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Poisson bracket

  1. Apr 27, 2010 #1
    How can I work out


    where {,} is a Poisson Bracket; π is the canonical momentum and φ is the spacial derivative of the field (ie. not including the temporal one).

    Basically the question boils down to (or atleast I think it does!), working out ∂(φ) /∂φ - ie. differentiating the spacial derivative φ wrt φ.

    Stupid question - but how to do this?

  2. jcsd
  3. Apr 27, 2010 #2
    You need to work out the commutator because {A,B} goes to [A,B](2pi/ih) in quantum mechanics.

    I would expect the momentum operator would commute with the spacial derivative since in position space they are one and the same.
  4. Apr 27, 2010 #3
    yes, that's what I would also think but if we have a term in a Hamiltonian which looks like this:


    ..after taking the PB (with the canonical momentum), it goes to:

    [tex]\partial_i \partial^i \phi[/tex] (which shows that they can't commute!).
  5. Apr 27, 2010 #4
    You lost me but I am no expert. Remember you are taking the commutator of the momentum operator with the derivative operator and not the derivative of the wave function.
  6. Apr 27, 2010 #5
    This is QFT, not QM. φ here is the operator (not a wavefuntion, like in QM), as is the canonical momentum.

    I am not sure if we can carry the QM result that the commutator is i times the PB over to QFT. I was thinking more along the lines that the PB is explicitly given by {A,B}=dA/dφ(x) *dB/dπ(y) - dB/dφ(y) *dA/dπ(x).

    If we work it out this way, you'll see that the PB in question reduces to ∂φ/∂φ
  7. Apr 27, 2010 #6
    I'm sorry, are you sure this is quantum mechanics ?
  8. Apr 27, 2010 #7


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    You know the fundamental Poisson bracket;

    [tex]\{\pi(\vec{y}),\phi(\vec{x}) \} = \delta^{3}(x-y)[/tex]

    Well, now differentiate both sides with respect to x.

  9. Apr 27, 2010 #8

    I am trying to work out {π,(φ)2}= 2{π,φ}φ

    Why would differentiating the expression you wrote, give me the PB on the RHS of the above?

    EDIT: I see what you#re suggesting - π is a function of y, so yes differentiating wrt x would give us the PB. How to differentiate the dirac delta function though?

    I'm also thinking along these lines:

    {π,(φ)2}= {π,(∂µφ∂µφ -∂t2φ)} = {π,∂µφ∂µφ}

    Might it be easier to work of the PB on the RHS of the above expression? How?

    Humanino: This isnt QM - It is Quantum Field Theory, where we promote fields themselves to operators.
    Last edited: Apr 27, 2010
  10. Apr 27, 2010 #9
    Everything you write is classical field theory, from my point of view. BTW, it's all very well explained by Susskind in a 2h or so lecture available freely, for instance if you're interested. It's an extremely important aspect of classical mechanics to be aware of before embarking on quantum field theory.
  11. Apr 28, 2010 #10
    Humanino, yes, sorry, ofcourse I am talking about classical field theory.

    thanks for the video - the guy explains things really well!

    Any thoughts on how I could work out {π,(φ)2}?
  12. Apr 29, 2010 #11
    If you worked out [tex]\{\Pi,\underline{\partial}\phi\}[/tex] you should be able to work this one with
    As in
    [tex]\{\Pi,\left(\underline{\partial}\phi\right)^2\} = \{\Pi,\underline{\partial}\phi\underline{\partial}\phi\} = \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} + \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} = 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}[/tex]

    How about the result of [tex]\{\Pi,\underline{\partial}\phi\}[/tex] ? It seems to me there are several ways. I have been wondering, was the previous question about [tex]\{\Pi,\phi\}[/tex] by any chance ?
    Last edited: Apr 29, 2010
  13. Apr 29, 2010 #12
    Yes, I get this.

    This is where I am stuck.

    Because we're dealing with the Hamiltonian density, we have to work out the integral:

    [tex]\int d^3x. 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}[/tex]

    If we use the result:

    \{\pi(\vec{y}),\phi(\vec{x}) \} = \delta(x-y)

    ..and pull the partial spacial derivative out of the PB to get:

    [tex]\{\Pi,\underline{\partial}\phi\}= \underline{\partial}\{\Pi,\phi\}[/tex] and intergrate, we find:

    [tex]\int d^3x. 2 \underline{\partial}\phi(\underline{x})\underline{\partial}\{\Pi(\underline{y}),\phi(\underline{x})\}=\int d^3x. 2 \underline{\partial}\phi(\underline{x})\underline{\partial}\delta^3(\underline{x}-\underline{y})[/tex]

    So the question really is, how to work out the above integral on the RHS?
  14. Apr 29, 2010 #13


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    \{\pi(y) . (\nabla \phi(x))^{2}\} = -2 \nabla \phi(x) . \nabla_{x} \delta^{3}(y-x)

    Are you ok with this?*

    Now integrate both sides over x and do integration by parts on the right hand side, you will get

    [tex]\{\pi(y) , (1/2) \int d^{3}x (\nabla \phi)^{2} \} = \int d^{3}x \nabla^{2}\phi(x) \delta^{3}(y-x) = \nabla^{2}\phi(y)

    Is this what you wanted?


    *Edit we have adifferent sign because we started with a wrong sign for [itex]\{\pi (y), \phi(x)\}[/itex]. this should have been [itex] -\delta^{3}(y-x)[/itex].
    Last edited: Apr 29, 2010
  15. Apr 29, 2010 #14


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    No this is wrong! the poisson bracket [itex]\{\pi, \partial_{t}\phi\}[/itex] is not zero! The field "velocity" [itex]\partial_{t}\phi[/itex] is a function of [itex]\pi , \nabla \phi[/itex] and [itex]\phi[/itex]

    Last edited: Apr 29, 2010
  16. May 2, 2010 #15
    yes, I see what you have done. Thanks!
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