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4-momentum of a massive scalar field in terms of creation and annihilation operators

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maverick280857
#1
Jun4-09, 10:44 PM
P: 1,780
Hi,

I'm trying to compute

[tex]P^{\mu} = \int d^{3}x T^{0\mu}[/tex]

where T is the stress energy tensor given by

[tex]T^{\mu\nu} = \frac{\partial\mathcal{L}}{\partial[\partial_{\mu}\phi]}\partial^{\nu}\phi - g^{\mu\nu}\mathcal{L}[/tex]

for the scalar field [itex]\phi[/itex] with the Lagrangian density given by

[tex]\mathcal{L} = \frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi - m^2\phi^2[/tex]

This is what I get

[tex]T^{\mu 0} = g^{0\mu}\mathcal{H}[/tex]

(using [itex]\mathcal{H} = \Pi\dot{\phi} - \mathcal{L} = \partial^{0}\phi\partial_{0}\phi - \mathcal{L}[/itex])

so

[tex]\int d^{3}x T^{0\mu} = g^{0\mu}H = \frac{1}{2}\int d^{3}p g^{0\mu}E_{p}[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)][/tex]

Now, the problem is that if we have

[tex]p^{\mu} = (E_{p}, \vec{p})[/tex]

then [itex]E_{p} = p^{0}[/itex], so

[tex]\int d^{3}x T^{0\mu} = \frac{1}{2}\int d^{3}p g^{0\mu}p^{0}[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)][/tex]

Is there some mistake here, because the answer should involve [itex]p^{\mu}[/itex]?

The correct answer is

[tex]\int d^{3}x T^{0\mu} = \frac{1}{2}\int d^{3}p p^{\mu}[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)][/tex]
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meopemuk
#2
Jun4-09, 11:06 PM
P: 1,746
Maveric,

I would like to draw your attention that from naive (common sense) considerations, the total energy of a system of several non-interacting particles is simply the sum of one-particle energies [tex] E(\mathbf{p}) [/tex]. In the creation-annihilation operator notation this means

[tex]E = \int d^{3}p E(\mathbf{p}) a^{\dagger}(\mathbf{p})a(\mathbf{p})[/tex]


Similarly, the total momentum is the sum of one-particle momenta

[tex]\mathbf{P} = \int d^{3}p \mathbf{p} a^{\dagger}(\mathbf{p})a(\mathbf{p})[/tex]

So, it seems that quantum field recipe outlined by you does not match exactly with common sense.
maverick280857
#3
Jun4-09, 11:19 PM
P: 1,780
Quote Quote by meopemuk View Post
Maveric,

I would like to draw your attention that from naive (common sense) considerations, the total energy of a system of several non-interacting particles is simply the sum of one-particle energies [tex] E(\mathbf{p}) [/tex]. In the creation-annihilation operator notation this means

[tex]E = \int d^{3}p E(\mathbf{p}) a^{\dagger}(\mathbf{p})a(\mathbf{p})[/tex]
Thank you for your reply meopemuk, but I am not sure how you can justify this rigorously. The expression I wrote comes from a direct (explicit) computation of the Hamiltonian of the Klein Gordon field. After normal ordering, the expression takes the form you wrote.

But maybe you have a different point?

Also, this question is from a book, and the answer I wrote down is the one given in the back of that book. I think there's a problem with the index raising/lowering.

EDIT -- Okay, I think I get the point of your post. Yes, offhand that is what the expression should be intuitively. But it isn't -- except in the normally ordered sense. One reason I can think of is the way the field is constructed...the second term in the ground state Hamiltonian is a momentum delta function evaluated at its singularity. To "remove" it, we define a normally ordered Hamiltonian. Is there a deeper reason? I'm new to QFT btw, so I would appreciate if you could dwell on the point you're trying to make further.

PS -- Please also take a look at my original question...I'm still stuck with the index ordering :-P

meopemuk
#4
Jun5-09, 12:32 AM
P: 1,746
4-momentum of a massive scalar field in terms of creation and annihilation operators

The point I am making does not answer your original question, but (I hope) it is not irrelevant.

