# Poisson Bracket for 1 space dimension field

1. ### Bobdemaths

3
Hi,

Suppose you have a collection of fields $\phi^i (t,x)$ depending on time and on 1 space variable, for $i=1,...,N$. Its dynamics is defined by the Lagrangian

$L=\frac{1}{2} g_{ij}(\phi) (\dot{\phi}^i \dot{\phi}^j - \phi ' ^i \phi ' ^j ) + b_{ij}(\phi) \dot{\phi}^i \phi ' ^j$

where $\dot{\phi}^i$ denotes the time derivative of the field ${\phi}^i$ and $\phi ' ^i$ denotes its space derivative, and where $g_{ij}(\phi)$ is a symmetric tensor, and $b_{ij}(\phi)$ an antisymmetric tensor.

One easily computes that the momenta conjugate to the fields $\phi^i (t,x)$ are $\pi_i = A_i + b_{ij} \phi ' ^j$, where $A_i = g_{ij} \dot{\phi}^j$.

Now I would like to show that the (equal time) Poisson Bracket $\{A_i,A_j\}$ is
$\{A_i(t,x),A_j(t,y)\}=(\partial_i b_{jk} + \partial_j b_{ki} + \partial_k b_{ij} ) \phi ' ^k \delta(x-y)$
using the canonical relation $\{\phi ^i(t,x) , \pi_j (t,y)\}=\delta_j^i \delta(x-y)$.

I tried to write $A_i = \pi_i - b_{ij} \phi ' ^j$, and then use $\{\phi ' ^i(t,x) , \pi_j (t,y)\}=\delta_j^i \delta ' (x-y)$. But then I can't get rid of the $\delta '$, and I don't get the $\partial_k b_{ij}$ term.

Am I mistaken somewhere ? Thank you in advance !

2. ### Bobdemaths

3
OK I found the answer. To get rid of the derivatives of Dirac deltas, I used the identity $f(x) \delta ' (x-y) = f(y) \delta ' (x-y) - f'(x) \delta (x-y)$ which follows from the convolution between delta and $(f \times \varphi)'$ ($\varphi$ is a test function).
This also produces the remaining wanted term.