# Spontaneous symmetry breaking from a QM perspective

• I
kelly0303
Hello! This questions might not make sense and I am sorry if that is the case (I am asking from a QM class perspective). I am a bit confused about the idea of spontaneously symmetry breaking (SSB), from the point of view of QM. I am talking here about the energy plot looking like a mexican hat in 2D (not the general Higgs one in the complex plane). I am mostly thinking of it by making an analogy with the ammonia molecule problem. The 2 lowest energy states in the SSB case don't respect the symmetry of the interaction, hence they can't be stationary states. Which means that they are not eigenstates of the 2 level Hamiltonian of the problem. Assuming that the energy of these 2 lowest energy levels is ##E_0## and the off diagonal term is ##A## (please see the link to Feynman lectures for more details about the notation), then, the actual stationary states, obtained by diagonalizing the hamiltonian (and which explicitly manifest the symmetry of the physical interaction) will have energies of ##E_0\pm A##. But this implies that there is an even lower in energy level, ##E_0-A##, than the 2 equal ones. Yet in the plot I mentioned above there is no lower energy than the 2 equal ones. Can someone explain to me what is wrong with my reasoning. Why don't we have that lower energy level? Thank you!

Gold Member
2022 Award
The important point of spontaneous symmetry breaking is that the ground states are not symmetric under the symmetry transformation. They are nevertheless eigenstates of the Hamiltonian and thus stationary states. That the Hamiltonian fulfills a symmetry but not the ground states implies that the lowest energy eigenvalues are degenerate, i.e., there is more than one linearly independent eigenvector to that eigenvalue.

A pretty intuitive example is a permanent magnet. Though the here relevant electromagnetic interactions fulfill rotational invariance, each of the ground states of the permanent magnet distinguish a direction by the finite magnetization of the magnet. The symmetry is still reflected in the fact that the energy is independent of the specific orientation of this magnetization.

• atyy
kelly0303
The important point of spontaneous symmetry breaking is that the ground states are not symmetric under the symmetry transformation. They are nevertheless eigenstates of the Hamiltonian and thus stationary states. That the Hamiltonian fulfills a symmetry but not the ground states implies that the lowest energy eigenvalues are degenerate, i.e., there is more than one linearly independent eigenvector to that eigenvalue.

A pretty intuitive example is a permanent magnet. Though the here relevant electromagnetic interactions fulfill rotational invariance, each of the ground states of the permanent magnet distinguish a direction by the finite magnetization of the magnet. The symmetry is still reflected in the fact that the energy is independent of the specific orientation of this magnetization.
I am still confused then, in relation to the ammonia example (not sure if you got to look over my link). In that case we have 2 states (with the N atom being on top or bottom of the plane created by the H atom) in which the system choses a preferential direction, but both having the same energy. Yet, none of these 2 states are stationary, and the stationary states are a real combination of them. Also these 2 states don't respect the symmetry of the original Hamiltonian. They have an electric dipole moment, which is not allowed by the QED hamiltonian describing the molecule interactions (ignore weak interaction here). How is the magnet choosing a preferential polarization different than the ammonia molecule choosing a preferential position of the N atom? Thank you!

kelly0303
The important point of spontaneous symmetry breaking is that the ground states are not symmetric under the symmetry transformation. They are nevertheless eigenstates of the Hamiltonian and thus stationary states. That the Hamiltonian fulfills a symmetry but not the ground states implies that the lowest energy eigenvalues are degenerate, i.e., there is more than one linearly independent eigenvector to that eigenvalue.

A pretty intuitive example is a permanent magnet. Though the here relevant electromagnetic interactions fulfill rotational invariance, each of the ground states of the permanent magnet distinguish a direction by the finite magnetization of the magnet. The symmetry is still reflected in the fact that the energy is independent of the specific orientation of this magnetization.
One more thing. I read this article and on page 2 it states: "the state of a system, if it is to be stationary, must always have the same symmetry as the laws of motion which govern it". But if I understand it right from your reply: "the ground states are not symmetric under the symmetry transformation. They are nevertheless eigenstates of the Hamiltonian and thus stationary states." you are stating exactly the opposite. And now I am really confused. Do stationary states have, or not, the same symmetry as the Hamiltonian of the system?

