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I have a QFT homework to do and there I should show for a given lagrangian

$\mathscr{L} = C_1 (\partial_{\nu} A_{\mu}) (\partial^{\nu} A^{\mu}) + C_2 (\partial_{\nu} A_\mu) (\partial^{\mu} A^\nu) + C_3 A_\mu A^\mu$

That the hamiltonian

$H = \int d^3x (\pi_\nu \partial_0A^\nu - \mathscr{L})$ with $\pi_\nu = \frac{\partial \mathscr{L}}{\partial \partial_0 A^\nu}$

is unbound from above and below if $C_3$ is not 0.

So I wanted to ask, how can one prove a statement like this?

Best Regards

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The Hamiltonian will be of the form $C^{\mu\nu\rho\sigma}\partial_\mu A_\nu \partial_\rho A_\sigma -C_3 A_\mu A^\mu$ for some rank-$4$ tensor $C^{\mu\nu\rho\sigma}$ you can work out. The last term is the potential energy in the Hamiltonian. Since $A_\mu A^\mu$ is unbounded, so is $-C_3 A_\mu A^\mu$.

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  • $\begingroup$ Oh Thanks so for the momentum I have something like $\pi_{\rho} = 2 C_1 \partial_0 A^{\rho} + 2 C_2 \partial_{\rho}A^0$ If I put that into the Hamiltonian equation, how do I integrate it then? $\endgroup$ – Armani42 Apr 21 '18 at 9:41
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    $\begingroup$ @Armani42 You don't need to compute the integral for this problem; the form of the Hamiltonian density is enough to explain the unboundedness. The other obvious use of the Hamiltonian, which you might need for something else, is to study on-shell behaviour. For that you'll need to ensure you remove each $\dot{A}^\mu$ using the formula for $\pi_\rho$. $\endgroup$ – J.G. Apr 21 '18 at 10:22
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    $\begingroup$ BTW it should be $\pi_\rho =2C_1\partial^0 A_\rho + 2C_2\partial_\rho A^0$. $\endgroup$ – J.G. Apr 21 '18 at 10:23
  • $\begingroup$ Yeah sorry I am sitting in the bus so I don't have my notices by me :) but thank you very much! $\endgroup$ – Armani42 Apr 21 '18 at 10:25

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