Spontaneous symmetry breaking from a QM perspective

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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!
 

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  • #2
vanhees71
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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.
 
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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!
 
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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?
 
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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.
 
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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##?
 
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The Mexican hat potential isn't one-dimensional, it states are a bit more complicated.
 
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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?
 
  • #9
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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.
 
  • #10
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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?
 
  • #11
vanhees71
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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.
 
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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!
 
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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.
 
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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##?
 
  • #15
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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.
 
  • #16
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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?
 
  • #17
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I didn't calculate the prefactor.
 
  • #18
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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.
 
  • #19
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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?
 
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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?
 
  • #22
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.
 
  • #23
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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.
 
  • #24
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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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  • #25
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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?
 

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