How to understand operators representing observables are Hermitian?

[shrodinger1226]

As we know, all operators representing observables are Hermitian. In my undersatanding, this statement means that all operators representing observables are Hermitian if the system can be described by a wavefunction or a vector in L2. For example, the momentum operator p is Herminitian, because for any states described by Ψ and Φ, we have

<Ψ|p|Φ> = ∫Ψ*(-ihdΦ/dx) dx = Ψ*(-ih)Φ + ∫(-ihdΨ/dx)*Φdx = <pΨ|Φ>

where Ψ*(-ih)Φ = 0 is employed at -∞ and +∞.

But the eigen function of p-operator (2πh)^-1/2exp(ipx/h) does not satisfy = 0 at -∞ and +∞, and the eigen function of the momentum operator, (2πh)^-1/2exp(ipx/h), is not a wave function describing quantum state. Am I right?

 

[WannabeNewton]

We had the exact same question being asked earlier today except it was with regards to the eigenfunctions of the position operator instead of those of the momentum operator. Regardless, see here: http://arxiv.org/pdf/quant-ph/0502053v1.pdf

 

[shrodinger1226]

... http://arxiv.org/pdf/quant-ph/0502053v1.pdf

Thanks, so the Hilbert space is replaced by rigged Hilbert space. It is a good paper and I will read it. One more question about the momentum operator.

In the 1D infinite well, 0<x<a, the momentum eigen equation is given by

-ih*dψ/dx=p*ψ, with ψ(0)=ψ(a)=0

and its solution is ψ=A*exp(ipx/h) which only has one constant "A" for the two boundary conditions ψ(0)=ψ(a)=0, and thus ψ=0 must hold. Does that mean the momentum operator p has no eigen wavefunctions in such a case? But according to the QE hypothesis, all operators representing observables are Hermitian, and their eigenfunctions constitute complete systems. How to understand this?

 

[bhobba]

But the eigen function of p-operator (2πh)^-1/2exp(ipx/h) does not satisfy = 0 at -∞ and +∞, and the eigen function of the momentum operator, (2πh)^-1/2exp(ipx/h), is not a wave function describing quantum state. Am I right?

It's the Rigged Hilbert Space formalism. The physically realizable states are taken as test functions (and they do the nice stuff like vanish at infinity, continuously differentiable etc etc) but for mathematical convenience it is extended by means of the Gelfland Triple to functions like the above and even weirder stuff like the Dirac Delta function:
http://homepage.univie.ac.at/roza.aceska/testfunctions.pdf
http://en.wikipedia.org/wiki/Rigged_Hilbert_space

Test functions are all that are considered physically realizable, but can approximate the functionals that are the superset of the Hilbert space in the Gelfland triple to any desired degree of accuracy. For example a wave that continues to infinity is not physically realizable, but is a good approximation to a wave that vanishes a long distance away. A position that is a Dirac Delta function doesn't exist but if it's known to a high degree of accuracy is a good approximation.

Thanks
Bill

 

[WannabeNewton]

In the 1D infinite well, 0<x<a, the momentum eigen equation is given by...

For a more detailed analysis of this system using the Rigged Hilbert space formalism see section 3 (p.10) of the paper.

EDIT: And see here for even more details: http://arxiv.org/pdf/quant-ph/0407195v1.pdf

But according to the QE hypothesis, all operators representing observables are Hermitian...

Yes. More precisely we want them to be self-adjoint so that the spectral theorem applies. To ensure this we must restrict the domains of operators (that correspond to dynamical variables) to appropriate dense subspaces of ##L^2## so that said domains agree with those of the adjoints.

For bounded operators there is really no issue with regards to any of this but we must be extremely careful with domains when it comes to unbounded operators; the property of being Hermitian is in a sense attached solely to the unbounded operator but the property of being self-adjoint is very sensitive to the domain because the domain of the adjoint has to match and so we must be cautious when dealing with unbounded operators (and by unbounded here I mean those operators which do not have a continuous extension from their dense domain of definition to all of ##L^2##).

and their eigenfunctions constitute complete systems

No. This is not part of the postulates of QM. Self-adjoint operators over infinite dimensional spaces can easily fail to have a complete set of eigenfunctions; the existence of a complete set of eigenfunctions for self-adjoint operators is not as fundamental as the spectral theorem. The spectral theorem is the main resort.

