For the step potential, $V=0$ for $x<0$ and $V=V_0$ for $x>0$, we have the following solution for a bounded state $E<V_0$: $$\psi_L=Ae^{ikx}+Be^{-ikx}\ \ \ \text{(left side)}$$ $$\psi_R=De^{-lx}\ \ \ \text{(right side)}$$ where $k^2=\dfrac{2mE}{\hbar^2}$ and $l^2=\dfrac{2m(V_0-E)}{\hbar^2}$. I understand how we reached these solutions, and I know by applying the proper continuity equations, we can determine $D$ and $B$ in terms of $A$. I've seen this a million times from everywhere but not once did I see the full time-dependent wavefunction. Is this because it is too hard to write down or something?

Normally, once we get the stationary states we take their linear combination and tack on the time-dependence factor to form the full wavefunction. I'm guessing I have to integrate over all $k's$ since $k$ is continuous and can range from $0$ to $\sqrt{2mE}/\hbar$. My guess is that the left wavefunction will look something like: $$\Psi_L(x,t)=\dfrac{1}{\sqrt{2\pi}}\int_0^{\sqrt{2mE}/\hbar} \phi_L(k)[Ae^{ikx}+Be^{-ikx}]e^{-i\hbar k^2 t/2m}\ dk$$ and $\Psi_R$ will take a similar form. Is my guess correct?

  • 1
    $\begingroup$ That's about the size of it - but you'll need to take into consideration the fact that the time domain is artificially split into before, during, and after the "collision phase," so you won't be able to write down a single expression for the wave function in each region. Wave packet scattering is covered in Shankar, as well as this paper for those who don't have a copy. $\endgroup$
    – J. Murray
    Sep 13, 2017 at 4:02

1 Answer 1


Since everything can be expressed in terms of the coefficient $A$ it is common to write $B=A R_+(k,\ell)$ and $D=A T_+(k,\ell)$. For $E>V_0$, one easily finds $$ R_+(k,\ell)=\frac{k-\ell}{k+\ell}\, ,\qquad T_+(k,\ell)=1+R_+(k,\ell)= \frac{2k}{k+\ell}\, . $$ with $$ k^2-\ell^2=\frac{2m V_0}{\hbar^2}\, . $$ For $E<V_0$ the coefficients $R_-(k,\ell)$ and $T_-(k,\ell)$ are different from $R_+(k,\ell)$ and $T_-(k,\ell)$: physically the solution on the right is a decaying exponentional rather than a plane wave. $R_-$ and $T_-$ can be found as usual by matching the wave function and its derivative at the discontinuity, or if you are secure in your math by analytic continuation of the $R_+$ and $T_+$ results. (The coefficients turn out to be complex.)

There is no reason to believe your integral over $k$ should be bounded or restricted to states having only energy either above or below $V_0$. Thus, your full solution ought to cover both regimes and so will be of the form \begin{align} \Psi_L(x,t)&=\frac{1}{\sqrt{2\pi}}\int_0^{\sqrt{2mV_0}/\hbar} dk\,\phi(k)[Ae^{ikx}+AR_-(k,\ell)e^{-ikx}]e^{-i\hbar k^2 t/2m}\\\ &+\frac{1}{\sqrt{2\pi}}\int_{\sqrt{2mV_0}/\hbar}^\infty dk\,\phi(k) [A e^{ikx} + A R_+(k,\ell)e^{-ikx}]^{-i\hbar k^2 t/2m} \end{align} with $\phi(k)$ your spectral function. Basically, for the range of $k$ corresponding to $E<V_0$ you use the reflection coefficient $R_-$ but for the range of $k$ corresponding to $E\ge V_0$ you use $R_+$.

On the right hand side the same idea holds: your solution will be in two parts. The first will cover $0<k<\sqrt{2mV_0}/\hbar$, will have transmission coefficient $T_-$ and an attenuation factor $e^{-\kappa x}$ since that part of the spectrum does not have enough energy to continue past the step as a plane wave. The second part will cover the range $\sqrt{2mV_0}/\hbar \le k <\infty$, will have transmission coefficient $T_+$, and will have the plane wave factor $e^{-i\ell x}$.

[Hopefully I didn't mess anything up too badly in the typesetting but watch for typos.]


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