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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

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# Relativistic correction of Schrödinger equation

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