 

[shrodinger1226]

Quote by shrodinger1226
and their eigenfunctions constitute complete systems
---
No. This is not part of the postulates of QM. Self-adjoint operators over infinite dimensional spaces can easily fail to have a complete set of eigenfunctions; the existence of a complete set of eigenfunctions for self-adjoint operators is not as fundamental as the spectral theorem. The spectral theorem is the main resort.

Thanks a lot. You corrected me a wrong idea. So in the 1D infinite well (discrete eigenvalue problem), the momentum operator p does not have eigenfunction, because its solution is ψ=A*exp(ipx/h) which only has one constant "A" for the two boundary conditions ψ(0)=ψ(a)=0, and thus ψ=0 must hold. Right?

 

[WannabeNewton]

Right.

 

[dextercioby]

[...] Does that mean the momentum operator p has no eigen wavefunctions in such a case? [...]

Due to the very strong boundary conditions, yes, the 0 vector is the only eigenfunction, thing which is trivial, as any operator when applied on the 0 vector turns the 0 vector times an arbitrary constant. So that for any operator, the 0 vector is a trivial eigenvector.

But according to the QE hypothesis, all operators representing observables are Hermitian, and their eigenfunctions constitute complete systems. How to understand this?

No, all observables are represented by self-adjoint operators. Nothing is being said on their spectral properties. For example, for the free galilean particle in 1D, the position operator is an observable, because it's represented on L^2(R) as a self-adjoint operator, however it has no eigenfunctions in L^2(R).

 

[shrodinger1226]

...
No, all observables are represented by self-adjoint operators. Nothing is being said on their spectral properties. For example, for the free galilean particle in 1D, the position operator is an observable, because it's represented on L^2(R) as a self-adjoint operator, however it has no eigenfunctions in L^2(R).

But how to understand the following statement in the book by Griffiths, [D. Griffiths, "Introduction to quantum mechanics", (Prentice Hall, NJ, 1995), p. 106, Chapter 3]:

"As we have seen, in the finite-dimensional case the eigenvectors of a Hermitian operator always span the space."

In my understanding, "always span the space" means the eigenvectors are complete. The 1D infinite-well case is of a "finite-dimensional case", and the momentum operator is Hermitian, but it has no eigenfunction. Seems not consistent with the above statement. Did I miss something? Thanks a lot.

 

[WannabeNewton]

The dimensionality there refers to the dimensionality of the state space, not that of the physical system. For example, the energy eigenstates in the coordinate representation of a particle confined to a 1D infinite square well with the usual boundary conditions are sinusoidal functions restricted to the width of the square well. These are clearly square integrable functions residing in ##L^2## which is infinite dimensional.

All operators over finite dimensional spaces are bounded hence being Hermitian is equivalent to being self-adjoint; furthermore over finite dimensional spaces all self-adjoint operators have a complete set of eigenvectors due to the finite dimensional spectral theorem as you probably know from your elementary linear algebra course(s). Over infinite dimensional spaces such as ##L^2## the situation is much more complicated as already noted above.

 

[dextercioby]

But how to understand the following statement in the book by Griffiths, [D. Griffiths, "Introduction to quantum mechanics", (Prentice Hall, NJ, 1995), p. 106, Chapter 3]:

"As we have seen, in the finite-dimensional case the eigenvectors of a Hermitian operator always span the space."

In my understanding, "always span the space" means the eigenvectors are complete. The 1D infinite-well case is of a "finite-dimensional case", and the momentum operator is Hermitian, but it has no eigenfunction. Seems not consistent with the above statement. Did I miss something? Thanks a lot.

Well, you chose your username a misspelling of Schrödinger, but the connection you look for resides however in the matrix mechanics invented by Born, Heisenberg and Jordan in 1925. There are no finite matrices in a formulation of the 1D infinite well case for any of the operators at stake: position, momentum, energy, so that your bolded comment has nothing to do with the quote from Griffiths (side note: there are probably at least 5 better textbooks to learn QM than Griffiths). So the red part above is true.

