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- Thread starter Domnu
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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.Just a quick question regarding Schroedinger's Equation... since we have [tex]\hat{H} \psi = \hat{p} \psi = E \psi[/tex],

Pete

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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?)

[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?)

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Fredrik

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Fredrik

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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.

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.

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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.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?)

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

[tex]\hat{p} \psi = p \psi[/tex]

- #7

reilly

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Regards,

Reilly Atkinson

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