Classical New E&M Text by Wald - Princeton Press 30% Off

[vanhees71]

Again: A self-adjoint operator represents (maybe!) an observable, but it is not the observable. A gauge-dependent expression cannot represent an observable, because it is not uniquely determined by any physical situation. E.g., the vector potential does not represent an observable local (vector-field like) quantity, because it's gauge dependent and thus not uniquely determined by the physical situation.

 

[Demystifier]

As a textbook, no. Weinberg and Landau give the student better physical insight to GR than Wald’s. However, if you require refined mathematic then go to Wald or (even better) Hawking & Ellis.

So by analogy, it's not unreasonable to expect that Wald's book on electrodynamics will give us a refined math that cannot be found in Landau and Jackson. And to expand the analogy, the analog of Hawking & Ellis could be Garrity or Hehl & Obukhov.
https://www.amazon.com/dp/B01K0TMP8K/?tag=pfamazon01-20
https://www.amazon.com/dp/B010WER8VW/?tag=pfamazon01-20

 

[martinbn]

As a textbook, no. Weinberg and Landau give the student better physical insight to GR than Wald’s. However, if you require refined mathematic then go to Wald or (even better) Hawking & Ellis.

Just out of curiosity, why do you think so?

 

[samalkhaiat]

But physical states in QED are gauge-independent, aren't they?

They are not. Gauge-dependent-object means that the object transforms under the gauge transformation. Gauge-independent-object means that the object is gauge invariant, i.e., it does not transform or (which is the same thing) it commutes with the generator of gauge transformation.

 

[samalkhaiat]

You assume that the generators satisfy some expected properties and then prove that they are in contradiction with gauge invariance.

The generators, J_{a}, of any symmetry group of the theory must satisfy the followings

1) Conservation: \frac{d}{dt}J_{a} = 0.
2) The closure of the Lie algebra: [J_{a} , J_{b}] = i C_{ab}{}^{c}J_{c} .
3) The correct infinitesimal transformation on local fields: \delta \varphi (x) = [iJ_{a} , \varphi (x)].
4) Consistency condition between 2 and 3: Do you know what it is?

So, if “your generators” fail to satisfy any one of the above, then you are in troubles because these “generators” have nothing to do with the symmetry of the theory and are not qualified to be called generators of the symmetry.

But you don't actually prove that the generators explicitly constructed from Belinfante really have those expected properties. Am I close?

The operators derived from the Belinfante expression don't satisfy 3 and/or 2. Now, to see the trouble with Belinfante expressions, say J_{\mu\nu} = \int d\sigma^{\rho}(x) M^{(Bel)}_{\rho \mu\nu}(x), I would ask you to calculate the commutator [J_{0i}, J_{ok}] and report your result to me.
Okay, close your eyes and ignore this difficulty, now go to see my proof bellow, to realize that it is completely irrelevant for the proof whether we use the canonical, Bellinfante or the super-duper-cucumber (if there is one ) versions.

 

[samalkhaiat]

The generators of the Poincare transformation are gauge invariant and the same for the Belinfante and the canonical expressions.

If this claim of yours is correct, then you should be able to prove it. So, show me your proof. In fact, I will now prove that the above quoted claim (of yours) is wrong.

Consider a gauge-invariant theory (such as QED), i.e., the theory is invariant under the infinitesimal c-number gauge transformation \delta_{\Lambda}A_{\nu} = \partial_{\nu} \Lambda \ \mbox{id} , where \Lambda is an arbitrary c-number field. Let Q_{\Lambda} be the generator of those local gauge transformations, so that \delta_{\Lambda}A_{\nu} = [iQ_{\Lambda} , A_{\nu}] = \partial_{\nu} \Lambda \ \mbox{id} \ \ (1) Let P_{\mu} be the total momentum operator, defined as the generator of space-time translations: \delta_{\mu}\varphi_{a} = [iP_{\mu} , \varphi_{a}] = \partial_{\mu}\varphi_{a} , \ \ \ \ \ \ (2) and let J_{\mu\nu} be the total angular momentum operator, defined as the generator of Lorentz transformation: \delta_{\mu\nu}\varphi_{a} = [iJ_{\mu\nu} , \varphi_{a}] = \left( x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu} \right) \varphi_{a} + \left( \Sigma_{\mu\nu}\right)_{a}{}^{b} \varphi_{b} . Claim: The operators (P_{\mu} , J_{\mu\nu}) (of our gauge-invariant theory) cannot be gauge-invariant operators, i.e., the followings are true: \delta_{\Lambda}P_{\mu} = [iQ_{\Lambda} , P_{\mu}] \neq 0 ,\delta_{\Lambda}J_{\mu\nu} = [iQ_{\Lambda} , J_{\mu\nu}] \neq 0 .

