# Poisson bracket

1. Apr 27, 2010

### vertices

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. Apr 27, 2010

### LostConjugate

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. Apr 27, 2010

### vertices

yes, that's what I would also think but if we have a term in a Hamiltonian which looks like this:

$$\int{d^3x[\frac{1}{2}(\partial_{space}\phi)^2}]$$

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

$$\partial_i \partial^i \phi$$ (which shows that they can't commute!).

4. Apr 27, 2010

### LostConjugate

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. Apr 27, 2010

### vertices

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. Apr 27, 2010

### humanino

I'm sorry, are you sure this is quantum mechanics ?

7. Apr 27, 2010

### samalkhaiat

You know the fundamental Poisson bracket;

$$\{\pi(\vec{y}),\phi(\vec{x}) \} = \delta^{3}(x-y)$$

Well, now differentiate both sides with respect to x.

sam

8. Apr 27, 2010

### vertices

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 isnt QM - It is Quantum Field Theory, where we promote fields themselves to operators.

Last edited: Apr 27, 2010
9. Apr 27, 2010

### humanino

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.

10. Apr 28, 2010

### vertices

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. Apr 29, 2010

### humanino

If you worked out $$\{\Pi,\underline{\partial}\phi\}$$ you should be able to work this one with
$$\{f,g\}=-\{g,f\}$$
$$\{f_1f_2,g\}=f_1\{f_2,g\}+f_2\{f_1,g\}$$
As in
$$\{\Pi,\left(\underline{\partial}\phi\right)^2\} = \{\Pi,\underline{\partial}\phi\underline{\partial}\phi\} = \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} + \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\} = 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}$$

How about the result of $$\{\Pi,\underline{\partial}\phi\}$$ ? It seems to me there are several ways. I have been wondering, was the previous question about $$\{\Pi,\phi\}$$ by any chance ?
$$\{\Pi,\phi\}=\delta$$

Last edited: Apr 29, 2010
12. Apr 29, 2010

### vertices

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:

$$\int d^3x. 2 \underline{\partial}\phi\{\Pi,\underline{\partial}\phi\}$$

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:

$$\{\Pi,\underline{\partial}\phi\}= \underline{\partial}\{\Pi,\phi\}$$ and intergrate, we find:

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

So the question really is, how to work out the above integral on the RHS?

13. Apr 29, 2010

### samalkhaiat

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

Are you ok with this?*

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

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

Is this what you wanted?

sam

*Edit we have adifferent sign because we started with a wrong sign for $\{\pi (y), \phi(x)\}$. this should have been $-\delta^{3}(y-x)$.

Last edited: Apr 29, 2010
14. Apr 29, 2010

### samalkhaiat

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

sam

Last edited: Apr 29, 2010
15. May 2, 2010

### vertices

yes, I see what you have done. Thanks!