Weyl exponential form of the Canonical Commutation Relations

Question

What is the physical meaning of the $c$-numbers $Q, P\in \mathbb{R}$ in the exponent of the Weyl system $\exp\left[\frac{i}{\hbar} Q \hat{p}\right]$ and $\exp\left[\frac{i}{\hbar}P\hat{q}\right]$? Here $\hat{q},\hat{p}$ are operators of position and momentum with canonical commutation relation $[\hat{q},\hat{p}]=i\hbar$.

Answer 1

There's a considerable danger of oversimplification here, but I find it helpful to think of the construction $e^{i\lambda \hat q}$ as an operator that generates the characteristic function of a probability distribution in a particular state. Being a little free with mathematical details, we can write the probability density of observing the value $q$ in a vector state $\left|\psi\right>$ as $\mathsf{Pr}(q)=\left<\psi\right|\delta(q-\hat q)\left|\psi\right>$, for which the characteristic function is the fourier transform $$C(\lambda)=\int\left<\psi\right|\delta(q-\hat q)\left|\psi\right>e^{i\lambda q}\mathrm{d}q= \left<\psi\right|e^{i\lambda \hat q}\left|\psi\right>.$$ We can inverse fourier transform this back to a probability density. What is commonly labeled $Q$, as you have here, is more helpfully labeled in a way that emphasizes its difference from the position operator and the values that position measurements may take. Mathematically, $\lambda$ is a linear dual of the position. It's very clear what a fourier transform of a probability density is algebraically, but perhaps not so clear what its physical meaning is. It's perhaps best to think of it as a formal device that uses $\lambda$ for keeping track of all the moments of the probabilities of different measurement results, which allows us to say that $\lambda^n$ is associated with the $n$-th moment. [You could do a lot worse, if you want to understand generating functions, though this is slightly idiosyncratic suggestion on my part, to read John Baez's recent series on Network Theory with an open mind; a Google search for "network theory (part" baez finds them. It may well twist your head a little, however, so if you want quick understanding this may not be for you.]

Everything is essentially the same for the momentum operator taken alone, but if we introduce both $\hat q$ and $\hat p$, so that we consider an object such as $$\tilde W(\lambda,\mu)=\int\left<\psi\right|\delta(q-\hat q)\delta(p-\hat p)\left|\psi\right> e^{i\lambda q+i\mu p}\mathrm{d}q\mathrm{d}p= \left<\psi\right|e^{i\lambda \hat q+i\mu\hat p}\left|\psi\right>,$$ and then inverse fourier transform this object, we obtain negative "probability densities" for some values of $p$ and $q$. This is the Wigner function, about which much has been written.

Answer 2

The Weyl system, $\exp\left[\frac{i}{\hbar} Q \hat{p}~\right]$ and $\exp\left[\frac{i}{\hbar}P\hat{q}~\right]$, comprise two "presentation" elements of the Heisenberg group.

To the extent $\hat{p}$ is a derivative with respect to position q, Q is but the shift amount that q in any function of it is translated by the action of $\exp\left[\frac{i}{\hbar} Q \hat{p}~\right]$; that is, $f(q) \mapsto f(q+Q)$.

The action of the group element $\exp\left[\frac{i}{\hbar}P\hat{q}~\right]$ on such functions is more prosaic: multiplication by $\exp\left[\frac{i}{\hbar}Pq\right]$, which is a bland rephasing.

What is special about the braiding Weyl relations, $\exp\left[\frac{i}{\hbar} Q \hat{p}~\right] \exp\left[\frac{i}{\hbar}P\hat{q}~\right]$ $= e^{iPQ/\hbar} \exp\left[\frac{i}{\hbar}P\hat{q}~\right]\exp\left[\frac{i}{\hbar} Q \hat{p}~\right]$ is that they are bounded and thus closer to their finite-dimensional analogs, and thus serve to illuminate the Heisenberg group and the Stone-von Neumann theorem much better than $\hat{q}$ and $\hat{p}$. In particular, they are the continuum limits of the substantially more tractable clock and shift matrices system (group).