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

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

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

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  If the ground states of interacting QFTs are so complicated, how did Nature find them?

+ 8 like - 0 dislike
3562 views

My question was inspired by trying to understand the paper Quantum Algorithms for Quantum Field Theories, by Jordan, Lee, and Preskill. The main result of that paper is that scattering experiments in one of the simplest-possible interacting quantum field theories, namely ϕ4 theory, can be efficiently simulated using a quantum computer (not just perturbatively, but in general).

Intuitively, that result sounds "obvious": how could it possibly be false? So what's surprising to me, and in need of explanation, is how much work goes into making it rigorous---and even more pointedly, the fact that JLP don't know how to generalize their algorithm to handle massless particles, let alone other aspects of the full Standard Model like chiral fermions. To some extent, these difficulties probably just reflect the complexity of QFT itself: when you're designing an algorithm (classical or quantum) to simulate something, and proving its correctness, you're not allowed to handwave any issue as "standard" or "well-known" or "who cares if it's not quite rigorous"! Indeed, for that very reason, I think understanding how one would go about simulating QFT on a quantum computer could provide an extremely interesting avenue to understanding QFT itself.

But there's one thing in particular that seems to cause JLP their main difficulty, and that's what leads to my question. Even for a "simple" interacting QFT like ϕ4, the ground state of a few well-separated particles in the vacuum is already complicated and (it seems) incompletely understood. So even for the very first step of their algorithm---preparing the initial state---JLP need to take a roundabout route, first preparing the ground state of the noninteracting field theory, then adiabatically turning on the interaction, adjusting the speed as they go to prevent level crossings. Here, they need some assumption that causes the spectral gap to remain reasonably large, since the inverse of the spectral gap determines how long the adiabatic process needs to run. And this, I think, is why they can't handle massless particles: because then they wouldn't have the spectral gap.

(Note that, at the end of the algorithm, they also need to adiabatically turn off the interaction prior to measuring the state. Another note: this adiabatic trick seems like it might have something to do with the LSZ reduction formula, but I might be mistaken about that.)

Here, though, I feel the need to interject with an extremely naive comment: surely Nature itself doesn't need to do anything nearly this clever or complicated to prepare the ground states of interacting QFTs! It doesn't need to adiabatically turn on interactions, and it doesn't need to worry about level crossings. To explain this, I can think of three possibilities, in rapidly-increasing order of implausibility:

(1) Maybe there's a much simpler way to simulate QFTs using a quantum computer, which doesn't require this whole rigamarole of adiabatically turning on interactions. Indeed, maybe we already know such a way, and the issue is just that we can't prove it always works? More specifically, it would be nice if the standard dynamics of the interacting QFT, applied to some easy-to-describe initial state, led (always or almost always) to the interacting ground state that we wanted, after a reasonable amount of time.

(2) Maybe the cooling that took place shortly after the Big Bang simulated the adiabatic process, helping to ease the universe into the "right" interacting-QFT ground state.

(3) Maybe the universe simply had to be "initialized," at the Big Bang, in a ground state that couldn't be prepared in polynomial time using a standard quantum computer (i.e., one that starts in the state |0...0⟩). If so, then the computational complexity class we'd need to model the physical world would no longer be BQP (Bounded-Error Quantum Polynomial-Time), but a slightly larger class called BQP/qpoly (BQP with quantum advice states).

I'll be grateful if anyone has any insights about how to rule out one or more of these possibilities.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson
asked Jan 2, 2013 in Theoretical Physics by ScottAaronson (795 points) [ no revision ]
Doesn't spontaneous symmetry-breaking mean that the universe has lots of different (but equivalent) ground states?

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Peter Shor
Thanks---yes, I suppose! So then I simply revise the question to, "how does Nature find A ground state?"

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson
An even trickier question is: "how did Nature find the same ground state throughout the observable universe?" And we're not quite in a ground state—there's that 3°K cosmic microwave background radiation.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Peter Shor
Essentially by quantum computing...trying every possibility at once. It is a little more clear in the language of path integrals.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user dmckee
dmckee: So, if that's correct, can you get rid of the "adiabatically turn on the interaction" step in computer simulations of QFT, or can't you?

