Vector Delta Function Identity

I'm trying to prove the the vector extension of the identity $$1 = \int \left|\sum_i\frac{ \partial g }{ \partial a }\big| _{a =a _i} \right| \delta ( g ( a ) ) da$$ where the sum is over all the zeros of $g$. The vector extension is: $$1 = \left( \prod _{i = 1 } ^n \int d a _i \right) \delta ^{ ( n ) } \left( {\mathbf{g}} ( {\mathbf{a}} ) \right) \det \left( \frac{ \partial g _i }{ \partial a _j } \right)$$ where ${\mathbf{g}}$ and ${\mathbf{a}}$ are $n$ dimensional vectors. This identity is used in Quantum Field Theory by Peskin and Schroeder (pg. 295).

I'm a physics graduate student and I don't know much formal mathematics. Any help on how to prove this would be greatly appreciated.

• The 1-dimensional version doesn't seem to work for $g(a)=-a$, since $\delta(-a) = \delta(a)$. Also the $n$-dimensional case needs a sum over roots too I think. – Bob Terrell Feb 12 '14 at 21:41
• @BobTerrell, Good point. I forgot an absolute value sign, thanks. I believe it should be okay now. I think the sum over the roots turns somehow into the determinate but I don't know how. The n-dimensional case expression I quoted agrees with the book I'm using. I think its correct. – JeffDror Feb 12 '14 at 21:48
• These are all simple roots, too, so none of them would have a multiplicity greater than 1... – Alex Nelson Feb 12 '14 at 22:28
• You ask about moving this question to phys.SE. It is not possible to move a question while a bounty is in progress. If at the end of the bounty you still haven't received a suitable answer, mention this again, and we'll see what we can do. – davidlowryduda Feb 17 '14 at 4:50
• Thanks, I didn't know how it worked. I will do that! – JeffDror Feb 17 '14 at 10:48

Here is a partial answer, in the case of a single root $\mathbf{a}_1$.

For each number $h>0$ you can define a function $d(\mathbf{x})$ to be zero outside the sphere $|\mathbf{x}|=h$, and 1/volume inside. When $h$ is small, $d$ is close to $\delta$ as distributions. Also, you can use an ellipsoid rather than a sphere.

$\mathbf{g}(\mathbf{a})$ is in such a disc if $\mathbf{a}$ is close enough to a zero of $\mathbf{g}$. Specifically, $$\mathbf{g}(\mathbf{a}) = D\mathbf{g}(\mathbf{a}_1)(\mathbf{a}-\mathbf{a}_1)$$ nearly, when $\mathbf{a}$ is near $\mathbf{a}_1$, where $D\mathbf{g}(\mathbf{a}_1)$ means the Jacobian matrix $\left(\frac{\partial g _i}{\partial a _j}\right)$ evaluated at the root.

The matrix maps any small sphere centered at $\mathbf{a}_1$ to some ellipsoid centered at $\mathbf{0}$, and the ellipsoid volume is $|\det(D\mathbf{g}(\mathbf{a}_1))|$ times the sphere volume. So, $$\delta^{(n)} \left( \mathbf{g}(\mathbf{a}) \right)$$ is well approximated by $$\frac{1}{|\det(D\mathbf{g}(a_1))|({\rm sphere vol.})}$$ when $\mathbf{a}$ is in any small sphere centered at $\mathbf{a}_1$ and zero outside the sphere. Then $$\delta^{(n)} \left( \mathbf{g}(\mathbf{a}) \right) |\det(D\mathbf{g}(\mathbf{a}_1))|$$ is nearly 1/sphere volume when $\mathbf{a}$ is inside, and 0 if outside. That is, it is nearly $\delta(\mathbf{x}-\mathbf{a}_1)$.

I don't see how to handle more than one root and I don't see why your authors don't have absolute values on the derivative.

• Thanks, this is more formal then I'm used to but I'll study your answer in detail soon. I see where you're coming from with regards to the absolute value signs... Maybe their is a typo. – JeffDror Feb 13 '14 at 3:29

The proof below uses only the chain rule (to change integration variables). It's what I, as a physics student, would write.