There are two ways to look at quantum field theory in general and at operators of observables in particular. One way is long and painful, and the other way is fast and easy.

The long and painful way is based on the idea of quantum field. Unfortunately, this way is presented in most QFT textbooks and fills many pages there. Roughly, it goes like this (in the case of a non-interacting field):

1. First we assume that there exists some (mysterious) substance called "field".
2. Then we postulate a certain Lagrangian and action for the field, and demand that this action must be minimized.
3. Then we apply Noether's theorem and derive field-based expressions for basic observables, such as total energy and momentum.
4. Then we derive a field equation (e.g., Klein-Gordon) by minimizing the action.
5. Then we solve this equation in the form of a "Fourier series".
6. Then we "quantize" this solution by converting coefficients to (creation-anihilation) operators with prescribed commutation relations.
7. Then we insert this solution for the "quantum field" in the formulas for the energy and momentum found in 3.
8. Then we see that obtained energy has a divergent term and artificially delete this term by normal ordering.
9. Finally, we arrive to the desired expressions

[tex] P = \int d^3p p a^{\dag}(p)a(p) [/tex].........(1)
[tex] E = \int d^3p E(p) a^{\dag}(p)a(p) [/tex]..........(2)


The other (fast and easy) way is based on the idea of particles. As far as I know the only major textbook that uses this path is Weinberg's "The quantum theory of fields", vol. 1.

The idea is that world is made of particles. If particles do not interact, then the total momentum and energy of any N-particle system are simply

P = p_1 + p_2 + ... + p_N.........(3)
E = e_1 + e_2 + ... + e_N.........(4)

Then we notice that operators (1) and (2) have exactly forms (3) and (4), respectively, in any N-particle Hilbert space (N-particle sector of the Fock space). So, quite naturally, we choose (1) and (2) as our total momentum and total energy operators.

As you can guess, I prefer the fast and easy way of working with QFT. So, I am not sure where you made a mistake in your algebra. You can compare your (long and painful) calculations with derivations of formulas (2.31) and (2.33) in M. E. Peskin and D. V. Schroeder "An introduction to quantum field theory".
maverick280857
#5
Jun5-09, 12:44 AM
P: 1,780
Thank you for your detailed reply.

Quote Quote by meopemuk View Post
As you can guess, I prefer the fast and easy way of working with QFT. So, I am not sure where you made a mistake in your algebra. You can compare your (long and painful) calculations with derivations of formulas (2.31) and (2.33) in M. E. Peskin and D. V. Schroeder "An introduction to quantum field theory".
However, I think you're referring to something else. The question asks to use the expression for the (conserved) stress energy tensor to explicitly compute the momentum this way. I don't see how it is long or painful (even for me :-D), and the problem is merely in the index raising/lowering in the final step...I believe I have the correct expression till the point before the final step.

Note that Peskin and Schroeder have already dropped the delta function term after equation (2.31), so their calculation is "slightly" different. I am not using the normally ordered notation yet, nor am I dropping the term..if you will, I am using equation (2.31) as the definition of the Hamiltonian in my calculation, without expressing the integrand in terms of the commutator -- this means I am using the 5th equation in my original post.
RedX
#6
Jun5-09, 12:58 AM
P: 969
Quote Quote by maverick280857 View Post
Hi,

[tex]T^{\mu 0} = g^{0\mu}\mathcal{H}[/tex]

(using [itex]\mathcal{H} = \Pi\dot{\phi} - \mathcal{L} = \partial^{0}\phi\partial_{0}\phi - \mathcal{L}[/itex])
The 2nd equation in parenthesis is correct. The 1st is wrong.

Just forget about the Hamiltonian and use the formula you have of the stress tensor in terms of the Lagrangian.
maverick280857
#7
Jun5-09, 01:03 AM
P: 1,780
Quote Quote by RedX View Post
The 2nd equation in parenthesis is correct. The 1st is wrong.

