# General self adjoint operator

1. Sep 6, 2005

### Palindrom

I'm asking from a mathematical aspect, please be as exact as possible.

Suppose A is an observable, and B a Borel set in R (real numbers). Without further assumptions on A, I would like a general expression for the probability of measuring one of the values from B for A.
The reason for my question is that it seems I'm stuck if A hasn't got any eigenvalues (like position X), and can't be associated with a wavefunction (like position X), or put in position representation.
Please notice that the two identical brackets have opposite interpretations.

*For example, for X, the answer would be the Lesbegues integral of $$$\left| \psi \right|^2$$$ over B.

Last edited: Sep 6, 2005
2. Sep 7, 2005

### Palindrom

Really? Nobody?
Is my question not clear enough?

3. Sep 7, 2005

### reilly

I'll take a crack: if some of the eigenvalues of an operaor are contained in B, then the measure (probability) will be |w(a)|**2 with no integral. For a discrete operator there should be no problem -- everything will be finite. And, as far as I can figure, the Lebesgue integral for any continuous spectra operator should be OK -- surely wavefunctions absolutely squared are Lebesque integrable. Technically a point in a compact Borel set (if I remember right, Borel sets are compact) has measure zero, so we'll have to be satisfied with the probability of finding a neighborhood of he appropriate point. For all pratical purposes, standard measure theoretic probability is perfectly adequate for your question.

Regards,
Reilly Atkinson

4. Sep 7, 2005

### Palindrom

Why should Borel sets be compact? R^n is Borel in R^n for all n, isn't it? (I'm not sure, this is an actual and not retorical question. haven't taken measure theory yet, only next semester)

Anyway, if A is an operator with no eigenvalues (like X), how do you associate it with a wavefunction? I'm just stuck in this specific technical point.
If the state is |psi>, what's the wavefunction associated to it given A as stated above?

5. Sep 7, 2005

### Hurkyl

Staff Emeritus
If we're being exact...

You should first recognize that the position operator X does not have any eigenvalues: there does not exist any λ and nonzero |ψ> such that X|ψ> = λ|ψ>.

The things we call its "eigenvectors" are really some icky objects that lie in some extension of the state space, but not in the state space itself.

I would like to assert that every self-adjoint operator has a complete set of eigenvectors in this generalized sense, but I don't know enough to assert that.

However, I can quote the spectral theorem, but first we need to learn about a spectral resolution.

A spectral resolution in a hilbert space H is a family of commuting projection operators {Eα} where α ranges over the real numbers. This family has the property that:

(1) If α < β then EαEβ = Eα.

Intuitively, this is just saying that Eα projects onto a smaller subspace than Eβ.

(2) The map α → Eα is continuous from the right.

(3) As α → -∞, Eα → 0, the zero operator.

(4) As α → +∞, Eα → 1, the identity operator.

Then, the spectral theorem says that every self-adjoint operator posesses a unique spectral resolution satisfying:

$$A = \int_{-\infty}^{+\infty} \alpha \, dE_{\alpha}$$

in particular, |ψ> is in the domain of A if and only if the integral in

$$<\psi|A|\psi> = \int_{-\infty}^{+\infty} \alpha \, d<\psi|E_{\alpha}|\psi>$$

converges. This integral is a Stieltjes integral.

Basically, we think of the Eα as being a "cumulative eigenspace" operator: it maps vectors onto the subspace spanned by all of the "eigenvectors" whose "eigenvalues" are less than or equal to α.

In fact, if we are working with a self-adjoint operator on a finite-dimensional space, we see that the above paragraph is literally true: we can remove the quotes.

Anyways, the connection to statistics is this: if we have a cumulative probability distribution f(a) = P(X <= a), then we would calculate the expected value of the random variable X by:

$$\bar{X} = \int_{-\infty}^{+\infty} x \, df(x)$$

The similarity should be obvious. In fact, the conditions on the definition of a spectral resolution are precisely the conditions needed for it to be a cumulative probability distribution function!

So, it is now clear how to answer your original question. The probability that the observable A lies in some region R is:

$$\int_R dE_{\alpha}$$

where {Eα} is the spectral resolution of A.

Last edited: Sep 7, 2005
6. Sep 7, 2005

### Palindrom

Wow. Thanks a million.
Only problem is, I'm not familiar yet with the integral you noted , and don't know its definition, so I got stuck at that point. I don't suppose it's a simple thing to explain?

A small question of pure curiosity: when you say, in (2), that the map needs be continuous from the right, with respect to what metric is that? Because it seems nothing promises you that E_a is bounded.

