Evaluating Gibbs state in the second quantized formalism First, let us fix our notation. If $A:\mathcal H \to \mathcal H$ is a linear operator on a single-particle Hilbert space $\mathcal H$, we can lift $A$ on the Fock space $\mathfrak F_{s/a}(\mathcal H)$ in the following two ways. Here the subscript indices $s/a$ stand for symmetrized/antisymmetrized versions corresponding to bosons/fermions respectively. Define $\Gamma, \text{d}\Gamma: \mathcal H^N_{s/a} \to \mathcal H^N_{s/a}$ on $N$-particle Hilbert spaces $\mathcal H^N_{s/a}$ by

*

*$\Gamma(A) := \bigotimes_{i=1}^N A$

*$\text{d}\Gamma := \sum_{j=1}^N 1\!\!1 \otimes...\otimes 1\!\!1 \otimes A \otimes 1\!\!1 \otimes ...\otimes 1\!\!1 \, ,$ where $A$ is in the $j$-th position, and $1\!\!1$ is the identity.

Note that $\text{d}\Gamma$ is a linear operator, whereas $\Gamma$ is not.
It follows from these definitions that if $U:\mathcal H \to \mathcal H$ is unitary, $$\Gamma(U)b^*(\varphi)=b^*(U\varphi)\Gamma(U) \tag 1 $$ where $b^*(\varphi): \mathcal H^{N-1}_{s/a} \to \mathcal H^N_{s/a}$ is the creation operator of state $\varphi \in\mathcal H\,.$
Moreover, $$\Gamma(e^{\alpha A})=e^{\alpha \ \text{d}\Gamma(A)} \,. \tag 2$$
Now, let $H$ be a self-adjoint Hamiltonian over the one-particle Hilbert space $\mathcal H$, governing the dynamics of a gas of non-interacting fermions at inverse temperature $\beta \in \mathbb R$ and chemical potential $\mu \in \mathbb R \,.$
If $K_\mu$ denotes the modified Hamiltonian
$$ K_\mu := \text{d}\Gamma(H - \mu 1\!\!1) = \text{d}\Gamma(H) - \mu \mathcal N \, , \tag3$$
then the Gibbs state is defined by
$$ \omega(A) := Z^{-1} \mathrm {Tr}(e^{-\beta K_\mu}A) \tag4$$
where we assume that $e^{-\beta K_\mu}$ is trace-class and in particular, $\mathrm {Tr} (e^{-\beta K_\mu}) =: Z$. Furthermore, $\mathcal N$ represents the number operator.
I want to know how the following two equalities follow from the above considerations.

*

*First, $$\boxed{ e^{-\beta K_\mu} b^*(\varphi) = \lambda \ b^*(e^{-\beta H}\varphi) e^{-\beta K_\mu}} \,, \tag5$$ where $\lambda := e^{\beta \mu}$ is the activity.


*Secondly, $$ \boxed{\omega \big( a^*(\varphi)a(\psi) \big) = \omega \big(a(\psi)a^*(\lambda e^{-\beta H}\varphi) \big)} \,. \tag6$$
I would appreciate if you do not skip any steps. For your reference, I am studying Operator Algebras and Quantum Statistical Mechanics II Proposition 5.2.22 (pp.46-48). The steps necessary to my understanding have been skipped from the book and I am struggling to connect the dots.
Kindly help.
 A: I know this a pretty old question, but I would like to give an incomplete answer, in the sense, that I will assume that $\mathcal{H}$ is finite-dim and that we are dealing with the anti-symmetrized Fock space so that everything is finite-dimensional. The generalization should be relatively straightforward, except that we would have to drown in details such as domains and convergence. I would love to see a complete answer though.
Also, let me assume that $\mu=0$ so that $K_\mu = d\Gamma(H)$, since the proof is pretty much the same with nonzero $\mu$.
Let $\varphi_n$ be an orthonormal basis of $\mathcal H$ and $a^*_n =a^*(\varphi_n)$. It's easy to check that $d\Gamma(\mathcal H)=\sum_n a^*(H\varphi_n) a(\varphi_n)$. Hence, by the Adjoint/adjoint representation, we see that
$$
e^{-\beta d\Gamma(H)} a^*(\varphi) e^{\beta d\Gamma(H)} = \exp{(-\beta [d\Gamma(H),\cdot])} a^*(\varphi) = a^*(e^{-\beta H}\varphi)
$$
where we used the fact that $[d\Gamma(H),a^*(\varphi)]=a^* (H\varphi)$. The second equality follows from the first equality and the fact that the trace is commutative.
