thermodynamics - Definition of equilibrium in statistical mechanics

Equilibrium statistical mechanics is (amongst other things) about deriving the equations of state of thermodynamic systems (in equilibrium) from a microscopic basis (i.e. starting with a microscopic Hamiltonian).

In order to do that, we observe the system over a very long time, which means taking the limit of time average and variance of a phase space function. For quasi-ergodic systems, this is equivalent to the (appropriate) ensemble average/variance. We get a very sharp peaked average value which is constant in time and reproduces the thermodynamic e.o.s for a system in equilibrium.

So far, so good.

How can one now define a 'system in equilibrium' in terms of statistical mechanics? Would it be convenient to define a subset of phase space in which the macroscopic variables (like total energy,...) differ only by a small value (eg. the variance) from the ensemble average and call all points of this subset equilibrium-states (and the other ones non-equilibrium states) of the system? Or is there another definition?

EDIT: Maybe this thought experiment will help to clearify my question. Let's assume that we have a small container filled with an (ideal) gas. The container itself is placed within another but much larger isolated container with no other gas in it. At time T1 we open the small container and simultanously measure the full microstate of the gas. Then we wait "long enough" and at T2 we measure the full microstate again. Intuitively, one would say that the system was out of equilibrium at T1 and in equilibrium at T2. Yet, both microstates are part of the microcanonical ensemble. If we would measure a macroscopic phase space function (where no particle is somehow favoured) at T1 we would propably get a different result compared to a measurement at T2 or the ensemble average of the function. Furthermore, because of the recurrence theorem, a state like at T1 will come again at some point in the future. So, how could one define equilibrium with this experiment in mind?

quantum field theory - On-shell symmetry from a path integral point of view

Normally supersymmetric quantum field theories have Lagrangians which are supersymmetric only on-shell, i.e. with the field equations imposed. In many cases this can be solved by introducing auxilary fields (field which don't carry dynamical degrees of freedom, i.e. which on-shell become a function of the other fields). However, there are cases where no such formulation is known, e.g. N=4 super-Yang-Mills in 4D.

Since the path integral is an integral over all field configurations, most of them off-shell, naively there is no reason for it to preserve the on-shell symmetry. Nevertheless the symmetry is preserved in the quantum theory.

Of course it is possible to avoid the problem by resorting to a "Hamiltonian" approach. That is, the space of on-shell field configurations is the phase space of the theory and it is (at least formally) possible to quantize it. However, one would like to have an understanding of the symmetries survival in a path integral approach. So:

How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?

"Hidden" theta-term in Hamiltonian formulation of Yang-Mills theory

I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $\theta$, because it can be absorbed in the electric field:

$$ \mathcal{H}=\frac{1}{g^2}\text{tr}(\mathbf{E}^2+\mathbf{B}^2)=g^2\text{tr}(\mathbf{\pi}-\frac{\theta}{8\pi^2}\mathbf{B})^2+\frac{1}{g^2}\text{tr}(\mathbf{B}^2). $$

Here, $g$ is the gauge coupling, $E_i=\dot{A}_i$ is the non-Abelian electric field, $B_i=-\frac{1}{2}\epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{\mu\nu}$ is the gluon field strength, and $$ \mathbf{\pi}=\frac{\partial \mathcal{L}}{\partial \mathbf{\dot{A}}}= \frac{1}{g^2}\mathbf{E}+\frac{\theta}{8\pi^2}\mathbf{B} $$ is the momentun conjugate to $\mathbf{A}$ (see pp. 39 and 40 of the lecture notes).

In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $\theta$-term: $$ \mathcal{L}= -\frac{1}{2g^2}\text{tr}(F^{\mu\nu}F_{\mu\nu})+\frac{\theta}{16\pi^2}\text{tr}(F^{\mu\nu}\tilde{F}_{\mu\nu})=\frac{1}{g^2}\text{tr}(\mathbf{\dot{A}}^2-\mathbf{B}^2)-\frac{\theta}{4\pi^2}\text{tr}(\mathbf{\dot{A}} \mathbf{B}), $$ where $\tilde{F}_{\mu\nu}$ is the Hodge dual of $F_{\mu\nu}$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.

How is this possible? Tong mentions that the $\theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $\theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $\theta$-dependence of Yang-Mills in this formalism?

quantum optics - Virtual photon description of B and E fields

I continue to find it amazing that something as “bulky” and macroscopic as a static magnetic or electric field is actually a manifestation of virtual photons.

