# Fourier series, Hermitian operators

(First of all I never saw Hilbert spaces in a mathematical class, only used it in intro QM so far, so please don't assume I know that much when answering.)

Let's consider the Hilbert space on the interval [a,b] and the operator ##\textbf{L} = \frac{d^{2}}{dx^{2}} ##. Then ##\textbf{L}## is hermitic only if for all functions ##f(x)## and ##g(x)## in the Hilbert space:

##<f| \textbf{L} g > = < \textbf{L} f |g > ##

Using partial integration one can find that:

##<f| \textbf{L} g > - < \textbf{L} f |g > = g'(b)f_{c}(b)-g'(a)f_{c}(a)-f_{c}'(b)g(b)+f_{c}'(a)g(a)##

( ' refers to the first derivative and subscript c to the complex conjugate )

In physics class now something weird happened. We said that the functions we will be working with will have certain boundary conditions we had discussed such that the right term in the last equation vanishes. And thus in our case the operator ##\textbf{L}## will be hermitian. This means we will use the eigenfunctions of ##\textbf{L}## as a basis for our space, saying that they can produce any square integrable function on [a,b].

This confuses me A LOT. I thought that the relationship above had to hold for ALL square integrable functions ##\textbf{L}## and only then the eigenfunctions could produce all the square integrable functions on [a,b].

So basically something fishy is happening in physics class and I'd like it explained on a non-math level since I have not taken a math course on Hilbert spaces and only have practical experiences of working with them in intro QM.

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strangerep
In physics class now something weird happened. We said that the functions we will be working with will have certain boundary conditions we had discussed such that the right term in the last equation vanishes. And thus in our case the operator ##\textbf{L}## will be hermitian. This means we will use the eigenfunctions of ##\textbf{L}## as a basis for our space, saying that they can produce any square integrable function on [a,b].
There are several theorems underpinning all this. Have you studied linear algebra? What you're seeing here is a generalization of the "spectral theorem", which says that eigenvectors of an Hermitian operator form a basis for the vector space.

This confuses me A LOT. I thought that the relationship above had to hold for ALL square integrable functions ##\textbf{L}##
##\textbf{L}## is an operator, not a function. Maybe you meant something else?

and only then the eigenfunctions could produce all the square integrable functions on [a,b]
It's not producing "all" the square integrable functions on [a,b], but rather all the square integrable functions applicable in the dynamics of the current problem.

So basically something fishy is happening in physics class and I'd like it explained on a non-math level since I have not taken a math course on Hilbert spaces and only have practical experiences of working with them in intro QM.
Well, there are severe limitations to how far one can go without involving a bit of math. You'll need at least some linear algebra. Also, if you can access a copy of Ballentine, see whether the material in his ch1 is within your grasp.

BTW, what textbooks have specified for your course?

Maybe because QM people tend have a little more sympathy for your predicament?

bhobba
bhobba
Mentor
Well, there are severe limitations to how far one can go without involving a bit of math. You'll need at least some linear algebra. Also, if you can access a copy of Ballentine, see whether the material in his ch1 is within your grasp.

All I can do is reiterate the suggestion.

Study some linear algebra then check out Chapter 1 of Ballentine.

Thanks
Bill

It's not producing "all" the square integrable functions on [a,b], but rather all the square integrable functions applicable in the dynamics of the current problem.

So if I say that I'm only going to consider all the functions that have the following boundary conditions in point ##a## and ##b##:

##\alpha f(a) - \beta \frac{df}{dx}(a) = 0 ##

##\gamma f(b) + \delta \frac{df}{dx}(b) = 0 ##

For these functions, operator ##L## is hermitian. What will the eigenvectors of this operator span then? It don't have to be ALL square integrable functions like you earlier said, so it must be all square integrable function THAT also agree with these boundary conditions?

So first of all, is the above true? If yes then let's consider the interval to be ##[0,2\pi]## and set the following boundary conditions:

##f(a) = \frac{df}{dx}(a) ## and ##f(b) = \frac{df}{dx}(b) ##

For these functions ##L## is hermitian and the eigenvectors will be ##sin(nx)## and ##cos(nx)##

However we know that those functions span all the square integrable functions on ##[0,2\pi]##, from the classical use of Fourier Series. So here, even if we do set constraining boundary conditions each square integrable function should be producable. Seems a little weird to me.

Edit: This particular thought proces didn't come up in QM. It's in a physicsy-math type of class where we learn hand-waving math methods to solve different kinds of physics problems like diffusion, wave equation, potential equation with source terms and so on. I am only familiar with Hilbert spaces from intro QM and here we suddenly use Hilbert spaces as well as if everyone knows all the finesses of it. I personally don't and would like to at least understand this particular reasoning to move on to the physics of the class. We don't use any particular textbook, I just use notes from the theory classes.

