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Poisson bracket

by vertices
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vertices
#1
Apr27-10, 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
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LostConjugate
#2
Apr27-10, 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.
vertices
#3
Apr27-10, 02:49 PM
P: 230
Quote Quote by LostConjugate View Post
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!).

LostConjugate
#4
Apr27-10, 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.
vertices
#5
Apr27-10, 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
humanino
#6
Apr27-10, 04:51 PM
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P: 2,828
I'm sorry, are you sure this is quantum mechanics ?
samalkhaiat
#7
Apr27-10, 05:54 PM
Sci Advisor
P: 903
Quote Quote by vertices View Post
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 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
vertices
#8
Apr27-10, 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,(∂φ∂φ -∂t2φ)} = {π,∂φ∂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.
humanino
#9
Apr27-10, 07:16 PM
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P: 2,828
Quote Quote by vertices View Post
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.
vertices
#10
Apr28-10, 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}?
humanino
#11
Apr29-10, 12:29 AM
humanino's Avatar
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]
vertices
#12
Apr29-10, 05:44 AM
P: 230
Quote Quote by humanino View Post
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]
Yes, I get this.

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\}=\underline{\partial}\phi[/tex]
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(\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?
samalkhaiat
#13
Apr29-10, 04:13 PM
Sci Advisor
P: 903
[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].
samalkhaiat
#14
Apr29-10, 04:35 PM
Sci Advisor
P: 903
I'm also thinking along these lines:
{π,(∂φ∂φ -∂t2φ)} = {π,∂φ∂φ}
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
vertices
#15
May2-10, 12:50 PM
P: 230
Quote Quote by samalkhaiat View Post
[tex]
Is this what you wanted?
yes, I see what you have done. Thanks!


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