# Real Scalar Field, Hamiltonian, Conjugate Momentum

1. Oct 29, 2016

### binbagsss

$L(x) = L(\phi(x), \partial_{u} \phi (x) ) = -1/2 (m^{2} \phi ^{2}(x) + \partial_{u} \phi(x) \partial^{u} \phi (x))$ , the Lagrange density for a real scalar field in 4-d, $u=0,1,2,3 = t,x,y,z$, below $i = 1,2,3 =x,y,z$

In order to compute the Hamiltonian I first of all need to compute the conjugate momentum:

$\Pi (t, x) = \frac{\partial L}{\partial \dot{\phi (x)}}$

I can see it's coming from the second term : $- \partial_{u} \phi(x) \partial^{u} \phi (x)) = - \partial_{0} \phi(x) \partial^{0} \phi (x)) - \partial_{i} \phi(x) \partial^{i} \phi (x))$, where I'm only interested in $\partial_{0} \phi(x) \partial^{0} \phi (x))$,

But I'm unsure how to deal with the one upper and one lower index.

The Hamiltonian is $H = \int d^{3} x \dot{\phi} \Pi - L(t, {x}) = \int d^{3} x (1/2m^{2}\phi^{2} + 1/2( \partial_{i} \phi )^{2} + 1/2\dot{\phi^{2}})$

I can see I clearly need to lower an index. So if do $g_{\alpha u} \partial ^{\alpha} \phi \partial_{u} \phi = (\partial _{u} \phi )^{2} = (\partial_{0} \phi(x))^{2} - (\partial_{i} \phi(x))^{2}$ , I get the correct answer that $\Pi = \dot{\phi}$

however then surely I have found $g_{\alpha u} \Pi$ and subbed in $g_{\alpha u} L$ in H, as a pose to $L$

Last edited: Oct 29, 2016
2. Oct 30, 2016

### binbagsss

anyone?

I believe the following is related:

Showing that $\partial^{u}\alpha(\phi^*\partial_{u}\phi)=\partial_{u}\alpha(\phi^*\partial^{u}\phi)$

I don't really know how to approach this, since I need to raise one index and lower the other, but they are the same index so I can't use the obvious like $g_{ab}x^{b}=x_{a}$ etc.

3. Nov 2, 2016

### binbagsss

it's okay got it , thanks for the help guys