Mentor
If they are degenerate then your choice is arbitrary. You can use states with the same symmetry but you do not have to. This means the off-diagonal term is zero, or at least so small that it's negligible in practice.

kelly0303
If they are degenerate then your choice is arbitrary. You can use states with the same symmetry but you do not have to. This means the off-diagonal term is zero, or at least so small that it's negligible in practice.
But my questions still remains. Do the stationary states have or not the same symmetry as the Hamiltonian (which means that in that basis the off diagonal terms are zero)? If they don't, then the 2 lowest energy states in the Mexican hat potential are not stationary (as they don't have the symmetry of the Hamiltonian), right? So, in order to get truly energy eigenstates i.e. stationary states you need to diagonalize the hamiltonian, but in that case you would get the stationary states with energy ##E_0\pm A##, where A is the off diagonal term. So as in my original post my main question is, what happens to the stationary, symmetric state with energy ##E_0-A##?

Mentor
The Mexican hat potential isn't one-dimensional, it states are a bit more complicated.

kelly0303
The Mexican hat potential isn't one-dimensional, it states are a bit more complicated.
Can't you make a 1D version? I remember this is how it was introduced first time in my particle physics class. Of course it has no counterpart in the real world, but it is a valid theoretical system, isn't it? But actually now I am confused about basic QM based on the first reply (regardless of the SSB). For a 2 level quantum mechanical system (or n-level in general), do the stationary states (which are exactly n) must have the same symmetry as the Hamiltonian (i.e. the interaction) or not?

Mentor
In a one-dimensional version: With a big hill in between the off-diagonal elements are negligible, if the hill isn't that big then you get two different energy levels close together, one will be symmetric and one will be antisymmetric.

kelly0303
In a one-dimensional version: With a big hill in between the off-diagonal elements are negligible, if the hill isn't that big then you get two different energy levels close together, one will be symmetric and one will be antisymmetric.
So in the one dimensional case, with a big hill in between, the 2 degenerate eigenstates are stationary states?

Gold Member
2022 Award
As is beautyfully explained in Feynman's lectures, the ammonia molecule's Hamiltonian is not degenarate due to tunneling. That's why the
But my questions still remains. Do the stationary states have or not the same symmetry as the Hamiltonian (which means that in that basis the off diagonal terms are zero)? If they don't, then the 2 lowest energy states in the Mexican hat potential are not stationary (as they don't have the symmetry of the Hamiltonian), right? So, in order to get truly energy eigenstates i.e. stationary states you need to diagonalize the hamiltonian, but in that case you would get the stationary states with energy ##E_0\pm A##, where A is the off diagonal term. So as in my original post my main question is, what happens to the stationary, symmetric state with energy ##E_0-A##?
The symmetry is explicitly broken here, because the two states have different energy, and the ground state is symmetric.

• atyy
kelly0303
As is beautyfully explained in Feynman's lectures, the ammonia molecule's Hamiltonian is not degenarate due to tunneling. That's why the

The symmetry is explicitly broken here, because the two states have different energy, and the ground state is symmetric.
Thank you for this! Ok so stuff that I am still confused: In the spontaneously symmetry breaking case, why can't the 2 states tunnel from one to another? I assume that the tunneling probability is very law, but I don't see why it would be zero. So if it is not zero, it means that there is an even lower energy level that those 2. The energy will be smaller by a very, very small amount, but based on the math of it that state should exist. So my question now is: is that state actually there? Or that lower energy state doesn't exist at all in the case of SSB? And if it doesn't why is that? Thank you!

• atyy
Mentor
Who says the system needs to be in one of the energy eigenstates (where the energies differ in theory only)? If the tunneling probability is negligible you can have a state that is in one valley only for all practical purposes.

kelly0303
Who says the system needs to be in one of the energy eigenstates (where the energies differ in theory only)? If the tunneling probability is negligible you can have a state that is in one valley only for all practical purposes.
Oh, no no, I totally agree. If the tunneling probability is very close to zero a state that is non-stationary in theory is basically stationary in practice for timescales as long as the age of the universe. My questions is purely theoretical now: as I said in my last post, if they are not stationary in theory, then it can be shown (as in the case of ammonia) that there are 2 other (stationary in theory, too) levels with energy ##E\pm A##. But in the plot for the spontaneously symmetry breaking (the Mexican hat potential), I don't understand where these 2 levels are located. Assuming that the 2 local minima have energy ##E##, where (on that plot) are these levels of energy ##E\pm A##?

