In one part of Marc Bee's book on Quasielastic Neutron Scattering, he defines the pair correlation function

$$ G(\textbf r,t) = \frac{1}{(2\pi)^3}\int I(\textbf Q,t)\text e^{-i\textbf Q.\textbf r}\ d^3 Q $$

where $$ I(\vec Q,t) = \frac{1}{N}\sum_{i,j} \left <\text e^{i\textbf Q.\hat{\textbf R}_i(t)}\text e^{-i\textbf Q.\hat{\textbf R}_j(0)}\right > $$ where $\hat{\textbf R}_i$ are Heisenberg operators. Now using this he evaluates to obtain the $G(\textbf r,t)$

$$ G(\textbf r,t) = \frac{1}{N}\sum_{i,j}\int\left<\delta(\textbf r -\textbf r'+\hat{\textbf R}_i(0))\delta(\textbf r' - \hat{\textbf R}_j(t))\right> d^3r' $$

Now my question is how is this evaluation done ? Most importantly, can operators be arguments to delta function ?


Yes, they can. It's not great form, but they can.

For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$, $$ \hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}). $$ This holds equally well if the potential is a delta function at $x_0$, $V(x)=\kappa \delta(x-x_0)$, in which case $$ \hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0). $$ Functions of an operator are usually defined by transplanting its power series, but the way it's really done is by acting on the eigenvalues, so for any potential $V(x)$ you have $$ V(\hat x)|x⟩=V(x)|x⟩ $$ regardless of whether the function $V(x)$ is analytic or not - including such "ugly" beasts as discontinuous potentials. The same holds true for our delta function: $$ \delta(\hat x-x_0)|x⟩=\delta(x-x_0)|x⟩, $$ though of course this is still problematic because the delta function is not actually a function. To get to what this really means, you apply the definition of the delta function: when integrated against an arbitrary function $f(x)$ it returns a specific value. Thus, you do $$ \int\mathrm d x\: f(x)\delta(\hat x-x_0)|x⟩=\int\mathrm d x\: f(x)\delta(x-x_0)|x⟩=f(x_0)|x_0⟩. $$ That is more OK, but it might still worry you. If you look closely, our operator $\delta(\hat x-x_0)$ has turned what might have been a perfectly reasonable state $\int f(x)|x⟩\mathrm dx$ into the singular state $f(x_0)|x_0⟩$, which isn't really a state at all - it sits outside of the normal Hilbert space. The way to resolve that is not to look at the eigenkets but at the eigenbras of the position operator, which is actually how functions of the position operator should work, $$ ⟨x|V(\hat x)=V(x)⟨x| $$ (because then the wavefunction of $V(\hat x)|\psi⟩$ is $⟨x|V(\hat x)|\psi⟩=V(x)⟨x|\psi⟩=V(x)\psi(x)$). Applying this to the delta function, the real definition of $\delta(\hat x-x_0)$ is that operator for which $$ \bbox[5px,border:1px solid black]{\int\mathrm d x\: f(x)⟨x|\:\delta(\hat x-x_0)=f(x_0)⟨x_0|} $$ for arbitrary (sufficiently well-behaved) functions $f$.

(In mathematical terms, this is no longer worrisome, as it takes linear functionals over $\mathcal H$, like $\int\mathrm d x\: f(x)⟨x|$, into other linear functionals, like $f(x_0)⟨x_0|$. Our worry from before (that $|x_0⟩$ wasn't a state) has now been de-fanged: the argument says that $\delta(\hat x-x_0)$ can take a bounded functional and return a non-bounded functional. That's perfectly fine - $\hat x$ itself does the same.)

In the example you quote it is hard to know what is and is not an operator without further context, but nothing looks like it needs to be an operator for the calculation to work. Unless you provide further context, all quantities you mention are perfectly valid c-numbers (i.e. not operators).

To do the calculation, you simply substitute for $I$ and invert the order integration and expectation value to get $$ G(\vec r,t) = \frac{1}{N}\sum_{i,j} \left < \frac{1}{(2\pi)^3} \int e^{i\vec Q·(\vec R_i(t)-\vec R_j(0)-\vec r)}\ d^3 \vec Q \right >. $$ The integral evaluates to a single delta function in $\vec r$: $$ G(\vec r,t) = \frac{1}{N}\sum_{i,j} \left < \delta\left(\vec R_i(t)-\vec R_j(0)-\vec r\right) \right >. $$ This coincides with the other result you give once you cancel out the $\vec r'$ integral, using the $\delta(\vec r' - \vec R_j(t))$ term, to substitute $\vec R_j(t)$ for $\vec r'$. It's hard to know why the author did this; presumably he had reasons to choose the form you give. Without more context it's hard (impossible) to tell.

  • $\begingroup$ My bad, I have edited the question to make it more evident of what are the operators there with a hat on those variables. I think with that thing there, I can't write it as a sum over the exponential as they don't commute. $\endgroup$ – user35952 Aug 13 '15 at 2:43
  • $\begingroup$ That's again not quite enough context to tell what the fix is, but there does seem to be done non kosher manipulation going on. $\endgroup$ – Emilio Pisanty Aug 13 '15 at 8:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.