# Composite system, rigged Hilbert space, bounded unbounded operator, CSCO, domain

1. Jan 31, 2013

### Petro z sela

Is something wrong in my assertions below?

Suppose we have two quantum systems $N$ and $X$. Let $N$ is described by discrete observable $\hat{n}$ (bounded s.a. operator with discrete infinite spectrum) with eigenvectors $|n\rangle$. Let $X$ is described by continuous observable $\hat{x}$ (unbounded s.a. operator with continuous spectrum) with generalized eigenvectors $|x\rangle$. Then:
1. physical states of $N$ lie in Hilbert space $H_{N}$;
2. $H_{N}$ is spanned by $|n\rangle$;
3. $|n\rangle$ lie $H_{N}$;
4. basis set $|n\rangle$ has cardinality aleph-null (countable);
5. system $X$ is considered in rigged Hilbert space $Ω_{X}\subset H_{X}\subset Ω^{\times}_{X}$;
6. physical states of $X$ lie in $Ω_{X}$;
7. $Ω_{X}$ is spanned by $|x\rangle$;
8. basis set $|x\rangle$ has cardinality aleph-one (uncountable);
9. $|x\rangle$ lie in $Ω^{\times}_{X}\backslash H_{X}$;
10. the complete set of commuting observables (CSCO) for composite system $NX$ is $\hat{n}$, $\hat{x}$;
11. composite system $NX$ is considered in rigged Hilbert space $Ω_{NX}\subset H_{NX}\subset Ω^{\times}_{NX}$;
12. $H_{NX}=H_{N}\otimes H_{X}$ (tensor product);
13. physical states of $NX$ lie in $Ω_{NX}$;
14. $Ω_{NX}$ is spanned by $|n,x\rangle$ ( $|n,x\rangle = |n\rangle\otimes|x\rangle$ );
15. basis set $|n,x\rangle$ has cardinality aleph-one (uncountable);
16. $|n,x\rangle$ lie in $Ω^{\times}_{NX}\backslash H_{NX}$;
17. operator $\hat{X}=\hat{1}\otimes\hat{x}$ is unbounded;
18. $\hat{X}$ has domain $Ω_{NX}$ and maps $Ω_{NX}$ into $Ω_{NX}$;
19. operator $\hat{N}=\hat{n}\otimes\hat{1}$ is bounded;
20. $\hat{N}$ has domain $H_{NX}$ and maps $H_{NX}$ into $H_{NX}$;
21. Suppose $NX$ is in the some state $ψ\inΩ_{NX}$. One has measured observables $\hat{X}$ and/or $\hat{N}$ in state $ψ$. After this procedure $ψ$ collapses to vector from $|n,x\rangle$ set, this vector $\notinΩ_{NX}$. It implies that 6 and 13 must be reformulated: physical states lie in $Ω_{NX}$ and in some subspace of $Ω^{\times}_{NX}\backslash H_{NX}$ ( subspace of generalized eigenvectors).
22. And what about $H_{NX}\backslash Ω_{NX}$ ? I can’t apply $\hat{X}$ to vector $\varphi$ from $H_{NX}\backslash Ω_{NX}$, because $||\hat{X}\varphi||\rightarrow∞$ and $\hat{X}\varphi\inΩ^{\times}_{NX}\backslash H_{NX}$ , i.e. I can’t measure observable $\hat{X}$ in states $H_{NX}\backslash Ω_{NX}$. But I can apply $\hat{N}$ to $\varphi$, because $\hat{N}\varphi\in H_{NX}$ and $||\hat{N}\varphi||<∞$ , i.e. I can measure observable $\hat{N}$ in states $H_{NX}\backslash Ω_{NX}$. But $\hat{N}$ and $\hat{X}$ form CSCO and they can be measured simultaneously. What can I do in this case? I can decrease domain of $\hat{N}$ from $H_{NX}$ to $Ω_{NX}$. Thus $\hat{N}$ and $\hat{X}$ will have common domain, but for $\hat{N}$ this domain is not invariant, because in general case $\hat{N}$ maps $Ω_{NX}$ into $H_{NX}$.

Last edited: Jan 31, 2013
2. Jan 31, 2013

### dextercioby

My comments are markes with caps lock in the quote.

21 & 22: My comment. I don't believe in von Neumann's measurement => collapse postulate. I don't know of a formulation of von Neumann's postulate for distribution spaces.

Last edited: Feb 1, 2013
3. Jan 31, 2013

### rubi

8, 15: This is actually neither true nor false. It's equivalent to the continnuum hypothesis, which is undecidable in ZFC. A true statement would be that the cardinality is beth-one.
9: This is true in my opinion.

21: There are two ways out:
a) You never really measure an exact value for a continuous variable and thus the state will not collapse to $|x\rangle$, but rather to something like $\int e^{-\frac{(y-x)^2}{\sigma^2}}|y\rangle dy$.
b) You are only in a distributional state for an infinitesimal amount of time, since the diffusive nature of the Schrödinger equation will quickly evolve the state into a physical state. This might not apply for every possible Hamiltonian however.

22: The states in $H_{NX}\backslash\Omega_{NX}$ are not physical. They correspond to things like infinite energy. A physical state must always give you finite values for all measurements. The fact that you can apply operators to a state doesn't imply that this state can be realized physically. The situation you described corresponds to something like a particle with infinite energy and spin up. You can easily write down a wavefunction for this situation, but that doesn't mean that this is realized in nature.

4. Feb 1, 2013

### dextercioby

9 is true, of course. I've corrected the point above.