In field theory we most of deal with theories whose Lagrangian densities are of the form (sticking to scalar fields for simplicity) $$\mathcal{L}= -\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - \frac{1}{2}m_{\phi}^{2}\phi^{2} + \cdots$$ where ##\partial := \frac{\partial}{\partial x^{\mu}}## and the dots represent any interaction terms.(adsbygoogle = window.adsbygoogle || []).push({});

Given this, one normally proceeds to define a conjugate momentum, ##\pi_{\phi}##, conjugate to the field ##\phi##. Now, in all the introductory texts that I've read, it is defined as $$\pi_{\phi} = \frac{\partial\mathcal{L}}{\partial\dot{\phi}} = \frac{\partial\mathcal{L}}{\partial(\partial_{0}\phi)}$$ However, this seems to be an abuse of notation, since ##\dot{\phi}## and ##\partial_{0}\phi## are not equal in general (after the former is a total time derivative, and the latter is a partial derivative with respect to time).

This being the case, how is one meant to interpret this definition? Is the conjugate momentum of the field equal to the partial derivative of the Lagrangian with respect to the total time derivative of the field, i.e. ##\dot{\phi}##, or with respect to the partial derivative of the field with respect to time, i.e. ##\frac{\partial\phi}{\partial t}:=\partial_{t}\phi##?

If it is the case that the conjugate momentum of the field equal to the partial derivative of the Lagrangian with respect to the partial derivative of the field with respect to time, then I fail to see how the formulation is consistent, since in the free-field case ##\phi## can be expressed as $$\phi(x)=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{(2E_{k})}\left(a(\mathbf{k})e^{-ik\cdot x}+a^{\dagger}(\mathbf{k})e^{ik\cdot x}\right)$$ where ##k\cdot x= k^{0}t -\mathbf{k}\cdot\mathbf{x}##.

This clearly has no explicit time dependence in it, which leaves me confused as to what's the correct way to proceed?!

Is the point that when we transition to field theory, the Lagrangian density is actually a function of ##\phi##, ##\dot{\phi}=\frac{d\phi}{dt}## and ##\nabla\phi##, i.e. $$\mathcal{L}=\mathcal{L}(\phi, \dot{\phi},\nabla{\phi})$$ and we then abuse notation, symbolically writing ##\partial_{\mu}\phi =(\dot{\phi},\nabla\phi)##, such that $$\mathcal{L}=\mathcal{L}(\phi, \dot{\phi},\nabla{\phi})=\mathcal{L}(\phi,\partial_{\mu}\phi)$$ Then, for example, in the Euler-Lagrange equation $$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=\left(\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{\phi}}\right),\nabla\left(\frac{\partial\mathcal{L}}{\partial(\nabla\phi)}\right)\right)$$ if this is the case, then it all makes sense to me (I think).

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# A Time dependence of field operators

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