Spectral decomposition for an arbitrary linear combination of position and momentum operators

Question

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

Q_i ψ(q₁,q₂,...,q_n) = q_i ψ(q₁,q₂,...,q_n)

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

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 Q_i to find the spectral family of projection operators E⁽ⁱ⁾_x is defined by

E⁽ⁱ⁾_x ψ(q₁,q₂,...,q_n) = ψ(q₁,q₂,...,q_n) if q_i < x and E⁽ⁱ⁾_x ψ(q₁,q₂,...,q_n) = 0 if q_i >= x

We then have

< ψ₁, Q_i 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 P_i is related to that for the Q's through the n-dimensional Fourier transform F. We have

< ψ₁, P_i 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 Q_i and P_i are very well understood and appear in lots of textbooks. My question concerns taking linear combinations of them.

We define φ = ∑_i=1..n a_i Q_i + b_i P_i for real coefficients a_i and b_i. 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 E_x that allow us to write

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

Answer 1

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?

Answer 2

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.).