1,288 reputation
1515
bio website
location
age
visits member for 2 years, 4 months
seen 2 mins ago

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.
Dec
4
comment Semantic problem about renormalizability
Thanks for the clarification :)
Dec
4
comment Semantic problem about renormalizability
Do you mean that QED (even though its perturbatively renormalizable) is more like an effective field theory because of "Landau pole" ? But even if its so, isn't it in principle possible that there may exist a continuum nonperturbative theory (with possibly some kind of phase transition) which is approximated by perturbative QED in the low energy limit ?
Dec
4
comment Semantic problem about renormalizability
I don't know much about Landau pole; but the statement that perturbative computations are not an approximation to non-perturbative (by which I guess you mean "exact"?) computations sounds very unintuitive and even wrong.
Dec
4
comment Semantic problem about renormalizability
Can you please elaborate on your statement "there is no underlying non-perturbative QED".
Dec
3
comment Can one show that ${\gamma^5}^\dagger = \gamma^5$ directly from the anticommutation relations?
@CrazyBuddy thanks :)
Dec
3
comment Can one show that ${\gamma^5}^\dagger = \gamma^5$ directly from the anticommutation relations?
@CrazyBuddy This question is not off-topic. Qmechanic also thinks the same. See his comment on this meta post.
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
Thanks for the reply David. From your edit I understand that one of the following must be true : Either i) the coherent states are not elements of the Hilbert space. OR ii) The Hilbert space is not over complex numbers, but the multiplication by grassmann numbers to also allowed (In which case its not a Hilbert space). I think the first statement is correct. Is that right?
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
sorry, I used the rules you mentioned for nonconstant functions which don't apply to the function 1 :) However now the same problem remains if we use function $exp(\psi \bar{\psi})$ in place of 1.
Nov
28
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
"The functions do not depend on $\bar{\psi}$" But since $a^{\dagger}_f 1 = \bar{\psi}$ ; so $\bar{\psi}$ should be linearly dependent upon 1 and $\psi$. However I can't see any way to write $\bar{\psi}$ in terms of 1 and $\psi$ using complex coefficients.
Nov
27
comment Error in books of conformal field theory?
@NowIGetToLearnWhatAHeadIs Yes these definitions are not equivalent. In active view two different metrics (namely the given metric and its pullback) are being compared at the same point; while in the passive view the coefficients of the same metric in two different coordinates are being compared (again at the same point).