But first, I believe that you have an error in your original posting. I believe that the 1D identity should be: \begin{align} \int \Big( \sum_{\substack{\textrm{roots $a_i$}\\ \textrm{ of $g$}}} \frac{1}{g'(a)} \Big)^{-1} \delta(g(a)) da = 1. \end{align}

I know that you asked about the multidimensional case. However, let's review the one-dimensional case first. They are extremely similar.

One-dimensional Case

\begin{align} \int_{\substack{\textrm{small $a$-region} \\ \textrm{ containing root $a_i$}}} \delta (g(a)) da = &\int_{\substack{\textrm{small $g$-region} \\ \textrm{ corresponding to} \\ \textrm{ root $a_i$}}} \delta(g) \frac{1}{g'(a)} dg \end{align} We knew $g$ as a function $a$ (i.e., $g(a)$), but we can also think of $a$ as a function of $g$. This works because in the small region near the root, $g(a)$ is invertible. So we can think of $\frac{1}{g'(a)}$ as a function of $g$. The delta function extracts the value of this function near $g=0$. But in the small region we care about, $g=0$ means $a=a_i$. Thus: \begin{align} \int_{\substack{\textrm{small $a$-region} \\ \textrm{ containing root $a_i$}}} \delta (g(a)) da = \frac{1}{g'(a_i)} \end{align} Integrating over all space instead of a small region gives: \begin{align} \int \delta(g(a)) da = \sum_{\textrm{roots $a_i$ of $g$}} \frac{1}{g'(a_i)} \end{align} which is a special case of an identity you can find at the Wikipedia article on delta functions

We can rewrite this to obtain: \begin{align} \int \Big( \sum_{\substack{\textrm{roots $a_i$}\\ \textrm{ of $g$}}} \frac{1}{g'(a)} \Big)^{-1} \delta(g(a)) da = 1. \end{align} Which is (except for the inversion) the 1D identity that you had. You can bring the factor $\left( \sum_{\textrm{roots$a_i$of$g$}} \right)^{-1}$ inside the sum because it's just a constant.

Multidimensional case The only difference in the multidimensional case is that when you change variables you need to use the Jacobian determinant (which Peskin and Schroeder call $\frac{\partial g_i }{\partial a_j}$). So, instead of $da = \frac{1}{g'(a)} dg$, we have \begin{align} \prod_k d a_k = \frac{1}{\det{ \frac{\partial g_i }{\partial a_j} } } \prod d g_k \end{align} Note that $\frac{\partial g_i }{\partial a_j}$ really stands for the matrix whose $ij$ entry is $\frac{\partial g_i }{\partial a_j}$''.

Hence the final identity from before becomes: \begin{align} \left( \prod_k \int d a_k \right) \Big( \sum_{\substack{\textrm{roots $\vec{a}_i$}\\ \textrm{ of $g$}}} \frac{1}{\det{ \frac{\partial g_i }{\partial a_j} } } \Big)^{-1} \delta^{(n)}(\vec{g}(\vec{a})) = 1. \end{align} where $\vec{a}_i$ are the roots of $\vec{g}(\vec{a})$ and each determinant is evaluated at the root $\vec{a}_i$.

If there's only one root , then the sum only has one term, and so inverting the terms is undone by inverting the sum: \begin{align} \left( \prod_k \int d a_k \right) \det{ \frac{\partial g_i }{\partial a_j} } \delta^{(n)}(\vec{g}(\vec{a})) = 1. \end{align} [Which is the identity on p295 of Peskin and Schroeder.] The fact that these two inversions cancel is nice, but I think it is what led you to the error in your original post.

• Perfect! Thanks for this very clear proof. You are right that this is exactly what I had in mind. I spoke with BobTerrell in person yesterday as we coincidentally work at the same university and I understood his answer first so I awarded the bounty to him, but thanks again! – JeffDror Feb 20 '14 at 14:37