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Assume that we have a Lagrange functional $L = L(\psi, \partial_t\psi,\partial_x\psi)$ with $\psi:(x,t) \rightarrow \psi(x,t)$.

From the this I want to calculate the Hamiltonian. I was wondering how here the generalized impulses look like? For $L(q,\dot{q})$ it would simply be $p = \frac{\partial L}{\partial \dot{q}}$. How do the generalised impulses look like if I have a field that also depends on the x-coordinate? Do I need two generalised impulses, one for $\partial_x \psi$ and one for $\partial_t \psi$?

Xander Henderson
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physicsGuy
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1 Answers1

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If you think $\mathcal{L}=\mathcal{L}(\psi,\partial_\mu\psi)$ like in classical field theory, you can generalize the old $\delta S=0$ condition with coordinate ($q,\dot{q}$) to the field version $$ \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\psi}\delta\psi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\psi}\delta\partial_\mu\psi=0 $$ and using an integration by parts for the second therm you obtain the Euler-Lagrange equation $$ \frac{\partial\mathcal{L}}{\partial\psi}=\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\psi}. $$ If you define the momenta $p^\mu=\frac{\partial\mathcal{L}}{\partial\partial_\mu\psi} $, the Hamiltonian has the form $$ \mathcal{H}=(p^\mu\partial_\mu\psi-\mathcal{L})_{|\partial_\mu\psi=\partial_\mu\psi(p^\mu)} $$ as usual. Your generalized momenta are $$ p^0=\frac{\partial\mathcal{L}}{\partial\partial_0\psi}=\frac{\partial\mathcal{L}}{\partial\dot{\psi}} $$ and $$ p^i=\frac{\partial\mathcal{L}}{\partial\partial_i\psi}=\frac{\partial\mathcal{L}}{\partial\nabla\psi}, $$ so yes, one momenta for the time derivative and three momenta for spatial derivatives.

yngabl
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