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson
@ScottAaronson If you have a quantum computer that can manage your problem. For the most part that is exactly what you don't have at the current state of the art.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user dmckee
dmckee: You seem to be answering a much easier question than the one I asked! I know there's an algorithm to efficiently simulate at least special QFTs on a quantum computer, since JLP showed it (check out their paper if you haven't yet). And yes, certainly you can describe such an algorithm in path-integral language. My question was about the step in their algorithm where you need to adiabatically prepare the interacting vacuum state. Can you avoid that step? If so how? If not, how does Nature do that step? (John Preskill has now partly answered this question; see below.)

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson

1 Answer

+ 7 like - 0 dislike

As you note, this is really a question about cosmology. There is a lot to say, but in the interest of brevity I will just address a part of the question.

Quantum chromodynamics is a strongly-coupled theory with a very complex vacuum state. The universe is not in the QCD vacuum today, but is pretty darn close to it, since the 2.7K background temperature is way below the mass gap of QCD.

At high temperature QCD is a lot simpler -- the quarks and gluons are weakly interacting (asymptotic freedom). The state of high temperature QCD is fairly simple (though not trivial); it's approximately an ideal gas of relativistic particles.

As the universe expanded and cooled adiabatically it passed through a (first order) phase transition from the "easier" high-temp state to the "harder" low temp state. Just below the phase transition the state was still pretty hot, but as the universe continued to expand adiabatically the state approached the QCD vacuum.

Of course, this would not have worked if preparing the QCD vacuum were QMA hard; the universe would have gotten stuck is some long-lived metastable state. That apparently did not happen, so we have cosmological evidence that preparing the QCD vacuum is easy, even if JLP have not proved it (yet).

You can ask further questions about how the hot initial state was prepared, why it was so homogeneous and isotropic, etc. which would lead us into a discussion about how cosmic inflation started and ended, but that's enough for now.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user John Preskill
answered Jan 2, 2013 by John Preskill (160 points) [ no revision ]
Most voted comments show all comments
@Hugh: This is off-topic, but I think that's completely wrong. If it were correct, then why shouldn't MWI believers simply shoot themselves with (very fast and effective) guns hooked up to quantum random number generators? Since they can never observe the gun being fired, obviously it will seem never to do so! ;-D Yet I don't see them lining up to try it (nor can I blame them, since I don't see anything in MWI that says we get to condition on remaining alive).

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson
@Hugh Allen: Do you also believe that you cannot fall completely asleep, because you cannot observe yourself do this?

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Mitchell Porter
@Scott Aaronson: obviously they don't do that because they have little to gain by it and everything to lose if mistaken, not to mention that the stunt would prove nothing to the scientific community, even if they were seen to survive against extreme odds. (the suicider gets to condition on remaining alive, but observers just see the expected outcome probabilities. So to an observer they were just lucky if they survive.)

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Hugh Allen
One minor comment: For realistic values of the quark masses the phase transition is a smooth crossover.

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Thomas
The main algorithm we have for find the ground state of an interacting QFT is Euclidean lattice field theory. Simulations indicate that ground state of QCD is complex, but not hard to find (in the sense that simple Metropolis algorithms converge quickly, even if initialized with very poor guesses of the ground state).

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Thomas
Most recent comments show all comments
BTW, I suppose we'd better HOPE that the early universe found the correct QCD ground state! I'd never quite put together that, if the known universe gets destroyed via tunneling to a lower-energy vacuum state, the Earth incinerated by a bubble expanding outward at the speed of light, then we ought to blame the failure of the quantum adiabatic algorithm to efficiently find global optima to NP- and QMA-hard optimization problems. :-)

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Scott Aaronson
@ScottAaronson There was an electroweak phase transition as well, when the universe cooled enough that the Higgs gained a nonzero vacuum expectation value. The transition is 2nd order in the standard model with the now measured Higgs mass, but some supersymmetric models can give a first order phase transition. I'm definitely not an expert on JLP, but my intuition is that these technical, thermodynamic details won't change the basic picture. Take my word with a big grain of salt though. :)

This post imported from StackExchange Physics at 2014-07-24 15:45 (UCT), posted by SE-user Michael Brown

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:
$\varnothing\hbar$ysicsOverflow
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).
Please complete the anti-spam verification




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

Your rights
...