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Aug 25, 2021 at 11:18 history closed ohneVal
Qmechanic
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Aug 25, 2021 at 11:17 history edited Qmechanic
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S Aug 20, 2021 at 12:37 history suggested user196948 CC BY-SA 4.0
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Aug 20, 2021 at 12:12 review Suggested edits
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Aug 20, 2021 at 11:51 history edited jng224 CC BY-SA 4.0
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Aug 20, 2021 at 11:28 comment added Aleksandr Artemev Let us continue this discussion in chat.
Aug 20, 2021 at 11:26 comment added phys_student Yes, you are right! But this is in contracdict with $\pi_0|0\rangle=0=\langle 0|\pi_0$ which seems also be right.
Aug 20, 2021 at 11:22 comment added Aleksandr Artemev $\langle 0 \mid \phi_0 \pi_0\mid 0 \rangle$ and $\langle 0\mid \pi_0 \phi_0 \mid 0 \rangle$ can not both be zero, since $[\phi_0, \pi_0] \neq 0$.
Aug 20, 2021 at 11:18 comment added phys_student Sorry, I made a typo in my last comments. Indeed, $a_0$ is identified with $\pi_0$, then the term like this $\langle 0|\phi_0 \pi_0|0\rangle=0= \langle 0| \pi_0\phi_0|0\rangle$ will not contribute, since $\pi_0|0\rangle=0=\langle0|\pi_0$.
Aug 20, 2021 at 10:53 comment added Aleksandr Artemev That's not exactly right. As you wrote correctly in the body of the question, a0 should be identified with pi0, not phi0. phi0 does not need to annihilate the vacuum, so the term pi0 phi0 contributes to the correlation function. The vacuum average of phi0^2 is what defines the (necessary for 2d boson correlator) infrared cutoff and is also non-zero.
Aug 20, 2021 at 9:58 comment added phys_student Thanks for you comments. I think phi0 is just a0, the latter acting on $|0\rangle$ gives 0, so phi0 will not contribute.
Aug 20, 2021 at 9:34 comment added Aleksandr Artemev I believe the problem is that terms with phi0 and pi0 should also contribute. By the way, you missed that they also commute nontrivially.
Aug 20, 2021 at 9:05 review Close votes
Aug 20, 2021 at 11:51
Aug 20, 2021 at 8:40 history asked phys_student CC BY-SA 4.0