I would like to show that a given solution really is a solution to a PDE. The discussion of this is from a book "Quanum Noise" by Gardiner and Zoller (around page 125).

The partial differential equation is (I've taken $\hbar$ to be 1):

$$\frac{\partial P(\alpha, \alpha^{*}, t) }{ \partial t} = i \left( -\omega \frac{\partial}{\partial \alpha} \alpha + \omega \frac{\partial}{\partial \alpha^{*}} \alpha^{*} -\lambda \frac{\partial}{\partial \alpha} + \lambda^{*} \frac{\partial}{\partial \alpha^{*}} \right) P(\alpha, \alpha^{*}, t) $$

It is "shown" that a solution to it is

$$P(\alpha, \alpha^{*}, t) = \delta^{(2)}[\alpha - \beta(t)] = \delta(Re(\alpha) - Re(\beta(t))) \delta(Im(\alpha) - Im(\beta(t)))$$

where $\alpha$ satisfies:

$$\dot \beta = i \left( \omega \beta + \lambda(t) \right)$$

I would like to be able to show that the function above for $P(\alpha, \alpha^{*}, t)$ written in terms of the $\delta$ functions really satisfies the PDE by directly substituting it in, but I can't quite do it. I've tried various things, and in particular some properties of the delta function like

$$\int \alpha \frac{\partial}{\partial \alpha} \delta(\alpha) d\alpha = - \int \delta(\alpha) d\alpha$$

but can't get RHS to agree with the LHS.

Does anyone know how one would do this?


  • $\begingroup$ I'm a little confused here. $\alpha - \alpha(t)$... so one is a constant, one is a function of $t$, but both have the same symbol? $\endgroup$ Aug 26, 2014 at 2:56
  • $\begingroup$ This is exactly how things are presented in the book, so for the sake of consistency I originally had it written the same way as them - I've now updated the question by renaming $\alpha(t)$ to $\beta(t)$ $\endgroup$ Aug 26, 2014 at 3:04

1 Answer 1


Lets check by direct computation. Let $\phi$ be some test function and consider $\partial_t P$ in the sense of distributions.

\begin{align*} (\partial_t P, \phi) = & - (P , \partial_t \phi) \end{align*}

Lets check the RHS first, since $P = \delta^{(2)} ( \alpha - \beta (t) )$, we have that

$$ - ( P, \partial_t \phi ) = - (\Im \partial_t \phi)( \alpha) (\Re \partial_t \phi)(\alpha) $$

where $\Im$ is the imaginary part, and $\Re$ is the real part. Now lets check the LHS by plugging in the PDE. $$(\partial_t P, \phi) = (- i\omega \partial_\alpha \alpha P, \phi) + (i \omega \partial_{\alpha^*} \alpha^* P, \phi ) - ( i \lambda \partial_\alpha P , \phi) + ( i \lambda^* \partial_{\alpha^*} P, \phi)$$

We can simplify things via product rule

\begin{align*} (\partial_t P, \phi) =& (-i \omega \alpha \partial_{\alpha}P, \phi) + (i \omega \alpha^* \partial_{ \alpha^*} P, \phi) - ( i \lambda \partial_\alpha P , \phi) + ( i \lambda^* \partial_{\alpha^*} P, \phi) \\ =& -([i \omega \alpha + i \lambda] \partial_a P , \phi ) + ([ i \omega \alpha^* + i \lambda ] \partial_{\alpha^*} P, \phi) \\ =& -( \partial_t \alpha \partial_\alpha P, \phi ) + ( \partial_t \alpha^* \partial_ {\alpha^*}P, \phi )\\ \end{align*}

Note that $$P (\alpha, \alpha^*,t) = \delta \left ( \frac{\alpha + \alpha^*}{2} - \frac{ \beta(t) + \beta^*(t)}{2} \right ) \delta \left ( \frac{\alpha - \alpha^*}{2i} - \frac{ \beta(t) - \beta^*(t)}{2i} \right ) $$

Now it's just a derivative calculation. Note that $( \delta' ,\phi) = -\phi'(0)$ (Sorry about switching between $\alpha$ and $\beta$ :s)

  • $\begingroup$ Thanks for taking the time to answer! Do you mind posting the last few lines? In particular how the ode for $\beta$ comes out of this PDE? - I'm not quite seeing it. Also, I was wondering if it's always necessary in a case like this to invoke the use of a test function? Thanks again. $\endgroup$ Aug 27, 2014 at 0:02
  • $\begingroup$ If you're working with distributions, then we have to use test functions! :P $\endgroup$
    – Jeb
    Aug 27, 2014 at 3:43

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