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

[samalkhaiat]

people want to somehow "interpret" gauge-dependent results as physical quantities.

One can easily prove the statement: “In a gauge-invariant theory (such as QED), the Poincare generators $(P_{\mu} , J_{\mu\nu})$ cannot be gauge invariant”. Aren’t these $(P_{\mu} , J_{\mu\nu})$ physical quantities?
I would also like to say that I don’t believe that Wald is able to provide better physical insight to the subject than Landau and Jackson.

 

[samalkhaiat]

You don't need to wait. This section can already be seen in the link in #4.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.

1) The tensor $F_{\mu\nu}$ is an observable physical field. However, as dynamical variables $F_{\mu\nu}$ gives incomplete description in the quantum theory.

2) The vector potential $A_{\mu}$ is not an observable. But, as dynamical variable, it was found to give a full (classical and quantum) description of the physical phenomena.
Indeed, this state of affair was demonstrated nicely by the Aharonov-Bohm effect:
Classical electrodynamics can be described entirely in terms of $F_{\mu\nu}$: Once the value of $F_{\mu\nu}(x)$ at a point $x$ is given, we know exactly how a charged particle placed at $x$ will behave. We simply solve the Lorentz force equation. This is no longer the case in the quantum theory. Indeed, in the A-B effect, the knowledge of $F_{\mu\nu}$ throughout the region traversed by an electron is not sufficient for determining the phase of the electron wave function, without which our description will be incomplete. In other words, $F_{\mu\nu}$ under-describes the quantum theory of a charged particle moving in an electromagnetic field. This is why we use the vector potential $A_{\mu}$ as dynamical variable (or primary field) in the A-B effect as well as in QFT. However, the vector potential has the disadvantage of over-describing the system in the sense that different values of $A_{\mu}$ can describe the same physical conditions. Indeed, if you replace $A_{\mu}$ by $A_{\mu} + \partial_{\mu}f$ for any function $f$, you will still see the same diffraction pattern on the screen in the A-B experiment. This shows that the potentials $A_{\mu}(x)$, which we use as dynamical variables, are not physically observable quantities. In fact, even the phase difference at a point is not an observable: a change by an integral multiple of 2π leaves the diffraction pattern unchanged.
3) The real observable in the A-B effect is the Dirac phase factor $$\Phi (C) = \exp \left( i e \oint_{C} dx^{\mu} A_{\mu}(x) \right) .$$ Just like $F_{\mu\nu}$, $\Phi (C)$ is gauge invariant, but unlike $F_{\mu\nu}$, it correctly gives the phase effect of the electron wave function.

 

[Demystifier]

I would also like to say that I don’t believe that Wald is able to provide better physical insight to the subject than Landau and Jackson.

For a comparison, would you say that Wald's book on general relativity provides better physical insight than Landau and Weinberg?

 

[Demystifier]

One can easily prove the statement: “In a gauge-invariant theory (such as QED), the Poincare generators $(P_{\mu} , J_{\mu\nu})$ cannot be gauge invariant”. Aren’t these $(P_{\mu} , J_{\mu\nu})$ physical quantities?

It depends on what one means by "physical". I think @vanhees71 meant measurable.

But perhaps the right question is this: What do these generators act on? If they act only on gauge invariant objects, then I would expect that they are themselves gauge invariant. If, on the other hand, the generators act also on the gauge non-invariant potential, then it's not surprising that they are not gauge invariant.

Another insight. The generators can be constructed from the energy-momentum tensor, but there are two energy-momentum tensors for EM field. The canonical one, which is not gauge invariant, and the symmetric one, which is gauge invariant. See Jackson, 3rd edition, Eqs. (12.104) and (12.113).

 

[martinbn]

For a comparison, would you say that Wald's book on general relativity provides better physical insight than Landau and Weinberg?

There are things in Wald that are not covered in Landau nor Weinberg.

 

[vanhees71]

One can easily prove the statement: “In a gauge-invariant theory (such as QED), the Poincare generators $(P_{\mu} , J_{\mu\nu})$ cannot be gauge invariant”. Aren’t these $(P_{\mu} , J_{\mu\nu})$ physical quantities?
I would also like to say that I don’t believe that Wald is able to provide better physical insight to the subject than Landau and Jackson.

In classcial relativistic field theory and local relativistic QFT the "Poincare generators" are built from local fields (i.e., not the gauge-dependent four-potentials), i.e., via the Belinfante energy-momentum tensor (gauge-invariant) and not the canonical one (gauge-dependent).

Landau and Jackson are of course hard to beat when it comes to clarity (in this order!).

 

[samalkhaiat]

For a comparison, would you say that Wald's book on general relativity provides better physical insight than Landau and Weinberg?

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.

I was trying to make the following: If you study Jackson & Landau (and do all the problems) your knowledge (about the EM phenomena) will (probably) be equivalent to that of Wald.

 

[samalkhaiat]

It depends on what one means by "physical". I think @vanhees71 meant measurable.

Physical quantity or measurable quantity, they should both mean the same thing. In QM, it is represented by Hermitian operator and it is called observable. In quantum gauge field theory (such as QED), It can be shown that a necessary condition for an operator $A$ to be observable is $$A |\Psi \rangle \in \mathcal{V}_{ph}, \ \ \ \forall |\Psi \rangle \in \mathcal{V}_{ph} , \ \ \ \ \ (1)$$ where $\mathcal{V}_{ph}$ is the physical vector space in which the scalar products are positive semi-definite. The positive definite Hilbert space is given by $$\mathcal{H}_{ph} = \mathcal{V}_{ph} / \mathcal{V}_{0},$$ where $\mathcal{V}_{0}$ is the subspace of $\mathcal{V}_{ph}$ consisting of zero-norm vectors. The essential point is that the condition Eq(1) does not necessarily require an observable operator to be gauge invariant, i.e., it does not need to commute with the generator of gauge transformation.

