Poisson bracket

[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

[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.

[vertices]

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:

\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!).

[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.

[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

[humanino]

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

[samalkhaiat]

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;

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

Well, now differentiate both sides with respect to x.


sam

[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.

[humanino]

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

[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}?

[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

[vertices]

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\}

Yes, I get this.

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\}=\underline{\partial}\phi

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?

[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).

[samalkhaiat]

I'm also thinking along these lines:
{π,(∂µφ∂µφ -∂t2φ)} = {π,∂µφ∂µφ}

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

[vertices]

[tex]
Is this what you wanted?


yes, I see what you have done. Thanks!