Proof: Consider the following Jacobi identity -\Big[ iQ_{\Lambda} , [iP_{\mu} , A_{\nu}] \Big] = \Big[ [Q_{\Lambda} , P_{\mu}] , A_{\nu} \Big] - \Big[iP_{\mu} , [iQ_{\Lambda} , A_{\nu}]\Big]. Now, evaluate the terms using (1) and (2) to obtain -\partial_{\mu}\partial_{\nu}\Lambda = \Big[ [Q_{\Lambda} , P_{\mu}] , A_{\nu} \Big] . Since \partial_{\mu}\partial_{\nu}\Lambda \neq 0, it follows that \delta_{\Lambda}P_{\mu} = [iQ_{\Lambda} , P_{\mu}] \neq 0. Thus the operator P_{\mu} is not gauge-invariant.
Now, your exercise is to complete the proof by showing \delta_{\Lambda}J_{\mu\nu} \neq 0.
Is it relevant for the proof whether we use the canonical or Bellinfante versions?

 

[samalkhaiat]

Just out of curiosity, why do you think so?

Because most physics graduates don’t understand Wald’s fancy math. Do the experiment

 

[samalkhaiat]

So by analogy, it's not unreasonable to expect that Wald's book on electrodynamics will give us a refined math that cannot be found in Landau and Jackson.

I very much doubt that because there aren't many un-answered (mathematical) questions left in EM.

 

[romsofia]

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The only other book I knew that did E+M in this formulation was written like 30 years ago: https://www.amazon.com/dp/0387964355/?tag=pfamazon01-20

And... it left me wanting more to say the least! I'm currently reading this, and wow, I'm hooked. Thanks for sharing!

 

[Demystifier]

@samalkhaiat I think I understand the origin of the difficulties. Let my try to express it in my own (less sophisticated) language. If we study classical symmetries, i.e. if relevant commutators are replaced by suitable Poisson brackets, then everything is fine and Poincare generators are gauge invariant. The problem is that replacing Poisson brackets with commutators may create some additional "anomalous" quantum terms. Furthermore, to write down the operator $A_{\mu}$, one must first fix a gauge, say a Lorentz gauge. It is known that quantization of $A_{\mu}$ is ambiguous, i.e. depends on the gauge. The measurable quantities (such as transition probabilities computed from the S-matrix) do not depend on the gauge, but here we are not talking about such measurable quantities. So it's not surprising that the anomalous terms in the commutators depend on the choice of gauge.

 

[vanhees71]

They are not. Gauge-dependent-object means that the object transforms under the gauge transformation. Gauge-independent-object means that the object is gauge invariant, i.e., it does not transform or (which is the same thing) it commutes with the generator of gauge transformation.

What's gauge invariant are, e.g., expectation values of gauge-invariant observables with respect to physical states. For a nice treatment of the Gupta-Bleuler formalism for QED, see

O. Nachtmann, Elementary Particle Physics - Concepts and
Phenomenology, Springer-Verlag, Berlin, Heidelberg, New
York, London, Paris, Tokyo (1990).

The operator approach to the non-Abelian case is more complicated.

 

[vanhees71]

If this claim of yours is correct, then you should be able to prove it. So, show me your proof. In fact, I will now prove that the above quoted claim (of yours) is wrong.

Consider a gauge-invariant theory (such as QED), i.e., the theory is invariant under the infinitesimal c-number gauge transformation \delta_{\Lambda}A_{\nu} = \partial_{\nu} \Lambda \ \mbox{id} , where \Lambda is an arbitrary c-number field. Let Q_{\Lambda} be the generator of those local gauge transformations, so that \delta_{\Lambda}A_{\nu} = [iQ_{\Lambda} , A_{\nu}] = \partial_{\nu} \Lambda \ \mbox{id} \ \ (1) Let P_{\mu} be the total momentum operator, defined as the generator of space-time translations: \delta_{\mu}\varphi_{a} = [iP_{\mu} , \varphi_{a}] = \partial_{\mu}\varphi_{a} , \ \ \ \ \ \ (2) and let J_{\mu\nu} be the total angular momentum operator, defined as the generator of Lorentz transformation: \delta_{\mu\nu}\varphi_{a} = [iJ_{\mu\nu} , \varphi_{a}] = \left( x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu} \right) \varphi_{a} + \left( \Sigma_{\mu\nu}\right)_{a}{}^{b} \varphi_{b} . Claim: The operators (P_{\mu} , J_{\mu\nu}) (of our gauge-invariant theory) cannot be gauge-invariant operators, i.e., the followings are true: \delta_{\Lambda}P_{\mu} = [iQ_{\Lambda} , P_{\mu}] \neq 0 ,\delta_{\Lambda}J_{\mu\nu} = [iQ_{\Lambda} , J_{\mu\nu}] \neq 0 .

