Suppose we have the Hilbert space L2(Rn) and we have n operators Qi and n operators Pi defined in the usual way by:

Qi ψ(q1,q2,...,qn) = qi ψ(q1,q2,...,qn)

Pi ψ(q1,q2,...,qn) = -i $\frac{\partial}{\partial q_i}$ ψ(q1,q2,...,qn)

for i=1,2,...,n. The Q's are the position operators and P's the momentum operators of usual quantum mechanics.

These each define self-adjoint operators (when restricted to the appropriate domains).

You can look up in a text book the spectral decomposition of the Qi to find the spectral family of projection operators E(i)x is defined by

E(i)x ψ(q1,q2,...,qn) = ψ(q1,q2,...,qn) if qi < x and E(i)x ψ(q1,q2,...,qn) = 0 if qi >= x

We then have

< ψ1, Qi ψ2 > = $\int_{-\infty}^\infty$ x d < ψ1, E(i)x ψ2 >

where <.,.> denotes the inner product in L2(Rn) and the integral is a Riemann-Stieltjes integral.

The spectral family for the Pi is related to that for the Q's through the n-dimensional Fourier transform F. We have

< ψ1, Pi ψ2 > = $\int_{-\infty}^\infty$ x d < ψ1, F-1 E(i)x F ψ2 > = $\int_{-\infty}^\infty$ x d < F ψ1, E(i)x F ψ2 >

The operators Qi and Pi are very well understood and appear in lots of textbooks. My question concerns taking linear combinations of them.

We define φ = ∑i=1..n ai Qi + bi Pi for real coefficients ai and bi. This operator is self-adjoint so should also have a spectral family that defines its spectral decomposition.

The question is what is the spectral family of φ? Is it related to the spectral families of the Q's and P's? What are the projection operators Ex that allow us to write

< ψ1, φ ψ2 > = $\int_{-\infty}^\infty$ x d < ψ1, Ex ψ2 > ?

I'm interested in this because I'm interested in defining functions of the φ operator (for example defining the operator exp(iφ2) or exp(iφ4). The only way I know how to do this is using the functional calculus defined through the spectral decomposition of φ. If this decomposition was known then we'd have (for example):

< ψ1, exp(iφ4) ψ2 > = $\int_{-\infty}^\infty$ exp(i x 4) d < ψ1, Ex ψ2 >

In any event I want to know how to calculate < ψ1, exp(iφn) ψ2 > say. i.e. I want a closed-form expression for what this number is.

If anyone knows how to do this or whether it's possible/impossible I'd really like to know.

PS - A good reference I've been using for this is "Linear Operators for Quantum Mechanics" by Thomas F. Jordan esp. chapter 3.

  • 2
    $\begingroup$ Why is $\varphi$ self-adjoint? Where is it even defined? $\endgroup$
    – Helge
    Jul 2 '10 at 13:39
  • 1
    $\begingroup$ For defining $\exp(i\phi)$, the Campbell-Baker-Hausdorff-Dynkin formula should give you a clue as to what it should be. Formally, at least. $\endgroup$ Jul 2 '10 at 15:15
  • 1
    $\begingroup$ This is an element of the Heisenberg Lie algebra in Stone-von Neumann representation, and it's conjugate to a single $P$ (or $Q$) under the metaplectic group. In particular, its spectral properties are the same. My first reaction would be to look it up in Folland, Harmonic analysis in phase space or Reid and Simon, Methods of mathematical physics, I have neither of them close at hand, unfortunately. $\endgroup$ Jul 2 '10 at 19:35
  • $\begingroup$ I'd be interested in hearing more about the Heisenberg Lie algebra/metaplectic group point of view - I don't have Follard's book yet but should be able to get it from my library in a few days (they're relocating the books at the moment!) From the Campbell-Baker-Hausdorff-Dynkin formula (and just trying to go from the Wikipedia page about this!) do we have $ \exp(i \phi) := \exp(i (a.Q) + i(b.P)) = \exp(i a.Q) \exp(i b.P) \exp(a.b/2)$ since the commutator satisfies: [a.Q,b.P] = i a.b ? Presumably a more complicated version holds for $\exp(i \phi^2)$, let along $\exp(i \phi^4)$! $\endgroup$
    – StevenJ
    Jul 5 '10 at 12:49
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    $\begingroup$ Update: In Follard's book (p21-22) he shows that [exp(i &phi;) f](q) = exp(i a.q + i a.b/2) f(x + b) which agrees with the C-B-H-D formula... $\endgroup$
    – StevenJ
    Jul 15 '10 at 15:34

Just to elaborate on my comment:

It is clear that your operator $\varphi$ is symmetric, in the sense that for $f,g$ smooth and compactly supported functions, one has $$ \langle f, \varphi g\rangle = \langle \varphi f, g\rangle. $$ However, this does not imply self-adjointness. Furthermore, the usual arguments fail why your sum should be well defined. In particular $Q_k$ and $P_k$ both are operators, whose spectrum is $\mathbb{R}$, so quadratic form methods are not available.

