Is every state an eigenstate of an observable?

[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.

No, I mean a pure state. I suppose my question is whether every state is an eigenvector of some observable.

This is true, but even the arbitrary vector is an eigenvector of some observable.

Is there a short proof, or can you point me to a reference? (Anyway, I do think it's not very relevant to the question, since I know this is true in many cases.)

A nonzero vector $|u>$ is an eigenvector of $|u><u|$.

Are all eigenvalues of $|u><u|$ real?

That's a projection operator, so it only has eigenvalues 0 and 1.

Note that if $P^2=P$ and $Px=\lambda x$ where $\lambda$ is a complex number and $\|x\|\neq 0$, we have $P^2x=P(Px)=P(\lambda x)=\lambda Px=\lambda^2x$, and also $P^2x=Px=\lambda x$. So $\lambda(\lambda-1)x=0$, and this implies that $\lambda$ is 0 or 1.

Ok, thanks. I have to say that it seems very unintuitive in the case of the wave function, but perhaps it's just very hard to measure such an observable, although it exists. (It seems quite intuitive to me for spin).

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?

 

[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.

 

[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.

 

[Fredrik]

I make a measurement and get a result of 0, what state has the system collapsed into?

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
$$\rho'=\frac{M_R(\Delta_a)\rho M_R(\Delta_a)}{\operatorname{Tr}\left(M_R(\Delta_a)\rho M_R(\Delta_a)\right)}.$$

 

[atyy]

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
$$\rho'=\frac{M_R(\Delta_a)\rho M_R(\Delta_a)}{\operatorname{Tr}\left(M_R(\Delta_a)\rho M_R(\Delta_a)\right)}.$$

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

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).

 

[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.

 

[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)?

 

[Fredrik]

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)?

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.

 

[strangerep]

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.

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

 

[jfy4]

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)?

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.

 

[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.

 

[micromass]

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

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.

 

[atyy]

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.

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 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf. 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?

 

[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).

 

[micromass]

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).

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.

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 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf. 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?

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.

 

[atyy]

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.

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

 

[Fredrik]

I guess I should let a physicist answer whether the eigenvectors should span the space or not.

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

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.

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

 

[micromass]

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

Yes. Sorry.

 

[micromass]

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.

But topology is so beautiful. Why ignore it

 

[atyy]

Yes. Sorry.

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 ...)

 

[micromass]

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 ...)

If you're really interested in the mathematics behind QM, then I highly recommend Kreyzig: https://www.amazon.com/dp/0471504599/?tag=pfamazon01-20
The book covers a lot of ground, but unlike most functional analysis books, it doesn't have many prerequisites.

Another nice book (and more physical) is Prugovecki: https://www.amazon.com/dp/0486453278/?tag=pfamazon01-20

The notion of closure is treated very well in both books.

 

[atyy]

If you're really interested in the mathematics behind QM, then I highly recommend Kreyzig: https://www.amazon.com/dp/0471504599/?tag=pfamazon01-20
The book covers a lot of ground, but unlike most functional analysis books, it doesn't have many prerequisites.

Another nice book (and more physical) is Prugovecki: https://www.amazon.com/dp/0486453278/?tag=pfamazon01-20

The notion of closure is treated very well in both books.

Thanks!

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

But topology is so beautiful. Why ignore it

BTW, when I asked my QM prof (Xiao-Gang Wen) whether I should take more maths classes, he said rigourous maths classes are mostly useless, because you only learn about things that don't exist, like functions that are everywhere continuous but nowhere differentiable. But he studies something called "topological order", so I think he likes topology.

Another prof said something like, he didn't know whether anything he told us was really mathematically correct, but it all matched experiment ...

 

[Fredrik]

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

Hilbert spaces are metric spaces. In the context of metric spaces, you can define "closed" like this: Let X be a metric space. A set $E\subset X$ is said to be closed if the limit of every convergent sequence with all its terms in E is in E. Then you can define the closure of an arbitrary $E\subset X$ as the smallest subset of X that contains E. Another way of saying that is that the closure of E is the the intersection of all the subsets of X that contain E. And yet another way is to say that the closure of E is the set of points that are limits of convergent sequences with all terms in E.

A subset of a Hilbert space that's also a vector space is called a linear subspace. It can be proved that a linear subspace of a Hilbert space is a Hilbert space if and only if it's a closed set. Hence the term closed linear subspace. I would prefer to call these things "vector subspaces" and "Hilbert subspaces", but that terminology seems less popular for some reason.

 

[micromass]

Maybe a more intuitive explanation (but also imprecise). If I say that the span of some vectors is the entire space, then that means that every vector is the linear combination of the vectors I start with. In particular, I can multiply with scalars and I can take finite sums.
If I take the closure of the space, then I also allow infinite sums to be taken.

 

[DrDu]

As Fredrik has pointed out already, the question is not one of mathematical rigor, as every self-adjoint operator has a spectral decomposition. The problem is rather how to realize a given operator as a measurement device.

 

[atyy]

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).

If one follows this approach, is every state also the eigenvector of some "mathematical observable" (ie. is the condition for a mathematical observable the same as in micromass's approach, that it just has to be self-adjoint, or are there additional conditions from the dynamical group)?

 

[strangerep]

If one follows this approach, is every state also the eigenvector of some "mathematical observable"

Given a state, one can always find a rank-1 operator, as others have pointed out above. That wasn't really the reason for my interjection though. I was really just curious why you're quite intensely focused on this question? Is there a specific case you have in mind?

(ie. is the condition for a mathematical observable the same as in micromass's approach, that it just has to be self-adjoint, or are there additional conditions from the dynamical group)?

The generators of the dynamical group are usually represented as self-adjoint operators (or perhaps a suitably extended version of "self-adjoint" in the case of rigged Hilbert space). More generally, the finite group elements should be represented as unitary operators.

(Strictly speaking, one can construct more general cases than Hermitian or self-adjoint, such as in the case of unstable systems -- there one can have non-Hermitian Hamiltonians and the imaginary part of the energy eigenvalues correspond to decay lifetimes.)

 

[atyy]

Given a state, one can always find a rank-1 operator, as others have pointed out above. That wasn't really the reason for my interjection though. I was really just curious why you're quite intensely focused on this question? Is there a specific case you have in mind?

It was just something I wasn't sure about because while in some systems like spin 1/2, I could see how rotating the apparatus would make any state an eigenvector, but I didn't see how to do this in the case of the wave function (or as jfy4 said - what apparatus would detect a half dead - half alive cat?).

So the answer mathematically is "yes". Physically it seems often to be "no", though I'm not sure if the limitation one of current technology or deeper than that.