# Wick's theorem and transverse field Ising model

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?

• Have you tried to express the operator $A_iA_j$ in terms of $\eta_q$ and $\eta_q^\dagger$ ? Maybe that solves your problem.
Commented May 27, 2016 at 7:25

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.

• 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? Commented May 28, 2016 at 1:31
• 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.
• 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.