# Phase velocity of a wave function across a potential energy step

I'm trying to do point c of problem 2.34 in Griffiths' book on QM:

It requires that I find the relative phase speed of the wave function outside and inside the potential. We're given the expression for the phase speed of the free particle in eq. 2.98 (p. 75), it is: $$v_{\text{phase}} = \frac{\hbar |k|}{2m} = \sqrt{\frac{E}{2m}}.$$

As the book explains, it derives from the expression of the wave function itself (eq. 2.95), $$Ψ(x,t) = Ae^{i \big( kx - \frac{\hbar k^2}{2m} t \big)},$$ by taking the ratio of the coefficients of $$t$$ and $$x$$. Inside the potential, the time-independent wave function has the form $$ψ(x) = Ce^{ilx},\qquad l = \frac{\sqrt{2m(E-V_0)}}{\hbar};$$

if I tack onto it the usual time dependence $$e^{-i\frac{E}{\hbar}t}$$, I get $$Ψ(x,t) = Ce^{i \big( lx - \frac{E}{\hbar}t \big)} = Ce^{\frac{i}{\hbar} \big( \sqrt{2m(E - V_0)} - Et\big)},$$ but this suggests that the wave function gains speed by crossing the step (increasing its potential energy), as the ratio of the coefficients of $$t$$ and $$x$$ here is $$\frac{1}{\sqrt{2m}}\frac{E}{\sqrt{E - V_0}}$$, while for $$V = 0$$ (from the equation above) it is $$\frac{1}{\sqrt{2m}}\frac{E}{\sqrt{E}}$$, and $$E > E - V_0$$. Where am I wrong?

The phase velocity has little physical significance in non-relativistic QM. You can change it at will by adding a position and time-independent potential or by includingthe $$mc^2$$ rest energy in $$E$$. The physically significant quantity is the group velocity
$$v_{\rm group}= \frac{\partial E}{\partial k}.$$ This is equal to $$\hbar k/m$$, which coincides with the classical veloity if $$\hbar k= mv_{\rm classical}$$ and is independent of any added constant potential. Consequently I do not see why Griffiths even asks the question --- never mind it seeming to give paradoxical answers.
Note that your formula $$k/{2m}=\sqrt{\frac e{2m}}$$ does not apply after the step.
• Thanks for your help. I've added a screenshot with the text of the problem. Anyway, I know that $k/2m = \sqrt{E/2m}$ only applies where V = 0, the thing that confuses me is that the phase velocity that I get for the wave inside the potential is $E/\sqrt{2m(E-V_0)}$, which is faster than outside the potential. – Arch Stanton Apr 19 '20 at 14:21
• Thanks fo thscreenshot. I see that he never asks you to compute $v_{\rm phase}$, only the probability current and that is $|\psi|^2 \hbar k/m$ not $|\psi|^2 v_{\rm phase}$. – mike stone Apr 19 '20 at 14:26
• Eq. 2.99 is about the wave velocity. It states, for the free particle, $v_\text{classical} = \sqrt{2E/m} = 2v_\text{quantum}$, where $v_\text{quantum}$ is the phase velocity of the wave function (then he derives $v_\text{group} = 2v_\text{phase}$). – Arch Stanton Apr 19 '20 at 14:32
• He even says that this is a paradox. The point is that his statement $2v_{\rm quantum}= v_{\rm classical}$ actually incorrect and nonsensical. The correct statement is that $dE/dk= v_{\rm classical}$. There is *no* physics in $v_{\rm phase}$. I expect he explain this later. I never did like Griffiths book. – mike stone Apr 19 '20 at 14:35