And, may I ask, what is your occupation? Into what field are you?

Thanks again.

7. Sep 7, 2005

### Hurkyl

Staff Emeritus
The norm of every nonzero projection operator is 1. It's an easy proof to work out (intuitively, at least), but I'll present it if you ask.

You can skip the integral, I suppose, and write down a direct correspondence to probability:

P(A <= α | ψ) = <ψ|Eα|ψ>

which, once you've chosen |ψ>, defines A as a random variable with the given cumulative probability distribution function.

The integral itself will make much more sense when you've learned measure theory. The notation is supposed to be suggestive, though: if f happens to be differentiable, then df(x) = f'(x) dx.

Oh, and I'm a mathematician.

8. Sep 7, 2005

### Hurkyl

Staff Emeritus
By the way, the reference I was using is:

Modern Physics for Mathematicians

which you can find by clicking through the "links" link from the top of the page, and a bit of digging. Appendix B is the appendix on Hilbert space, and it talks about this stuff.

9. Sep 7, 2005

### Palindrom

Well, you've pretty much solved me the problem that's been bothering me for two weeks.

Another (related) question: in probability theory, are these kinds of structures being studied somewhere? I mean, Hilbert spaces where hermitean operators are associated with random variables, and basically all that theory used in quantum mechanics.
What should I look for in a course sylabii if I wanna take that in the mathematics department? (Which I do)

10. Sep 7, 2005

### Hurkyl

Staff Emeritus
Quantum physics uses all sorts of crazy mathematics from all over the board! I'm probably not the right person to ask, but I'd imagine that a good place to start is with some sort of Real Analysis course, mainly just to get a good foundation for what's to come.

Though it could very well get close to some of the topics of interest... for instance my Real Analysis text has a section on real Hilbert space in a section on Banach spaces (A Hilbert space is a type of Banach space), and it talks a good bit about bounded linear operators on a Banach space.

11. Sep 7, 2005

### David

Courses will vary in what they teach exactly and will be called different things in different places. For example, the first course I ever did that was called real analysis did not go into details about Banach spaces, etc. It pretty much started with set theory and topology, and worked through to vector spaces, using only the basic machinery needed for analysis (effectively integration) on the reals. WIth luck, your course catalog has some decent descriptions of content. Otherwise, find out what textbooks are recommended and wander through your university's bookstore and browse the books.

The actual mathematics you NEED to do quantum physics depends on what exactly you want to do with it. For day-to-day use, you might not need anything substantial. But if you really want to get into the foundations, then you'll need some probability and measure theory, for example. The measure theory will get you into the other types of integrals. For certain purposes, you'll want some group theory (especially when it comes to understanding angular momentum as a first stop).

One thing to note is that you might learn the mathematics in courses that never make the connections to quantum physics, and the quantum physics courses you take might not ever require the mathematics you learned in the other courses. This basically reflects the fact that to just "use" quantum physics, you don't necessarily need much of the mathematical machinery that sits behind it.

12. Sep 8, 2005

### Palindrom

Thank you both, and David, you've just just stated more or less what's bothering me in Physics courses.

13. Sep 8, 2005

### reilly

With all due respect: The math of quantum mechanics is by no means crazy -- unless you are restrictive in the type of spaces and operators you consider. There's a huge mathematical literature based on quantum theory or dealing with similar issues.`. Some examples:

Compact Quantum Metric Spaces by Prof. Marc Rieffel of Berkeley, my college roommate.
http://lanl.arxiv.org/abs/math.OA/0308207 (has many references)

George Mackey's classic work on unitary representations and commutation rules.

The entire field of Distribution Theory, one way to deal with unbounded operators.

A good bit of functional analysis

Of course X has eigenvalues(characteristic values), and eigenstates (characteristic functions). And the mathematical literature is full of references to eigenstates, and eigen values. What makes them "icky"? X has no eigenvalues? Where does this come from?

To make life simple and direct, I'll quote from Hille and Phillips, Functional Analysis and Semi-groups (1953, American Mathematical Society) -- I learned my functional analysis from Prof. Phillips.

In Chap. XIX, Translations and Powers, they examine, among other things, translations in the complex plane (hence along the real line). the infitesimal generator of such translations is d/dt, as in dz(t)/dt= Lz(t), or in finite form T(a)z(t)=z(t+a) ---- think of z as a wave function, if you want, and t as time, or space, or ....

