From de Broglie's postulates to Shrödinger's equation

[quasar987]

TL;DR Summary

Trying to make sense of a vague comment by Susskind

So I was watching one of Susskind's lecture on Youtube* and about the 37:00 mark he has this equation relating the frequency of a matter wave to its wavelength:

$$\nu = \frac{h}{2m\lambda^2}.$$

This is arrived at by assuming that matter has a wavelike nature and that the energy and momentum formula for the photon ($E=h\nu$ and $p=h/\lambda$) still hold true for the matter wave. Then Susskind said that this is essentially just Shr¨ödinger's equation, or at least that Schrödinger was looking for an equation which would result in this relation between $\nu$ and $\lambda$. But if I set $\psi(x,t) = e^{i(x-vt)}$ and use $v=\lambda\nu$, then feeding $\psi$ into Shrödinger's equation

$$\frac{1}{2m}\frac{\partial^2}{\partial x^2}\psi = -\frac{i}{h}\frac{\partial}{\partial t}\psi,$$

I find instead

$$\nu = \frac{h}{2m\lambda}.$$

What am I missing??

*

 

[etotheipi]

The Schrödinger equation for a free particle will read (I think you have dropped a factor of $2\pi$)$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$Then if you let $\psi(x,t) = Ae^{i(kx-\omega t)}$, you have $\frac{\partial \psi}{\partial t} = - i \omega \psi$ and $\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi$. If you plug those results in, and use that $\hbar k = p$, you will notice that$$\hbar \omega = \frac{\hbar^2 k^2}{2m} \implies h\nu = \frac{p^2}{2m}$$Now the velocity of a particle is identified as the group velocity, $v_g = \frac{\partial \omega}{\partial k}$, which is twice the phase velocity, $v_p = \frac{\omega}{k}$, i.e. $v_g = 2v_p = 2\nu \lambda$, which means that$$h\nu = \frac{(mv_g)^2}{2m} = \frac{4\nu^2 \lambda^2 m^2}{2m} = 2\nu^2 \lambda^2 m$$ $$\nu = \frac{h}{2m \lambda^2}$$which is the equation in the video.

 

[quasar987]

etotheipi, thanks for your answer! You made me realize that I made two mistakes. The first was to start from $\psi(x,t) = e^{i(x-vt)}$ and not realizing this implies $\lambda=2\pi$ and the other was to forget a factor of $2\pi$ in the SE. These mistakes "cancel out" in a sense because the computations with the correct factor of $2\pi$ lead to $$v = \frac{h/2\pi}{2m\lambda}$$ but since $2\pi = \lambda$ we get the desired form $$\nu = \frac{h}{2m\lambda^2}.$$ Or, in other words, if we (correctly) start from the more general expression$\psi(x,t) = \exp\left[i\frac{2\pi}{\lambda}(x-vt)\right]$ then the SE implies $$v=\frac{2\pi\hbar}{2m \lambda}=\frac{h}{2m \lambda}$$ and finally the fundamental relation $v=\lambda\nu$ gives the desired result.

But you also make the observation that if we trust that $p=h/\lambda$ holds for massive particles as it does for photons, then we get $$v=\frac{p}{2m}=\frac{v_{particle}}{2}.$$ I.e. the velocity of the plane wave "attached" to the particle is half that of the particle itself!

 

[vanhees71]

This is the phase velocity. If you take the group velocity, you'll get $v=p/m$, as expected. That makes some sense, because the velocity of the particle is rather the velocity of the center of the wave packet rather than the phase velocity.