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
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.
yes, that's what I would also think but if we have a term in a Hamiltonian which looks like this: [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!).
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.
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
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. sam
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.
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. NhNBW8a8-lI[/youtube]
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}}?
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]
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: [tex] \{\pi(\vec{y}),\phi(\vec{x}) \} = \delta(x-y) [/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(\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?
[tex] \{\pi(y) . (\nabla \phi(x))^{2}\} = -2 \nabla \phi(x) . \nabla_{x} \delta^{3}(y-x) [/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}(y-x) = \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}(y-x)[/itex].
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] sam