On the definition of the bosonic one-particle reduced density matrix

Question

Consider the bosonic/fermionic Fock space $F^\pm:=\bigoplus\limits_{N=0}^\infty H_N^{\pm}$, where $H^+_N:=\vee^N \mathfrak h$ and $H^-_N:=\wedge^N \mathfrak h$ for some (complex, separable) one-particle Hilbert space $\mathfrak h$ and $H^\pm_0 :=\mathbb C$.

Define

$$ \langle f,\gamma_\rho g\rangle_{\mathfrak h}:=\mathrm{Tr}_{F^\pm} \rho\, a^\dagger_{\pm}(g)a_\pm(f)\tag{1}\quad , $$

for $f,g\in \mathfrak h$ and some density operator $\rho$ on $F^\pm$ with $\mathrm{Tr}_{F^\pm}\rho N <\infty$. Here, $a^\dagger_\pm$ and $a_\pm$ denote the usual creation and annihilation operators on $F^\pm$ and $N$ denotes the number operator.

Question: Under what conditions is the RHS of $(1)$ well-defined for the bosonic case? And under what conditions can we prove the existence of a positive semi-definite, trace-class operator $\gamma_\rho$ on $\mathfrak h$ with trace $\mathrm Tr_{\mathfrak h} \gamma_\rho = \mathrm{Tr}_{F^\pm}\rho N$ fulfilling $(1)$? Are these conditions the same?

More concretely, I am worried since in the bosonic case, the creation and annihilation operators are unbounded and hence the trace on the RHS of $(1)$ might be ill-defined from the very beginning. Is the condition that $\mathrm{Tr}_{F^+} \rho N <\infty$ sufficient to ensure that $(1)$ is well-defined?

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I think that for the fermionic case this can be done as follows: We note that $a^\dagger_-(g)$ and $a_-(f)$ are bounded operators on $F^-$ (defined on the whole space) and from the fact that $\rho$ is of trace-class we find that $\rho\, a^\dagger_-(g) a_-(f)$ is trace-class as well and thus the RHS of $(1)$ is a well-defined sesquilinear form for all $f,g\in \mathfrak h$. Further, it is bounded, because

$$ |\mathrm Tr_{F^-} \rho \,a^\dagger_- (g)a_- (f)| \leq ||\rho|| \,||a_-^\dagger(g)||\, ||a_-(f) ||\, = ||\rho|| \, ||g||_\mathfrak h \,|| f||_\mathfrak h\quad . \tag{2} $$

Now the existence of a bounded operator on $\mathfrak h$ denoted by $ \gamma_\rho$, fulfilling $(1)$ follows. Positive semi-definiteness and the trace normalization are more or less trivial, from which in turn it follows that it is of trace-class, too.

However, the bosonic creation and annihilation operators are not bounded, and as such the above cannot be applied.

Answer 1

I do not have a definite answer but some remark. First of all notice that if $A$ is selfadjoint and $Ran(B) \subset D(A)$ with $B$ everywhere defined and bounded, then $AB$ is everywhere defined an bounded as well. It easily flows from the closed graph theorem ($A$ is closed because selfadjoint). On the other hand, if $T$ is trace class and $S$ is bounded and everywhere defined, then $ST$ is trace class as is well known. Therefore, if $\rho$ is a mixed state such that $\sqrt{\rho}$ is still trace class and $A$ is selfadjoint with $Ran(\sqrt{\rho}) \subset D(A)$, then $A\rho$ is trace class even if $A$ is unbounded.

For instance, if $H=H^\dagger$ is bounded below with discrete specrum such that $e^{-\beta H}$ is trace class for positive $\beta$, then every operator $H^n e^{-\beta H}$ is trace class for $n>0$ integer.