But perhaps the right question is this: What do these generators act on? If they act only on gauge invariant objects, then I would expect that they are themselves gauge invariant. If, on the other hand, the generators act also on the gauge non-invariant potential, then it's not surprising that they are not gauge invariant.

The generators are Hermitian operators. They act on (gauge-dependent) states not “potentials”.

The canonical one, which is not gauge invariant, and the symmetric one, which is gauge invariant. See Jackson,

Hehehe, Jackson is for students not for me. I spent big chunk of my academic life working with those two tensors. But any way, see my reply to vanhees71 bellow.

 

[samalkhaiat]

In classcial relativistic field theory and local relativistic QFT the "Poincare generators" are built from local fields (i.e., not the gauge-dependent four-potentials), i.e., via the Belinfante energy-momentum tensor (gauge-invariant) and not the canonical one (gauge-dependent).

Are you aware about the problems with the Belinfante’s expressions for the "generators"? Well, 1) the algebra closes only on-shell, or 2) they fail to generate the correct Poincare transformations on local fields. But, even if we ignore this difficulty and declare that $(P_{\mu} , J_{\mu\nu})_{Bel} = (P_{\mu} , J_{\mu\nu})_{Can}$ , we can still prove (on general ground) that $(P_{\mu}, J_{\mu\nu})_{Bel}$ cannot be gauge invariant. I can show you the easy proof if you want.

 

[Demystifier]

Are you aware about the problems with the Belinfante’s expressions for the "generators"? Well, 1) the algebra closes only on-shell, or 2) they fail to generate the correct Poincare transformations on local fields. But, even if we ignore this difficulty and declare that $(P_{\mu} , J_{\mu\nu})_{Bel} = (P_{\mu} , J_{\mu\nu})_{Can}$ , we can still prove (on general ground) that $(P_{\mu}, J_{\mu\nu})_{Bel}$ cannot be gauge invariant. I can show you the easy proof if you want.

I would like to see the proof. In fact, I cannot imagine how generators based on Belinfante can fail to be gauge invariant.

 

[Demystifier]

The generators are Hermitian operators. They act on (gauge-dependent) states not “potentials”.

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

 

[vanhees71]

It seems to me that you don't distinguish two different meanings of the word "observable".

In one meaning it is an adjective, meaning the same as measurable. This meaning probably cannot be made mathematically precise. It has more to do with experimental physics than with mathematical physics.

In the other meaning it is a noun, meaning the same as a self-adjoint operator (i.e., obeying your Eq. (1)). This meaning is mathematically precise.

A gauge non-invariant quantity can be observable as a noun, but it is not observable as an adjective. No experimentalist has ever measured a gauge non-invariant quantity in the laboratory.

It's a question of physics, not grammar. An observable is never a self-adjoint operator. It is represented by a self-adjoint operator in the mathematical formalism of QT. Only things that are observable (and in physics it should even be quantifiable, i.e., measurable) are relevant for the physics. A gauge-dependent quantity cannot be observable by definition, because it is not unique, given a physical situation.

 

[vanhees71]

Are you aware about the problems with the Belinfante’s expressions for the "generators"? Well, 1) the algebra closes only on-shell, or 2) they fail to generate the correct Poincare transformations on local fields. But, even if we ignore this difficulty and declare that $(P_{\mu} , J_{\mu\nu})_{Bel} = (P_{\mu} , J_{\mu\nu})_{Can}$ , we can still prove (on general ground) that $(P_{\mu}, J_{\mu\nu})_{Bel}$ cannot be gauge invariant. I can show you the easy proof if you want.

The generators of the Poincare transformation are gauge invariant and the same for the Belinfante and the canonical expressions. It's about the local densities (energy, momentum, stress, angular momentum density) of the corresponding observables, which are not unique and not a priori observable. For that you need gauge invariant expressions.

It is also easy to see that the usual gauge invariant densities, defined with the field rather than the potential are what's measurable. E.g., the energy density or the energy-current density of the electromagnetic field are measurable, and if you analyze how they are measured, i.e., via the interaction of the em. field with matter (e.g., the photoeffect for photomultipliers, CCD cams etc.), and there you get via the usual standard theory of the photoeffect (1st-order time-dependent perturbation theory, again under careful consideration of gauge invariance, as famously worked out by Lamb) that you measure the standard gauge-invariant densities and not some gauge-dependent canonical one.

 

[Demystifier]

It's a question of physics, not grammar. ... A gauge-dependent quantity cannot be observable by definition, because it is not unique, given a physical situation.

It is a matter of precise language, to avoid confusion. A gauge-dependent quantity cannot be measurable, but it can satisfy the formal definition of "observable" as an arbitrary self-adjoint operator.

 

[Demystifier]

we can still prove (on general ground) that $(P_{\mu}, J_{\mu\nu})_{Bel}$ cannot be gauge invariant. I can show you the easy proof if you want.

Let me guess, the proof is axiomatic, not constructive. You assume that the generators satisfy some expected properties and then prove that they are in contradiction with gauge invariance. But you don't actually prove that the generators explicitly constructed from Belinfante really have those expected properties. Am I close?

 

[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]

https://www.amazon.com/dp/B010WER8VW/?tag=pfamazon01-20

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.