Why is the Lagrangian density for fields treated as a functional in QFT?

[spookyfish]

This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field,

$$\partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0$$

i.e. $\cal L$ is treated like a functional (seen from the $\delta$ symbol). But why would it be a functional? Functonals map functions into numbers, and in our case $\cal L$ is a function of the fields (and their derivatives).

 

[fzero]

If you want to be completely rigorous, the action is the true functional. The variational derivatives of the Lagrangian (density) should be considered distributions.

 

[spookyfish]

But why should there be a functional derivative of $\cal L$? we have $\cal L$ which is a function of $(\phi, \partial_\mu \phi)$ and we differentiate (as a function) with respect to $\partial_\mu \phi$

 

[Bill_K]

This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field,

$$\partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0$$

i.e. $\cal L$ is treated like a functional (seen from the $\delta$ symbol). But why would it be a functional? Functonals map functions into numbers, and in our case $\cal L$ is a function of the fields (and their derivatives).

If they wrote it that way it's a misprint. The derivatives should be ∂'s, not δ's.

 

[fzero]

But why should there be a functional derivative of $\cal L$? we have $\cal L$ which is a function of $(\phi, \partial_\mu \phi)$ and we differentiate (as a function) with respect to $\partial_\mu \phi$

It's common to abuse the notation and use $\delta$ for these derivatives in order to distinguish them from the coordinate derivatives $\partial_\mu$.