Proof: Consider the following Jacobi identity -\Big[ iQ_{\Lambda} , [iP_{\mu} , A_{\nu}] \Big] = \Big[ [Q_{\Lambda} , P_{\mu}] , A_{\nu} \Big] - \Big[iP_{\mu} , [iQ_{\Lambda} , A_{\nu}]\Big]. Now, evaluate the terms using (1) and (2) to obtain -\partial_{\mu}\partial_{\nu}\Lambda = \Big[ [Q_{\Lambda} , P_{\mu}] , A_{\nu} \Big] . Since \partial_{\mu}\partial_{\nu}\Lambda \neq 0, it follows that \delta_{\Lambda}P_{\mu} = [iQ_{\Lambda} , P_{\mu}] \neq 0. Thus the operator P_{\mu} is not gauge-invariant.
Now, your exercise is to complete the proof by showing \delta_{\Lambda}J_{\mu\nu} \neq 0.
Is it relevant for the proof whether we use the canonical or Bellinfante versions?

What I was referring to is of course the Gupta-Bleuler formalism. The operators themselves are not gauge invariant, but the physical observables are, i.e., expectation values of gauge-invariant operators wrt. physical states or the S-matrix elements for physical scattering processes. It's lengthy to post the proof here. A nice treatment can be found in

O. Nachtmann, Elementary Particle Physics - Concepts and
Phenomenology, Springer-Verlag, Berlin, Heidelberg, New
York, London, Paris, Tokyo (1990).

I've written it up only in German here (Sect. 1.4):

https://itp.uni-frankfurt.de/~hees/faq-pdf/qft.pdf

 

[samalkhaiat]

What I was referring to is of course the Gupta-Bleuler formalism. The operators themselves are not gauge invariant,

My approach is just little bit more general than the Gupta-Bleuler approach: \mathcal{L}_{QED} = \mathcal{L}_{class} + \mathcal{L}_{gf}, where \mathcal{L}_{class} = -\frac{1}{4}F^{2} + \frac{1}{2} \left( \bar{\psi} \left( i\gamma^{\mu}\partial_{\mu} - m + e\gamma^{\mu}A_{\mu}\right) \psi + \mbox{h.c.}\right), and \mathcal{L}_{gf} = - B(x) \partial^{\mu}A_{\mu} + \frac{1}{2} \lambda B^{2} , where B(x) is the gauge fixing field, whose positive frequency part defines the physical states by B^{+}(x)|\Psi \rangle = 0 , and setting \lambda = 1 gives you the Gupta-Bleuler formalism. In this formalism, and in the Gupta-Bleuler formalism, (as I said that in #43 ) an operator O is an observable, if, for all |\Psi \rangle \in \mathcal{V}_{ph}, O|\Psi \rangle is itself a physical state, i.e. if B^{+}(x)\left( O| \Psi \rangle \right) = 0 . This can be rewritten as \big[ B^{+}(x) , O \big]|\Psi \rangle = 0 . \ \ \ (1) All (gauge-dependent or gauge-invariant) operators satisfying (1) are observables with physical eigenstates. For example, consider our friend the momentum operator P_{\mu}, we already proved that P_{\mu} is not gauge-invariant, however, it is an observable with physical eigenstates because \big[B^{+}(x) , P_{\mu} \big]|\Psi \rangle = i\partial_{\mu}B^{+}(x)|\Psi \rangle = 0 . Of course, what we measure in the lab are the matrix elements of operators (between physical states) not the operators themselves. And these matrix elements are indeed gauge invariant. In fact, for any (\Phi, \Psi ) \in \mathcal{V}_{ph}, one can show that \langle \Phi |P_{\mu}| \Psi \rangle \ \ \mbox{and} \ \ \langle \Phi |J_{\mu\nu}| \Psi \rangle , are gauge-invariant even though the operators (P_{\mu},J_{\mu\nu}) are not gauge invariant.

What I wanted to demonstrate is that this lack of gauge-invariance-of-operators is of no physical significance.
And happy new year to you and to the others

 

[Demystifier]

Of course, what we measure in the lab are the matrix elements of operators (between physical states) not the operators themselves.

