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I am trying to understand calculation of correlation function in the ground state of the Transverse Field Ising model, from the following book, which is freely available: http://link.springer.com/book/10.1007/978-3-642-33039-1

The calculations can be found in Chapter 2 of the book. I shall follow the notation from this book and try to describe most of the steps.

Set up:

Consider a spin chain with $N$ sites. The hamiltonian for transverse field Ising model is (Page $17$ of the book) $$H= -\sum_i S^z_i - \lambda\sum_i S^x_i\otimes S^x_{i+1}.$$ Now, the book follows the well known process of using Jordan Wigner transformation to map Pauli operators ($S^x_i,S^y_i,S^z_i$) to fermionic operators $c_i, c^{\dagger}_i$. After this, a fourier transform is performed (equation $2.2.7$), defining new operators $c_q, c^{\dagger}_q$, which are the Fourier transforms of original $c_i,c^{\dagger}_i$ and now the hamiltonian looks like (equation $2.2.8$): $$H=N-2\sum_q(1+\lambda \cos(q))c^{\dagger}_qc_q - \lambda\sum_q(e^{-iq}c^{\dagger}_qc^{\dagger}_{-q}-e^{iq}c_qc_{-q}).$$

Then a Bogoliubov transformation is performed, which is the source of my confusion. They define operators $\eta_q,\eta^{\dagger}_q$ in the following way (equation $2.2.11$): $$\eta_q = u_qc_q + iv_qc^{\dagger}_{-q} , \quad \eta^{\dagger}_q = iv_qc_q + u_qc^{\dagger}_{-q}.$$

This transformation diagonalizes the hamiltonian $H$, with appropriate choice of $u_q,v_q$ and one infers that the ground state $|\psi_0\rangle$ is the state annihilated by all $\eta_q$: $\eta_q|\psi_0\rangle = 0.$

Main question:

Now in appendix $2.A.3$ (Page $42$), correlation function $\langle \psi_0|S^x_iS^x_{i+n}|\psi_0\rangle$ is computed. This is a complicated expression when written in terms of operators $c_i, c^{\dagger}_i$ and for this Wick's theorem is used. But, as can be seen in equation $2.A.30$, calculation is done as if $|\psi_0\rangle$ is annihilated by $c_i$ themselves. Whereas, we saw that $|\psi_0\rangle$ is actually annihilated by $\eta_q$, which is a mixture of both $c_i$ and $c^{\dagger}_i$.

In fact, all the further calculations appear to be done in same manner, assuming that $|\psi_0\rangle$ is annihilated by $c_i$. I traced equation $2.A.32$ to the following reference: http://pcteserver.mi.infn.it/~molinari/NOTES/Wick.pdf

In this reference, wick's theorem has been stated as Theorem $IV.4$ (Page $4$). Equation $2.A.32$ (of the book) looks very similar to corollary $IV.6$ (of the reference). But the corollary is true only if $|\psi_0\rangle$ has $0$ expectation value with all normal-ordered operators.

So how can $|\psi_0 \rangle$ have $0$ expectation value with normal-ordered form of a product of $c_i,c^{\dagger}_i$? Shouldn't this be true only with $\eta_q,\eta^{\dagger}_q$? Is there a underlying principle here, that expectation values do not change under Bogoliubov transformation?

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  • $\begingroup$ Have you tried to express the operator $A_iA_j$ in terms of $\eta_q$ and $\eta_q^\dagger$ ? Maybe that solves your problem. $\endgroup$ – Adam May 27 '16 at 7:25
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Eq. 2.A.30 is a somewhat non-trivial identity for the ground-state $|\psi_0\rangle$ which only uses the ground-state property that $\eta_q|\psi_0\rangle =0$. (Of course, as the OP has noted, $c_i|\psi_0\rangle \neq 0$.)

What we need to show is that $$I=\langle \psi_0 | (c_j+c_j^\dagger)(c_i+c_i^\dagger)|\psi_0\rangle =\delta_{ij}.$$

Using Eq. 2.A.37a, we find $$I=\frac1N\sum_{q,q'}e^{-i q R_i +i q' R_j}(u_q+i v_q)(u_{q'}-iv_{q'})\langle \psi_0 | \eta_q \eta_{q'}^\dagger|\psi_0\rangle ,$$ which is simplified, using $u_q^2+v_q^2=1$, into $$ I = \frac1N \sum_q e^{-i q (R_i-R_j)}=\delta_{ij}.$$

This is indeed the identity we wanted to prove.

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  • $\begingroup$ Adam, that is a good point. But my main question is connected to how to derive Equation 2.A.32 . The way I see it, it appears to be using the following statement: Any normal ordered product of c^{\dagger}_i and c_i also has zero expectation value with ground state. Is that right? If yes then how do you see it? $\endgroup$ – anurag anshu May 28 '16 at 1:31
  • $\begingroup$ No, it is using the fact that the Hamiltonian is quadratic in $\eta$ and $\eta^\dagger$ (or equivalently in $c$ and $c^\dagger$), which implies that the expectation values of any operator can be simplified using the Wick theorem. What you then need is to know the averages of some operators, either $\eta^\dagger \eta$, or equivalently that of the $A$ and $B$ operators. $\endgroup$ – Adam May 28 '16 at 8:47
  • $\begingroup$ Or, stated otherwise, the normal ordering and all that is done with respect to the $\eta$'s, that indeed annihilate the groundstate. In any case, to me, the easiest proof of the Wick theorem is to use the path-integral formalism, that avoids all this boring discussion of normal ordering. $\endgroup$ – Adam May 28 '16 at 8:54
  • $\begingroup$ Adam , can you point to a reference for the proof of your following statement?: "No, it is using the fact that the Hamiltonian is quadratic in ηη and η†η† (or equivalently in cc and c†c†), which implies that the expectation values of any operator can be simplified using the Wick theorem. " Specially for fermions, where commutation rule makes things more complicated. Thanks. $\endgroup$ – anurag anshu May 29 '16 at 3:53
  • $\begingroup$ See for instance here. It is proved using the path integral formalism, and the properties of the grassmann integrals lptmc.jussieu.fr/files/chap_fi.pdf $\endgroup$ – Adam May 29 '16 at 12:16

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