# Moving electron - finding the wavefunction

1. Aug 7, 2013

### 71GA

1. The problem statement, all variables and given/known data
On our modern physics class e did a problem:
At first i said: "Oh i know this!" and solved the case like this.

2. Relevant equations
Lorentz invariant: $E=\sqrt{{E_0}^2 + p^2c^2}$
Schrodinger equation where $V(x)=0$.

3. The attempt at a 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 afterall defined using the full energy and not kinetic energy...

2. Aug 7, 2013

### vela

Staff Emeritus
You're wrong. Both your expression for L and the Schrodinger equation are non-relativistic. It's not appropriate to use the relativistic expression for the energy of the electron here.

3. Aug 7, 2013

### 71GA

Thank you.

But can you explain to me why don't we generaly use the rest energy when dealing with a classical approximations. In books everyone explains the rest energy but what it realy is? I mean i know how to calculate it and whatsoever but when does it appear and why do we have to take it into the calculation?

I am confused only because of that Lorentz invariance which says that if a particle is mooving slowly most of its energy is the rest energy. On the other hand we just neglect it like it isn't there... It seems to me like a contradiction...

4. Aug 7, 2013

### vela

Staff Emeritus
It's just the way you're doing the classical approximation isn't correct. A plane wave is given by $$\psi = e^{i(kx-\omega t)}$$ where $p= \hbar k$ and $E = \hbar \omega$. For the spatial dependence, you need to find the classical approximation for the momentum p:
\begin{align*}
pc &= \sqrt{E^2 - (mc^2)^2} \\
&= \sqrt{(mc^2+K)^2 - (mc^2)^2} \\
&= \sqrt{2mc^2 K + K^2} \\
&\cong \sqrt{2mc^2 K}
\end{align*} where K is the kinetic energy. The rest energy term cancels out.

The other way to look at it is that $k = \sqrt{\frac{2mE}{\hbar^2}}$ is equivalent to the relationship $E = \frac{(\hbar k)^2}{2m} = \frac{p^2}{2m}$. This latter expression is clearly the classical quantity for the kinetic energy of a particle of mass m, not the total energy.