What if we restrict the domain of the operators, i.e. define the operators to be objects that act only on physical states? In that case, are the operators themselves gauge invariant?

 

[samalkhaiat]

What if we restrict the domain of the operators, i.e. define the operators to be objects that act only on physical states? In that case, are the operators themselves gauge invariant?

1) I don’t know what you mean by “act only”? We want all operators to act on the physical states.
2) Physical states are in the domain of observable operators.
3) Observable operators may or may not be gauge invariant.
4) An operator O is gauge invariant if and only if, it commutes with Q_{\Lambda}, the generator of local gauge transformation: \delta O = [iQ_{\Lambda},O] = 0.

If I give you the QCD Lagrangian, which is the sum of the following parts: \mathcal{L}_{class} = - \frac{1}{4} \mathrm{tr} \left(F^{2}\right) + \mathcal{L} (\psi , \nabla_{\mu}\psi ) ,\mathcal{L}_{gf} = - (\partial_{\mu}B^{a}) A_{\mu}^{a} + \frac{\lambda}{2} B^{a}B_{a} ,\mathcal{L}_{FP} = - i \partial_{\mu} \bar{C}_{a} \left( D^{\mu}C \right)^{a} , can you construct any useful gauge-invariant operators other than those included in the \mathcal{L}_{class}? And what is the meaning of gauge invariance for this QCD Lagrangian? Do you even have an analogue to the Q_{\Lambda} of QED?
I don’t understand this obsession with “gauge-invariant” operators.

 

[Demystifier]

I don’t understand this obsession with “gauge-invariant” operators.

I'm not obsessed, I just want to understand it properly. So let me check whether I do. The generators such as $P^{\mu}$ are at least weakly gauge invariant, in the sense that $\langle\psi_1|P^{\mu}|\psi_2\rangle$ is gauge invariant for any physical states $|\psi_1\rangle$, $|\psi_2\rangle$. On the other hand, the gauge potential operator $A^{\mu}$ is not even weakly gauge invariant, i.e. $\langle\psi_1|A^{\mu}|\psi_2\rangle$ is not gauge invariant. Is that right?

 

[vanhees71]

It's not obsession but just the necessity to define what's observable, and what's observable can be calculated from gauge-invariant operators only, where gauge invariance is "weakly defined" as @Demystifier said in the previous posting.

 

[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 $|\psi\rangle$, $|\psi'\rangle$, which implies that $\delta P^{\mu}|\psi'\rangle$ is orthogonal to any physical state $|\psi\rangle$. This implies that $\delta P^{\mu}$, viewed as operator acting in the physical Hilbert space, is not self-adjoint, despite the fact that $P^{\mu}$ is self-adjoint. I think it's rather surprising.

 

[vanhees71]

But $\delta P^{\mu}$ is 0 on the physical Hilbert space, because
$$\langle \psi|\mathrm{i} (Q_{\Lambda} P^{\mu}-P^{\mu} Q_{\Lambda})|\psi' \rangle=0$$
for all $|\psi \rangle$ and $|\psi' \rangle$ in the physical Hilbert space. The 0-operator is for sure self-adjoint!

 

[Demystifier]

But $\delta P^{\mu}$ is 0 on the physical Hilbert space, because
$$\langle \psi|\mathrm{i} (Q_{\Lambda} P^{\mu}-P^{\mu} Q_{\Lambda})|\psi' \rangle=0$$
for all $|\psi \rangle$ and $|\psi' \rangle$ in the physical Hilbert space. The 0-operator is for sure self-adjoint!

But it is a non-zero operator on some bigger space, i.e. by acting on a physical space it goes out of this physical space. I think this means that it is not self-adjoint.

 

[vanhees71]

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.

 

[Demystifier]

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]

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.

 

[samalkhaiat]

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}.

 

[atyy]

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?

 

[vanhees71]

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.

 

[Demystifier]

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.

 

[vanhees71]

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

 

[martinbn]

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.

 

[vanhees71]

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

 

[martinbn]

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.

 

[vanhees71]

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.

 

[martinbn]

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?

 

[vanhees71]

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

 

[Demystifier]

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.

 

[Demystifier]

The "Schrödinger field" has no classical interpretation

Actually it has, in the macroscopic Ginzburg-Landau theory of superconductivity.

 

[martinbn]

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

 

[Demystifier]

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.

 

[Frabjous]

This showed up today in the today’s mail.

 

[ergospherical]

This showed up today in the today’s mail.

Cool! What are your first impressions of the text?

 

[Frabjous]

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.

 

[hutchphd]

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.

 

[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

 

[Demystifier]

Interestingly, the book has no references.