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A Time dependence of field operators

  1. Jun 7, 2016 #1
    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.

    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).
     
    Last edited: Jun 7, 2016
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  3. Jun 7, 2016 #2

    Orodruin

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    I do not think this is much of an abuse. How would you define the total derivative? You would need the field along a path, which is generally not what you have. In field theory we often simply define our notation as ##\dot\phi = \partial\phi/\partial t##.
     
  4. Jun 7, 2016 #3
    Yes, I appreciate that it's not much of an abuse of notation, but I just want to check that I've understood it all correctly. Would what I put in my last paragraph be correct at all?

    My problem with it is that ##\partial_{t}\phi## will be zero unless ##\phi## has explicit time dependence, however, if it has implicit time dependence, then ##\frac{d\phi}{dt}## (which classical one would denote as ##\dot{\phi}## would be non-zero.
     
  5. Jun 7, 2016 #4

    Orodruin

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    The last expression in your OP does not make sense. You have written that a divergence, which is a scalar, is equal to some sort of vector.

    You will generally not see d/dt in field theory. I do not think there is any room for confusion.

    So what if ##\phi## is not time dependent? This should be determined from the field equations.
     
  6. Jun 7, 2016 #5
    Yes, apologies for that. I meant to write $$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)= \frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{\phi}}\right) + \nabla_{i}\left(\frac{\partial\mathcal{L}}{\partial(\nabla_{i}\phi)}\right)$$
    Where ##\nabla_{i}=\frac{\partial}{\partial x^{i}}=\partial_{i}##

    Fair enough, but how then does one determine the conjugate momentum ##\pi =\frac{\partial\mathcal{L}}{\partial(\partial_{0}\phi)}## of a field ##\phi## expressed as $$\phi(x)=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{(2E_{k})}\left(a(\mathbf{k})e^{-ik\cdot x}+a^{\dagger}(\mathbf{k})e^{ik\cdot x}\right)$$ Am I just being silly here and the explicit time dependence is contained in the exponentials ##e^{\pm ik\cdot x}##, i.e. ##\partial_{0}(e^{\pm ik\cdot x})=\pm ik^{0}te^{\pm ik\cdot x}##?
     
  7. Jun 8, 2016 #6

    vanhees71

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    I don't understand the confusion. A field is a function of space and time arguments which are independent variables. In the Lagrange density you have the fields and their first partial derivatives (or covariant derivatives in General Relativity) as independent variables.

    Then you can transform to the Hamiltonian formulation with the canonical field momenta defined as
    $$\Pi=\frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}$$
    and quanize the theory assuming the canonical equal-time commutators (or anti-commutators) for bosons (fermions). This leads to the time dependence according to the Heisenberg picture of the time evolution (i.e., operators defining observables carry the full time dependence.
     
  8. Jun 8, 2016 #7

    Apologies, I think I was just being really stupid here. Of course, in the Heisenberg picture the field operators have explicit time dependence (due to the explicit time dependence contained in the evolution operator ##\hat{U}(t)=e^{-i\hat{H}t}##).

    Is what I put in the last part of my previous post correct then (i.e. the explicit time dependence in ##\hat{\phi}(x)## is contained in the exponentials, such that ##\partial_{0}(e^{\pm ik\cdot x})=\pm ik^{0}e^{\pm ik\cdot x}##)?!
     
  9. Jun 8, 2016 #8

    vanhees71

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    It's much simpler. For the KG field you wrote you have (with your sign convention, i.e., east coast)
    $$\mathcal{L}=-\frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi+V(\phi)=\frac{1}{2} (\dot{\phi}^2-(\vec{\nabla} \phi)^2)+V(\phi).$$
    Then the canonical field momentum is
    $$\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}.$$
    For the free field you get the momentum-mode decomposition you've written down, and the space-time dependence is of course in the exponentials.
     
  10. Jun 8, 2016 #9
    Yes, apologies. That is what I was trying to get at, but in a bit of a round-about way :sorry:
     
  11. Jun 8, 2016 #10
    Just a quick follow-up question (which is I guess essentially what my original question boils down to), do the field operators ##\hat{\phi}(x)## in the Heisenberg picture evolve as $$ \dot{\hat{\phi}}(x) = i\left[\hat{H},\hat{\phi}(x)\right] + \partial_{t}\hat{\phi}(x)$$ (in units where ##\hbar = c = 1##), or as $$ \dot{\hat{\phi}}(x) = i\left[\hat{H},\hat{\phi}(x)\right]$$ If the latter is true, what is the reasoning for why the field operator ##\hat{\phi}(x)## (and its conjugate momentum ##\hat{\pi}_{\phi}(x)##) contain no explicit time dependence (i.e. why we can take ##\partial_{t}\hat{\phi}(x) = 0##)? Is it simply because, quantizing in the Schrödinger picture, the field operators have no time dependence and thus, since $$\partial_{t}\hat{\phi}_{H}(t,\mathbf{x}) =\hat{U}^{\dagger}(t)\partial_{t}\hat{\phi}_{S}(\mathbf{x})\hat{U}(t)$$ where the subscripts denote the two different pictures. Clearly, ##\partial_{t}\hat{\phi}_{S}(\mathbf{x})=0## and so ##\partial_{t}\hat{\phi}_{H}(t,\mathbf{x}) =0##.
     
    Last edited: Jun 8, 2016
  12. Jun 9, 2016 #11

    vanhees71

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    By definition the fields obey
    $$\dot{\hat{\phi}}(x) = \mathrm{i} \left[\hat{H},\hat{\phi}(x)\right].$$
    Explicit time dependence comes in if you have functions of the "fundamental operators" of the theory, and the "fundamental operators" here is the field. So by definition its only time dependence in the Heisenberg picture is due to the equation of motion and not additionally "explicit" somehow.
     
  13. Jun 9, 2016 #12
    So it's by construction then?! Is there any physical intuition for why they shouldn't have explicit time dependence?

    By this, do you mean that ##\hat{\phi}## and ##\hat{\pi_{\phi}}## are in a sense, the fundamental variables of the theory and so any explicit time can only arise through functions of these fundamental variables, i.e. ##f(\hat{\phi},\hat{\pi_{\phi}},t)##?
     
  14. Jun 10, 2016 #13

    vanhees71

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    Yes, in QFT the field operators are the fundamental operators out of which you build all observables, and by definition the fundamental operators "move" with some time-local operator (which one you choose defines the picture of time evolution; for the Heisenberg picture it's ##\hat{H}##, for the Schrödinger picture it's ##0##, for the interaction picture it's the non-interacting part of the Hamilltonian, ##\hat{H}_0##, etc.). So your last sentence nails it to the answer to your question :-).
     
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