Suppose we have a finite-dimensional Hilbert space $H = \mathbb{C}^s$ (for a natural number s) and we construct the symmetric (or bosonic) 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.

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 || ||n denotes the appropriate norm.

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.

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: 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

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 $\sqrt{2}$).

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?

  • $\begingroup$ You a.s. know this, but $a^*a$ is the number operator. $\endgroup$ Mar 16, 2010 at 15:15
  • $\begingroup$ Bratteli and Robinson volume 2 is also a good reference: books.google.com/…. Both volumes are available to download from Bratteli's webpage: folk.uio.no/bratteli. $\endgroup$ Mar 17, 2010 at 1:38

1 Answer 1


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 $\mathbb C^s$ (with respect to the Gaussian measure) defined in the classical paper:

Bargmann, V., On a Hilbert space of analytic functions and an associated integral transform, Commun. Pure Appl. Math. 14, 187-214 (1961). ZBL0107.09102.

An introduction to the Bargmann space may be found in chapter 4 of the book by Uri Neretin

Neretin, Yurii A., Lectures on Gaussian integral operators and classical groups, EMS Series of Lectures in Mathematics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-080-7/pbk). xii, 559 p. (2011). ZBL1211.22001.

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.

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:

Stochel, Jerzy B., Representation of generalised creation and annihilation operators in Fock space, Zesz. Nauk. Uniw. Jagiell. 1208, Univ. Iagell. Acta Math. 34, 135-148 (1997). ZBL0949.47027.

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.

  • $\begingroup$ Thank you for your answer. Using the Bargmann space sounds like a good idea. If we represent the a(f) and a*(f) as operators on the Bargmann space then can we get an explicit form for the spectral decomposition of a(f)+a*(f)? As far as I understand it we'd want something like $$\langle \psi_1 |(a(f) + a*(f)) |\psi_2 \rangle = \int_{-infty}^\infty \alpha d\langle \psi_1| E_\alpha | \psi_2\rangle $$ for some spectral projectors $E_\alpha$ depending on the real number $\alpha$. Is it possible to work out the $E_\alpha$ explicitly? $\endgroup$
    – StevenJ
    Mar 19, 2010 at 15:01
  • $\begingroup$ The answer is yes, but there is a complication because the eigenvectors of this operator lie outside the Fock space. (As a simple exercise, the solution of the eigenvalue problem of this operator for H = C^1, produces eigenfuntions which are not square integrable). One way to do that is to use the unitary isomorphism between the bargmann-Fock space and L^2(R^s) given by the Bargmann transform. Continued in the following comment $\endgroup$ Mar 21, 2010 at 16:27
  • $\begingroup$ In L^2(R^s), this operator is proportional a component of the position operator, i.e., the multiplication operator by a linear combination of the coordinates of R^n. Once the eigenvectors or the spectral projectors are found, one can use the bargmann transform to transform them back to the Bargmann space. The following review article by N.P. Landsmann: Continued in the next comment $\endgroup$ Mar 21, 2010 at 16:29
  • $\begingroup$ math.ru.nl/~landsman/HSQM2006.pdf actually describes two methods for the construction of the spectral projectors, the first is by approximating the eigenvectors with square integrable functions (page 42), or by restriction of the Hilbert space to the Domain of the operator (page 52, 53). $\endgroup$ Mar 21, 2010 at 16:29

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