ at $t = 0$ and it must be the case that:

$$\psi\left(x,0\right) = \int^{\infty}_{-\infty} \psi\left(k\right) a\left(k\right) dk$$

I.e., it is formed only by scattered eigenstates ( not super sure if this is actually true). Would this wave packet then be completely reflected?

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    $\begingroup$ There will not be any bound states. This is sort of physically obvious since they would have energy $\ge 0$. $\endgroup$ Aug 2, 2017 at 0:41
  • 1
    $\begingroup$ However, the "physically obvious" is wrong. Von Neumann and Wigner in 1929 showed that bound states in the continuum are possible. As I understand it, their (positive) potential (in 3D, I think) is oscillatory. I speculate that there would not be bound states with your monotonic potential. $\endgroup$ Aug 2, 2017 at 5:22
  • $\begingroup$ Landau&Livshits QM has some interesting exact solutions under the guise of problems, including scattering from the infinite potentials. $\endgroup$
    – Roger V.
    Jul 2, 2021 at 7:15

1 Answer 1


Integrate the Schrödinger equation by time and you obtain

$\psi(x,t) = \psi(x,0) + \frac{1}{i \hbar} \int_0^t\left[- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial^2 x} + V(x)\right]\psi(x,t')dt'$.

Every time we can make a mode Expansion of $\psi(x,t)$ in Terms of $e^{i\frac{p}{\hbar}x}$. If this wave is reflected, momentum $p$ goes to $-p$ and thus for the reflected wave function $\psi_r$ we have the relation $\psi_r(x,t)=\psi(-x,t)$, because the mode Expansion is in Terms of $e^{i\frac{-p}{\hbar}x}$.

You obtain the Amplitude for reflection (square of it is reflection probability) by computing the overlap of $\psi(x,t)$ with the Initial, but reflected wave function $\psi_r(x,0)$, i.e. the reflection Amplitude is

$r = \int d^3x \psi_r^*(x,0) \psi(x,t)$.

We overlap with the reflected Amplitude at $t=0$, because the outcome of the scattering will be the Initial state, but reflected.

Assume the wave function as

$\psi(t) = \sum_{n=0}^\infty a_n(t)e^{i\frac{p_nx-E_nt}{\hbar}} + \int d^3k a(k,t)e^{ikx-iE_kt}$

to eliminate the kinetic energy term in time integrated Schrödinger equation. After multiplication of it with the complex conjugate of reflected Initial wave function and Integration over space you will get for the discrete sum

$r = \int_0^t dt'\frac{1}{i \hbar}\int d^3x \sum_{n,m}a_n^*(t=0)a_{m}(t=0)V(x)e^{i \frac{(p_n+p_m)x-(E_m-E_n)t'}{\hbar}} \mapsto \sum_{n,m} (a_n^*a_m)(t=0)\frac{1}{i \hbar}F_V(p_n+p_m)\delta(E_n-E_m)$.

and for the continuous term a very similar result. Reflection Amplitude depends strongly Fourier-transformed potential $F_V$, e.g. if $F_V \approx \delta(p_m+p_n)$ and $a_m = a_{-n}$ then by orthonormality it would hold $r=1$, i.e. a total reflection.

In General, the incoming wave packet will be scattered to different ones, but one expects that in the Limit $t \mapsto \infty$, all These (modified) wave packets are also reflected off this potential wall.

  • $\begingroup$ Thanks for your comment @kryomaxim, could you give me some more intuition for why $r = \int d^{3}x \psi_{r}^{\star}(x,0) \psi(x,t)$ is true ?, also are you taking $t \rightarrow \infty$?, the limit of long times. $\endgroup$
    – ace7047
    Aug 3, 2017 at 2:51
  • $\begingroup$ We can express every wave function in Terms of orthonormal Basis functions. $\psi_r(t,0)$ is one of such a Basis function. The Limit $t \mapsto \infty$ is an asymptotic Limit, it is taken for having energy conservation in the scattering (note the uncertainty between energy and time). $\endgroup$
    – kryomaxim
    Aug 3, 2017 at 6:41

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