- #1
- 124
- 0
Hi everyone,
This summer (it's summer in Australia) I have been studying quantum mechanics from a mathematician's perspective, and the physical interpretation has become a little more difficult as the theory has become more in-depth.
Do we have a particular method for choosing which Hilbert space we use for a given system?
There seem to be two answers to my question: (let's just neglect spin here)
1) No. The underlying space is always chosen to be [tex]L^2(\mathbb{R}^{mn})[/tex], where n is the number of particles and m is the number of physical degrees of freedom each particle has. Alternatively we can use some closed subspace of [tex]\mathbb{R}[/tex] in the case of a finite-'volume' system, but that is sort of not an issue here.
2) Yes - the Hilbert space we work in is the closed linear span of the set of eigenvectors.
In the first case, my notes say that we consider that each [tex]\phi[/tex] satisfying [tex] ||\phi || = 1 [/tex] to be a pure state of the system, where pure states are unique only up to a numerical factor of modulus one.
If we take this approach, there exist pure states that are not in the domain of the position operators. Indeed, since the position operators are the same for every physical problem, there will *always* be states (supposed to be physical) on which the position operators cannot be applied. This would seem to bring interpretational difficulties - can we exclude these states *just because*?
Introducing the continuity/differentiability requirement of course solves this problem, but from a mathematical perspective this is a cop out, especially since the theory itself shows why wavefunctions need to be continuous and differentiable.
The problem is, you can't avoid this difficulty by switching to interpretation (2) - if you do, you lose a lot of physical systems, for example the unbound states in a Coulomb potential. Also, to my mind you lose quite a bit of beauty in the theory - by using a large Hilbert space, you can decompose a system into it's components mathematically as a direct sum of closed subspaces. For example, you can write the larger Hilbert space as the direct sum of the spaces of bound and unbound states.
Is there a rigourous explanation resolving the difficulties I've posed here, or has it been largely ignored? Interpretation (2) seems to be the one used by physicists when they are just considering specific parts of a system, but I'm more concerned about the whole range of states.
Cheerio,
Kane
This summer (it's summer in Australia) I have been studying quantum mechanics from a mathematician's perspective, and the physical interpretation has become a little more difficult as the theory has become more in-depth.
Do we have a particular method for choosing which Hilbert space we use for a given system?
There seem to be two answers to my question: (let's just neglect spin here)
1) No. The underlying space is always chosen to be [tex]L^2(\mathbb{R}^{mn})[/tex], where n is the number of particles and m is the number of physical degrees of freedom each particle has. Alternatively we can use some closed subspace of [tex]\mathbb{R}[/tex] in the case of a finite-'volume' system, but that is sort of not an issue here.
2) Yes - the Hilbert space we work in is the closed linear span of the set of eigenvectors.
In the first case, my notes say that we consider that each [tex]\phi[/tex] satisfying [tex] ||\phi || = 1 [/tex] to be a pure state of the system, where pure states are unique only up to a numerical factor of modulus one.
If we take this approach, there exist pure states that are not in the domain of the position operators. Indeed, since the position operators are the same for every physical problem, there will *always* be states (supposed to be physical) on which the position operators cannot be applied. This would seem to bring interpretational difficulties - can we exclude these states *just because*?
Introducing the continuity/differentiability requirement of course solves this problem, but from a mathematical perspective this is a cop out, especially since the theory itself shows why wavefunctions need to be continuous and differentiable.
The problem is, you can't avoid this difficulty by switching to interpretation (2) - if you do, you lose a lot of physical systems, for example the unbound states in a Coulomb potential. Also, to my mind you lose quite a bit of beauty in the theory - by using a large Hilbert space, you can decompose a system into it's components mathematically as a direct sum of closed subspaces. For example, you can write the larger Hilbert space as the direct sum of the spaces of bound and unbound states.
Is there a rigourous explanation resolving the difficulties I've posed here, or has it been largely ignored? Interpretation (2) seems to be the one used by physicists when they are just considering specific parts of a system, but I'm more concerned about the whole range of states.
Cheerio,
Kane