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Jan
15
comment Mode expansions of fields
Thanks to both ACM and AFT
Jan
15
comment Mode expansions of fields
Ok I think I see so since nabla is a differential operator, it is defined like $\partial_i = \nabla$?
Jan
15
comment Mode expansions of fields
@ACuriousMind: Thanks, but I don't understand: Why does $\mathbf k$ have components $k^i$? On the lhs I have $\nabla \phi = \partial_i \phi$. If what you say is correct then $\nabla = \partial^i$ and the relative minus will cancel between the two sides.
Jan
15
asked Mode expansions of fields
Nov
21
accepted Derivation of momentum in QFT - from Energy-Momentum Tensor
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Perfect, thanks :)
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
But why would I need to work with $-\frac{\partial}{\partial x^i}$? I just worked with $\partial^i$ and when taking derivatives, used the fact that $$\pm i \partial^i (p \cdot x) = \pm i \partial^i (p_ot - p^jx^j) = \mp i p^i.$$ Proceeding with this I get the minus error in the end. Thanks!
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Thanks! Sorry to be a pain but could you perhaps edit your answer to also include the starting point of computing $P^i$? As I wrote in my opening post, I have that $$P^i = \int d^3 x T^{0i} = \int (\partial^0 \psi^{\dagger} \partial^i \psi + \partial^i \psi^{\dagger}\partial^0 \psi)d^3 x$$ Going through with this computation I get minus what the answer should be. This minus I just don't know how to get rid of it and have been stuck with it for three days now. Thanks!
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Ah I see thanks! So, from this argument, $a^{\dagger}_{p}a_{p'}|u_p, u_p'\rangle $ is a state with 2 particles of momentum $p$. Since $|u_p, u_p'\rangle$ is also a state with 2 particles, they are the same eigenvectors of N and thus do not vanish. So the terms remaining must just be $a^{\dagger}a$ and $b^{\dagger}b$. Is that correct?
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
So do you mean that $a_p^{\dagger}b_{p'}^{\dagger}|0\rangle = |u_p, \bar v_{p'}\rangle$? But in what way will the eigenstate differ if we apply this operator onto the state again? Won't it just produce a state with two particles of momentum $p$ and two antiparticles of momentum $p'$?
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
To those who decided to close my thread, please consider the edit I have now made to reopen it.
Nov
21
revised Derivation of momentum in QFT - from Energy-Momentum Tensor
deleted 126 characters in body
Nov
21
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Thanks for your answer. Could you please explain though why 'the expectation values of the latter ones in any momentum eigenstate is obviously zero'? is it because those terms come with an oscillating exponential and therefore vanish over all p?
Nov
20
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Just wondering, what starting expression did you use for computing $P^i$? I had $$P^i = \int (\partial_o \psi^{\dagger} \partial^i \psi + \partial^i \psi^{\dagger} \partial_o \psi) d^3 x$$ which follows exactly from the definition of the stress energy tensor yet does not give the right answer. I thought I had $P^i$ but I noticed a sign error in what I had previously.
Nov
20
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Yes I agree with you for the evaluation of $P^i$. In that case there is component $g^{0i}$ multiplying $\mathcal L$ so indeed it vanishes. However, in the case of $P^0$ this is not true and is the case I am having difficulty with. Although, I did derive it using the observation by ACuriousMind but I still would like to see why my method is not giving me what I want.
Nov
20
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
Thanks for your response. Two things I don't understand: It is obvious (by inspection) that the lagrangian density vanishes for field configurations that solve the Dirac equation, but this is not so clear for the lagrangian for a complex scalar field. Also, as far as I can see, $\mathcal L$ for a complex scalar field does contain a term of the form $\partial_{\mu}\psi^{\dagger}$ (As it must otherwise $\mathcal L$ is not real). Please tell me if you agree with this.
Nov
20
comment Derivation of momentum in QFT - from Energy-Momentum Tensor
@Noiralef: Yes exactly, I use the energy momentum relationship to write $p_0^2 - p_i^2 = m^2$ which allows me to combine the terms that did not depend on m. Basically I have $$\frac{p_0^2 - p_i^2}{w(p)^2} + \frac{m^2}{w(p)^2} = 2\frac{m^2}{w(p)^2}$$ as written in the last equation in my post. This is where I am stuck.
Nov
20
asked Derivation of momentum in QFT - from Energy-Momentum Tensor
Nov
1
asked IBP Identities to solve differential equation
Oct
30
comment Non-symmetry of a lagrangian
@Qmechanic: the notes are hand written from my lectures.