Let me elaborate the question by using 2D Ising model without external magnetic field. When we lower the temperature and pass $T_c$ a little bit, the theory of spontaneous symmetry breaking tells us that the system will settle in one of the two degenerate ground states with non-zero magnetization.

My question is why cannot the system be in a superposition of the two states and thus the average magnetization still remains zero. People may say that at low $T$ the system does not have enough energy to tunnel through the barrier in the middle of the double-well free energy landscape. Well, this might be true for relatively high barrier, but when $T\rightarrow T_c^-$ the barrier can have a height that's close to 0, in which case the superposition of states seems to be possible and the average magnetization should be zero if that is so, which in turn contradicts the results from symmetry breaking.

Is it because the system is supposed to be coupled with a heat bath such that quantum decoherence destroys the superposition? But What about spontaneous symmetry breaking in quantum phase transitions?


Ising model is at his core a classical model. Meaning that we don't consider the possibility of a "superposition of states" in the quantum sense. If you had an experiment recreating the Ising model (a sort of magnet, I guess), that kind of quantum effect would appear. But a model is not an experiment and we voluntary don't take into account quantum effects.

But, "Quantum" Ising model do exists, and have exactly the sort of behaviour you seems to expect, if there is multiple ground states possibles, the system will be in a superposition of states. (see by example https://en.wikipedia.org/wiki/Heisenberg_model_(quantum))

As for symmetry breaking, it's still here even in the quantum world, as soon as you measure the state, you will fall in one of the ground state.

  • 1
    $\begingroup$ even though sometimes measurement can induce symmetry breaking, but my intuition tells me that measurement induced quantum state collapse is essentially different from symmetry breaking $\endgroup$ – M. Zeng Oct 18 '16 at 15:24
  • 1
    $\begingroup$ Yes the two concepts are different. What I do mean is that given a completely symmetric hamiltonian without external forces there wouldn't be a symmetry breaking. But quantum collapse would still act (if there is a measurement, else the system would just keep both ground states). While for a classical system purely symmetric without external forces, the solution would stay undefined (and we would say that it isn't a physical situation). $\endgroup$ – Jeannette Oct 18 '16 at 16:02
  • $\begingroup$ could you elaborate a bit on the quantum phase transition part that has superposition of two degenerate ground states and also the behaviour of the order parameter? $\endgroup$ – M. Zeng Oct 19 '16 at 2:03
  • $\begingroup$ The ground state of your system is the eigenvector associated with the smallest eigenvalue of the hamiltonian (think $|0>$). But that eigenvalue can be degenerate (two non parallel eigenvectors), then your low temperature phase will be a superposition of both of those states. Order parameters behave like classically $\endgroup$ – Jeannette Oct 19 '16 at 6:39
  • $\begingroup$ what do you mean by order parameter behaves classically? $\endgroup$ – M. Zeng Oct 19 '16 at 9:31

This is an extremely deep question that still isn't fully understood. Such a "macroscopic superposition" $|\uparrow \uparrow \uparrow \uparrow \dots \rangle + |\downarrow \downarrow \downarrow \downarrow \dots \rangle$ is a perfectly valid state in the Hilbert space, and yet we never see it experimentally (at least not without a lot of careful work to isolate the system). Such a state is called a "cat state" (after Schrodinger's cat) and you are essentially asking about the paradox of why we never see Schrodinger's cats walking around.

More precisely, a cat state is a state $|\psi\rangle$ that fails to obey the "cluster decomposition property," which says that for all local operators $\hat{A}(x)$ and $\hat{B}(y)$, $$\lim_{x - y \to \infty} \left[ \langle \psi | \hat{A}(x) \hat{B}(y) | \psi \rangle - \langle \psi | \hat{A}(x) | \psi \rangle \langle \psi | \hat{B}(y) | \psi \rangle \right] = 0,$$ i.e. expectation values of far-separated observables factorize. You can easily check that your proposed superposition does not satisfy the cluster property, while the physically realistic symmetry-broken states do.

The most widely accepted resolution is that as you suggest, cat states are unstable to decoherence. Before you measure the system, it can indeed be in such a macroscopic superposition. But when you do measure it, the measuring device itself acts as a thermal bath that produces the random jitters that destroy the coherent superposition! A measuring device is always macroscopic and can act as an entanglement bath. (Indeed, under the current way of thinking, the property of being macroscopic is what makes it a measuring device! Of course, a "measuring device" doesn't need to have a conscious human watching it or anything like that - random air molecules bouncing off the system and so on can "measure" the system and collapse/decohere any macroscopic superpositions.)

  • $\begingroup$ So, a Schrödinger cat is not a symmetry-broken state?! The cat is indeed "symmetric" untill a measure is performed? $\endgroup$ – AndreaPaco Feb 16 '18 at 19:38
  • $\begingroup$ @AndreaPaco Well, for the actual Schrodinger cat setup the Hamiltonian does not have an exact symmetry that takes the live cat state to the dead cat state and vice versa, so there's no symmetry to be broken. But in the Ising model analogy, yes, under the decoherence interpretation the system is the symmetric state (which is analogous to the Schrodinger's cat macroscopic superposition) until the measurement is performed. $\endgroup$ – tparker Feb 17 '18 at 3:50
  • $\begingroup$ @AndreaPaco But you have to understand that the decoherence time scales are incredibly short for a macroscopic object, like $10^{-26}$ s or something. So realistically the environment will "measure" the cat long before you actually open the box. $\endgroup$ – tparker Feb 17 '18 at 3:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.