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I thought it is due to spontaneous symmetry breaking. But isn't that because we never observe the superposition states, then we claim that there is spontaneous symmetry breaking. It looks like spontaneous symmetry breaking is an assumption.

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As the size of the lattice gets large, the superposition

$$\vert \psi \rangle = \frac{1}{\sqrt{2}}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle + \vert \downarrow \downarrow \downarrow \cdots \rangle\right)$$

in the Ising model becomes unstable to environmental perturbation. Just like Schrodinger's cat (a superposition $\vert \mathrm{live} \rangle + \vert \mathrm{dead} \rangle$), the Ising ground state is formally possible but in practice impossible to create and preserve because of decoherence. So-called "pure decoherence" happens when the unitary system-environment interaction takes the form $$U_{\mathcal{SE}} = \sum_i \vert S_i \rangle \langle S_i \vert \otimes U_{\mathcal{E}}^{(i)}$$ where $\vert S_i \rangle$ is some basis of orthogonal states of the system (the "pointer states") and the $U_{\mathcal{E}}^{(i)}$ are conditional unitaries on the environment. For the Ising model, the pointer states would be $\vert \uparrow \uparrow \uparrow \cdots \rangle$ and $\vert \downarrow \downarrow \downarrow \cdots \rangle$. In this idealized case, the superposition $\vert \psi \rangle$ will decay into a mixed state

$$\rho = \frac{1}{2}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle \langle \uparrow \uparrow \uparrow \cdots \vert + \vert \downarrow \downarrow \downarrow \cdots \rangle \langle \downarrow \downarrow \downarrow \cdots \vert \right)$$

exponentially fast, with a rate proportional to the number of lattice sites (under a reasonable noise model). Pure decoherence is minimal in the sense that it has "no classical effect" on the system, in the narrow sense that pure decoherence doesn't change the probabilities of various outcomes for a hypothetical measurement of the system in the pointer basis.

Now in general, perturbations from the environment can act in much more complicated ways, and often will have classical effects in addition to pure decoherence. For instance, when you open the box containing Schordinger's cat, the air currents in the room will rustle that cat's hair (meaning the'll be in a different position state). Likewise the environment's effect on $\psi$ will generally cause it to decay into a mixed state that has non-zero over lap with states that are a short Hamming distance away from $\vert \uparrow \uparrow \uparrow \cdots \rangle$ and $ \vert \downarrow \downarrow \downarrow \cdots \rangle$.

The actual dynamics taken is a complicated question that depends sensitively on how the environment couples to the system, and the environment's initial state. However, it's intuitive that long range entanglement is unusually sensitive for the following reason. If a single particle in the environment interacts with a single spin in the system, then this will usually reduce the entanglement by a factor of order unity. (The factor is called a decoherence factor, and it's given by the inner product between $U_{\mathcal{SE}} \vert \uparrow \rangle \vert E_0 \rangle$ and $U_{\mathcal{SE}} \vert \downarrow \rangle \vert E_0 \rangle$, even if the decoherence isn't pure, where $\vert E_0 \rangle$ is the initial state of the particle in the environment.) As multiple particles interact, the decoherence factors combine multiplicatively, and the entanglement is destroyed exponentially quickly in the size of the environment and system.

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  • $\begingroup$ Could you explain why it will decay to the mixed state of up and down, not other mixed state? Does the up and down ferromagnetic states have special properties? $\endgroup$ – Tim Sep 3 '14 at 16:29
  • $\begingroup$ The up-down state does have special properties (long range entanglement), but it's also true that more realistic sources of environmental perturbations will do more than simple decohere. I've revised my responde, so let me know if it answers your question. $\endgroup$ – Jess Riedel Sep 8 '14 at 18:49

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