Wednesday, December 10, 2008

Symmetry Breaking in Quantum Mechanics

Today, in our Mathematical Quantum Mechanics lecture, I put my foot in my mouth. I claimed that under very general circumstances there cannot be spontaneous symmetry breaking in quantum mechanics. Unfortunately, there is an easy counter example:

Take a nucleus with charge Z and add five electrons. Assume for simplicity that there is no Coulomb interaction between the electrons, only between the electrons and the nucleus (this is not essential, you can instead take the large Z limit as explained in this paper by Frieseke. The only way the electrons see each other is via the Pauli exclusion principle. The Hamiltonian for this system has an obvious SO(3) rotational symmetry. The ground state, however is what chemists would call 1S^2 2S^2 2P^1. That is, there is one electron in a P-orbital and in fact this state is six-fold degenerate (including spin). Of course, there is a symmetric linear combination but in that six dimensional eigen-space of the Hamiltonian there are also linear combinations that are not rotationally invariant. Thus, the SO(3) symmetry here is in general spontaneously broken.

This is in stark contrast to the folk theorem that for spontaneous symmetry breaking you need at least 2+1 non-compact dimensions. This for example is discussed by Witten in lecture 1 of the IAS lectures or here and is even stated in the Wikipedia.

Witten argues using the Stone-von Neumann theorem on the uniqueness of the representation of the Weyl group (the argument is too short for me) and explicitly only discusses the particle on the real line with a potential (the famous double well potential where the statement is true). In 1+1 dimensions, there is the argument due to Coleman, that in the case of symmetry breaking you would have a Goldstone boson. But free bosons do not exist in 1+1d since the 2-point function would be a log which is in conflict with positivity.

I talked to a number of people and they all agreed that they thought that "in low dimensions (QM being 0+1d QFT), quantum fluctuations are so strong they destroy any symmetry breaking". Unfortunately, I could not get hold of Prof. Wagner (of the Mermin-Wagner theorem) but maybe you my dear reader have some insight what the true theorem is?

Anonymous said...

Witten's argument only applies to Bosons.

Take a peek a Witten in the book "Quantum Fields and Strings: A Course for Mathematicians Volume II." His arguments are around page 1126.

Basically the argument is that for Bosons in N dimensions the Heisenburg algebra of p's and q' is realized with a Hilbert space of square integrable functions on R^N. This representation is irreducible.

Any symmetry group G of the potential also acts projectively on this Hilbert space, and hence, symmetry breaking does not occur.

Robert said...

Thanks for the hint but that book is exactly what I used (the "IAS lectures").

I don't see how the difference between bosons and fermions would make a difference here. In quantum mechanics (where there are no field operators and no spin statistics theorem), fermions just have anti-symmetric wave functions but are acted upon by the same p and q operators so that argument should apply as well.

In your paragraph 2 and 3 just replace "Bosons" by "Fermions" and it should hold equally well. Or am I missing somehting?

Anonymous said...

You mean continuous symmetries, of course. Discrete symmetries can be spontaneously broken in 2D, e.g. the Ising model. Abelian continuous symmetries are a borderline case, cf the Kosterlitz-Thouless transition

Anonymous said...

If, as you say, in quantum mechanics there is no spin statistics theorem. Then there need not be a Pauli exclusion principle in quantum mechanics for electrons and your original counter example is suspect. For without a Pauli exclusion principle why should the electrons not all pile into 1S?

Quantum mechanics usually says there is this magical thing called a spin statistics theorem that you just have to take on faith. Once you take that on faith, then you can build atoms as you do in your counter example.

So, as one can see, Witten's argument only applies to Bosons.

Robert said...

There is no spin statistics theorem in the sense that spin (since there is no internal SO(d) group only defined in terms of the dimension of the spin label) is not liked to statistics and nothing stops you from having bosons with two internal states or fermions without internal degree of freedom.

But of course you can specify that your multi-particle wave function is anti-symmetric under permutations of the particles. In this "fermionic Hilbert space" you can ask for the ground state and this is what I am doing in the example.

Anonymous said...

I would guess then that the quantum mechanics that you speak of, RQM for short, is different from the quantum mechanics that Witten speaks of.

For example in RQM the terms Boson and Fermion are ill-defined. What one needs to do in RQM is to define four species of particles: Boson-Boson, Boson-Fermion, Fermion-Fermion, and Fermion-Boson. The first name indicates how the particle transforms under rotations while the second name indicates how the particle transforms under particle exchange.

For example, a Fermion-Boson will have a wave function that changes sign under a 360 degree rotation and the wave function will not change sign under particle exchange.

I doubt this is the theory Witten is speaking of. He mentions nothing about Bosons-Bosons or Boson-Fermions or...

So, I can only assume, without proof, that Witten assumes spin-statistics holds and Bosons and Fermions retain their standard meaning and his statement applies only to Bosons.

However, as far as I know, RQM is a self consistent non-relativistic theory. It's just not what we see in nature.

Anonymous said...

Perhaps you should have your PhD rescinded! Or, maybe you should have to complete extra coursework.

NS
http://sciencedefeated.wordpress.com

Anonymous said...

Quite the contrary.

What he proposes is a study of QM in absence of any knowledge of QFT. This, while not physical, may yield physical insights, much as the study of topological quantum field theory yields insights into physical systems. Think Chern Simons theory and quantum computers.

Matthias said...

See this for a bunch of arguments why exchange and rotation by 360° are the same, i.e. that there are only fermionic-fermionic and bosonic-bosonic particles.

Anonymous said...

Robert, there is a good discussion in Weinberg QFT book of what precisely is (global) symmetry breaking, and why it requires a strict infinite volume limit. The issue is not so much finding the true ground state, but ensuring that local perturbations have only small effect (cluster decomposition). I'm not sure this coincides with Witten's argument.

Anonymous said...

Matthias, I don't want to put words in John Baez's mouth. But, he never claims spin statistics does not require QFT. All the proofs sketched there require QFT. What Robert is proposing is to forget QFT. If one forgets QFT, there is no spin statistics theorem, and there is no reason Boson-Fermions, for example, can not exist.

Anonymous said...

Hi Robert. Your counterexample is not a counterexample. In quantum mechanics you have to average over the possible ground states. As explained in Witten's book, if I remember correctly (it's been a while) you need to take an infinite volume limit to get superselection sectors that can decouple in the path integral. I.e. in the infinite volume limit you can suppress tunneling.

Robert said...

Sean, this is exatly the point: Why should I have to average for the finite volume system? I could not find the answer in Witten's lecture.

For the anharmonic oscillator with the double well potential you just compute that the symmetrised state has the lowest energy, while the anti-symmetric state is lifted in energy by an amount computable in terms of an instanton sum. Anything localised in one of the minima is not a statinary state and eventually oscillates to the other minimum and back.

This is different for this fermionic atom: There, in the limit I consider, all these states are degenerate in energy and I cannot see any (energy) "advantage" of the symmetric state over the others.

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