Transformations on correlation functions

Question

Let $\hat{U}$ be a unitary transformation that acts on fermionic fields (defined on a lattice) as $$\hat\psi_j \rightarrow \hat{U}\hat\psi_j\hat{U}^{-1},$$ $$\hat\psi_j^\dagger \rightarrow \hat{U}\hat\psi_j^\dagger\hat{U}^{-1}.$$

I consider it to be anti-unitary, just because it makes more explicit my argument, so $$i\rightarrow \hat{U}\hat i\hat{U}^{-1}=-i,$$ where $i$ is the imaginary unit.

Now I want to understand how this transformation modify correlation functions of the type $\left<\hat\psi_i\hat\psi_j^\dagger\right>$. I can think about this in two ways, but I don't really know which one is the correct approach:

1. Since $\left<\hat\psi_i\hat\psi_j^\dagger\right>$ is just a scalar, one way of proceeding is to only use the anti-unitary property of the operator, i.e. $$\left<\hat\psi_i\hat\psi_j^\dagger\right>\rightarrow \hat{U}\left<\hat\psi_i\hat\psi_j^\dagger\right>\hat{U}^{-1}=\left<\hat\psi_i\hat\psi_j^\dagger\right>^{*},$$ so the transformation only complex-conjugate the correlation function. This does not look good to me, because e.g. for a Gaussian states different transformations which only share the anti-unitary property would modify the state, determined thy the correlation matrix, in the same way, and there would be no difference between e.g. time-reversal and sublattice symmetry.

2. The other, more sensible, approach, would be to get the operators $\hat{U}$ into the braket, so $$\left<\hat\psi_i\hat\psi_j^\dagger\right>\rightarrow \hat{U}\left<\hat\psi_i\hat\psi_j^\dagger\right>\hat{U}^{-1}=\left<\hat{U}\hat\psi_i\hat\psi_j^\dagger\hat{U}^{-1}\right>=\left<\hat{U}\hat\psi_i\hat{U}^{-1}\hat{U}\hat\psi_j^\dagger\hat{U}^{-1}\right>.$$ This looks better, but still does not satisfy me, because with these approach the state wouldn't be affected by whether the operator $\hat{U}$ is unitary or anti-unitary, in the sense that the anti-unitarity in this calculation doesn't complex conjugate anything.

So in conclusion, neither of this approaches satisfy me, as the first one is insensitive to the specific properties of the transformation, while the second one is insensitive to the unitary or anti-unitary character of it. Is any of this approaches correct, or maybe some intermediate approach?

Answer 1

Let $(~,~)$ denote the inner product on a Hilbert space, so that $$ (\Psi_1,\Psi_2)^* = (\Psi_2 , \Psi_1) , \qquad (\Psi_1,a \Psi_2)= a (\Psi_1,\Psi_2) , \qquad (a \Psi_1,\Psi_2) = a^* (\Psi_1,\Psi_2) . $$

Unitary, linear operators $U$ satisfy $$ ( U \Psi_1 , U \Psi_2 ) = ( \Psi_1 , \Psi_2 ) , \qquad ( \Psi_1 , U^\dagger \Psi_2 ) = ( U \Psi_1 , \Psi_2 ) , \quad U(a \Psi) = a ( U \Psi ) . $$ Anti-unitary, anti-linear operators $T$ satisfy $$ ( T \Psi_1 , T \Psi_2 ) = ( \Psi_1 , \Psi_2)^* , \qquad ( \Psi_1 , T^\dagger \Psi_2 ) = ( T \Psi_1 , \Psi_2 )^* , \qquad T(a \Psi) = a^* ( T \Psi ) . $$

Let us now consider your correlator. In this language, it is given by $$ \langle {\hat \psi}_i {\hat \psi}_j \rangle = \langle 0| {\hat \psi}_i {\hat \psi}_j | 0\rangle = ( \Psi_0 , {\hat \psi}_i {\hat \psi}_j \Psi_0 ) $$ We then have $$ ( \Psi_0 , {\hat \psi}_i {\hat \psi}_j \Psi_0 ) = ( T \Psi_0 , T {\hat \psi}_i {\hat \psi}_j \Psi_0 )^* = ( T \Psi_0 , T {\hat \psi}_i T^{-1} T {\hat \psi}_j T^{-1} T \Psi_0 )^* $$ We now assume the vacuum state is invariant under $T$, i.e., $T \Psi_0 = \Psi_0$. Then, $$ ( \Psi_0 , {\hat \psi}_i {\hat \psi}_j \Psi_0 ) = ( \Psi_0 , T {\hat \psi}_i T^{-1} T {\hat \psi}_j T^{-1} \Psi_0 )^* $$ This implies $$ \langle {\hat \psi}_i {\hat \psi}_j \rangle = \langle T {\hat \psi}_i T^{-1} T {\hat \psi}_j T^{-1} \rangle^* $$