Moving electron - finding the wavefunction

Question

On our modern physics class my professor did a problem:

Write down a wavefunction of an electron which is moving from left to right and has an energy $100\text{ eV}$.

At first i said: "Oh i know this!" and solved the case like this.

My solution:

The energy $100eV$ must be the kinetic energy of the electron. So i said ok this kinetic energy is very small compared to the rest energy and i can say that $pc \ll E_0$ which means i have a classical limit where:

\begin{align} E=\sqrt{{E_0}^2 + p^2c^2}\\ E=\sqrt{{E_0}^2 + 0}\\ \boxed{E=E_0} \end{align}

So now i can write the general wavefunction for a free right -mooving particle like this:

$$\psi=Ae^{iLx}\quad L=\sqrt{\tfrac{2mE}{\hbar^2}}$$

So if i want to get the speciffic solution i need to calculate the constant $L$ and then normalise the $\psi$. Because $E=E_0$ i calculated the constant $L$ like this:

$$L=\sqrt{\frac{2mE_0}{\hbar^2}}$$

while my professor states that I should do it like this:

$$L=\sqrt{\frac{2mE_k}{\hbar^2}},$$

where $E_k$ is the kinetic energy of the electron. Who is wrong? I mean whaaaaat? The constant $L$ is after all defined using the full energy and not kinetic energy...

Answer 1

Well, your proposed wavefunction has the same spatial dependence no matter what the particle's velocity/momentum. What you should observe is a change in wavelength, and therefore a change in the wavefunction, as you change the momentum. Your professor is right.

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Edit in response to comment.

No. A theory is Lorentz invariant but quantities like momentum and energy are covariant. That means that you need to change $E$ and $p$ when you change frames. $E=E_0$ holds only in the rest frame when $p=0$; otherwise you need to take the full $E=\sqrt{E_0^2+p^2 c^2}$. If you are in the classical limit, then you do need to work to second order in momentum, so that $$E\approx E_0+p^2c^2/2E_0=E_0+p^2/2m=E_0+E_k.$$ Then $E_0$ is a constant and can be dropped.

Let me explain in a bit more detail how that works. You are postulating some wavefunction $\psi=e^{iLx}$ as a solution to Schrödinger's equation $$\hat{H}\psi=E\psi.$$ You then have two choices: count $E_0$ in, or drop it. If you do not drop it, then you will have $E=E_0+E_k$ as above but you also need to include $E_0$ in the hamiltonian, so $\hat H=E_0+\frac{\hat{p}^2}{2m}$. The terms in $E_0$ will then cancel out, and you will be left with the same equation for $\psi$ and $L$ as you would have if you had dropped $E_0$ from the beginning: $$\frac{\hbar^2L^2}{2m}\psi=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}=E_k\psi.$$

As you can see, $L=\sqrt{2mE_k/\hbar^2}=p/\hbar$ depends on the momentum. This is absolutely necessary, as $L=2\pi/\lambda$ is inversely proportional to the wavelength, and this must depend on momentum because of de Broglie's relation, which embodies the canonical commutation relations; an answer which doesn't have this must be discarded as wrong before it leaves the table. Your initial mistake, on the other hand, was dropping $E_0$ from one side of the relation $H=E$ and not from the other.