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Apr
2
comment Why do we use functional integration in QFT?
@AlexNelson Its true that we choose a time coordinate but in all the expressions time and spatial coordinates appear on equal footing. There are no expressions such as Exp(itH) in which time 'appears' to be different from spatial coordinates.
Apr
2
comment Why do we use functional integration in QFT?
@AlexNelson You may be right; I am not sure. However its true that in path integral formalism all the spacetime coordinates t,x,y,z are treated on equal footing which is more in the spirit of relativity. In particular, path integral formalism doesn't require the spacetime to be of the form Space X Time (as is necessarily required in Hamiltonian formalism) and hence can be used to define theories on more general manifolds including compact ones.
Apr
2
comment Why do we use functional integration in QFT?
Some of the benefits of path integral formulation are - i) Perturbation series is easy to derive compared to Hilbert space formalism ii) All objects in path integral formulation are classical; one doesn't need to introduce quantum operatorts iii) In Hilbert space formalism one needs to choose a time slice which breaks the manifest Lorentz invariance; this is not a problem in path integral formalism iv) After Wick rotation to Euclidean space, the path integral formalism is completely equivalent to the partition function formalism of stat mech; hence use of RG methods etc is more intuitive.
Feb
17
awarded  Nice Question
Jan
21
comment How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?
@joshphysics - There is a finite multiplicative group contained in the Clifford algebra generated by gamma matrices. E.g. in two dimensions and with metric of signature (1,1), this group consist of elements $1$, $-1$, $\gamma^0$, $-\gamma^0$, $\gamma^1$ ,$-\gamma^1$, $\gamma^0 \gamma^1$ and $-\gamma^0 \gamma^1=\gamma^1 \gamma^0$.
Dec
28
revised Irreducible Representations Of Lorentz Group
added 51 characters in body
Dec
12
comment A key relation in Di Francesco's book on Conformal Field Theory
You are right about the second term on r.h.s of equation 6.174. From the standard definition of a primary field this term should be $(n+1)h z^n$ instead of $(n+1)h$. However, I am not sure how the author replaced $L_n \phi$ with $[L_n,\phi]$ in the second step in equation 6.174 ? This can be done for n>0, but for n<0, $L_n$ will not annihilate the state $|h,\bar{h}>$.
Dec
12
answered Proof that higher genus surface admits a metric of negative Ricci scalar everywhere
Dec
10
comment Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
@David Z (Comment 2, after 21 hours) I am still waiting for any response from you. Can this question be considered for reopening ?
Dec
10
comment Angular momentum conservation in pion decay?
Thanks for the clarification :)
Dec
10
comment Angular momentum conservation in pion decay?
I was imagining bound state of quark and antidown quark as forming a kind of atom (like the hydrogen atom) with various angular momentum levels of which pion is one state. But I am not sure if this picture make sense for strong interation
Dec
10
comment Angular momentum conservation in pion decay?
Hmm.. but does the state l=1, S=0 exist for a bound state of up and anti-downquark ?
Dec
10
comment Angular momentum conservation in pion decay?
Since pion is a bound state, so could it be that the pion first goes to a state of angular momentum 1 before going to the final state via W+ channel?
Dec
9
comment Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
@David Z can it be considered for reopening? I mean its not so trivial homework problem, and the OP has also shown some effort
Dec
9
revised Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
changed the picture, changed some notation
Dec
9
suggested suggested edit on Where does the divergence in the $g\phi^3$ $d=4$ 3 point one loop diagram (three external legs) come from?
Dec
9
comment How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?
You can look at any of the references mentioned in this post
Dec
9
comment How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?
The proof that any finite dimensional representation of a finite group can be chosen to be unitary is not diffcult. Let $V$ be a finite dimensional complex representation of a finite group $G$. Let ( , ) be any Hermitian product on $V$. Define a new Hermitian product as $(x , y)' = \displaystyle\sum_{g\in G} (gx,gy)$. To see that the new product is unitary note that for any $h\in G$, $(hx,y)'=\displaystyle\sum_{g\in G} (ghx,gy) =\displaystyle\sum_{g\in G} (ghx,gh h^{-1}y)=(x,h^{-1}y)'$.
Dec
9
comment How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?
Use the fact that Gamma matrices are all unitary i.e. $(\gamma^{\mu})^{\dagger}=(\gamma^{\mu})^{-1}$ (they can be chosen to be so since they form a representation of a finite group). Now using this fact along with the commutation relations we find that $(\gamma^0)^{\dagger}=\gamma ^0$ and $(\gamma ^i)^{\dagger}=-\gamma^i$. Again using the commutation relations we have $\gamma^0 \gamma^i \gamma^0 = -\gamma^i =(\gamma^i)^{\dagger}$. Also $\gamma^0 \gamma^0 \gamma^0=\gamma^0 =(\gamma^0)^{\dagger}$
Dec
9
comment XX coupling vs Heisenberg coupling
The difference is in the definition of the Hamiltonian as is mentioned in the paper that you linked to. In particular Heisenberg interaction is more symmetric since it involves all the spins democratically. I am not sure of the physical consequences though.