Particle in a box in momentum basis

  • Thread starter Ravi Mohan
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  • #26
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Post #13 cites a very good source which demonstrates that computation (page 9-14).
 
  • #27
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Yeah ,right.
So, the momentum operator being Hermitian means that its eigenvalues are real But, what does it mean if it's also self-adjoint? I mean, in my introductory QM course, they just said that if an operator is Hermitian, it has real eigenvalues and orthogonal eigenfunctions. So, what does the lack of self-adjointness mean in general? I can see that there are problems with the momentum operator in this case, but how can we state more generally what the lack of self-adjointness means?
 
  • #28
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A quote from that article:
Fran ̧cois Gieres said:
One may wonder whether it is possible to characterize in another way the little “extra” that a Hermitian operator is lacking in order to be self-adjoint. This missing item is exhibited by the following result which is proven in mathematical textbooks. If the Hilbert space operator A is self-adjoint, then its spectrum is real [6, 8][13]-[18] and the eigenvectors associated to different eigenvalues are mutually orthogonal; moreover, the eigenvectors together with the generalized eigenvectors yield a complete system of (generalized) vectors of the Hilbert space4 [19, 20, 8]. These results do not hold for operators which are only Hermitian. As we saw in section 2.1, this fact is confirmed by our previous example: the Hermitian operator P does not admit any proper or generalized eigenfunctions and therefore it is not self-adjoint (as we already deduced by referring directly to the definition of self-adjointness)

And that is what you need for the observable postulate of Quantum Mechanics. The observables should correspond to Self-Adjoint operators and not to the operators that are only Hermitian. Only-Hermitian operators don't have complete set of eigenvectors.

You may want to read a blog post I wrote few years ago https://ravimohan.net/2013/07/25/particle-in-a-box-infinite-square-well/
 
  • #29
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A quote from that article:


And that is what you need for the observable postulate of Quantum Mechanics. The observables should correspond to Self-Adjoint operators and not to the operators that are only Hermitian. Only-Hermitian operators don't have complete set of eigenvectors.

You may want to read a blog post I wrote few years ago https://ravimohan.net/2013/07/25/particle-in-a-box-infinite-square-well/
Oh, I missed that. So, by not being self-adjoint, it does not have proper eigenfunctions. But, this is valid for this particular case. Because for the free particle for example, the momentum operator has well-defined eigenfuctions(plane waves). So, what's up with that?

P.S. Your blog seems useful! Thanks
 
  • #31
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Oh, I missed that. So, by not being self-adjoint, it does not have proper eigenfunctions. But, this is valid for this particular case. Because for the free particle for example, the momentum operator has well-defined eigenfuctions(plane waves). So, what's up with that?

P.S. Your blog seems useful! Thanks
Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.
 
  • #32
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For a more formal treatment, see

http://arxiv.org/abs/quant-ph/0103153
Please correct me if I am wrong, but from the paper that you provided a link to, I concluded that the self-adjointness of the momentum operator depends on the problem that we examine, right? Because the momentum operator is hermitian(and this is independent of the problem-again correct me if I am wrong) and for it to be self-adjoint it's domain must be the same as the domain of its hermitian conjugate. In the case of the free particle, their domains are the same, so the momentum operator is also self-adjoint, right?
 
  • #33
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Actually the existence of the well defined eigenfunctions is not enough. They have to form a complete set and satisfy the boundary conditions. The enginefunctions of momentum operator in this problem don't form a complete set. You may want to read post #4.
Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?
 
  • #34
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Yes, I understand that, but in the case of the free particle, all these conditions are satisfied, so the momentum operator is self-adjoint. So, am I right to conclude that the self-adjointness of an operator depends on the given problem?
Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.
 
  • #35
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Ah, I am sorry. I didn't read the "free particle". In that case you are correct. Also, as you said, the nature of an operator depends on the given physical system.
If you have some extra time, could you please show me how to conclude if the momentum operator is self-adjoint or not in the case of the harmonic oscillator? Is it easy to show if it's self-adjoint? I am guessing that it is since there are no boundary conditions, so again the domains of P and P+(dagger) are the same, right?
 
  • #36
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I will have to think and work it out. But essentially, as you say, the vectors in the domain of ##\hat{P}^\dagger## and ##\hat{P}## need not satisfy any extra boundary condition (save square integrability and vanishing at infinity), their domain should turn out same.
 
  • #37
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I will have to think and work it out. But essentially, as you say, the vectors in the domain of ##\hat{P}^\dagger## and ##\hat{P}## need not satisfy any extra boundary condition (save square integrability and vanishing at infinity), their domain should turn out same.
Great, I think I got it! Thanks!
Also, your blog is excellent!
 

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