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Rest energy in Schrodinger's equation

  1. Jan 18, 2012 #1
    Why is the rest energy usually ignored in Schrodinger’s equation? (I am aware of Dirac’s later relativistic equation.) What is the justification? Wouldn’t it change the nature of the solutions to the last equation below if it were included?

    Well, ok, it won't copy my Word equations. Why isn't "E" in Schrodinger's spatial equation actually (E-mc^2)?
     
  2. jcsd
  3. Jan 18, 2012 #2

    jtbell

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    First, Schrödinger's equation is non-relativistic to begin with.

    Second, introducing [itex]m_0 c^2[/itex] as a constant energy term has the same effect as shifting the potential energy function by a constant. Namely, it changes the overall phase of the wave function by a (spacewise) constant factor, which doesn't change any physical results. It always cancels out, for example in the probability density:

    [tex]P(x,t) = \Psi^*(x,t) \Psi(x,t) = \Psi_0^*(x,t)e^{+im_0 c^2 t / \hbar} \Psi(x,t)
    e^{-im_0 c^2 t / \hbar} = \Psi_0^*(x,t) \Psi_0(x,t)[/tex]
     
  4. Jan 18, 2012 #3
    Thanks for that answer. I guess I understand that changing the reference point for the potential has no net effect, as that changes both "V" and "E" in opposite directions. But it doesn't seem that including the rest energy affects both like that? i.e., It changes the sum?

    Isn't the following equation correct? (I'll try "tex", it's been a really long time since I typed with that stuff!)

    [tex]\grad^2 \Psi + 2 m (E - V - m c^2) / \hbar^2 = 0[/tex]

    Well I can't remember how to get the Laplacian, but you can see what I mean.
     
  5. Jan 18, 2012 #4
    Your equation looks fine except I guess there should be a psi in the second term. But in any case from your equation you can see that including the mc^2 term is exactly the same as shifting V to V+mc^2. If you solve the time-independent Schrodinger equation with this mc^2 term included you will get the same energy eigenstates and the same energy eigenvalues except that the eigenvalues will all be shifted up by mc^2.

    As another way of thinking about it you could absorb the mc^2 by defining F = E-mc^2. This gives you the regular time-independent Schrodinger equation with F in place of E. If you now solve for the allowed values of the "energy eigenvalue" F you will get the regular result you would have gotten neglecting the mc^2 term. Now E = F+mc^2, so you can see the energy eigenvalues are now just shifted up by mc^2.
     
  6. Jan 18, 2012 #5
    Historically Schrodinger originally tried to find an equation that was relativistic and ended up with what is now called the Klein-Gordon equation. However, as an equation describing a particle with mass and spin this equation gave a number of absurd answers so he abandoned it (it's actually the correct equation for a spin-0 system). He then abandoned relativity altogether and tried his eponymous equation which gave him extremely accurate predictions for atomic spectra. Dirac later figured out the relativistic form for an electron. Anyway, the point is the Schrodinger's equation is intentionally non-relativistic (time and position aren't even on the same footing) so cobbling some random aspect of SR (the rest mass) on to it doesn't make a lot of sense. If you want to include relativity and the possibility of particle production then you simply have to abandon Schrodinger's equation and move to quantum field theory (which is basically quantum mechanic+special relativity).
     
  7. Jan 19, 2012 #6

    tom.stoer

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    This argument fails when not considering a static potential but e.g. interacting particles of different masses. Write down the Schrödinger equation for a two-particle system (e.g. the hydrogen atom taking the proton as qm d.o.f. into account) with e²/r potential. How would you then introduce the rest energy for proton and electron with two different masses? Would you simply add mec² and mpc². That seems to be rather strange.
     
  8. Jan 19, 2012 #7
    (I'll have to study the new answers above - thanks!)

    Yes thanks, I missed that last psi before the equal sign. But I am still confused, sorry.

    That first answer up there, that including the rest energy is equivalent to just changing the reference point for the potential: I can understand how that might have no effect on the spatial equation, since it doesn't change (E - V). But wouldn't it change (E) in the time equation, exp[-i E t / h-bar]? After the separable assumption that breaks the partial equation into space-only and time-only sides, that each must therefor equal "some constant" (E) that is not a function of space or time, but that constant cannot be arbitrary can it?

    And I don't see how that (just changing the potential reference point) comes from the application of the non-relativistic expansion of the general Hamiltonion to Schrodinger's equation.

    H|psi> = i h-bar partial .wrt. time |psi>

    There must be some reason to add the rest energy constant to the right side as well as the left side??

    Thanks again for any help.
     
  9. Jan 19, 2012 #8
    Relativistic quantum mechanics (RQM) doesn't work. That's what QFT was developed. A wave equation CANNOT be made consistent with relativity. So I'm not quite sure what this topic is about.
     
  10. Jan 19, 2012 #9

    atyy

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    If you add a constant to the LHS (without adding it to the RHS), it automatically appears in the the evolution of the wavefunction as an overall phase, exactly as jtbell wrote it. An overall phase does not change physical predictions, since it cancels out when the wavefunction is "squared" to get probabilities.

    As jtbell and others have mentioned, adding a constant to the Schroedinger equation does not make it relativistic. Rather, from the relativistic equations (relativistic quantum field theory), we can derive Schroedinger's equation plus a constant in the non-relativistic limit. Since in this limit physical predictions are not changed by omitting such a constant, we go ahead and do so.
     
  11. Jan 19, 2012 #10

    tom.stoer

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  12. Jan 21, 2012 #11
    Furthermore, just as the probability density P is laid out, would it true that P is always equal to 1?
    For example, Psi = cos(kx-wt)+isin(kx-wt), while Psi* = cos(kx-wt)-isin(kx-wt). Once you multiply the two, you get 1.

    Thanks!
     
  13. Jan 22, 2012 #12
    Try making it mesh with

    [tex] \Delta E \Delta t > h [/tex]

    and

    [tex]E = m c^2 [/tex]

    (i.e. particle number being a matter of time scales).
     
  14. Jan 22, 2012 #13

    vanhees71

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    Of course, a single-particle interpretation of the various types of relativistic wave equations for interacting particles does not work as it does in non-relativistic physics with the Schrödinger or Pauli equation. The reason is that local relativistic field theories always imply the possibility of creation and annihilation of particles. The Dirac equation has to be interpreted as an equation for field operators ("2nd quantization") to establish a many-particle theory in terms of a quantum field theory with the corresponding Fock space as the Hilbert space of a system with arbitrary particle numbers, admitting the creation and destruction of particles.

    A one-particle interpretation is, however, possible in the non-relativistic limit, where the presence of antiparticles and the creation and annihilation of particles can be neglected. The paradigmatic example is the relativistic hydrogen atom, which can in first approximation be described as the motion of an electron in the Coulomb field of the proton (classical-field approximation through soft-photon resummations, see Weinberg QT of Fields, Vol. 1). On top of this approximation one can do perturbation theory, leading to one of the great achievements of modern QED by explaining the Lamb shift very accurately.
     
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