Considering an Ising model in the quantum scenario in quantum spatial dimension d=1 (that corresponds to classical D=2=d+1 dimension). Starting with the Ising model hamiltonian under the approximation of nearest neighbor and uniform interaction between spins and also in the limit $h\longrightarrow 0$ and under the assumption of finite lattice sites ($M$ finite) $$H=-h\sum_{i=1}^M(\hat{\sigma_i^x})-J\sum_{i=1}^M(\hat{\sigma_i^z}\hat{\sigma_{i+1}^z})$$ there are 2 degenerate ground states $$|\uparrow,\uparrow,\dots, \uparrow, \dots\rangle$$ $$|\downarrow,\downarrow,\dots, \downarrow, \dots\rangle$$ And the true ground state must be the superposition of these 2, i.e.$$\frac{1}{\sqrt{2}}\lbrace |\uparrow,\uparrow,\dots, \uparrow, \dots\rangle+|\downarrow,\downarrow,\dots, \downarrow, \dots\rangle\rbrace$$ Why the superposition state has a lower energy than the single state where all the spins are alligned up or down? I read that this is due to the tunneling.
So, my question is: Why the tunneling (if the lattice is finite) lowers the energy?
If the lattice approaches $\infty$ ($M\longrightarrow \infty$) the tunneling energy required also tends to $\infty$ and the true ground state is not the superposition anymore and the system shows spontaneous symmetry breaking.
I guess that i've not get the meaning of tunneling in this context.