Coming to the question: I would say there is no such description. It is well known that the operators $- \frac{d^2}{dx^2}$ and multiplication by $x^2$ both have spectrum $[0, \infty)$, but there sum has spectrum $2n + 1$ (if I remember correctly). Furthermore, the eigenvalues of this sum, the harmonic oscillator, are the Hermite polynomials, and not simple combinations of $\delta_x$ or $x \mapsto \sin(k x)$, which are the generalized eigenfunctions of $x^2$ respectively $-\frac{d^2}{dx^2}$.

However, if your operator is self-adjoint you should be able to write down the spectral projections. First your problem factors into problems on $L^2(\mathbb{R})$. Second, I think one can explicitly solve the ODE $u' + a xu = z$, which should allow one to write down the spectral projections ...

Update, July 5th: So I make fewer typos, denote by $H$ some self-adjoint operator. Define $$ R(z) = (H - z)^{-1} , \quad im(z) > 0 $$ Then one can write the spectral projection $E(a,b)$ of the interval $(a,b)$ as $$ E(a,b) = \lim_{\varepsilon\to 0} \frac{1}{\pi} \int_{a}^{b} im( R(t + i \varepsilon)) dt $$ with some subtelitties because of point mases, I choose to ignore. So in order to understand $E(a,b) f$, one can now try to understand $$ \lim_{\varepsilon\to 0} \frac{1}{\pi} \int_{a}^{b} im( R(t + i \varepsilon) f) dt $$ Since, we can write down what $R(t + i \varepsilon) f$ using the ODE, we can compute this (with some effort). However, one first need to write down the domain of $H$ so, one knows what $R$ is.

Further Update

Let $f$ be compactly supported and smooth. Introduce $$ K_{a, z}(x) = \exp\left(\frac{ia}{2} x^2 + z x\right). $$ A direct computation shows that $i K_{a,z}'(x) + (a x - z) K_{a,z}(x) = 0$. So if we define for $f$ compactly supported and smooth $$ R_{z} f (x) = K_{a,z}(x) \cdot \left(\int_{-\infty}^{x} K_{a,z}(t) f(t) dt + C\right), $$ then we have $i R_{z} f'(x) + (ax -z) R_{z} f(x) = f(x)$. Now, if we want to have $R_{z} f \in L^2(R)$, we must have $C \equiv 0$ by looking at $x \to - \infty$ asymptotics and also $$ \int_{-\infty}^{\infty} K_{a,z}(t) f(t) dt =0 $$ by looking at the $x \to \infty$ asymptotic.

This worries me, since $R_{z}$ should be defined for all $f \in L^2$. Does anybody see my mistake?

  • 1
    $\begingroup$ This is a linear problem in the Heisenberg Lie algebra, so the properties of quadratic operators may be misleading (since $P$ and $Q$ do not commute). $\endgroup$ Jul 2 '10 at 19:37
  • $\begingroup$ I pretty sure &phi; is self-adjoint (or at least has a unique self-adjoint extension or closure). To see this we can think of writing the P and Q operators in terms of creation and annihilation operators a and $a^\dagger$. When we do this we find that &phi; becomes the Segal field operator discussed in Reed and Simon Methods of Modern Mathematical Physics Vol II p210 Thm X.41. There they show that that the Segal field operator is essentially self-adjoint - i.e. its closure is self-adjoint. How would I be able to use the 1st order ODE given by Helge to get the spectral projections? $\endgroup$
    – StevenJ
    Jul 5 '10 at 12:16
  • $\begingroup$ I expanded my post. Is this already helpful? $\endgroup$
    – Helge
    Jul 5 '10 at 14:12
  • $\begingroup$ Thanks, that sounds like an interesting way to think about it... $\endgroup$
    – StevenJ
    Jul 5 '10 at 16:27
  • $\begingroup$ Having continued my computation, I believe even less that this operator is self-adjoint ... ;-) The word 'algebra' in this context always seem worrisome, since the product of unbounded operators is something nasty (for example not always defined ...) $\endgroup$
    – Helge
    Jul 5 '10 at 21:05

I'm going to give a remarkably unhelpful answer. Namely: shouldn't this be a straightforward (although rather ugly) calculation using the techniques of quantum optics? Unfortunately, I don't know quantum optics well enough to solve this easily, nor do I know of any references on quantum optics which are written for mathematicians (Please, if anybody knows one, tell me. If not, somebody please write one.).


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