Just forget about the Hamiltonian and use the formula you have of the stress tensor in terms of the Lagrangian.
Thanks RedX. I'll check this out.

EDIT: This was my working:

[tex]T^{0\mu} = \partial^{0}\phi\partial^{\mu}\phi - g^{0\mu}(\partial^{0}\phi\partial_{0}\phi-\mathcal{H})[/tex]

Is this correct?
RedX
#8
Jun5-09, 01:08 AM
P: 969
yeah that's correct.
maverick280857
#9
Jun5-09, 01:14 AM
P: 1,780
Quote Quote by RedX View Post
yeah that's correct.
Ok, so now

[tex]T^{0\mu} = \partial^{0}\phi\partial^{\mu}\phi - g^{0\mu}(\partial^{0}\phi\partial_{0}\phi-\mathcal{H})[/tex]

gives me

[tex]T^{0\mu} = \partial^{0}\phi\partial^{\mu}\phi - g^{0\mu}\partial^{0}\phi\partial_{0}\phi + g^{0\mu}\mathcal{H}[/tex]

I get my mistake: I canceled the first two terms -- clearly a wrong thing to do, since there is no repeated index which would convert the second term into the first one. But I retained the Hamiltonian since I know how to write it in terms of [itex]a(p)[/itex] and [itex]a^{\dagger}(p)[/itex]. I'll now try to solve it without expressing it in terms of the Hamiltonian.
meopemuk
#10
Jun5-09, 01:19 AM
P: 1,746
Quote Quote by maverick280857 View Post
I don't see how it is long or painful (even for me :-D), and the problem is merely in the index raising/lowering in the final step...

By "long and painful" I meant the full 9-step procedure of introducing quantum fields in QFT. My goal was to draw your attention to the alternative (particle-based) approach to QFT. Sorry for hijacking your thread.
RedX
#11
Jun5-09, 01:28 AM
P: 969
Quote Quote by maverick280857 View Post
Ok, so now

[tex]T^{0\mu} = \partial^{0}\phi\partial^{\mu}\phi - g^{0\mu}\partial^{0}\phi\partial_{0}\phi + g^{0\mu}\mathcal{H}[/tex]

I get my mistake: I canceled the first two terms -- clearly a wrong thing to do, since there is no repeated index which would convert the second term into the first one. But I retained the Hamiltonian since I know how to write it in terms of [itex]a(p)[/itex] and [itex]a^{\dagger}(p)[/itex]. I'll now try to solve it without expressing it in terms of the Hamiltonian.
I don't know if it's too late and you calculated everything, but you can see from the expression that if mu is not equal to zero, then only the first term survives. So you really only need to calculate the first term.
maverick280857
#12
Jun5-09, 01:36 AM
P: 1,780
Quote Quote by RedX View Post
I don't know if it's too late and you calculated everything, but you can see from the expression that if mu is not equal to zero, then only the first term survives. So you really only need to calculate the first term.
Yeah, saw that. Thanks. (PS -- I'm in a different time zone..its 5 minutes past noon here, so its certainly not late )
maverick280857
#13
Jun5-09, 01:41 AM
P: 1,780
Quote Quote by meopemuk View Post
By "long and painful" I meant the full 9-step procedure of introducing quantum fields in QFT. My goal was to draw your attention to the alternative (particle-based) approach to QFT. Sorry for hijacking your thread.
Hi meopemuk, thanks for your insight...you certainly did not hijack the thread! The 9 step procedure you described put everything into perspective for me actually. As I said I am new to QFT, but I wanted to go through that 9 step process since I have not really done relativistic QM or classical field theory "formally". So I reckon before I can get to the shorter method (ala Weinberg) I need to get my hands dirty in all this algebra..I'm still fairly new to such manipulations and interpretations.

Thanks a ton for your help again, and hope I run into you more often on this forum...I'm still on Chapter 2 of most QFT books


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