1
$\begingroup$

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 $\omega(aa^{*})\geq 0$?

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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: $$\omega(A^*A)=\langle A\Omega_{\mu_c},A\Omega_{\mu_c}\rangle_{\Gamma_s(\mathscr{H}_{\mu_c})}\; .$$

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.