QFT Commutator (momentum and Hamiltonian) Issue

[orentago]

Hi,

I haven't posted this in the homework section, as I don't really see it as homework as such.

I'm trying to derive the Heisenberg equations of motion for the Klein Gordon field (exercise 2.2 of Mandl and Shaw).

I'm trying to derive the commutator of the Hamiltonian and canonical momentum, i.e. [H,\pi(x)]

For Klein Gordon, we have:H={1 \over 2}\int\mathrm{d}^3\mathbf{x}\left[c^2\pi^2(x) + (\mathbf{\nabla}\phi)^2+\mu^2\phi^2\right]

Inserting this into the commutator give several sub-commutators:
[c^2\pi^2(x),\pi(x)]
[(\mathbf{\nabla}\phi)^2,\pi(x)]
[\mu^2\phi^2,\pi(x)]

I have evaluated the first and last of these commutators, but I'm really struggling with the second. The first is of course zero, and the last can be evaluated practically directly using the canonical commutation relations to give:
[\mu^2\phi^2,\pi(x)]=2\mathrm{i}\hbar\mu^2\delta(x-x')\phi(x)

My attempt at evaluating the second gives: [(\mathbf{\nabla}\phi)^2,\pi(x)]=2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi

However, the result given indicates that [(\mathbf{\nabla}\phi)^2,\pi(x)]=-2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi. I'm not sure where I've gone wrong. I've used the commutation relation:
[AB,C]=A[B,C]+[A,C]B
to expand the commutator as much as possible, but I'm not sure this is the best method.

Does anyone know how to derive this commutator?

Any help is most appreciated.

EDIT: I'm using the commutation relation: [\phi(x),\pi(x)]=\mathrm{i}\hbar\delta(x-x')

 

[Avodyne]

Neither of the commutators you give is correct; one of the gradients must act on the delta function. Then, because you want to integrate over x', you can integrate by parts to move the gradient off the delta function.

 

[strangerep]

I'm trying to derive the commutator of the Hamiltonian and canonical momentum, i.e. [H,\pi(x)]

For Klein Gordon, we have:H={1 \over 2}\int\mathrm{d}^3\mathbf{x}\left[c^2\pi^2(x) + (\mathbf{\nabla}\phi)^2+\mu^2\phi^2\right]<br />

[...]

I'm using the commutation relation: [\phi(x),\pi(x)]=\mathrm{i}\hbar\delta(x-x&#039;)

Hint #1: be more careful with your variable names. Either use [H,\pi(y)] or
change the dummy integration variable in H to something else. And your CR above is
missing a prime somewhere. (That's one reason I prefer x,y rather than x,x' -- it's less
easy to forget a prime.

[...] several sub-commutators:
[c^2\pi^2(x),\pi(x)]
[(\mathbf{\nabla}\phi)^2,\pi(x)]
[\mu^2\phi^2,\pi(x)]
My attempt at evaluating the second gives: [(\mathbf{\nabla}\phi)^2,\pi(x)]=2\mathrm{i}\hbar\delta(x-x&#039;)\nabla^2\phi

However, the result given indicates that [(\mathbf{\nabla}\phi)^2,\pi(x)]=-2\mathrm{i}\hbar\delta(x-x&#039;)\nabla^2\phi. I'm not sure where I've gone wrong. I've used the commutation relation:
[AB,C]=A[B,C]+[A,C]B
to expand the commutator as much as possible, but I'm not sure this is the best method.

Yes, using Leibniz is fine. Then, if you clarify your variable names, and (Hint #2) perform
an integration by parts, you'll get the right answer in a couple of lines (at least, I did).

The trick about "integration by parts" when you have a delta distribution, but no explicit
integral, is to remember that the delta only makes sense under an integral -- hence you can
kinda imagine there's an integral, perform integral-like manipulations and then forget
the imagined integral.

HTH.