Mentor
What is the "energy of a minimum"? The levels will be very close to the energy of the ground state each potential well on its own would have. If that is +-A or +-2 A or whatever: Didn't check.

kelly0303
What is the "energy of a minimum"? The levels will be very close to the energy of the ground state each potential well on its own would have. If that is +-A or +-2 A or whatever: Didn't check.
I am not sure I understand what you mean. Check what?

Mentor
I didn't calculate the prefactor.

Staff Emeritus
Can't you make a 1D version?

This is where the thread went off the rails. The 1D version is not a case of SSB. The problem is symmetric and so is the ground state wavefunction.

We can discuss this, if you want, but it doesn't have anything to do with SSB.

kelly0303
This is where the thread went off the rails. The 1D version is not a case of SSB. The problem is symmetric and so is the ground state wavefunction.

We can discuss this, if you want, but it doesn't have anything to do with SSB.
Thank you for this! I was actually confused by that. As I mentioned before the SSB was introduced in my particle physics class as a 1D (I think it is in Griffiths, too). Then we extended it to 2D. I didn't see any flaws in the calculations (but I was just introduced to it), so what is exactly not working with the 1D?

Staff Emeritus
so what is exactly not working with the 1D?
The problem is symmetric and so is the ground state wavefunction.

kelly0303
I am not sure I understand. How is the 1D more symmetric than the 2D? The local maxima looks symmetric in both cases, while the valleys don't. Why do we have SSB in 2D?

Dr.AbeNikIanEdL
You keep forgetting that you have to look at the ground state. A state where the wave function is localized at one of the minima might not be symmetric, but it is also not an eigenstate of the hamiltonian, much less the ground state, so looking at them is irrelevant. You have to look at the lower energy state of the two states you mention in your first post (and later). This is indeed symmetric the same way as the hamiltonian.

Staff Emeritus
he local maxima looks symmetric in both cases, while the valleys don't. Why do we have SSB in 2D?

You have an idea in your head and you think we all can see it. We don't.

Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about. You will be tempted to write down more words. Please don't. Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about.

kelly0303
You have an idea in your head and you think we all can see it. We don't.

Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about. You will be tempted to write down more words. Please don't. Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about.

Griffiths introduces a 1D spontaneous symmetry breaking, describes it and then moves to the 2D case. So I am confused why is this 1D case not a real SSB?

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kelly0303
You keep forgetting that you have to look at the ground state. A state where the wave function is localized at one of the minima might not be symmetric, but it is also not an eigenstate of the hamiltonian, much less the ground state, so looking at them is irrelevant. You have to look at the lower energy state of the two states you mention in your first post (and later). This is indeed symmetric the same way as the hamiltonian.
That's exactly what I am saying. In SSB, the 2 states (please see the link in the above post) that are symmetric with respect to the center should not be eigenstates, as they don't respect the symmetry of the Hamiltonian (hence the name of SSB). Yet in that plot (on page 362) I see no energy lower than those 2, yet a lower energy level should exist, shouldn't it?

Staff Emeritus
Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about. You will be tempted to write down more words. Please don't. Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about.

• vanhees71 and weirdoguy
Thank you for this! Ok so stuff that I am still confused: In the spontaneously symmetry breaking case, why can't the 2 states tunnel from one to another?

