# Functional derivative and units

The both sides of below equation don't give the same units, e.g. $$\frac{\delta}{\delta \phi (\tau)}\int_a^b \phi (\tau') d\tau'=1\;.$$ where $a<\tau<b$. To show this assume that the field $\phi$ has "$j$" unit and $\tau$ has "$s$" unit then $$[\frac{1}{j} j s] \neq [ 1 ].$$ What's wrong with my reasoning?

-
The identity looks wrong. Or maybe the $1$ in RHS is not dimensionless, i.e. It might be giving the value $1$ for some specific $\phi (\tau)$, which doesn't imply it is dimensionless. – udiboy1209 Aug 8 '13 at 9:53
Where did you find this equation? In what context? – Ali Aug 8 '13 at 9:58
This equation I wrote as an example of more complex case. Especially I think that sometimes people forget about planck constant in the action $e^{-S/\hbar}$ but they don't forget about it in another places of the same given equation (when there is no assumption that $\hbar=1$). – memat Aug 8 '13 at 10:06

The integral has units $js$ as you write, using your notation, but the functional derivative has the compensating units $1/(js)$ so the units cancel.
To see that dimension of the functional derivative is $1/(js)$, one may use a more general "defining" identity for the functional derivatives $$\frac{\delta}{\delta \phi(\tau)} \phi(\tau') = \delta(\tau-\tau')$$ which is the continuous-index counterpart of $\partial / \partial x_i (x_j) = \delta_i^j$. The units of the functional derivative would be just $1/j$ – canceling that of $\phi$ itself – if the delta-function on the right hand side were dimensionless, like the Kronecker delta. But it's not. The delta-function may be written as the $\tau$-derivative of the dimensionless step function (zero for negative arguments, one for positive arguments), so its units are $1/s$ (the inverse units to the units of the argument, in this case $\tau$), which means that the units of $\delta / \delta \phi(\tau)$ have an extra $1/s$ as well, to get the total units $1/(js)$.