2
$\begingroup$

In Quantum Mechanics, the quantum state of the physical system lives in an infinite-dimensional Hilbert space and can be written in terms of two different bases, the position basis (uncountably infinite) and the energy basis (countably infinite). Apparently, the two bases are of different cardinalities, which violates a theorem in Linear Algebra that all bases of a vector space must be of the same cardinality. How to explain this confusion?

$\endgroup$
0
4
$\begingroup$

Note that the "position basis" is not a basis, because its elements $|x\rangle$ are not elements of the Hilbert space. It is a useful structure to work with, but making mathematically rigorous statements about it can be quite tricky.

$\endgroup$
4
  • $\begingroup$ Hello, so I assume that the OP's argument is wrong for two different reasons: the reasons in my answer and the reason of your answer. Am I right ? $\endgroup$ – Amr Oct 30 '20 at 12:59
  • 1
    $\begingroup$ @Amr Yes. The "position basis" to which the OP is referring is not a basis at all, and the energy basis is a Schauder basis (in fact, an orthonormal basis of the type you mention). $\endgroup$ – J. Murray Oct 30 '20 at 13:18
  • $\begingroup$ Aha, I see. Thank you so much. $\endgroup$ – Amr Oct 30 '20 at 13:40
  • $\begingroup$ @J.Murray Thanks for your answer. Where can I read more about this? $\endgroup$ – Ahmad Haitham Ibrahim Oct 30 '20 at 15:04
2
$\begingroup$

I am a mathematician and not a physicist, so I don't know the physical context and other physicists here are very welcome to correct my answer if it is wrong.

I think your confusion comes from the unfourtantate terminology of "basis" in the context of Hillbert spaces. There are two different concepts both happen to be called basis:

  1. Hamel Basis: That's just a basis in the sense of linear algebra. ie A linearly independent subset such that any vector can be written as a finite linear combination of the members of your basis.

  2. Hillbert Orthonormal basis: That's a linearly independent orthonormal subset of your Hilbert space such that any vector in your vector space can be approximated with arbitrary precision using finite linear combinations of your basis. It follows mathematically that any vector in your hillbert space can be expressed as an infinite sum of members of your basis.

I don't know the physical context, but I suspect that what you call an "energy basis" is the second type of basis I talked about above and not the first one(ie Hillbert Orthonormal basis and not Hamel Basis). However, the theorem of linear algebra which you quote refers to Hamel Basis and not Hillbert orthonormal basis. Hence, you are applying the linear algebra theorem in a wrong way/invalid context and that should solve the confusion.

See this link of Wikipedia:

https://en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases

$\endgroup$
8
  • $\begingroup$ Although the terminology you speak about is correct it doesn't correspond to the "position basis" as mentioned in the other comments and answers, because the position kets do not formally belong to the Hilbert Space. $\endgroup$ – ohneVal Oct 30 '20 at 12:59
  • $\begingroup$ @ohneVal Yes I don't know the physical context so I was not in a position to judge the validity of the OP's decision to count "position basis" as a basis (I just don't know what that is) $\endgroup$ – Amr Oct 30 '20 at 13:00
  • $\begingroup$ @ohneVal As a person who never studied QM, Am I right to say that the OP's argument is wrong because of these two reasons: 1) Energy basis is not a basis in the sense of Linear algebra 2)Position basis is not actually a subset of the Hillbert space so not actually a basis ? $\endgroup$ – Amr Oct 30 '20 at 13:02
  • $\begingroup$ You can think of them as Dirac distributions centered at the point $x$. If you collect them all you will have a "sort of basis". $\endgroup$ – ohneVal Oct 30 '20 at 13:02
  • $\begingroup$ The energy eigenstates basis IS a Hilbert basis (which in physics we just call basis). The important part here is actually just point 2 I believe $\endgroup$ – ohneVal Oct 30 '20 at 13:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.