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Question regarding Schroedinger's Equation

  1. Jun 17, 2008 #1
    Just a quick question regarding Schroedinger's Equation... since we have [tex]\hat{H} \psi = \hat{p} \psi = E \psi[/tex], does this mean that the operators [tex]\hat{H}[/tex] and [tex]\hat{p}[/tex] are compatible? In other words, does this mean that the operators commute and that the total energy and momentum of the particle system can be observed simultaneously? This would make sense, assuming we knew the mass of the system...
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  3. Jun 17, 2008 #2
    Woa there. Where did you get the idea that this was true? As you wrote it you have the energy eigenvalues being the same as the momentum eigenvalues. It is true that the energy operator and the momentum operator have the same set of eigenkets and, as I recall (I'll have to double check to make sure) for that reason these two operators commute.

  4. Jun 17, 2008 #3
    Um, well, you know that Schroedinger's time dependent equation is just

    [tex]-\frac{\hbar^2}{2m} \nabla^2 \psi + V \psi = i \hbar \frac{d \psi}{dt}[/tex]

    and that

    [tex]\hat{H} = -\frac{\hbar^2}{2m} \nabla^2+ V[/tex],
    [tex]\hat{p} = i \hbar \frac{d}{dt}[/tex].

    So, you know that they can simultaneously be measured (they have the same set of eigenkets, and they are thus commutative... so they are measurable at the same time?)
    Last edited: Jun 17, 2008
  5. Jun 17, 2008 #4


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    No, [itex]p=-i\hbar\nabla[/itex]. The operator you call p is actually H, written in an alternative form.
  6. Jun 17, 2008 #5


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    One answer to your original question is that in relativistic quantum mechanics, H is actually the 0th component of the four-momentum. All the components commute with each other, [itex][P^\mu,P^\nu]=0[/itex]. Also, the operator [itex]P^2=-H^2+\vec P^2[/itex] commutes with all of the generators of (inhomogeneous) Lorentz transformations (momentum, spin and boost operators), which means that all the other inertial observers will agree with you about its value. The value is written as [itex]-m^2[/itex], and m is not surprisingly called the mass.

    I should add that the above is only true when we're dealing with a Hilbert space of one-particle states. It obviously isn't true when our state vectors represent some complicated multi-particle system with interactions.
    Last edited: Jun 17, 2008
  7. Jun 17, 2008 #6
    Yes. But you have it written as having the same eigenvalues, which they don't. The eigenvalues of the Hamiltonian operator are different than the eigenvalues of the momentum operator even though the eigenket is the same. I.e.

    [tex]\hat{H} \psi = E \psi[/tex]

    [tex]\hat{p} \psi = p \psi[/tex]
  8. Jun 22, 2008 #7


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    p and H do not commute if there is a spatial interaction -- then, dp/dt=F = - grad(V). For a free particle, H and p do commute, thus they share eigenstates, as do any set of commuting operators. They share common eigenvalues only for massless particles, and then only for the magnitude of p.
    Reilly Atkinson
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