# New E&M text by Wald

• Classical
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Well, yes, but on the bigger space you don't have a proper scalar product either. I'm not sure, whether it makes sense to define self-adjointness wrt. such a non-Hilbert space to begin with.

Gold Member
Well, yes, but on the bigger space you don't have a proper scalar product either. I'm not sure, whether it makes sense to define self-adjointness wrt. such a non-Hilbert space to begin with.
It makes sense to define non-self-adjointness in this way. If an operator does not satisfy a definition of self-adjointness (see e.g. Ballentine, the stipulation after Eq. (1.21)) then this operator is not self-adjoint.

Furthermore, often (but not always) when an operator is not self-adjoint, it is possible to make a self-adjoint extension, i.e. to define a new bigger Hilbert space on which the operator is self-adjoint. Your argument above suggests that in our case even a self-adjoint extension is impossible, which, if true, is even more surprising (and hence even more interesting).

But this is not as crazy as it may look. For instance, even in ordinary QM, there is no self-adjoint extension of the momentum operator on a half-line. In other words, if a particle in one dimension lives at ##x>0##, then it's not possible to choose boundary conditions at ##x=0## such that the momentum is a self-adjoint operator.

vanhees71
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2022 Award
But there you have at least a proper Hilbert space you deal with. It's of course clear that an essentially self-adjoint operator is only defined on a restricted domain and co-domain.

Demystifier
One additional insight. If ##P^{\mu}## is not gauge invariant as @samalkhaiat said, then $\delta P^{\mu}=[iQ_{\Lambda},P^{\mu}] \neq 0$. But I think we also have $\langle\psi |\delta P^{\mu}|\psi'\rangle=0$ for any physical states $|\psi\rangle, \ |\psi'\rangle$
True.
which implies that $\delta P^{\mu}|\psi'\rangle$ is orthogonal to any physical state $|\psi\rangle$.
Correct. And this means that the state $| \chi \rangle \equiv \delta P^{\mu}|\Psi \rangle$ is a zero-norm vector belonging to the subspace $\mathcal{V}_{0}$ which is orthogonal to the whole physical space $\mathcal{V}_{phy}$. The (completion of) quotient space $$\mathcal{H}_{phy} = \overline{\mathcal{V}_{phy}/ \mathcal{V}_{0}},$$ is the (positive metric) Hilbert space of the theory. So, all zero-norm states (including $\delta_{\Lambda}P^{\mu}|\Psi \rangle$) are excluded from the physical Hilbert space $\mathcal{H}_{phy}$. This means that $\mathcal{H}_{phy}$ is unstable under the action of the “operator” $\delta_{\Lambda}P_{\mu}$.

vanhees71
But as you know, the Aharonov-Bohm gauge-invariant observable is expressed in terms of the potential, not in terms of the magnetic field, provided that you insist on a local description. It all boils down to the fact that the integral ##\int dx^{\mu}A_{\mu}## is gauge invariant, so it's not really necessary to deal with ##F_{\mu\nu}## in order to have a gauge-invariant quantity.
Is this true even for classical EM?

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There is no Aharonov-Bohm effect in classical physics, and also in QT and with the AB effect not the potential is observable but the corresponding non-integrable phase, which is a gauge-invariant quantity, i.e., the magnetic flux through any surface with the integration path as its boundary.

hutchphd
Gold Member
Is this true even for classical EM?
Wald points out that AB effect is classical, in the sense that it exists even for a classical charged field coupled to EM field.

atyy
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I've no clue, what he means by this. Which "charged field" has a classical meaning?

I've no clue, what he means by this. Which "charged field" has a classical meaning?
I think he by classical he means not quantum, not that it actually exists in what we observe.

atyy and Demystifier
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If there is no such field, then there's also no classical AB effect.

If there is no such field, then there's also no classical AB effect.
Mathematically there are such fields. I might be wrong, but I think that the point is that it is not an effect of the quantum theory.

Gold Member
2022 Award
We are talking about physics, not math. The "Schrödinger field" has no classical interpretation, and that's why the AB effect doesn't refer to anything that can be described within classical physics.

hutchphd
We are talking about physics, not math. The "Schrödinger field" has no classical interpretation, and that's why the AB effect doesn't refer to anything that can be described within classical physics.
You may be talking about physics, but what is Wald talking about in this specific instance?

Gold Member
2022 Award
I don't know, as the book is not out yet in conventiently readable form.

atyy
Gold Member
We are talking about physics, not math. The "Schrödinger field" has no classical interpretation, and that's why the AB effect doesn't refer to anything that can be described within classical physics.
Wald perhaps has a different definition of the difference between physics and math. We don't see classical charged fields in actual experiments, but in theory, before performing quantization of a complex scalar field, one can study its classical properties. One of those classical properties is classical interference of classical waves, which includes interference around solenoids. The latter is the theoretical classical AB effect.

vanhees71
Gold Member
The "Schrödinger field" has no classical interpretation
Actually it has, in the macroscopic Ginzburg-Landau theory of superconductivity.

hutchphd, vanhees71 and atyy
All this makes me want to get the book. Is any of you Wald in disguise? (Or may be the publisher.)

Last edited:
kith, vanhees71 and atyy
Gold Member
All this makes me want to get the book. Is any of you Wald in disguise? (Or may be the publisher.)
I'm Wald's ex wife, but he does not longer speak to me since I told him that Carroll's book on GR is better than his.

atyy and vanhees71
Gold Member
This showed up today in the today’s mail.

vanhees71, Demystifier, Hamiltonian and 2 others
ergospherical
This showed up today in the today’s mail.
Cool! What are your first impressions of the text?

Gold Member
Cool! What are your first impressions of the text?
It is clearly a physics text, not a mathematical methods book. My one word description is “Clean”. He knows the path he wants to tread and he does not wander from it. It’s short with 225 pages of text, but it is more than just a set of lecture notes. If you do not know the math he uses, you will need supplementary material. That being said, the math doesn’t look scary. I think I will need to refresh my knowledge of Green’s functions. After thumbing through it, I am still looking forward to reading it.

atyy, hutchphd and vanhees71
Homework Helper
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Mine came about an hour ago. It seems very well formatted and logically presented. My first impression is that for me this is a perfect recapitulation. This means that there is not very much new to me (an old dog) yet I am certain I will learn a lot. Just like Christmas morning. If I find any major warts I will bring them up.

vanhees71
dude662
It is clearly a physics text, not a mathematical methods book. My one word description is “Clean”. He knows the path he wants to tread and he does not wander from it. It’s short with 225 pages of text, but it is more than just a set of lecture notes. If you do not know the math he uses, you will need supplementary material. That being said, the math doesn’t look scary. I think I will need to refresh my knowledge of Green’s functions. After thumbing through it, I am still looking forward to reading it.
It is literally lecture notes though. I was one of his students when he was making the book

atyy and vanhees71