 

[tom.stoer]

... and of course you can write formally

[\partial_x A(x), B(y)] = \partial_x[A(x), B(y)] = -[A(x), B(y)]\partial_x

the last step is valid in case of a partial integration w.r.t. x

 

[orentago]

Thanks for all the help. I still have one question. When I expand the commutators I end up with a term [\mathbf{\nabla},\pi(x)]. Have I expanded correctly? If so, how do I evaluate this commutator?

 

[orentago]

I've had a go, using your suggestions. I'm still running into a few problems, probably due to ignorance on my part.

Here's my derivation:

\int\mathrm{d}^3\mathbf{x}[(\mathbf{\nabla}\phi)^2,\pi]=\int\mathrm{d}^3\mathbf{x}\left\{\mathbf{\nabla}\phi\cdot[\mathbf{\nabla}\phi,\pi]+[\mathbf{\nabla}\phi,\pi]\cdot\mathbf{\nabla}\phi\right\}
=\int\mathrm{d}^3\mathbf{x}\left\{\mathbf{\nabla}\phi\cdot\mathbf{\nabla}[\phi,\pi]+\mathbf{\nabla}\phi\cdot[\mathrm{\nabla},\pi]\phi+\mathbf{\nabla}[\phi,\pi]\cdot\mathbf{\nabla}\phi+[\mathbf{\nabla},\pi]\phi\cdot\mathbf{\nabla}\phi\right\}

Now, I've assumed that any two terms either side of a dot product commute, so:

\int\mathrm{d}^3\mathbf{x}[(\mathbf{\nabla}\phi)^2,\pi]=\int\mathrm{d}^3\mathbf{x}\left\{2 \mathbf{\nabla}\phi\cdot\mathbf{\nabla}[\phi,\pi]+2 \mathbf{\nabla}\phi\cdot[\mathrm{\nabla},\pi]\phi\right\}

I'm not sure how the nabla commutes with the momentum, hence my previous post, so for now I'll concentrate on the first term in the integral:

2\int\mathrm{d}^3\mathbf{x} \mathbf{\nabla}\phi\cdot\mathbf{\nabla}[\phi,\pi]=2\mathrm{i}\hbar\int\mathrm{d}^3\mathbf{x} \mathbf{\nabla}\phi\cdot\mathbf{\nabla}\delta(x-x&#039;)=2\mathrm{i}\hbar\left[\delta(x-x&#039;)\mathbf{\nabla}\phi-\int\mathrm{d}^3\mathbf{x}\delta(x-x&#039;)\nabla^2\phi\right]
=2\mathrm{i}\hbar\left[\delta(x-x&#039;)\mathbf{\nabla}\phi-\nabla^2\phi\right]

And then I get stuck, both because I don't know how that nabla commutes with the canonical momentum, and I don't know how to get rid of that Dirac function in the result immediately above. Perhaps the problems are linked, but I'm not sure.

Any hints?

 

[tom.stoer]

Please apply the rule

[\partial_x A(x), B(y)] = \partial_x[A(x), B(y)]

I have written down before.

For every commutator, every field operator and every derivative you have to write down the coordinate - either x or y! You haven't done that.
Let all derivatives act on the whole commutator; A and B must not contain derivatives!
If you do that you will never ever have to calculate a commutator of a derivative with a field operator.

Remark: a coordinate x in quantum field theory is nothing else but a "continuous index". Think about ordinary quantum mechanics. Think about the harmonic oscillator. You have operators, e.g. a_i and a_k^\dagger and - working in three dimensions - you have an index i, k = 1 \ldots 3. You would never ask how to calculate the commutator [i, a_k].

 

[orentago]

Ok I'll give that a go. To be thorough, do you know where I could find a proof of the identity you provided?

 

[tom.stoer]

Honestly? No! I always used it and it worked.

It should follow from some basic rules regarding operator valued distributions (there may be problems with regularization in case of operator products evaluated at the same spacetime point).

