- #1
Jonsson
- 78
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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
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: