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Relativistic correction of Schrödinger equation

  1. Mar 15, 2015 #1
    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. jcsd
  3. Mar 15, 2015 #2

    jtbell

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    Staff: Mentor

    I'm suspicious of this, because it's based on the non-relativistic relationship between momentum and kinetic energy.
     
  4. Mar 15, 2015 #3
    Hello jtbell, thanks for your response. Are you able to define the word «suspicious»?

    Thank you for your time
     
  5. Mar 15, 2015 #4

    Vanadium 50

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    I'm suspicious of the E=hbar omega part.
     
  6. Mar 16, 2015 #5
    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?
     
  7. Mar 16, 2015 #6
    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
  8. Mar 16, 2015 #7

    Vanadium 50

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    One thing you can do is to apply the second term as a (1st order) perturbation on the Hamiltonian and then use perturbation theory to work everything out.
     
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