In the examples that we have, quantum spontaneous symmetry breaking involves two limits that do not commute (these limits are not present in the ammonia molecule example). In the thermodynamic limit, the tunneling rate goes to zero, and the states are exactly degenerate.
Spontaneous Symmetry Breaking in Quantum Mechanics by Jasper van Wezel, Jeroen van den Brink

You can also see the interesting comment by Xiao-Gang Wen that in the transverse Ising model, as the B field goes to zero one can have spontaneous symmetry breaking (if one also takes two non-commuting limits), but it is not trivial to demonstrate, since for finite B field, the ground state respects the symmetry.
https://physics.stackexchange.com/q...ntaneous-symmetry-breaking-in-quantum-systems (see comment by Xiao-Gang Wen; see also the paper by Munoz et al pointed out by Peter Morgan)
https://physics.stackexchange.com/q...perposition-state-is-unstable-in-spontaneousl (see tparker's answer, especially the part where he again mentions that in this example, we usually take two limits that do not commute to get the spontaneous symmetry breaking)
https://blogs.umass.edu/rvasseur/files/2019/02/Transverse-Field-Ising-Chain.pdf (explicit working of the example by Romain Vasseur, showing how the tunneling rate and degeneracy go to zero in the thermodynamic limit)

As an aside, the Higgs mechanism is not an example of spontaneous gauge symmetry breaking, although it is called by that name.
Is electromagnetic gauge invariance spontaneously violated in superconductors? by Martin Greiter
http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism

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Dr.AbeNikIanEdL
Yet in that plot (on page 362) I see no energy lower than those 2, yet a lower energy level should exist, shouldn't it?

There are no energy levels of any states plotted on that page...

In SSB, the 2 states (please see the link in the above post) that are symmetric with respect to the center should not be eigenstates, as they don't respect the symmetry of the Hamiltonian (hence the name of SSB).

If states are not eigenstates of the hamiltonian, they are irrelevant for the question whether the system exhibits SSB. The only relevant question is whether the ground state shares the symmetries of the hamiltonian. Eigenstates do not have to share the symmetries of the hamiltonian, not even the ground state has to, and the consequences of this are the entire point of SSB.

I can only second this:

Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about.

kelly0303
Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about. You will be tempted to write down more words. Please don't. Please write down the Hamiltonian you are talking about and the ground state wavefunction you are talking about.
It's equation 11.108 in the link I sent (I guess you're fine with me not literally copy-pasting that here).

kelly0303
There are no energy levels of any states plotted on that page...

If states are not eigenstates of the hamiltonian, they are irrelevant for the question whether the system exhibits SSB. The only relevant question is whether the ground state shares the symmetries of the hamiltonian. Eigenstates do not have to share the symmetries of the hamiltonian, not even the ground state has to, and the consequences of this are the entire point of SSB.

I can only second this:
The Hmiltonian is in the link I sent (well the Lagrangian)... It's equation 11.108.

• weirdoguy
Staff Emeritus
This is not what I asked for. It is not the Hamiltonian for the 1D situation. It is not the Hamiltonian for the 2D situation. It is not even a Hamiltonian. There is no wavefunction.

Here's what I think. I think you are in way over your head here and just don't have the background to understand what is going on with SSB, since you are not answering even basic QM questions about the system. In lieu of being able to calculate, you are forced to rely on stream of words that explain the outcome of the calculation you are not able to do.

You have a choice - you can go back to the QM and solidfy your background, or you can hope that someone will hit on the right arrangement of words.

• weirdoguy
One more thing. I read this article and on page 2 it states: "the state of a system, if it is to be stationary, must always have the same symmetry as the laws of motion which govern it". But if I understand it right from your reply: "the ground states are not symmetric under the symmetry transformation. They are nevertheless eigenstates of the Hamiltonian and thus stationary states." you are stating exactly the opposite. And now I am really confused. Do stationary states have, or not, the same symmetry as the Hamiltonian of the system?

The ferromagnet cited by @vanhees71 is a classic example of spontaneous symmetry breaking. However the "More is Different" article is by Anderson, who atypically does not consider the ferromagnet to be an example of spontaneous symmetry breaking. https://condensedconcepts.blogspot.com/2011/06/ferromagnetism-is-not-spontaneously.html

For the ferromagnet, even a finite system will demonstrate spontaneous symmetry breaking in the sense that @vanhees71 mentioned, because the finite system has ground states that are exactly degenerate - and stable - they don't tunnel into each other.

However, in most examples of quantum spontaneous symmetry breaking, there is only approximate ground state degeneracy that becomes more and more exact as the system becomes larger, ie. the thermodynamic limit is required for spontaneous symmetry breaking.

