
#1
Apr2710, 02:17 PM

P: 230

How can I work out
{π,∂φ} 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? Thanks 



#2
Apr2710, 02:32 PM

P: 842

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. 



#3
Apr2710, 02:49 PM

P: 230

[tex]\int{d^3x[\frac{1}{2}(\partial_{space}\phi)^2}][/tex] ..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!). 



#4
Apr2710, 03:33 PM

P: 842

Poisson bracket
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.




#5
Apr2710, 04:20 PM

P: 230

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
Apr2710, 05:54 PM

Sci Advisor
P: 817

[tex]\{\pi(\vec{y}),\phi(\vec{x}) \} = \delta^{3}(xy)[/tex] Well, now differentiate both sides with respect to x. sam 



#8
Apr2710, 06:50 PM

P: 230

sam:
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,(∂_{µ}φ∂^{µ}φ ∂_{t}^{2}φ)} = {π,∂_{µ}φ∂^{µ}φ} 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. 



#9
Apr2710, 07:16 PM

P: 2,828





#10
Apr2810, 05:37 PM

P: 230

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}}? 



#11
Apr2910, 12:29 AM

P: 2,828

If you worked out [tex]\{\Pi,\underline{\partial}\phi\}[/tex] you should be able to work this one with
[tex]\{f,g\}=\{g,f\}[/tex] [tex]\{f_1f_2,g\}=f_1\{f_2,g\}+f_2\{f_1,g\}[/tex] 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 ? [tex]\{\Pi,\phi\}=\delta[/tex] 



#12
Apr2910, 05:44 AM

P: 230

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: [tex] \{\pi(\vec{y}),\phi(\vec{x}) \} = \delta(xy) [/tex] ..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(\under line{y}),\phi(\underline{x})\}=\int d^3x. 2 \underline{\partial}\phi(\underline{x})\underline{\partial}\delta^3(\un derline{x}\underline{y})[/tex] So the question really is, how to work out the above integral on the RHS? 



#13
Apr2910, 04:13 PM

Sci Advisor
P: 817

[tex]
\{\pi(y) . (\nabla \phi(x))^{2}\} = 2 \nabla \phi(x) . \nabla_{x} \delta^{3}(yx) [/tex] 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}(yx) = \nabla^{2}\phi(y) [/tex] Is this what you wanted? sam *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}(yx)[/itex]. 



#14
Apr2910, 04:35 PM

Sci Advisor
P: 817

sam 



#15
May210, 12:50 PM

P: 230




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