So putting on your QFT spectacles, look closely at the space near the pole of a powerful magnet – virtual photons! Now look between the plates of a charged capacitor – virtual photons again!

But if it’s all virtual photons, how do we get the difference between a magnetic and electric field?

[This question needs the tag "magnetostatics" as well].

Answer

the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of $E$ and $B$ vectors at each point. The probability density that a photon is found at a particular point is proportional to the energy density $(E^2+B^2)/2$ at this point. But again, the interpretation of $B,E$ for a single photon has to be changed.

So whether the field around an object is electric or magnetic or both is encoded in the "polarization" of the virtual photons.

You may imagine that the photon has 6 possible polarizations or so, identified with the components of $E$ and $B$. Well, for a particular direction, it is really just the $E+iB$ combination that acts as the wave function, so there are only three polarizations for a given direction - and one of them (the longitudinal) is forbidden, too. ;-) But the qualitative point that there are many polarizations is correct.

However, as emphasized repeatedly, you shouldn't imagine that a virtual photon is a real particle that can be counted. That's a reason why QGR's answer is pretty much irrelevant for your question because there is no operator counting virtual photons at all - so it makes no sense to ask whether it commutes with other operators. QGR may have thought about real photons but he hasn't answered your question, anyway.

By the way, static fields correspond to a vanishing frequency - because everything with a non-vanishing frequency will go like $\exp(i\omega t)$ or $\cos(\omega t)$. So if you want to describe the fields of electric sources and magnets as a collection of virtual photons, you must realize that the static nature of the field implies that the relevant fields will have the energy equal to zero. But the momentum is nonzero because the field depends on space - because of the sources. Such virtual photons are very far from being on-shell - they're very virtual, indeed. It is not too helpful to talk about virtual photons with particular frequencies and wavenumbers if there are electric sources in the middle of the region you want to describe. The Fourier analysis is only helpful for photons in a pretty much empty space.

But you could calculate the probabilities of various outcomes for a charged particle in an external electric or magnetic field, produced e.g. by many spinning electrons, using Feynman diagrams - where the virtual photons are the internal lines. The Feynman diagrams would be able to calculate the force acting on the probe particle. Some terms in the force wouldn't depend on the velocity - the electric forces - while others would depend on the velocity - the magnetic ones. These different terms would always come from the "same type" of virtual photons but all these photons depend on the sources of the field, so you would of course get different results for electric and magnetic fields.

All this stuff is confusing and really unnecessary. If you worry that quantum electrodynamics won't reproduce basic properties of electromagnetism - such as the difference between electricity and magnetism; or the difference between attractive and repulsive forces - then you shouldn't worry. It can be easily demonstrated that in the classical limit - e.g. for strong enough fields with a low enough frequency - the quantum electrodynamics (and the quantum field) directly reduces to the right classical limit, the classical electrodynamics (and the classical fields). Virtual photons are just a very helpful tool to study all kinds of processes similar to scattering. Their maths can be deduced from quantum fields - not the other way around - and these virtual photons don't happen to be useful to describe your kind of highly classical situations.

Best wishes Luboš

general relativity - I think I am misunderstanding Einstein's equivalence principle and his elevator

I'm having difficulties understanding why a gravitational acceleration can be guaranteed to be locally equivalent to an accelerating frame. Doesn't it matter on how the force is being applied? If the floor of the elevator is exerting a force on me (due to some external force accelerating it) then this would be very different from a gravitational acceleration that would accelerate each part of my body equally. If I held a string up during my acceleration in the elevator and I let go, the string should fall, while in the gravitational case, it would remain the same, or am I missing something?

Answer

I'm having difficulties understanding why a gravitational acceleration can be guaranteed to be locally equivalent to an accelerating frame.

Actually, it can't. See section 20 of Relativity: the Special and General Theory where Einstein said this:

“We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes”.

You can't transform away a real gravitational field. So the room you’re in is not exactly equivalent to the room in the rocket. See this article and how it refers to an infinitesimal region? That's a "local" region of zero size. That's no region at all. The principle of equivalence was "Einstein's happiest thought", but it's just a principle, it doesn't mean being in a real gravitational field is exactly the same as being in an accelerating rocket.

Doesn't it matter on how the force is being applied? If the floor of the elevator is exerting a force on me (due to some external force accelerating it) then this would be very different from a gravitational acceleration that would accelerate each part of my body equally.