Fact: any finite dimensional complex Hilbert space is isomorphic to C^n, for some n a natural number. C^n is the setting for the theory of square matrices with n² entries.

 

[shrodinger1226]

WannabeNewton and dextercioby, I got it, thanks a lot.

 

[shrodinger1226]

So in the 1D infinite well (discrete eigenvalue problem), the momentum operator p does not have eigenfunction, because its solution is ψ=A*exp(ipx/h) which only has one constant "A" for the two boundary conditions ψ(0)=ψ(a)=0, and thus ψ=0 must hold. Right?
---------
Right.

I still have a question. The momentum operator p does have any eigenfunctions in the 1D infinite well, since zero vector cannot be counted as a eigenvector. No eigenvectors, no eigenvalues. Then how to understand the postulate: A measurement of observable p is certain to return one of the eigenvalues of the operator p?

 

[dextercioby]

I still have a question. The momentum operator p does have any eigenfunctions in the 1D infinite well, since zero vector cannot be counted as a eigenvector. No eigenvectors, no eigenvalues. Then how to understand the postulate: A measurement of observable p is certain to return one of the eigenvalues of the operator p?

Extremely simple: if the wavefunction is required to be nil at the 2 points forming the boundary of the box, then momentum is not an observable, hence no contradiction with the measurement postulate.

 

[shrodinger1226]

Extremely simple: if the wavefunction is required to be nil at the 2 points forming the boundary of the box, then momentum is not an observable, hence no contradiction with the measurement postulate.

Sorry, I got confused a little bit.

(1) In principle, the momentum operator is Hermitian, observable operator. Why not observable?

(2) In the book by Griffiths [D. Griffiths, Introduction to quantum mechanics, (Prentice Hall, NJ, 2005), 2ed, p. 38, Chapter 2, Problem 2.5 (d) ], there is an exercise for calculating the expectation value of p, <p>, in the 1D infinite well; the answer is <p>=(8h/3a)sin(3ωt). This also indicates that p should be observable. Am I right?

 

[dextercioby]

For the point 1), well said 'in principle'. That leaves the door open for exceptions such as the momentum operator in a box = 1D infinite square well.
2) I don't see how that follows. You're merely computing <psi, p psi> for a given psi. How does that tell you it's an observable ?

 

[WannabeNewton]

I still have a question. The momentum operator p does have any eigenfunctions in the 1D infinite well, since zero vector cannot be counted as a eigenvector. No eigenvectors, no eigenvalues. Then how to understand the postulate: A measurement of observable p is certain to return one of the eigenvalues of the operator p?

This is a very nice question my friend !

Take a look at example 4 in page 6 of this paper, see if you can answer the question yourself, and then go to page 39 of said paper for the resolution: http://arxiv.org/pdf/quant-ph/9907069.pdf

The moral of the story is that Dirac's braket notation, if taken more seriously than being a mere calculational facility, will lead to a lot of mathematical contradictions in functional analysis.

 

[shrodinger1226]

For the point 1), well said 'in principle'. That leaves the door open for exceptions such as the momentum operator in a box = 1D infinite square well.
2) I don't see how that follows. You're merely computing <psi, p psi> for a given psi. How does that tell you it's an observable ?

Physically, in my undersatanding, <p> means that many measurements of p for an ensemble of the wells, we get an average value, <p>. Is that right?

 

[shrodinger1226]

This is a very nice question my friend !

Take a look at example 4 in page 6 of this paper, see if you can answer the question yourself, and then go to page 39 of said paper for the resolution: http://arxiv.org/pdf/quant-ph/9907069.pdf

The moral of the story is that Dirac's braket notation, if taken more seriously than being a mere calculational facility, will lead to a lot of mathematical contradictions in functional analysis.

Interesting! I am asking the same question as the paper does. But I don't understanding some words:

(1) "P does not admit any eigenvalues. Nevertheless, the spectrum of P is the entire complex plane [8] and P does not represent an observable."
The first statement seems not consistent with the second.

(2) "Thus, we established that P is Hermitian, but not self-adjoint."
In my understanding, "Hermitian" means "self-ajoint", <Ψ|p|Φ> = <pΨ|Φ>.

