# Relativistic correction of Schrödinger equation

1. Mar 15, 2015

### Jonsson

Hello there,

I've been given the relativistic correction of the Schrödinger equation for a free particle:

$$- \frac{\hbar^2}{2m} \frac{\partial ^2\Psi}{\partial x^2} - \frac{\hbar^4}{8m^3c^2} \frac{\partial ^4\Psi}{\partial x^4} + E_0 \Psi = i \hbar \frac{\partial \Psi}{\partial t}$$

How we derived this correction is not important. It is important to note that $E_0 = mc^2$. I am asked to prove that $\Psi(x,t) = e^{i(kx-\omega t)}$ with $k = (2mE)^{1/2}/ \hbar$ is a solution. We have also been given that this is the same $\Psi$ as is the solution of the ordinary Schrödinger equation for free particle

The problem is that I think I can prove that this is not a solution, which is controversial, because it contradicts what my professor is asking. I want to show you my proof before I present it to him, in case I made some silly mistake. Can you please scrutinize it?

We assume for contradiction that $\Psi(x,t) = e^{i(kx-\omega t)}$ is a solution. We try by inserting $\Psi$ into the equation. We get:
$$\frac{\hbar^2 k^2}{2m} \Psi - \frac{\hbar^4k^4}{8m^3c^2} \Psi + E_0\Psi = \hbar \omega \Psi.$$
Since $k = (2mE)^{1/2}/ \hbar$, that implies $E = \hbar^2 k^2/(2m)$ which by using de Broglie identity $E=\hbar \omega$ we deduce that $\hbar^2 k^2/(2m)$. Inserting this into the above equation, we obtain
$$\hbar \omega \Psi - \frac{\hbar^2 \omega^2}{2mc^2} \Psi + E_0\Psi = \hbar \omega \Psi \implies \frac{\hbar^2 \omega^2}{2mc^2} \Psi = mc^2 \Psi \implies E^2 = 2 (mc^2)^2 = 2 E_0^2$$
That means for all free particles which are described by $\Psi$ we have:
$$2^{1/2} = \frac{E}{E_0} = \gamma,$$
which means that $\gamma$ is equal to root two, a constant, which is impossibly correct for all free particles described by $\Psi$.

Are you able to find mistakes in the above which destroys the argument? What are your views? Thank you for your time.

Kind regards,
Marius

Last edited: Mar 15, 2015
2. Mar 15, 2015

### Staff: Mentor

I'm suspicious of this, because it's based on the non-relativistic relationship between momentum and kinetic energy.

3. Mar 15, 2015

### Jonsson

Hello jtbell, thanks for your response. Are you able to define the word «suspicious»?

4. Mar 15, 2015

Staff Emeritus
I'm suspicious of the E=hbar omega part.

5. Mar 16, 2015

### PhilDSP

Hello Marius,

I'm also suspicious that $k = (2mE)^{1/2}/ \hbar$ is a solution.

If I've figured correctly, the Fourier transform of your given equation is $\frac{\hbar^4}{8m^3c^2}k^4 - \frac{\hbar^2}{2m}k^2 - E_0 = -\hbar \omega$

Doesn't a solution need an $\omega$ term also?

6. Mar 16, 2015

### Jonsson

Thanks guys. It is obviously non-relativistic. But that doesn't matter. Why? The expression I used for $k$ is given to me implicitly by my professor, so the mistake is on his behalf (phew). So it doesn't matter. I will present this proof to him after I have worked out what he is really after. Meanwhile, I will have to find another interpretation of the assignment. Thank you for your time.

Last edited: Mar 16, 2015
7. Mar 16, 2015