In LQFT, a state, $\omega$, is a linear map $\omega:A=:CCR({\cal{S}},\Omega)\rightarrow \mathbb{C}$ satisfying:

  • $\omega(aa^{*})\geq 0$ for all $a\in A$.
  • $\omega(I)=1$ where $I$ denotes the identity element of $A$.

where $CCR({\cal{S}},\Omega)$ is the Weyl-algebra that comes from the symplectic vector space $({\cal{S}},\Omega)$.

Now let $\mu:{\cal{S}}\times{\cal{S}}\rightarrow\mathbb{R}$ be an arbitrary (real) inner product on $\cal{S}$ that satisfies \begin{equation} \frac{1}{4}\Omega(u^{1},u^{2})\le\mu(u^{1},u^{1})\mu(u^{2},u^{2}) . \end{equation}

where $\Omega$ is a symplectic structure.

Then $\omega_{\mu}: A\rightarrow\mathbb{C}$ defined by

\begin{equation} \omega(W(u))=e^{-\frac{\mu(u,u)}{2}} \end{equation}

for all $u\in {\cal{S}}$, is a called a quasifree state.

Why does this state satisfy $\omega(aa^{*})\geq 0$?

  • $\begingroup$ I suppose you meant to write $\omega(aa^\ast)\geq 0$ and $A= \mathrm{CCR}(\mathcal{S},\Omega)$. You also didn't define what $\langle W(u)\rangle_{u\in\mathcal{S}}$ is supposed to be, or how $\omega_\mu$ is a functional on $A$. $\endgroup$ – ACuriousMind May 29 '16 at 18:18
  • $\begingroup$ Thanks for the comment. I have edited the question now. If it is not clear, I think using linearity and continuity one can extend the state from the domain of definition (elements of the form $W(u)$) to the whole $A$. $\endgroup$ – yess May 29 '16 at 19:43

The condition on the modulus of $\Omega$ ensures that the complex bilinear form $\mu_c$ defined as $\mu_c(\cdot,\cdot)=\mu(\cdot,\cdot)+i\Omega(\cdot,\cdot)$ is a scalar product. Therefore $(\mathscr{S},\mu_c)$ is a complex pre-Hilbert space. Denoting by $\mathscr{H}_{\mu_c}$ its completion, it is then possible to define the symmetric Fock space $\Gamma_s(\mathscr{H}_{\mu_c})$ in the usual fashion: $$\Gamma_s(\mathscr{H}_{\mu_c})=\bigoplus_{n=0}^\infty \mathscr{H}_{\mu_c}^{\otimes_s n}\; .$$ In addition, $$\omega(W(u))=e^{-\frac{1}{2}\langle u,u\rangle_{\mathscr{H}_{\mu_c}}}\; .$$ This is the generating functional of the vector state $\lvert \Omega_{\mu_c}\rangle\langle\Omega_{\mu_c}\rvert$, where $\Omega_{\mu_c}\in \Gamma_s(\mathscr{H}_{\mu_c})$ is the vacuum vector. This can be easily seen since $(\Gamma_s(\mathscr{H}_{\mu_c}),\pi_{\mu_c})$ with $$\pi_{\mu_c}(W(u))=e^{\frac{i}{\sqrt{2}}\bigl(a^*_{\mu_c}(u)+a_{\mu_c}(u)\bigr)}$$ is a representation of the CCR on the Fock space ($a^{\#}_{\mu_c}$ are the usual creation/annihilation operators).

Now since $\omega$ is a vector state, then it is automatically positive, in fact: $$\omega(A^*A)=\langle A\Omega_{\mu_c},A\Omega_{\mu_c}\rangle_{\Gamma_s(\mathscr{H}_{\mu_c})}\; .$$

  • $\begingroup$ I am having problem to prove it for linear combinations. Would you mind doing that step explicitly? $\endgroup$ – yess May 29 '16 at 19:44
  • $\begingroup$ Having thought about it, there is a much smarter way to prove the assertion; I will edit the answer accordingly ;-) $\endgroup$ – yuggib May 30 '16 at 7:34

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