Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Stability of the vacuum state of interacting quantum fields

+ 3 like - 0 dislike
20 views

"Stability" is generally taken to be the justification for requiring that the spectrum of the Hamiltonian should be bounded below. The spectrum of the Hamiltonian is not bounded below for thermal sectors, however, but thermal states are nonetheless taken to be stable because they satisfy thermodynamic constraints. In classical Physics, we would say that the thermal state has lowest free energy, which is a thermodynamic concept distinct from Hamiltonian operators that generate time-like translations.

The entropy component of free energy, meanwhile, is a nonlinear functional of the quantum state (presuming that the definition of entropy in quantum field theory would be at least this much like von Neumann's definition in terms of density operators), so we can reasonably expect the sum of the energy and entropy components to have a minimum in the state of greatest symmetry, as we see for thermal states. [It seems particularly notable in this context that the entropy is not an observable in the usual quantum mechanical sense of a linear functional of the quantum state.]

The presence of irreducible randomness in quantum mechanics presumably puts quantum field theory as much in the conceptual space of thermodynamics as in the conceptual space of classical mechanics, despite the quasi-functorial relationship of "quantization", so perhaps we should expect there to be some relevance of thermodynamic concepts.

Given this background (assuming, indeed, that no part of it is too tendentious), why should we think that requiring the Hamiltonian to have a spectrum that is bounded below should have anything to do with stability in the case of an interacting field? The fact that we can construct a vacuum sector for free fields in which the spectrum of the Hamiltonian is bounded below does not seem enough justification for interacting fields that introduce nontrivial biases towards statistically more complex states.

This question is partly motivated by John Baez' discussion of "quantropy" on Azimuth. I am also interested in the idea that if we release ourselves from the requirement that the Hamiltonian of interacting fields must have a spectrum that is bounded below, then we will have to look for analogues of the KMS condition for thermal states that restore some kind of analytic structure for interacting fields.

I asked a related Question here, almost a year ago. I don't see an answer to the present Question in the citations given in Tim van Beek's Answer there.

This post has been migrated from (A51.SE)
asked Jan 12, 2012 in Theoretical Physics by Peter Morgan (1,075 points) [ no revision ]
Most voted comments show all comments
You might enjoy reading the work of Hollands & Wald on QFT on curved spacetimes. They suggest an axiomatic alternative to the usual stability condition in curved spacetimes, where one doesn't necessarily have Hamiltonians or preferred vacuum states.

This post has been migrated from (A51.SE)
If a free Hamiltonian spectrum being bounded below is OK with you, then a temporary interaction can only change the occupation numbers in the system. If your interaction is permanent, then your system must be represented as a set of free quasi-particles with a free quasi-particle Hamiltonian to which the previous reasoning is applicable. The latter case may require efforts to obtain a physical Hamiltonian.

This post has been migrated from (A51.SE)
If the Hamiltonian is unbounded from below in the vacuum sector then there is no thermal sector: there in no thermal equilibrium. Another POV is that in finite volume the Hamiltonian has to bounded from below in all sectors. I think it should imply the bound for the vacuum sector in infinite volume since negative energy states survive transition to finite volume

This post has been migrated from (A51.SE)
@Squark, By unbounded below, I mean, loosely, that the energy of every state in the Hilbert space is infinitely positive (as is the case for a thermal sector), that there is no state of lowest energy in the Hilbert space, not that there are states of negative energy. The states in the Hilbert space are constructed by the action of the *-algebra of observables on a given(constructed) Lorentz and translation invariant state (together with closure in the GNS-norm that results from that state). Managing the infinities in some explicit way to construct a thermal sector of course may not be trivial.

This post has been migrated from (A51.SE)
This is not possible in the vacuum sector because of Poincare invariance. The vacuum state has to be Poincare invariant hence its energy is exactly zero. If the Hamiltonian is unbounded from below it means that in the spectral decomposition we have representations that live in the negative light cone

This post has been migrated from (A51.SE)
Most recent comments show all comments
@Squark, the state constructed is a mixture, not pure. It can be rejected out of hand because it doesn't satisfy the Wightman axioms, however on the same grounds one would have to reject thermal states and the corresponding sectors. Changing $N$ to $H/\mathsf{kT}$ results in a thermal state, so that $\mu/2$ in the expression for the inner product $(f,g)_\mu$ changes to $\hbar k_0/2\mathsf{kT}$. The Poincare invariant fluctuations of the new state are greater than the fluctuations of the conventional vacuum state, but the field (and the annihilation/creation operator) algebra is unchanged.

This post has been migrated from (A51.SE)
Peter my point is that the Hamiltonian only has to be bounded below in vacuum sectors. Such sectors are by definition generated by pure Poincare invariant states

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...