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4momentum of a massive scalar field in terms of creation and annihilation operators 
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#1
Jun409, 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] 


#2
Jun409, 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 noninteracting particles is simply the sum of oneparticle energies [tex] E(\mathbf{p}) [/tex]. In the creationannihilation 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 oneparticle 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. 


#3
Jun409, 11:19 PM

P: 1,780

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 


#4
Jun509, 12:32 AM

P: 1,746

4momentum 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 noninteracting 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 fieldbased expressions for basic observables, such as total energy and momentum. 4. Then we derive a field equation (e.g., KleinGordon) 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 (creationanihilation) 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 Nparticle 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 Nparticle Hilbert space (Nparticle 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". 


#5
Jun509, 12:44 AM

P: 1,780

Thank you for your detailed reply.
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. 


#6
Jun509, 12:58 AM

P: 969

Just forget about the Hamiltonian and use the formula you have of the stress tensor in terms of the Lagrangian. 


#7
Jun509, 01:03 AM

P: 1,780

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? 


#8
Jun509, 01:08 AM

P: 969

yeah that's correct.



#9
Jun509, 01:14 AM

P: 1,780

[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. 


#10
Jun509, 01:19 AM

P: 1,746

By "long and painful" I meant the full 9step procedure of introducing quantum fields in QFT. My goal was to draw your attention to the alternative (particlebased) approach to QFT. Sorry for hijacking your thread. 


#11
Jun509, 01:28 AM

P: 969




#12
Jun509, 01:36 AM

P: 1,780




#13
Jun509, 01:41 AM

P: 1,780

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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