# Is every state an eigenstate of an observable?

1. Mar 30, 2013

### atyy

I started a new thread from a side discussion in https://www.physicsforums.com/showthread.php?t=681625&page=2, since it seems very off topic, but I still had questions.

Is there a requirement for an operator that corresponds to an observable to be part of a complete set, ie. its eigenvectors span the vector space?

Would $|u><u|$ be part of a complete set?

2. Mar 31, 2013

### micromass

Not sure what you mean with complete set. Are you just asking whether the eigenvectors of $|u><u|$ span the complete space?

This is certainly true. If we calculate the eigenvectors then we get that $U=\{\lambda |u>|~\vert~\lambda\in \mathbb{C}\}$ and all multiples are eigenvectors with eigenvalue $<u|u>$ (except the zero vector). And the orthogonal complement $U^\bot=\{|v>~\vert~<v|u>=0\}$ are eigenvectors with eigenvalue $0$ (except the zero vector).

Now, it is easy to see that $H=U\oplus U^\bot$. Indeed, any $|w>\in H$ can be written as $|w> = \frac{<u|w>}{<u|u>} |u> + ( |w> - \frac{<u|w>}{<u|u>}|u>)$
and we see that the first term is in $U$ and the second term is in $U^\bot$.
It is also true and easy to see that $U\cap U^\bot=\{0\}$, but this is not necessary here.

It is not true in general that Hermitian operators always have a collection of eigenvectors that spans the space. Indeed, some Hermitian operators have no eigenvectors at all! But for some Hermitian operators, we are indeed guaranteed that such a collection exists. In this case, we see that the range of $|u><u|$ is finite-dimensional (and even one-dimensional). This guarantees the existence of a orthonormal basis of eigenvectors. Limits of Hermitian operators with finite-dimensional range also are guaranteed an orthonormal basis of eigenvectors. Such operators are called compact.

3. Mar 31, 2013

### atyy

Yes, I meant that the eigenvectors should span the space. I was thinking complete set as in "complete set of commuting observables" I'm not sure what all the requirements for an observable are, but in this case, if I understand you correctly, several eigenvectors have the eigenvalue 0. So if I make a measurement and get a result of 0, what state has the system collapsed into? Usually I imagine that if I get a result λ, then the system has collapsed into the eigenstate corresponding to the eigenvalue λ.

4. Mar 31, 2013

### Fredrik

Staff Emeritus
If the eigenspace isn't 1-dimensional, I think all you can say is in general is that the new state vector will be a member of the eigenspace (i.e. the subspace spanned by all the eigenvectors with eigenvalue 0). So this sort of measurement doesn't prepare a pure state. If the eigenspace is finite-dimensional, I would guess that we can take the state to be P/d where P is the projection operator of the eigenspace, and d is the dimension of the eigenspace. I haven't really thought that through. I just noticed that P/d seems to satisfy the definition of a state operator. I don't know what we'd do if the eigenspace is infinite-dimensional.

Edit: Chapter 9 of Ballentine discusses things like this. If I understand him correctly after only having a quick look, he considers a measurement of an observable R, followed by a filtering where we discard the particle if the result isn't in an interval he denotes by $\Delta_a$. He denotes the projection operator associated with the subspace spanned by eigenvectors of R with eigenvalue in $\Delta_a$ by $M_R(\Delta_a)$. Formula (9.28) says that the state after this "filtering-type measurement" is
$$\rho'=\frac{M_R(\Delta_a)\rho M_R(\Delta_a)}{\operatorname{Tr}\left(M_R(\Delta_a)\rho M_R(\Delta_a)\right)}.$$

Last edited: Mar 31, 2013
5. Mar 31, 2013

### atyy

Thanks, let me think about it. Funnily, googling around gives

http://arxiv.org/abs/1208.0773

http://philosophyfaculty.ucsd.edu/faculty/wuthrich/PhilPhys/EarmanJohn2002IntStudPhilSci_TRI.pdf [Broken]

The former is about T-violation by weak interactions, and mentions a state that is filtered (paragraph after Eq 3).

The latter also mentions T-violation and mentions mixed states (p259).

Last edited by a moderator: May 6, 2017
6. Mar 31, 2013

### jfy4

For a finite dimensional vector space, I think that for a Hermitian operator, the eigenvectors of that operator form a complete set. But for and infinite dimensional vector space, It isn't the case.

7. Apr 1, 2013

### atyy

So it does look like every state is an eigenvector of an observable, but the observable may have more than one eigenvector for a given eigenvalue.

What experimental setup would measure this observable for a particle with wave function ψ(x)?

8. Apr 1, 2013

### Fredrik

Staff Emeritus
I don't think that there's a way to assign measuring devices to each self-adjoint operator. However, if a measuring device is represented by a self-adjoint operator B, and b is an eigenvalue of B with a 1-dimensional eigenspace, then |b><b| is simply the measuring device that you get by taking the first measuring device and relabeling the results, so that the result b is now thought of as result 1 and all other results are now thought of as result 0.