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So if I say that I'm only going to consider all the functions that have the following boundary conditions in point ##a## and ##b##:

##\alpha f(a) - \beta \frac{df}{dx}(a) = 0 ##

##\gamma f(b) + \delta \frac{df}{dx}(b) = 0 ##

For these functions, operator ##L## is hermitian. What will the eigenvectors of this operator span then? It don't have to be ALL square integrable functions like you earlier said, so it must be all square integrable function THAT also agree with these boundary conditions?

So first of all, is the above true? If yes then let's consider the interval to be ##[0,2\pi]## and set the following boundary conditions:

##f(a) = \frac{df}{dx}(a) ## and ##f(b) = \frac{df}{dx}(b) ##

For these functions ##L## is hermitian and the eigenvectors will be ##sin(nx)## and ##cos(nx)##

However we know that those functions span all the square integrable functions on ##[0,2\pi]##, from the classical use of Fourier Series. So here, even if we do set constraining boundary conditions each square integrable function should be producable. Seems a little weird to me.

You won't get all square integrable functions, since in general a square integrable function is not necessarily differentiable. A big mistake that beginners make is that an operator ##L## is everywhere defined, but this needs not be true and is usually not true.

The argument with the Fourier series would be the following. You can indeed express any square integrable function as ##f(x) = \sum_{n\in \mathbb{Z}} c_n e^{inx}##. But what you want to do is to set

$$L(f)(x) L\left(\sum_{n\in \mathbb{Z}} c_n e^{inx}\right) = \sum_{n\in \mathbb{Z}} c-n L(e^{inx})$$

and then conclude that ##L## makes sense on ##f##. The error here is that the second step is not allowed. You cannot in general switch an infinite sum and a linear operator. Yes, ##L## is linear but that only allows you to switch finite sums, not infinite ones. You would be allowed to switch sums and the operator if ##L## were continuous, but it usually isn't.

bhobba
Mentor
You won't get all square integrable functions, since in general a square integrable function is not necessarily differentiable. A big mistake that beginners make is that an operator ##L## is everywhere defined, but this needs not be true and is usually not true.

That's where generalised functions come into it:
https://terrytao.wordpress.com/tag/schwartz-functions/

In quantum mechanics that has been further generalised to what's called Rigged Hilbert Spaces. If you read Ballentine you will get a quick and dirty introduction to it. To start with that's pretty much all you need to know - later you can delve into the details.

But generalised functions are so important in not just physics, but applied math in general, knowledge of it should really be part of undergrad math education. Here is my goto book on it:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill

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I believe the point of your problem is to prove that d^2/dx^2 is Hermitian. You only need to substitute and show that the difference <f|Lg> - <Lf|g> is zero using integrals. All functions in quantum physics should be square integrable because if they weren't, then the inner product, and hence your probabilities would not exist. The operator d^2/dx^2 should infact be Hermitian because it is related to the Energy observable p^2/2m where the p is the momentum operator. That momentum operator is -(hbar)^2*d^2/dx^2. All your observables like position, momentum, energy, etc are represented by Hermitian operators.

Sorry the square of the momentum operator is -(hbar)^2*d^2/dx^2

I believe the point of your problem is to prove that d^2/dx^2 is Hermitian. You only need to substitute and show that the difference <f|Lg> - <Lf|g> is zero using integrals. All functions in quantum physics should be square integrable because if they weren't, then the inner product, and hence your probabilities would not exist. The operator d^2/dx^2 should infact be Hermitian because it is related to the Energy observable p^2/2m where the p is the momentum operator. That momentum operator is -(hbar)^2*d^2/dx^2. All your observables like position, momentum, energy, etc are represented by Hermitian operators.

This isn't actually a QM question, I know it's in the QM subsection but it got moved here by the mods without my doing.

So I'm talking about functions on a closed interval [a,b], and there clearly exist functions that are well behaving and all but for which the operator ##\textbf{L}## is not hermitian.

You won't get all square integrable functions, since in general a square integrable function is not necessarily differentiable. A big mistake that beginners make is that an operator ##L## is everywhere defined, but this needs not be true and is usually not true.

The argument with the Fourier series would be the following. You can indeed express any square integrable function as ##f(x) = \sum_{n\in \mathbb{Z}} c_n e^{inx}##. But what you want to do is to set

$$L(f)(x) L\left(\sum_{n\in \mathbb{Z}} c_n e^{inx}\right) = \sum_{n\in \mathbb{Z}} c-n L(e^{inx})$$

and then conclude that ##L## makes sense on ##f##. The error here is that the second step is not allowed. You cannot in general switch an infinite sum and a linear operator. Yes, ##L## is linear but that only allows you to switch finite sums, not infinite ones. You would be allowed to switch sums and the operator if ##L## were continuous, but it usually isn't.