With the ammonia molecule, there is a high tunneling rate between the approximately degenerate states, which is why the ammonia molecule is usually not considered an example of spontaneous symmetry breaking. In contrast, with the best examples of quantum spontaneous symmetry breaking, tunneling between approximately degenerate low-energy states is increasingly suppressed as the system gets larger.

You can see more discussion on why the ferromagnet is atypical as an example of quantum spontaneous symmetry breaking in these papers:
Spontaneous Symmetry Breaking in Quantum Mechanics by Jasper van Wezel, Jeroen van den Brink (see footnote 11)
An Introduction to Spontaneous Symmetry Breaking by Aron J. Beekman, Louk Rademaker, Jasper van Wezel (see Exercise 2.7 on p56)

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• vanhees71
Gold Member
2022 Award
There are few topics which are confusing for students because of sloppy jargon of physicists than spontaneous symmetry breaking. The reason may be that it took some time to get a clear understanding about the different cases.

The ammonia molecule cannot be an example for spontaneous symmetry breaking, because by definition if there is spontaneous symmetry breaking, the ground state must be degenerate. If the broken (global!) symmetry is a Lie symmetry, then you have a Goldstone boson, i.e., a massless excitation. The ammonia molecule is rather an example for explicit symmetry breaking, and indeed there's no Goldstone boson.

What was not clear for a long time also is that if the symmetry is a local gauge symmetry, there cannot be spontaneous symmetry breaking, though almost all textbook describe it as such. I'd rather talk about the "Anderson-Higgs mechanism". It has been discovered by Anderson in a famous paper on superconductivity and was fully understood by Higgs as not only a mechanism to describe massive gauge (non-Abelian) gauge bosons without breaking the local gauge symmetry but also that there must be (in the non-Abelian case) necessarily also a massive scalar boson, the Higgs boson. The formalism used to write down a Anderson-Higgsed gauge model indeed at the first glance looks a lot like the way you write down a model with a spontaneously broken global symmetry. The crucial discovery by Anderson however was that in fact here one has no spontaneously broken symmetry and also no degenerate ground state, because the apparent degenerate ground states are all connected by a local (and that's the crucial difference to the case of a spontaneously broken global symmetry!) gauge transformation and thus to be identified as the same state. The immediate implication is the also crucial point that the field degrees of freedom which would describe massless Goldstone modes in the global-symmetry case do not occur in the physical spectrum of the Anderson-Higgsed gauge theory but (in the socalled unitary gauge) are completely absorbed into the gauge-boson fields providing the additional degree of freedom of a massive vector boson as compared to a massless one. For more general gauges (particularly the important 't Hooft ##R_{\xi}## gauges) you get apparent would-be Goldstone field degrees of freedom, which however in such gauges are "ghosts" in the same sense as you have to introduce Fadeev-Popov ghosts, i.e., to cancel out the contributions from unphysical field degrees of freedom, leading to gauge invariant description of observable quantities like cross sections.

• atyy
Another note on the ammonia molecule. As stated above, the ammonia molecule is usually not considered an example of spontaneous symmetry breaking. Spontaneous symmetry breaking is most commonly demonstrated with a thermodynamic limit. However, there is a way to start from the ammonia molecule and modify the potential to get a an example of spontaneous symmetry breaking - like the thermodynamic limit, the modification involves a strong non-analyticity, but it is a simpler modification than the thermodynamic limit.

The ammonia molecule can be modeled as a double-trough potential. If one now replaces the troughs with rectangular wells, and makes the barrier between the two wells infinitely high, then one can get spontaneous symmetry breaking. This is discussed qualitatively in Anderson's More is Different, and mentioned in Weinberg's Quantum Mechanics (p179-181): "These considerations point up a general feature of spontaneous symmetry breaking: It is always associated with systems that in some sense are very large. It is only the very large barrier in molecules like proteins and sugars that allow these molecules to have a definite handedness. In quantum field theory, it is the infinite volume of the vacuum state that allows other symmetries to be spontaneously broken."

Similar discussions:
https://arxiv.org/abs/1205.4773
https://thiscondensedlife.wordpress.com/2016/05/12/broken-symmetry-and-degeneracy/.

• vanhees71
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