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  Can a theory gain symmetries through quantum corrections?

+ 10 like - 0 dislike

It is well known that not all symmetries are preserved when quantising a theory, as evinced by renormalising composite operators or by other means, which show that quantum corrections may alter a conservation law, such as with the chiral anomaly, or 'parity' anomaly of gauge fields coupled to fermions in odd dimensions.

However is the reverse possible: can a theory after quantisation gain a symmetry? Or if not, can it gain a 'partial symmetry'?

(For example invariance under $x\to x+a$ for any $a$ is translation symmetry, and invariance under $x\to x+2\pi$ would be said to be a partial symmetry. My question concerns whether a theory can gain a full symmetry, or a partial one at least through being quantised.)

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user user2062542
asked Oct 4, 2017 in Theoretical Physics by user2062542 (90 points) [ no revision ]
Most voted comments show all comments
One example could be Liouville CFT. In the Lagrangian description there is a single coupling constant $b$. Upon quantising the theory one finds a symmetry $b\to 1/b$ which was not manifestly present in the Lagrangian description. However, it's important to bare in mind that there are generally many ways to specify the same QFT, consequently symmetries that may be manifest in one description may not be manifest in another.

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user Tom Bourton
en.m.wikipedia.org/wiki/Accidental_symmetry ... also many forms of enhanced or emergent symmetry

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user Mitchell Porter
See my answer to this question for an example

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user jpm
I know from my lattice simulations of $\lambda\phi^4$ theory that quantum corrections can restore discrete symmetries. For example, you can start with a potential $-\frac{u^2\phi^2}{2} + \frac{\lambda\phi^4}{24}$, which has two global minima at $\phi =\pm \sqrt{\frac{6u^2}{\lambda}}$, which one would naively expect to be the vacuum expectation value of the theory. What I found is that for sufficiently small lattice spacing (sufficiently high momentum cutoffs) quantum corrections result in the VEV being zero.

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user chuckstables

"Normally" the zeroth-order approximation is simpler than the exact theory. I have taken the word "normally" in quote marks because often it is not only simpler, but also very "distant" from the exact physical solutions; thus infinite corrections.

But let us take a look at the Hydrogen atom. As we know a non-relativistic exact formulation is $O(4)$-invariant, whereas the zeroth-order may have a different level of symmetry.

Most recent comments show all comments
There is a thing that has been studied in the past which is called "order-by-(quantum)disorder" that seems to be exactly what you are looking for. As far as I remember, it is discussed in the book "quantum field theory in condensed matter theory" by Tsvelik.

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user Fabian
@AccidentalFourierTransform maybe Chern-Simons theory is an ok example (I realize that is not 100% what OP is looking for, but still, tecnhically, it is not classicaly gauge-invariant, but is quantum-mechanically gauge invariant for integer levels $k$).

This post imported from StackExchange Physics at 2019-01-11 12:59 (UTC), posted by SE-user Solenodon Paradoxus

1 Answer

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Here is one example, but it's probably remote from the systems you have in mind. Consider a model of spins/moments  $S^{\alpha=x,y,z}_j$ living on a lattice with sites $j \in \mathbb{Z}$, subject to a Hamiltonian of form

$H = \sum_{j}[ J_j S^z_j S^z_{j+1}+ \epsilon h_j S^x_j + g_j S^z_j] $

where $J_j,h_j,g_j$ are chosen randomly in $[-1,1]$ and $ \epsilon $ is a small number (but non-zero). Now if you quantize this model, and promote $S^\alpha$ to spin-1/2 operators, one can show that the model has an infinite set of (quasi)local conserved quantities. Thus, in a sense it has an infinite set of emergent conservation laws/symmetries. 

Some points: 1)There's a proof of the above statement, with one caveat which most people don't worry about (see also this). 2) Quasi-local = the conserved quantities are local, but have very small support on larger operators; the weight of this support on larger operators decays exponentially with the size of the operator. 3) There's a wide class of models which exhibit this sort of behavior; they constitute a phase of matter known as many-body localized (MBL) phases. 

Now, in the classical version of the same model, the above construction fails. Indeed, it is believed that the classical version of the model does not possess the local conservation laws I mentioned. This statement is connected to the fact that the KAM theorem for classical systems becomes less tenable as the number of degrees of freedom in a system increases. 

answered Aug 5, 2020 by Curt vK (50 points) [ revision history ]
edited Aug 6, 2020 by Curt vK

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