Calc Poisson Bracket: {π,∂φ} Calculation

In summary: Poisson bracket of the canonical variables).In summary, the question is how to work out {π,(∂φ)2} in classical field theory, which can be done by using the fundamental Poisson bracket and differentiating both sides with respect to x. This leads to the result that the Poisson bracket in question reduces to ∂(∂φ)/∂φ. Integrating this result over x and using integration by parts, we can obtain the desired outcome of ∂^2φ(y).
  • #1
vertices
62
0
How can I work out

{π,φ}

where {,} is a Poisson Bracket; π is the canonical momentum and φ is the spatial 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 spatial derivative φ wrt φ.

Stupid question - but how to do this?

Thanks
 
Physics news on Phys.org
  • #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 spatial derivative since in position space they are one and the same.
 
  • #3
LostConjugate said:
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 spatial 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!).
 
  • #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.
 
  • #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
 
  • #6
I'm sorry, are you sure this is quantum mechanics ?
 
  • #7
vertices said:
How can I work out

{π,φ}

where {,} is a Poisson Bracket; π is the canonical momentum and φ is the spatial 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 spatial 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
 
  • #8
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,(∂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 isn't QM - It is Quantum Field Theory, where we promote fields themselves to operators.
 
Last edited:
  • #9
vertices said:
Humanino: This isn't 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]
 
  • #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}?
 
  • #11
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]
 
Last edited:
  • #12
humanino said:
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 spatial 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?
 
  • #13
[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].
 
Last edited:
  • #14
I'm also thinking along these lines:
{π,(∂µφ∂µφ -∂t2φ)} = {π,∂µφ∂µφ}

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
 
Last edited:
  • #15
samalkhaiat said:
[tex]
Is this what you wanted?

yes, I see what you have done. Thanks!
 

1. What is a Poisson bracket?

A Poisson bracket is a mathematical operator used in classical mechanics to determine the time evolution of a physical system. It is denoted by {A, B} and is defined as the product of the partial derivatives of A and B with respect to the system's variables, multiplied by the system's Hamiltonian.

2. How is a Poisson bracket calculated?

To calculate a Poisson bracket, you first need to determine the Hamiltonian of the system. Then, take the partial derivatives of the two functions inside the bracket with respect to the variables of the system. Finally, multiply these derivatives by the Hamiltonian and add them together to get the final result.

3. What is the significance of a Poisson bracket in physics?

The Poisson bracket is significant in classical mechanics because it allows us to determine the time evolution of a physical system by quantifying the relationship between different physical quantities. It also helps us to understand the symmetries and conserved quantities of a system.

4. How is a Poisson bracket used in practical applications?

Poisson brackets are used in various applications in classical mechanics, such as in the study of fluid dynamics, electromagnetism, and celestial mechanics. They can also be used to solve Hamilton's equations of motion and determine the stability of a system.

5. Are there any limitations to using a Poisson bracket?

While the Poisson bracket is a useful tool in classical mechanics, it has some limitations. It can only be applied to systems that follow Hamilton's equations of motion, and it does not work for quantum systems. Additionally, it may not always provide an accurate representation of a system's behavior in non-linear or chaotic systems.

Similar threads

  • Electromagnetism
Replies
2
Views
785
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Quantum Physics
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
908
  • Quantum Physics
Replies
3
Views
7K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Classical Physics
Replies
6
Views
5K
Replies
4
Views
1K
Back
Top