It's different, but generally speaking, you can't tell the difference between the floor pressing up on you or you pressing down on the floor. Like David Hammen said, in practice you can't distinguish between the two using local experiments. Gennaro Tedesco said much the same re if you just look at the dynamics But note what CuriousOne said: you wouldn't be able to tell, to first order. If you had super-precise measuring equipment, such as NIST optical clocks, you could tell. Especially if your room had a high ceiling.

If I held a string up during my acceleration in the elevator and I let go, the string should fall, while in the gravitational case, it would remain the same, or am I missing something?

I think you're missing something I'm afraid. In both cases the string falls down. Or maybe I'm missing something!

The Reeh-Schlieder theorem and quantum geometry

There have been some very nice discussions recently centered around the question of whether gravity and the geometry and topology of the classical world we see about us, could be phenomena which emerge in the low-energy limits of a more fundamental microscopic theory.

Among these, @Tim Van Beek's reply to the question on "How the topology of space [time] arises from more fundamental notions" contains the following description of the Reeh-Schlieder theorem:

It describes "action at a distance" in a mathematically precise way. According to the Reeh-Schlieder theorem there are correlations in the vacuum state between measurements at an arbitrary distance. The point is: The proof of the Reeh-Schlieder theorem is independent of any axiom describing causality, showing that quantum entanglement effects do not violate Einstein causality, and don't depend on the precise notion of causality. Therefore a change in spacetime topology in order to explain quantum entanglement effects won't work.

which is also preceded with an appropriate note of caution, saying that the above paragraph:

... describes an aspect of axiomatic quantum field theory which may become obsolete in the future with the development of a more complete theory.

I had a bias against AQFT as being too abstract an obtuse branch of study to be of any practical use. However, in light of the possibility (recently discussed on physics.SE) that classical geometry arises due to the entanglement between the degrees of freedom of some quantum many-body system (see Swingle's paper on Entanglement Renormalization and Holography) the content of the Rees-Schilder theorem begins to seem quite profound and far-sighted.

The question therefore is: Does the Rees-Schlieder theorem provide support for the idea of building space-time from quantum entanglement? or am I jumping the gun in presuming their is some connection between what the theorem says and the work of Vidal, Evenbly, Swingle and others on "holographic entanglement"?

Answer

No. The Reeh-Schlieder theorem is a consequence of the Wightman axioms and of the Haag-Kastler axioms, both of which assume a Minkowski space, complete with the trivial metric structure, underlies everything. Variations of these axiomatic systems either allow or do not allow an analogue of the Reeh-Schlieder theorem to be derived. The idea that we might build space-time from quantum entanglement, while conceivable, presumably introduces either an axiomatic or some less well-defined mathematical structure, which will then either allow or not allow a Reeh-Schlieder-like theorem to be derived. Some thought this afternoon has suggested to me no way in which the Reeh-Schlieder theorem might of itself suggest a particular underlying mathematical structure. I take the lack of other Answers to suggest that no-one else immediately thought of such a thing either.

I personally take the Reeh-Schlieder theorem to say that there is a sense in which modulations of vacuum fluctuations are nonlocal, in a way that is weakly analogous to the nonlocality of analytic complex functions, for which the equivalent to the Reeh-Schlieder theorem is the possibility of analytic continuation. Analyticity, in a relatively sophisticated sense, is fundamental to the proof of the Reeh-Schlieder theorem. This is talk rather than mathematics, but to me it suggests the negative reply that I began with.

quantum field theory - What is the issue with interactions in QFT?

I've started studying QFT this year and in trying to find a more rigorous approach to the subject I ended up find out lots of people saying that "there is no way known yet to make QFT rigorous when there are interactions".

As for the textbook approach, even without interactions it already seems not much rigorous, still, approaching it in the right way it seems to be possible to make it precise.

Now, the rigour issue with interactions in QFT isn't explained in the books I'm using, and I confess I still didn't get it.

I mean: some people say the problem are the Dyson's series that in QFT wouldn't converge, some people say the issue is that the Fock space representation cannot be built with interactions and hence particles don't even exist in this case. Some people even say that it is not even possible to describe the theory with Hilbert spaces. And there are quite a few more points people make on this matter.

My question here isn't "how to solve these issues" because it seems to me that up to this day no one knows this yet. My question is: what really is the problem in more concrete terms.

What are the problems that make QFT with interactions be non rigorous? How interactions causes these problems in contrast to free QFT?