9. Apr 2, 2013

### strangerep

Actually, there are generalized spectral theorems for quite a few inf-dim cases.

10. Apr 2, 2013

### jfy4

Being called an observable means "measurable," at least indirectly. Consider a system in a superposition state that isn't classical, for instance, being half dead and alive. You will not find a measuring device which registers that state. Here is a possible state to write down, yet this is not an eigenstate of an observable.

11. Apr 2, 2013

### atyy

jfy4, that's an intuitive way of saying what I think Fredrick also said.

Fredrik and jfy4, just to make sure I understand, is this a correct paraphrase:

1) Mathematically, for any state we can always construct a mathematical observable - an operator with real eigenvalues, and whose eigenvectors span the Hilbert space.

2) Physically, there may not be any "easily constructed" experimental apparatus that corresponds to such an observable. Presumably there is no proof of this statement (maybe they'll detect zombie cats in 3001), but experience indicates this to be the case.

12. Apr 2, 2013

### micromass

I don't think this is an accurate definition of an observable. An observable should be Hermitian (this is stronger than just saying that it has real eigenvalues). And I don't think there are any conditions on the eigenvectors at all.

13. Apr 2, 2013

### atyy

Why is being Hermitian stronger than just requiring real eigenvalues? (Maybe this would be a good time to also clarify what the difference is between Hermitian and self-adjoint).

By the eigenvectors spanning the space, I mean the eigenvectors form a basis for the Hilbert space. One place that gives this condition is Preskill's notes. He says an observable is self-adjoint, and that such an operator has eigenstates that form a complete orthonormal basis. He says there's some subtlety for unbounded operators in infinite dimensional spaces. Is that the reason you are saying the eigenvectors aren't required to span the space?

Last edited: Apr 2, 2013
14. Apr 2, 2013

### strangerep

atyy,

Not sure whether I should get involved in this thread, but I get the feeling you're looking at all this in a cart-before-horse manner...

To construct a quantum model of a particular physical scenario, one starts with a physically-meaningful dynamical group (which maps solutions of the equations of motion into other solutions), and attempts to construct a unitary representation thereof (i.e., a representation of the group as operators on a Hilbert space, or generalization thereof). One way to do this is to find the group's Casimirs, and a maximal commuting set of generators within the dynamical algebra of group generators, determine their joint spectrum, and construct a Hilbert state space from that spectrum (verifying of course that the rest of the generators and the whole group are also satisfactorily represented on the state space so-constructed).

[Edit: In more general cases, it may be more physically-sensible to construct a POVM on the spectrum, since a POVM is closer to what real apparatus does. Cf. http://en.wikipedia.org/wiki/POVM]

So what I'm trying to say is: the dynamical group comes first, followed by the states which are (in a sense) derived from the dynamical group and the requirement of unitary representation. However, in this thread, it seems you're trying to do it the other way around, therefore encountering some difficulty with the physical interpretation.

BTW, Ballentine chapters 1-3 give a better introduction to this way of looking at things than the Preskill notes (imho).

Last edited: Apr 2, 2013
15. Apr 2, 2013

### micromass

This is even true in finite-dimensions. For example,

$$\left(\begin{array}{cc} 1 & 1\\ 0 & 2\end{array}\right)$$

is a matrix that induces an operator $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$. This operator will have real eigenvalues (namely $1$ and $2$), but it won't be a Hermitian operator (as you can see from the matrix).

The difference between hermitian and self-adjoint is one that arises in unbounded operators. I don't think the difference is very important here. Basically, the difference has to do with the domain of $T^*$ with respect to $T$. So if for a densely defined operator $T$, we can find an operator $T^*$ such that $Tx=T^*x$ for all $x$ in the domain of $T$, then $T$ is called hermitian or symmetric. This implies that the domain of $T$ is included in the domain of $T^*$. If the domain are actually equal, then the operator is called self-adjoint.

I guess I should let a physicist answer whether the eigenvectors should span the space or not. In my (extremely limited) experience, you only demand that the operator is hermitian.

You have to be careful though. Saying that the eigenvectors span the space is very useless in infinite dimensions. You will find almost no examples of this. What you should say is that the closure of the span of the eigenvectors is the entire space. So it is important to take the closure. In infinite dimensions, this is unnecessary though.

16. Apr 2, 2013

### atyy

In the last sentence, did you mean to say "In finite dimensions, this is unnecessary though"?

17. Apr 2, 2013

### Fredrik

Staff Emeritus
As a mathematician, you should know that physics is far too difficult to be left to the physicists.

Many standard QM books (e.g. Sakurai) never use concepts from topology. That's why our understanding of these things is so poor.

18. Apr 2, 2013

### micromass

Yes. Sorry.

19. Apr 2, 2013

### micromass

But topology is so beautiful. Why ignore it

20. Apr 3, 2013

### atyy

OK, good - now I need to find out what "closure" means

(Thanks for all your detailed answers, maybe you can point me to some good book to read ...)