# What does it mean for a quantum spin chain to be “low energy”?

Is it another way of saying the ground state?

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basically, although sometimes people also intend for it to include the low-lying collective excitations as well. –  wsc Jan 10 '12 at 15:56
@wsc thanks, do you mean like spin waves? If the spin chain is in equilibrium, if it is low energy, does it have to mean that it's in its ground state? (Since spin waves are, according to wikipedia, propagating disturbances in magnetic materials). –  Space boy Jan 11 '12 at 14:55
Like spin waves, yes. For the spin-1/2 Heisenberg chain there is no gap to make very long wavelength excitations -- in fact, this means that for any temperature $T>0$ the expected number of spin waves in equilibrium diverges. This is (part of) the Mermin-Wagner theorem. –  wsc Jan 11 '12 at 15:37
So for low dimensional systems, ground states are of course interesting, but people very much like to know about the low-lying excitations, and whether they destroy the thermal stability of that ground state. Generally one won't talk of a low-energy chain, but a low-energy model or theory, and then you know for sure they intend to include certain excitations. –  wsc Jan 11 '12 at 15:41
@wsc Thank you. You say that the spin waves in equilibrium diverge.. do you mean that starting off in a spin chain that's not in equilibrium, as it settles into equilibrium, the number of spin waves diverges? If you write up your comments in an answer, I'll accept it! –  Space boy Jan 13 '12 at 13:43