- #26

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Post #13 cites a very good source which demonstrates that computation (page 9-14).

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- 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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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?

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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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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?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/

P.S. Your blog seems useful! Thanks

- #30

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- #31

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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.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

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- #33

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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?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.

- #34

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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.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?

- #35

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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?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.

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- #37

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Great, I think I got it! Thanks!

Also, your blog is excellent!

- #38

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https://www.physicsforums.com/threa...r-the-free-particle-position-operator.880324/

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