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Why is the hamiltonian density defined as:

$$\mathcal{H}=\dot{\phi}\pi-\mathcal{L}$$

Where $\pi \equiv \frac{d\mathcal{L}}{d\dot{\phi}}$ and $\mathcal{L}(\dot\phi, \nabla \phi, \phi)$ is a function of the field and it's derivatives.

The thing that bothers me is that we favour the time coordinate over other coordinates such that we define $\pi \equiv \frac{d\mathcal{L}}{d\dot{\phi}}$ and not for example $\pi \equiv \frac{d\mathcal{L}}{d\partial_x \phi}$, which is what I would expect for a relativistic theory.

In case such thing exist, I hypothesize that it would be of the form:

$$\mathcal{H} = \partial_{\mu}\phi \frac{d\mathcal{L}}{d\partial_\mu \phi}-\mathcal{L}$$

Does such a thing exist? If not then what is the reason?

Habouz
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1 Answers1

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My question is why the hamiltonian density is defined as the first equation?

In classical mechanics of many particle systems: $$ \frac{\partial H}{\partial p_n} = \frac{d x_n}{dt} $$ and $$ \frac{\partial H}{\partial x_n} = -\frac{d p_n}{dt}\;, $$ where we see that time is singled out as special (since it is the only parameter really, the positions are dynamical variables).

In quantum mechanics the separation of time is even more prominent. Time is still a parameter and the positions are operators.

So in relativistic quantum mechanics or QFT (which I guess is what you are concerned with) we either have to promote time to an operator (which doesn't work for various reasons) or demote position to a parameter (which is what we do).

But, regardless, time still holds a special position in the Hamiltonian formalism, which is why we see the explicit time derivatives.

hft
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