Thursday, 4 June 2020

quantum field theory - Multiple vacua vs. vev's in qft


Take a (possibly supersymmetric) relativistic quantum field theory: when we construct it, we suppose that there is a unique vacuum state $|0\rangle$ which is Lorentz invariant, vector of some Hilbert space and annihilated by all annihilation operators of the theory.


First question: is this true only for free theories, for the perturbative sector, or also for the full non-perturbative story? in particular, unicity is one of Wightman axioms: is this also the case in any qft?


From this also follows that vacuum is a ground state, i.e. state of minimal energy: it is in this respect that we can consider more than one state of minimal energy, even though the vacuum is still unique: is that correct? And it is in this sense that we speak of a moduli space of vev's, where we have modded out equivalent states connected by gauge transformations.


Now a second question arises: when we say that some vev's (calculated on the unique vacuum of our Hilber space) minimize the potential, are we referring to expectation values calculated on the unique vacuum or to expectation values calculated in some other ground state? This question is motivated (other than by my ignorance) from two facts:


i. it is meaningful to ask in a susy theory if $Q_\alpha|\Psi\rangle=0,$ where $|\Psi\rangle$ can be the vacuum state or some other (ground-)state, to see if susy is broken or not



ii. when a symmetry is broken, as discussed in Spontaneous symmetry breaking: How can the vacuum be infinitly degenerate? and in http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism we are facing vacua which are in different Hilbert spaces.


Bonus question: when we move to string theory, do vacua landscape still have the same meaning as above, as long as one labels vacua by momentum index, say $|0;p\rangle$?



Answer



Quantum field theories usually have a unique ground state – by the ground state, I mean the Hamiltonian eigenstate corresponding to the lowest energy eigenvalue.


This may be demonstrated in various ways, depending on the assumptions we're allowed to make. For example, if a quantum field theory is a free field theory, the ground state may be constructed simply as the ground state of an infinite-dimensional harmonic oscillator – because the whole field or collection of fields is an infinite-dimensional harmonic oscillator. The ground state is effectively the same thing as the tensor product of ground states of one-dimensional harmonic oscillators – which we know to be unique, too.


This uniqueness holds for perturbations of free theories, too – for weakly interacting quantum field theories etc.


We expect uniqueness of the ground state much more generally whenever the ground state may be expected to be "really empty". If it is "really empty", it should transform as a singlet (trivial representation) under all the symmetries, so the ground state's subspace of the Hilbert space is composed of one-dimensional representations. This condition typically singles out discrete values of energy and it is infinitely unlikely for two such energy eigenvalues to match exactly – it is unlikely for the two levels to coincide.


In conformal field theories, we may describe the ground state as the state corresponding to the identity operator $1$ in the state-operator correspondence. The identity operator always exists, and that's why the ground state exists and is unique, too. There also exist other conditions that are enough to prove the uniqueness of the ground state.


In condensed matter physics, most "phases" also have a unique ground state. So it's true not only in the vacuum, it's also true for materials. The uniqueness of the ground state is expressed by nothing else than the third law of thermodynamics: the entropy goes to zero for $T=0$. Well, one may design systems for which the ground state is degenerate but they're contrived in some sense.


A different issue are of course moduli spaces. If there are moduli i.e. many allowed values of scalar fields that lead to the same potential energy, then the vacuum states are numerous i.e. degenerate. But each of them typically produces its own superselection sector that are called "the Hilbert space" and separated from each other.



The terms "vacuum" and "ground state" are synonyma – they label energy eigenstates with the lowest energy eigenvalues – unless one of them is locally given a different definition. But such a modified definition would depend on the author and there's no generally accepted meaning in which "vacuum" and "ground state" would mean something else in a field theory.


You ask which vevs of the potential are minimized for the true vacua in the presence of a potential: the vev of the unique vacuum, or the vev of some other states? This is a really confusing question because the answer must be self-evident. The state $|0\rangle$ is the true ground state if $\langle 0|V|0\rangle$ is minimal among all the vevs in this and other states $\langle G|V|G\rangle$. But in the previous sentence, I am just clarifying what the word "minimization" means in general. Minimization means that one has many "competitors" to start with and one looks for one of them (or many of them) for whom some quantity is minimal. So it's both true that one needs the vev in many states to talk about minimization at all; and that the vev in the chosen true ground state plays a preferred role because it's the "winner".


Your question sounds identical to me as if you asked the question: When we say that a runner won an Olympic sprint race, do we talk about his time or about the time of others? Well, we must talk about the times of all of them if we want to determine who is the winner but once we determine who is the winner, it's his time that matters most. Sorry if I misunderstood your question but I really see nothing in the wording that would go beyond this triviality.


SUSY. It is possible to ask whether a state $|\Psi\rangle$ itself is supersymmetric. This is equivalent to the question whether there exists at least one combination of supercharges $Q=\sum_a c_a Q^a$ such that $Q|\Psi\rangle=0$. When it's true, we are still interested in the number of linearly independent supergenerators that annihilate the state. If there are such states, it doesn't imply that $|\Psi\rangle$ is the ground state. There can be objects etc. that are supersymmetric but not ground states, the so-called BPS states. Quite generally, they have to carry some charges that also appear on the right hand side of the SUSY algebra along with the momentum vector.


A spontaneously broken symmetry indeed tends to produce several different superselection sectors built upon different minima etc. However, only global symmetries may "really" be spontaneously broken. The breaking of the gauge symmetries is just a matter of a simplified yet misleading terminology. They're never really broken. They're just nonlinearly realized. And that also means that the vacua remain unique – the electroweak theory is an example. All the other new minima of the Mexican hat potential are equivalent to each other via gauge transformations but gauge transformations are fundamentally redundancies so the states related by them to others must be considered the same state.


The vacua in string theory are treated just like vacua in a complicated field theory with many scalars that have many local minima (one may imagine the effective low-energy field theory approximating string theory even though one must realize that the relevant effective field theory depends on the "neighborhood" of the landscape; as we move across the landscape, the subset of fields whose particles are light is changing). There are some papers questioning whether it's a legitimate way to proceed or whether string theory forces us to change the attitude – most famously, from my ex-adviser Tom Banks (who likes to suggest that attempts to physically connect the different vacua are foiled by the production of black holes and other things) – but it's surely how most practitioners approach the vacua in the stringy landscape.


There isn't any momentum labeling (spacetime) vacua in the string theory (landscape). You have probably confused the second-quantized vacua in the spacetime (and it's those that are referred to when we talk about the landscape) with states of a single string i.e. with the ground state of the theory on the world sheet (instead of the spacetime). The Hilbert space of a single string, i.e. the first-quantized Hilbert space i.e. the world sheet theory, may have states such as $|0;k\rangle$ that describe the tachyon with spacetime momentum $k$. But these states don't make up the landscape. They're just states of a single string in a "particular string theory" (using the pre-1995 terminology) – i.e. states of a single string in a particular vacuum of the string theory landscape (using post-1995 terminology). They are the stringy counterpart/extension of the Hilbert space of one particle – whose basis of the Hilbert space is just given by $|\vec x\rangle$ vectors – when we talk about point-like particle quantum field theory and focus on the one-particle excitations in them (e.g. because we discuss the non-relativistic limit).


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