 

[DrFaustus]

The "proof" of the above "rule" can be found in any calculus book. The derivative wrt x is a partial one so it does not act on B(y). In other words, B(y) is treated as a constant when it comes to partial differentiation wrt x. Expand the commutator and see the magic unveil...

All the hints you've received should be enough for you to do it, but you have to be careful with the notation. Write the explicit integration variables and the variables of your fields - this is a source of major confusion in your calculation. If you keep struggling, I will write you the solution, but it's better if you work it out yourself.

 

[tom.stoer]

The "proof" of the above "rule" can be found in any calculus book.

In principle you have to verify that the rules remain valid when applied to (products of) operator-valued distributions.

 

[DrFaustus]

tom.stoer -> QFT is full of mathematical problems, but this one most definitely is not one of them. Field operators depend on a continuous parameter, the spacetime coordinates, and taking derivatives with respect to one really is just what one would expect it to be. At least if we are talking about the "taking the derivative out of the commutator" rule. For the partial integration things are a bit more subtle, but they do hold, rigorously, "in the sense of distributions".

 

[tom.stoer]

I agree. I used these rules for years and they always worked w/o problems :-) so I didn't care about a proof. Btw.: I am pretty sure that I am not able to proof (in a mathematical sense) that 1+1=2.

 

[orentago]

Thanks guys, I've had an crack, but I'm still getting snagged on that rogue Dirac delta, i.e.:

\int\mathrm{d}^3\mathbf{x}[(\mathbf{\nabla}_x\phi(x))^2,\pi(y)]=2\int\mathrm{d}^3\mathbf{x}\mathbf{\nabla}_x\phi(x)\cdot[\mathbf{\nabla}_x\phi(x),\pi(y)]=2\int\mathrm{d}^3\mathbf{x}\mathbf{\nabla}_x\phi(x)\cdot\mathbf{\nabla}_x[\phi(x),\pi(y)]
=2\mathrm{i}\hbar\int\mathrm{d}^3\mathbf{x}\mathbf{\nabla}_x\phi(x)\cdot\mathbf{\nabla}_x\delta(x-y)=2\mathrm{i}\hbar\left[\delta(x-y)\mathbf{\nabla}_x\phi(x)\right-\int\mathrm{d}^3\mathbf{x}\delta(x-y)\nabla^2_x\phi(x)]
=2\mathrm{i}\hbar\left[\delta(x-y)\mathbf{\nabla}_x\phi(x)\right-\nabla^2_x\phi(x)]

I must be overlooking something huge, I just don't know what it is.

 

[DrFaustus]

orentago -> Your first line seems completely wrong (typos?), but the last equality in the first line would seem what you need. Move the \nabla_x acting on the commutator to the field by partial integration, you pick a minus and a \nabla_x^2 \phi(x). Then evaluate the commutator, you get a delta. Then integrate against the delta and you should get what you are looking for.

 

[orentago]

But integrating that expression by parts gives a delta function that is outside of an integral :-(. I'm rather confused. Do I just ignore that and only use the second part of integral?

PS I've corrected my previous post.

 

[DrFaustus]

orentago -> Now I get it where your confusion is! When integrating by parts you always assume (because it's convenient) that the relevant quantities decay to zero at infinity. Or in this case, that the "surface term" vanishes. Remember that when integrating by parts the term "outside the integral" has to be evaluated between two values. Refresh your memory on integration by parts and check out the higher dimensional analogue.

So yes, you do ignore that delta "outside the integral", and no you do not "ignore" that term but it vanishes.

PS Your previous post still looks confusing...

 

[orentago]

So essentially:

\left[\delta(x-y)\mathbf{\nabla}_x\phi(x)\right]^\infty_{-\infty}=0

And this is because:

\lim_{x\rightarrow \infty}\phi(x)=0

 

[Avodyne]

This expression is zero, but it's because we assume y is finite, and \delta(x-y) vanishes unless x=y.