Thanks for the answer. The following question does not realte to your answer directly but I'm still trying to put it all correctly to shelves in my mind and since you replied I assume you don't mind me asking on.

Question: Do different Hilbert spaces exist on the interval [a,b]? So basically, when talking about the Hilbert space of functions with certain boundary conditions, or the Hilbert space of the square integrable functions ##\textbf{L}^{2} [a,b] ## ( note , this L doesn't relate to my operator L ) we are talking about different spaces?

Question: Do different Hilbert spaces exist on the interval [a,b]?

Yes.

So basically, when talking about the Hilbert space of functions with certain boundary conditions, or the Hilbert space of the square integrable functions ##\textbf{L}^{2} [a,b] ## ( note , this L doesn't relate to my operator L ) we are talking about different spaces?

The space of all functions with boundary conditions is not necessarily a Hilbert space and usually is not. The issue is a mathematical term called completeness. You need to be aware that the domain of ##L## (the set of all functions on which ##L## works) is not a Hilbert space at all, but merely a subspace of a Hilbert space.

Yes.

The space of all functions with boundary conditions is not necessarily a Hilbert space and usually is not. The issue is a mathematical term called completeness. You need to be aware that the domain of ##L## (the set of all functions on which ##L## works) is not a Hilbert space at all, but merely a subspace of a Hilbert space.

Then, returning to the previous post you have commented on ( when I said that we are going to work with functions with the general boundary conditions I mentioned) it is not correct to say that when working with these functions we are working in a Hilbert space? Damn, the confusion keeps increasing. I have read the first chapted of the book recommended by the people above me but it only very lightly touches upon the crucial part I'm interested in, namely how to think about boundary conditions in the context of Hilbert spaces. Too bad can't afford the time to learn Hilbert space theory thoroughly since I have a busy physics semester.

Then, returning to the previous post you have commented on ( when I said that we are going to work with functions with the general boundary conditions I mentioned) it is not correct to say that when working with these functions we are working in a Hilbert space? Damn, the confusion keeps increasing. I have read the first chapted of the book recommended by the people above me but it only very lightly touches upon the crucial part I'm interested in, namely how to think about boundary conditions in the context of Hilbert spaces. Too bad can't afford the time to learn Hilbert space theory thoroughly since I have a busy physics semester.

Yes, we are working in a Hilbert space. The operator ##L## takes in values of a Hilbert space. But the issue is that ##L## is not defined for all the elements of the Hilbert space.

Yes, we are working in a Hilbert space. The operator ##L## takes in values of a Hilbert space. But the issue is that ##L## is not defined for all the elements of the Hilbert space.

What can we then say about the eigenvectors of ##\textbf{L}## for the functions with the forementioned boundary condition? How do those eigenvectors relate to the Hilbert space / L²[a,b]?

What can we then say about the eigenvectors of ##\textbf{L}## for the functions with the forementioned boundary condition? How do those eigenvectors relate to the Hilbert space / L²[a,b]?

They are elements of the Hilbert space.

They are elements of the Hilbert space.

They don't form a basis for anything?

They don't form a basis for anything?

Not necessarily. But if you take the generalized eigenvectors (that are: eigenvectors that are not in the Hilbert space), then you do get a basis

bhobba
bhobba
Mentor
Just to expand on the the above that's where Rigged Hilbert Spaces come into it.

Here is a paper about it:
http://arxiv.org/pdf/quant-ph/0502053v1.pdf

Unfortunately at your current level it will likely be goobly gook.

It will be clearer when you read the first chapter of Ballentine.

Don't get despondent. You are obviously a thinking student. Unfortunately some of the fine points glossed over in beginner or even intermediate texts in QM require a sojourn into some tricky advanced areas of math. Most students don't get too concerned - but if you think hard about it, it is an issue. It means however you will need to do some supplemental reading to sort them out.

The same thing happened to me. I am self taught in physics but have a degree in applied math where I studied Hilbert spaces and such. I read Dirac's famous book on QM (Principles of QM) - elegant and beautiful but mathematically it left a lot to be desired - but it was what physicists used. Then I read Von-Neumanns book (Mathematical Foundations of QM) - equally elegant and beautiful - and mathematically rigorous - but it was not what physicists used. I was pulling my hair out so decided to get to the bottom of it.

I did - but it was a long sojourn into generalised functions, Nuclear Spaces and Rigged Hilbert spaces - quite advanced esoterica. I came out the other end knowing what's going on but it didnt help my understanding of QM at all. Because of that my advice is not to get caught up too much in it - Ballentine provides what you need for the physics.

Later you can delve into the detail if it interests you. The math is actually quite profound and beautiful - but unfortunately an advanced area of functional analysis.

Thanks
Bill

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strangerep