 

[WannabeNewton]

It's very hard to explain these things if you don't know basic functional analysis. When we are talking about operators over infinite dimensional spaces, the spectrum of the operator contains more than just the eigenvalues of the operator. See here: http://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)

Hermitian does not mean self-adjoint and I've explained why previously in the thread. An operator can be Hermitian but if it's domain does not agree with the domain of it's adjoint then clearly it cannot be self-adjoint. See post #5.

If you want, I can recommend some functional analysis texts to get you started.

 

[Jilang]

Interesting! I am asking the same question as the paper does. But I don't understanding some words:

(1) "P does not admit any eigenvalues. Nevertheless, the spectrum of P is the entire complex plane [8] and P does not represent an observable."
The first statement seems not consistent with the second.
.

I think the issue is that you are confusing an eigenvalue with an observable. You can measure P and find a value for it, but if you measure it again you won't probably get the same value for it. So P is not an eigenvalue for the state, it's not fixed.

 

[shrodinger1226]

It's very hard to explain these things if you don't know basic functional analysis. When we are talking about operators over infinite dimensional spaces, the spectrum of the operator contains more than just the eigenvalues of the operator. See here: http://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)

Hermitian does not mean self-adjoint and I've explained why previously in the thread. An operator can be Hermitian but if it's domain does not agree with the domain of it's adjoint then clearly it cannot be self-adjoint. See post #5.

If you want, I can recommend some functional analysis texts to get you started.

Thanks, I don't want to go further right now in this regard because it's too hard for me. Your corrections are very important to me. Thanks again.

 

[WannabeNewton]

Thanks, I don't want to go further right now in this regard because it's too hard for me. Your corrections are very important to me. Thanks again.

Anytime bud.

 

[shrodinger1226]

Quote by shrodinger1226
Interesting! I am asking the same question as the paper does. But I don't understanding some words:

(1) "P does not admit any eigenvalues. Nevertheless, the spectrum of P is the entire complex plane [8] and P does not represent an observable."
The first statement seems not consistent with the second.

I think the issue is that you are confusing an eigenvalue with an observable. You can measure P and find a value for it, but if you measure it again you won't probably get the same value for it. So P is not an eigenvalue for the state, it's not fixed.

I don't think I am confusing it. For an eigen state, a measurement has to get the eigen value. For a superposition state, a meausrement is certain to return one of the eigenvalues with a certain probability. In my undersatanding, <p> means that many measurements of p for an ensemble of the wells, we get an average value, <p>. From this, if <p> can be defined, then p should be an observable. Right?

According to the QE hypothesis, all operators representing observables are Hermitian operators. In my impression, the momentum seems to be observable. But I really don't know what is the exact mathematical definition of an observable (measurable quantity). What is called "an measurable quantity"?

 

[WannabeNewton]

An observable has to correspond to a self-adjoint operator. A Hermitian operator is not necessarily self-adjoint as already explained throughout this thread. The momentum operator associated with the infinite square well problem is not self-adjoint even though it's Hermitian thus not an observable.

 

[shrodinger1226]

An observable has to correspond to a self-adjoint operator. A Hermitian operator is not necessarily self-adjoint as already explained throughout this thread. The momentum operator associated with the infinite square well problem is not self-adjoint even though it's Hermitian thus not an observable.

They say "a square matrix is Hermitian (or self-adjoint) if it is equal to its Hermitian conjugate" (Griffiths book, 1ed, p.83). This means Hermitian = self-adjoint for matrix. You mean it holds only for a Hermitian matrix, but it does not hold for an Hermitian operator. Right?

 

[WannabeNewton]

Read post #10 again and keep in mind that matrix representations of operators are only well-behaved in finite dimensions.

By the way, as a tip, if you want to understand the formal mathematics behind QM then I wouldn't suggest using Griffiths as your reference. In fact I wouldn't suggest using any standard classroom QM text for such a purpose because they all butcher the beautiful intricacies of functional analysis.

If you really want to understand the mathematical foundations of QM then you're going to have to learn functional analysis and subsequently read a text on QM for mathematicians.

