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I am unable to derive the Hamiltonian for the electromagnetic field, starting out with the Lagrangian $$ \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-\frac{1}{2}\partial_\nu A^\nu \partial_\mu A^\mu $$ I found: $$ \pi^\mu=F^{\mu 0}-g^{\mu 0}\partial_\nu A^\nu $$ Now $$ \mathcal{H}=\pi^\mu\partial_0 A_\mu-\mathcal{L} $$ Computing this, I arrive at: $$ \mathcal{H}=-\frac{1}{2}\left[\partial_0 A_\mu\partial_0 A^\mu+\partial_i A_\mu\partial_i A^\mu\right]+\frac{1}{2}\left[\partial_i A_i\partial_j A_j-\partial_j A_i\partial_i A_j\right] $$ The right answer, according to my exercise-sheet, would be the first to terms. Unfortunately the last two terms do not cancel. I have spent hours on this exercise and I am pretty sure, that I did not commit any mistakes arriving at this result as I double checked several times. My question now is: did I start out right and am I using the right scheme? Is this in principle the way to derive the Hamiltonian, or is it easier to start out with a different Lagrangian maybe using a different gauge? Any other tips are of course also welcome. Maybe the last two terms do actually cancel and I simply don't realize it. Texing my full calculation would take a very long time so I am not going to post it, but as I said, it should be correct. But if everything hints at me having committed a mistake there, I will try again.

Edit:

After reading and thinking through Stephen Blake's answer, I realized that one can get rid of the last two terms in $H$, even though they do not vanish in $\mathcal{H}$. This is done by integrating the last term by parts and dropping the surface term, leaving $A_i\partial_j\partial_i A_j$. One can now proceed to combine the last two terms: $$ \partial_i A_i\partial_j A_j+A_i\partial_j\partial_i A_j=\partial_i(A_i\partial_j A_j) $$ This can be converted into a surface integral in $H$ which can be assumed to vanish, leaving us with the desired "effective" $\mathcal{H}$.

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Why do you think the Lagrangian should be that to begin with? Shouldn't it be $- F^{\mu \nu} F_{\mu \nu} / 4 - J^\mu A_\mu$? –  Eric Angle Jan 3 at 19:06
    
Possible duplicate: physics.stackexchange.com/q/90224/2451 –  Qmechanic Jan 3 at 20:25
    
@EricAngle We were working with a source-free EM field in Lorentz gauge, which allowed us to add the second term to the Lagrangian. I do, however, not quite understand why we did it. –  user35915 Jan 4 at 8:00
    
I see. Since the Lorentz gauge has $\partial_\mu A^\mu = 0$, I guess you're free to add any function of $\partial_\mu A^\mu$ to the Lagrangian. –  Eric Angle Jan 4 at 12:41
    
Or, at least a power of $\partial_\mu A^\mu$. –  Eric Angle Jan 4 at 13:57

1 Answer 1

up vote 1 down vote accepted

There doesn't appear to be anything wrong with user35915's calculation. However, in order to get the desired answer, the canonical momenta needs to be different. Starting from user35915's action, $$ S=\int d^{4} x\left( -\frac{1}{4}F_{\mu\lambda}F^{\mu\lambda}-\frac{1}{2}A^{\mu}_{,\mu}A^{\lambda}_{,\lambda}\right) $$ change the second term by integrating by parts and chuck the surface term away to get, $$ S=-\frac{1}{4}\int d^{4} x( F_{\mu\lambda}F^{\mu\lambda}-2A^{\mu}A^{\lambda}_{,\lambda\mu}) \ . $$ Now expand the electromagnetic field tensor $F_{\mu\lambda}=A_{\lambda,\mu}-A_{\lambda,\mu}$ and do a bit of swopping dummy indices to get, $$ S=-\frac{1}{2}\int d^{4}x \eta^{\mu\rho}\eta^{\lambda\sigma}(A_{\lambda,\mu}A_{\sigma,\rho}-A_{\lambda,\mu}A_{\rho,\sigma}-A_{\rho}A_{\lambda,\sigma\mu})\ . $$ The last two terms can be combined into a surface integral which vanishes at infinity and the final form of the action is only the first term in the last line. The Lagrangian is now, $$ L=-\frac{1}{2}\int d^{3}x \eta^{\mu\rho}\eta^{\lambda\sigma}A_{\lambda,\mu}A_{\sigma,\rho}=-\frac{1}{2}\int d^{3}x A_{\mu,\lambda}A^{\mu,\lambda}\ . $$ The reason for getting the Lagrangian in this form is because it looks like the Lagrangian for four scalar fields. The canonical momenta are now, $$ \pi^{\mu}=-A^{\mu}_{,0}=-\frac{\partial A^{\mu}}{\partial t} $$ which look like the momenta for four scalar fields. Now, it's straightforward to go over to the desired Hamiltonian, $$ H=-\frac{1}{2}\int d^{3}x (\pi^{\mu}\pi_{\mu}+A^{\mu}_{,r}A_{\mu,r}) $$

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If I might add one thing to this, it's that there is an ambiguity for the momenta. It is perfectly possible to have momentum as a generator of flows in one of the four independent parameters $(t, \vec{x})$. Marsden did some mathematical work on "multisymplectic field theories" in the past, but I don't know if it's ever really been translated from math-talk to physics-talk. –  webb Jan 3 at 22:58
    
@webb: I think that it's not an ambiguity in the momenta, the two momenta are, I guess, related by a canonical transformation (symplectomorphism). –  Stephen Blake Jan 4 at 0:40
    
@StephenBlake Thank you very much. I guess that is how we were supposed to solve the exercise. A hint might have been helpful...after reading your answer, I realized, that the two terms troubling me can also be combined into a (spacial) surface integral which vanishes in $H$ so I might as well leave them out of $\mathcal{H}$. –  user35915 Jan 4 at 7:43

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