They say; "....the spectrum of d/dt considered as an operator on C[0, infinity], is precisely the half-plane R(L) <=0, the points on the imaginary axis except for L=0 are in the continuous spectrum and the rest of the half plane is in the point spectrum. To each characteristic value L corresponds essentially only one characteristic function, x(t)=exp(Lt)."
(page 532)
(C stands for comples, R for real, and x=Re(z) )

My comment about the compactness of Borel sets is wrong. Borel sets obey a sigma algebra, which is closed under finite and countably infinite unions and intersections. Measures are then defined on sigma algebras, which guarantee the additivity property necessary for measures and probability. Typically in physics we deal with Hausdorff spaces, and compact sets in Hausdorff spaces are necessarily closed, while sigma algebras contain both open and closed sets.

The classic work on measure theoretic probability theory is Cramer's Mathematical Methods of Statistics, which is nucely summarized in; ttp://cscs.umich.edu/~crshalizi/reviews/cramer-on-math-stat/

The math of QM, in my opinion is very elegant, cf Dirac, and, to be sure, it often is based on intuition. But good physics creates good math. The problems of unbounded operators and such were largely solved 70-80 years ago. More recently physicist have pushed the boundaries of topology, and given mathematicians grist for their mill.

If you look carefully, hurkyl and I end up in the same place.

Again, the math of QM is anything but crazy -- mind bending as it may be sometimes. (To see this in detail requires the intro course in real variables, and at least one more in functional analysis to get the tools necessary to make you feel sane.)

Regards,
Reilly Atkinson

14. Sep 8, 2005

### Hurkyl

Staff Emeritus
I feel compelled to say that I didn't mean "crazy" as any sort of sleight: I was trying to refer to just how much high-powered mathematics is used to put various techniques used in QM on a rigorous foundation.

15. Sep 8, 2005

### Palindrom

Which field in particular deals with such an approach to physics? That is, a mathematical rigorous approach? Is that exactly what mathematical physics is all about?

16. Sep 9, 2005

### Ratzinger

so what is now with the position operator and eigenvalue problem?..as far as my (poor) understanding goes it's like this:

In a time-independent problem, we are looking for energy eigenstates, given the Hamilton operator. There are operators that commute with the Hamilton (that have common eigenstates with H). X does not. So there is no eigenvalue problem with the X operator and statevectors as its eigenkets.

17. Sep 9, 2005

### Hurkyl

Staff Emeritus
There's a bit of semantics going on here.

Normally, we speak of the states of a system living in a Hilbert space.

For example, we might say the state space is the set of all wavefunctions.

Now, the position operator X does not have any eigenvectors in this Hilbert space. (And thus, X has no eigenvalues when operating on it)

The things that we like to call the eigenvectors of X actually live in a messy thing called a rigged Hilbert space, which is your Hilbert space plus some additional things that are a pain to rigorously define. Operators on the Hilbert space can be extended to operate on the rigged Hilbert space, and when we extend X in this way, we find that it does have some eigenvectors in the rigged Hilbert space.

But, the rigged Hilbert space is a much more difficult thing to treat rigorously, so if you can avoid invoking it, you're often better off.

18. Sep 9, 2005

### Palindrom

A couple of questions:
1. Is the uncertainty principle valid for states in the rigged Hilbert space? (There's a good reason for the question)

2. To repeat my question from the precedent page, what specific area treats mathematical rigorous approach to physics? This is exactly what I'd like to apply to in Graduate School, when the happy day comes.

19. Sep 9, 2005

### Hurkyl

Staff Emeritus
The uncertainty principle doesn't really care about states: it's essentially just a theorem about operators.

However, the very thought about trying to apply it when we're considering the rigged Hilbert space makes me uneasy, because I don't know how to make sense of <φ|ψ> when |φ> and |ψ> are arbitrary elements of the rigged Hilbert space.

I think I suggested a real analysis course before... but as was mentioned, what is considered "real analysis" varies. The main topics I think you would like to have in such a course for your purposes are generic Banach spaces and generic measures. (Not just the Lesbegue measure!) Beyond that as a starting point, I don't really know. (Certainly some functional analysis, I suppose... that's a whole course on its own isn't it?)

But as I mentioned, I'm not really the person to ask. Most of what I know about quantum stuff is self-taught, and that includes a good deal of the mathematics too.

Last edited: Sep 9, 2005
20. Sep 9, 2005

### Palindrom

I didn't mean what courses, I meant what area in physics or mathematics- like mathematical physics, or applied mathematics.

Why does the uncertainty principle not care about states? Given two observables A and B, and a state |psi>, we have:
stdev(A)*stdev(B)>|<psi|[A,B]|psi>|

I can prove that if the uncertainty principle is valid, then X has no eigenvectors.