 

[R136a1]

They say "a square matrix is Hermitian (or self-adjoint) if it is equal to its Hermitian conjugate" (Griffiths book, 1ed, p.83). This means Hermitian = self-adjoint for matrix. You mean it holds only for a Hermitian matrix, but it does not hold for an Hermitian operator. Right?

A matrix corresponds to an operator on finite dimensional spaces. In these contexts, hermitian and self-adjoint are the same. However, if you do QM, you need to consider infinite dimensional spaces sooner or later. On these spaces, the notion of a matrix is ill-behaved, so it's rarely used. The notion of operator is important, but self-adjoint no longer is the same as hermitian.

 

[dextercioby]

And one more piece of advice: you have questions, but you're not prepared to receive the answers. To be honest, grasping the mathematics behind a physical theory is not for everyone, only for the ambitious. So read and learn mathematics, if you want to understand the answers to your questions.

 

[shrodinger1226]

And one more piece of advice: you have questions, but you're not prepared to receive the answers. To be honest, grasping the mathematics behind a physical theory is not for everyone, only for the ambitious. So read and learn mathematics, if you want to understand the answers to your questions.

Thanks for your advice. I have questions, and of course, I will receive correct answers when I understand.

In the QE books, they say
(1) "all operators representing observables are Hermitian".

From you and WannabeNewton, I think such a statement is confusing. It should be
(2) "all operators representing observables are self-adjoint".

"self-adjoint" is "Hermitian", but "Hermitian" is not necessarily "self-adjoint".
Of course, there is nothing wrong with (1), but really confusing, easy to get misunderstanding. Right?

However in the book [Steven Weinberg, Lectures on Quantum Mechanics, (Cambridge, New York, 2013), p. 61], Prof. Weiberg says:
-----
3.3 Observables

According to the second postulate of quantum mechanics, observable physical quantities like position, momentum, energy, etc., are represented as Hermitian operators on Hilbert space. An Hermitian operator is one that is linear and self adjoint. So before we spell out what this postulate means, we need to consider what is meant by operators in general, by linear operators in particular, and by the adjoint of an operator.
-----
Does Prof. Weinberg really mean "an Hermitian operator is self adjoint"?

 

[George Jones]

When first learning quantum mechanics, I recommend against delving too deeply into these issues.

For the difference between "selfadjoint" and "hermitian", see

https://www.physicsforums.com/showthread.php?p=1887619#post1887619

 

[dextercioby]

The book written by the Nobel Prize winner is intended as a gentle presentation of the formalism/content of the theory and its main results. It's not about the mathematical foundations/formulation of quantum mechanics. That's why no distinction is made between hermitean and self adjoint. A distinction exists and is made at a higher level of rigor than the books by Ballentine, Grifftihs, Weinberg, Sakurai, Merzbacher, Messiah, Cohen-Tannoudji, etc.

(2) as you wrote is the second postulate (the 1st postulate regards the states, the second is about observables, the third is usually the Born rule and the 4th is the evolution equations for states/observables).

 

[bhobba]

When first learning quantum mechanics, I recommend against delving too deeply into these issues.

I fully concur.

When I first learned QM properly I became caught up with this damnable Dirac Delta function. It took me on a long detour into Rigged Hilbert spaces and such. I emerged with my questions answered but the cost was my understanding of QM took a hiatus.

Much better in my view is put such things aside at the moment, knowing they have an answer, but it requires a more advanced background. You can get that background by reading up on some functional analysis. I have a few texts, but one I like is Applied Analysis by John Hunter.

While doing that Ballentine is reasonably mathematically complete and would give you a good foundation in the physics.

Thanks
Bill

 

[shrodinger1226]

When first learning quantum mechanics, I recommend against delving too deeply into these issues.

For the difference between "selfadjoint" and "hermitian", see

https://www.physicsforums.com/showthread.php?p=1887619#post1887619

Excuse me, in your post #6 https://www.physicsforums.com/showthread.php?p=1887619#post1887619, you said "all Hermitian operators are self-adjoint, but not all self-adjoint operators are Hermitian. "

However my understanding is the opposite: "all self-adjoint operators are Hermitian, but not all Hermitian operators are self-adjoint". Did I get it wrong?

 

[George Jones]

Yes.