1
$\begingroup$

I was reading a book on QFT and the example he uses to make an analogy of spontaneous symmetry breaking in QFT is that of the rod where a compressive force is applied on the top surface of the rod directed down the axis of the rod in line with the rod's axis.

1

The rod bends and is no longer rotationally symmetric. I don't really understand how it is "spontaneous". I did a little more digging and someone else who uses this example states that SSB does not require an external force and as an example he showed if you apply another force tangent to the top of the rod, this also breaks rotational symmetry but isn't spontaneously broken. But both of these examples use a force to break symmetry of the rod. Why is one not spontaneous symmetry breaking but the other is spontaneous symmetry breaking, since they are both the same thing - applying a force to break symmetry?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Which book is it? $\endgroup$ – QuantumEyedea Jan 20 at 6:04
  • $\begingroup$ Cases analogous to tangential force are referred to as explicit symmetry breaking. In this rod example, the direction of the force explicitly selects out a preferred direction that the rod will bend. $\endgroup$ – kaylimekay Jan 20 at 6:07
  • $\begingroup$ @QuantumEyedea just for reference, Ryder (from the edit history) $\endgroup$ – Nihar Karve Jan 20 at 6:14
2
$\begingroup$

I very much recall that example (many years ago!), and at the time it did confuse me too.

The most important thing to take from that example, is the fact that in the case of the force being applied in the longitudinal direction of the rod (from top pressing down) as in diagram (c) you have above, is that the rod can bend in any direction. That is, there are multiple different directions the rod could bend, meaning the final state of the rod, or the "ground state", is "degenerate".

This is not possible in the situation depicted in part (b) of the diagram, since even though rotational symmetry is broken, there is only one direction it will bend - in the direction of the force - and so this represents explicit symmetry breaking and there is only one possible "ground state".

Having degenerate ground states, or multiple vacuum states in QFT, is what defines the idea of spontaneous symmetry breaking in QFT.

It is also worth noting two properties of this spontaneously broken state. First, as already discussed, is that the direction in which the rod bends is arbitrary and therefore there are several "ground states" that could all be equally occupied: the ground states are degenerate.

The second thing to note is that the different directions of bending (the different ground states) are related by (the original) rotational symmetry. The symmetry has thus not "disappeared" but is "hidden" in the relation between the ground states. This reflects the fact that the equations of motion still obey the symmetry, although the ground states do not, and sometimes spontaneous symmetry breaking is referred to as a hidden symmetry.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Oh wow! It's so obvious! Thank so much Drjh. I finally get it. $\endgroup$ – user285893 Jan 20 at 5:07
  • 1
    $\begingroup$ No worries. You should have added pictures of the rod and before/after so that people without the book? you used can see what's happening. It is a good example. $\endgroup$ – joseph h Jan 20 at 5:10
  • $\begingroup$ I can probalby add it to the question if you think more people can learn form it. $\endgroup$ – user285893 Jan 20 at 5:12
  • 1
    $\begingroup$ I think it could help. Thanks. $\endgroup$ – joseph h Jan 20 at 5:13
  • $\begingroup$ I added a diagram so more can understand. $\endgroup$ – user285893 Jan 21 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy