Spectral theory for self-adjoint field operators on a symmetric Fock space - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T00:39:50Zhttp://mathoverflow.net/feeds/question/18368http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/18368/spectral-theory-for-self-adjoint-field-operators-on-a-symmetric-fock-spaceSpectral theory for self-adjoint field operators on a symmetric Fock spaceStevenJ2010-03-16T12:57:46Z2010-03-18T10:22:41Z
<h2>Background</h2>
<p>Suppose we have a finite-dimensional Hilbert space $H = \mathbb{C}^s$ (for a natural number s) and we construct the symmetric (or <em>bosonic</em>) Fock space built from it: $$F(H):= \mathbb{C} \oplus H \oplus S(H \otimes H) \oplus S(H \otimes H \otimes H) \oplus \ldots$$
where S is the symmetrising operator.</p>
<p>Vectors in F are sequences of vectors $\psi = (\psi_0, \psi_1,\psi_2,\ldots)$ such that
$\psi_0 \in \mathbb{C}$, $\psi_1 \in H$, $\psi_2 \in S(H \otimes H)$ etc such that
$\sum_{n=0}^\infty ||\psi_n||_n^2 < \infty$ where || ||<sub>n</sub> denotes the appropriate norm.</p>
<p>For any vector f $\in$ H we can define a pair of unbounded densely defined operators $a^\dagger(f)$ and $a(f)$ acting on F. These are called the "creation and annihilation operators". They are mutually adjoint and satisfy a commutation relation of the form: $$a(f) a^\dagger(g) - a^\dagger(g) a(f) = \langle f, g\rangle $$
where $\langle f, g\rangle $ is the inner-product of f, g $\in$ H.</p>
<p>The best reference for all this is M. Reed, B. Simon, "Methods of Mathematical Physics, Vol 2", section X.7 p207-212. This is partially available on Google books here: <a href="http://books.google.co.uk/books?id=Kz7s7bgVe8gC&lpg=PA141&dq=reed%20and%20simon%20x.7&client=firefox-a&pg=PA210#v=onepage&q=&f=false" rel="nofollow">http://books.google.co.uk/books?id=Kz7s7bgVe8gC&lpg=PA141&dq=reed%20and%20simon%20x.7&client=firefox-a&pg=PA210#v=onepage&q=&f=false</a></p>
<p>The sum $\phi(f) = a(f) + a^\dagger(f)$ is self-adjoint (more properly the closure of their sum is self-adjoint) and is called the Segal quantisation of f (up to a factor of $\sqrt{2}$).</p>
<blockquote>
<p>Since $\phi(f)$ is self-adjoint we can apply the spectrum theorem to it. The question is, what is its spectral decomposition? Or more loosely, what are its eigenvalues and eigenvectors? or what can we tell from about its spectral decomposition?</p>
</blockquote>
http://mathoverflow.net/questions/18368/spectral-theory-for-self-adjoint-field-operators-on-a-symmetric-fock-space/18478#18478Answer by David Bar Moshe for Spectral theory for self-adjoint field operators on a symmetric Fock spaceDavid Bar Moshe2010-03-17T11:20:05Z2010-03-17T11:20:05Z<p>One convenient way to do analysis on the symmetric Fock space is to use its isomorphism to
the Bargmann (reproducuing Kernel Hilbert) space (sometimes called the Bargmann-Fock pace)
of analytic functions on C^s (with respect to the Gaussian measure) defined in the classical paper:</p>
<p>Bargman V. On a Hilbert space of analytic functions and associated integral transform I,
Pure Appl. Math. 14(1961), 187-214. </p>
<p>An introduction to the Bargmann space may be found in chapter 4 of the
<a href="http://www.mat.univie.ac.at/users/neretin/public_html/lectures/book.pdf" rel="nofollow">book</a> by Uri Neretin </p>
<p>On the Bargmann space the creation and anihilation operators
are just the multiplication a_j = z_j and the derivation a*_j = d/dZ_j
and consequently, the theory of several complex variables can be used for the analysis on this space,
for example the trace of (a trace class) operator can be represented as an integral on its symbol.</p>
<p>Remark: The isomorphism between the symmetric Fock and Bargmann spaces is not proved in the Book. It can be found for example in the references of the following <a href="http://www.emis.de/journals/UIAM/actamath/PDF/34-135-148.pdf" rel="nofollow">article</a>:</p>
<p>Regarding the question about a(f)+a*(f), it is proportional to the position operator of quantum mechanics.
This is an unbounded operator, its spectrum is the whole real line, but it does not have
eigenfunvectors within the Fock space (Loosly speaking, they are Dirac delta functions), however one can find a series of vectors which approximate arbitrarily closely its eigenvectors. Using the corresponding projectors, one can approximate the spectral decomposition of this operator.
The case of the momentum operator i(a(f)-a*(f)) is used more frequently, a possible choice of the approximate eigenvectors is by means of wave packets.</p>