reference for wick product - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:01:23Z http://mathoverflow.net/feeds/question/37037 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37037/reference-for-wick-product reference for wick product BigBill 2010-08-29T09:37:31Z 2010-11-10T13:49:12Z <p>Let $H$ be a real Hilbert space with complexification $H_{\mathbb{C}}$. We denote by $\mathfrak{F}$ the antisymmetric Fock space over $H_\mathbb{C}$ ("fermions"). A creation operator is denoted by $c(f)$. I need a reference for the calculus of $$&lt;\Omega ,\big(c(f_1)+c(f_1)^*\big)...\big(c(f_{2k})+c(f_{2k})^*\big)\Omega>_{\mathfrak{F}}$$ where $\Omega$ the unit vector, called vacuum.</p> <p>Thank you.</p> http://mathoverflow.net/questions/37037/reference-for-wick-product/37047#37047 Answer by JosÃ© Figueroa-O'Farrill for reference for wick product JosÃ© Figueroa-O'Farrill 2010-08-29T12:29:10Z 2010-08-29T12:29:10Z <p>The original reference for Wick's theorem is, not surprisingly, Wick's original 1950 paper: <a href="http://prola.aps.org/abstract/PR/v80/i2/p268_1" rel="nofollow"><em>The Evaluation of the Collision Matrix</em></a> published in the Physical Review <strong>80</strong> (2) pp. 268-272. He also shows how to compute it and it is surprisingly readable 60 years on.</p> <p>Of course, depending on your background, this may be too physical. A more mathematical reference are the Bombay Lectures by Kac and Raina <em><a href="http://books.google.co.uk/books?id=SZVqYvvD8rQC" rel="nofollow">Highest-weight representations of infinite-dimensional Lie algebras</a></em>, particularly the 5th lecture on the Bose-Fermi correspondence.</p> <p>The basic idea is to think of $\mathfrak{F}$ as the space of semi-infinite forms. The vacuum vector would be given by $$\Omega = f_1^* \wedge f_2^* \wedge \cdots$$ and $c(f_i)^*$ acts by wedging with $f_i^*$ whereas $c(f_i)$ acts by contracting with $f_i$.</p> http://mathoverflow.net/questions/37037/reference-for-wick-product/37414#37414 Answer by BigBill for reference for wick product BigBill 2010-09-01T19:49:30Z 2010-09-01T19:49:30Z <p>The answer is provided by the article</p> <p><a href="http://www.pnas.org/content/100/15/8629.full" rel="nofollow">http://www.pnas.org/content/100/15/8629.full</a></p> <p>However, the authors work in the context of $q$-fock space. I does not know if there exists an older paper which provides the answer in the less general context of antisymmetric Fock space (i.e. q=1).</p> http://mathoverflow.net/questions/37037/reference-for-wick-product/37492#37492 Answer by David Bar Moshe for reference for wick product David Bar Moshe 2010-09-02T12:14:59Z 2010-09-02T12:40:14Z <p>The following <a href="http://www.ams.org/journals/proc/1998-126-12/S0002-9939-98-05028-X/S0002-9939-98-05028-X.pdf" rel="nofollow">article</a> by E. R. Negrin provides the required formula for the antisymmetric Fock space in the corollary on page 3644.</p> <p>I want to point out that the Wick products (for the antisymmetric Fock space) can be constructed from a Gaussian generating function which is Gaussian in (real) Grassmann variables, which is given for the case presented in the question by:</p> <p>$G(\mathbf{\xi}) = exp((\Sigma_{i=0}^{2k} \xi_i f_i, \Sigma_{j=0}^{2k} \xi_j f_j))$</p> <p>where $( , )$ denotes the Hilbert sapce $H_\mathbb{C}$ inner product.</p> <p>The required Wick product is obtained as the coefficient of $\xi_1 \xi_2 . . .\xi_{2k}$.</p> http://mathoverflow.net/questions/37037/reference-for-wick-product/45551#45551 Answer by Marcel Bischoff for reference for wick product Marcel Bischoff 2010-11-10T13:49:12Z 2010-11-10T13:49:12Z <p>One should be able to obtain the formula from the appendix of: </p> <p><a href="http://arxiv.org/abs/math-ph/0204029" rel="nofollow">http://arxiv.org/abs/math-ph/0204029</a> </p> <p>They have a formula for all vectors, to the vacuum expectation just the summand with $2p=n$ contributes. They using Arakis self dual CAR algebra, and if you consider $a(f)$ for $f=\Gamma f$ it should equal your $c(f)+c(f)^\ast$. </p>