Diagonal operators in the coherent state basis

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I asked a question back in November on Physics.StackExchange, but that didn’t attract any interest from anyone. I started thinking about it again recently and figured out a good solution. The question and answer are explained below.I posted the answer on Physics.SE too since they encourage the answering of one’s own question. How lonely is that?!?a  

Q: Is there a good notion of a “diagonal” operator with respect the overcomplete basis of coherent states?
A: Yes. The operators that are “coherent-state diagonal” are those that have a smooth Glauber–Sudarshan P transform.

The primary motivation for this question is to get a clean mathematical condition for diagonality (presumably with a notion of “approximately diagonal”) for the density matrix of a system of a continuous degree of freedom being decohered. More generally, one might like to know the intuitive sense in which X, P, and X+P are all approximately diagonal in the basis of wavepackets, but RXR^\dagger is not, where R is the unitary operator which maps

(1)   \begin{align*} \vert x \rangle \to (\vert x \rangle + \mathrm{sign}(x) \vert - x \rangle) / \sqrt{2}. \end{align*}

(This operator creates a Schrodinger’s cat state by reflecting about x=0.)

For two different coherent states \vert \alpha \rangle and \vert \beta \rangle, we want to require an approximately diagonal operator A to satisfy \langle \alpha \vert A \vert \beta \rangle \approx 0, but we only want to do this if \langle \alpha \vert \beta \rangle \approx 0. For \langle \alpha \vert \beta \rangle \approx 1, we sensibly expect \langle \alpha \vert A \vert \beta \rangle to be within the eigenspectrum of A.

One might consider the negativity of the Wigner-Weyl transformCase has a pleasingly gentle introduction.b   of the density matrix (i.e. the Wigner phase-space quasi-probability distribution aka the Wigner function) as a sign of quantum coherence, since it is known that coherent superpositions (which are clearly not diagonal in the coherent state basis) have negative oscillations that mark the superposition, and also that these oscillations are destroyed by decoherence. However, this is not quite right since the Wigner function always is negative somewhere for any state besides perfect Gaussian states. Additionally, negativity is not useful for identifying more general diagonal operators (like the position operator) which are negative over half their domain.

We get much closer to the answer by recalling that the frequency of the oscillations observed in the Wigner function of superpositions are “sub-Planckian” in the sense that they fit insides a phase-space area smaller than \hbar (which obviously does not mean below the Planck length or Planck momentum scales). But it’s not transparent exactly what this has to do with diagonality, and exactly how small the oscillations should be.

Let’s take a naive stab at what an operator would look like if it were diagonal in the basis of coherent states \vert \alpha \rangle, where \alpha = (x,p) is a variable ranging over phase space. First, note that we can write the position operator in the position basis as

(2)   \begin{align*} \hat{X}=\int \mathrm{d}x \, x \vert x \rangle \langle x \vert. \end{align*}

So, given an arbitrary real-valued (classical) function f(\alpha) on phase-space (e.g. position, momentum, or energy) consider the following method of defining a corresponding operator:The integral on the right should be normalized by a factor of 1/2 \pi but I’m going to ignore normalization in everything I have to say.c  

(3)   \begin{align*} \hat{\Omega}[f] := \int \mathrm{d}\alpha\, f(\alpha) \vert \alpha \rangle \langle \alpha \vert. \end{align*}

Intuitively, this operator acts on a state vector by “measuring it” in the wavepacket basis and weighting the corresponding components by the value of the function at that point in phase space. In fact, one can check that \hat{\Omega}[X] = \hat{X}, where X(\alpha) = X(x,p) = x is the position function on phase space.

How does this relate to the Wigner-Weyl transform? Well, the W transform of \hat{\Omega}[f] is

(4)   \begin{align*} W\left[\hat{\Omega}[f]\right](x,p) &= \int \mathrm{d}\Delta x \, e^{i p \Delta x} \langle x + \frac{\Delta x}{2} \vert \hat{\Omega}[f] \vert x - \frac{\Delta x}{2} \rangle \\ &= \int \mathrm{d}\Delta x \int \mathrm{d}\alpha\, e^{i p \Delta x} f(\alpha) \langle x + \frac{\Delta x}{2} \vert \alpha \rangle \langle \alpha \vert x - \frac{\Delta x}{2} \rangle.\\ &= \int \mathrm{d}\alpha\, e^{-\left[\alpha -(x,p)\right]^2} f(\alpha).\\ &= (f \circ g) (x,p) \end{align*}

where \circ denotes the convolution and g(\alpha) = e^{-\alpha^2} is a Gaussian kernel. Another way to write this is to use the convolution theorem:

(5)   \begin{align*} \tilde{W}\left[\hat{\Omega}[f]\right](\xi) = \tilde{f}(\xi) \tilde{g}(\xi) \end{align*}

where the tilde denotes the Fourier transform, so that \tilde{g}(\xi) = e^{-\xi^2/4}, and \tilde{W} is the characteristic function associated with W.Usually the “symplectic Fourier transform” is used for the characteristic function, but that’s differs from the above only by reparameterizing as \xi=(\xi_1,\xi_2)\to(-\xi_2,\xi_1). Note also that the factors like the 4 in \tilde{g}(\xi) = e^{-\xi^2/4} are dependent on the choice of Fourier convention, and whether \hbar = 1/2,1, or 2.d  

So we see that the W transform of a “diagonal” operator like \hat{\Omega}[f] is just f smoothed by a convolution with the Gaussian kernel g. But that’s exactly what makes f the P transform of \hat{\Omega}[f].See Wikipedia for a summary of the connection between the W, P, and Husimi Q transforms.e   Indeed, the definition of \hat{\Omega}[f] given above is the same as the defining equation for a P transform for a density matrix, only generalized to operators that aren’t necessarily density matrices.

Now, all smooth operators on the Hilbert space will have W transforms associated with them, but not all such W transforms can be de-convolved to a smooth P transform. In particular, if we move to Fourier space, de-convolution is just multiplication by e^{+\xi^2/4} which, in general, will produce a function on which the inverse Fourier transform does not converge.There are ways to try construct a P transform in this case, but they are highly singular, involving infinite derivatives of Dirac delta functions. See Wikipedia and references therein.f   Operators that have smooth P transforms will have the nice expansion given above in terms of coherent states, and are reasonably interpreted as diagonal in this overcomplete basis. For technical discussion of smoothness and convergence issues, see Bonifacio et al. and Sudarshan.

Proving that this implies all the sensible properties expected of diagonality is left as an exercise…

Footnotes

(↵ returns to text)

  1. I posted the answer on Physics.SE too since they encourage the answering of one’s own question. How lonely is that?!?↵
  2. Case has a pleasingly gentle introduction.↵
  3. The integral on the right should be normalized by a factor of 1/2 \pi but I’m going to ignore normalization in everything I have to say.↵
  4. Usually the “symplectic Fourier transform” is used for the characteristic function, but that’s differs from the above only by reparameterizing as \xi=(\xi_1,\xi_2)\to(-\xi_2,\xi_1). Note also that the factors like the 4 in \tilde{g}(\xi) = e^{-\xi^2/4} are dependent on the choice of Fourier convention, and whether \hbar = 1/2,1, or 2.↵
  5. See Wikipedia for a summary of the connection between the W, P, and Husimi Q transforms.↵
  6. There are ways to try construct a P transform in this case, but they are highly singular, involving infinite derivatives of Dirac delta functions. See Wikipedia and references therein.↵
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