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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).
Nov
27
comment Error in books of conformal field theory?
@Prahar Some authors also use the second definition when $x'$ is an active transformation.
Nov
27
revised Error in books of conformal field theory?
added 19 characters in body
Nov
27
answered Error in books of conformal field theory?
Nov
27
comment Proof that eigenvalues of Fermionic creation/annihilation operators are Grassman numbers
"It's stated probably in all textbooks.." Can you give some particular reference which states this ?
Nov
26
revised Is Einstein-Hilbert action the unique action whose variation gives Einstein's field equations?
added 12 characters in body; edited title
Nov
26
asked Is Einstein-Hilbert action the unique action whose variation gives Einstein's field equations?
Nov
24
comment C, T, P transformation mistakes in ``Peskin&Schroeder's QFT''?
I guess the reason that both the ways of transformation are valid is the following. $C,P,T$ are discrete operators. Moreover $C^2$~$I$,$P^2$~$I$,$T^2$~$I$. So even if (e.g.) you use $C^{-1}$ instead of $C$ to transform a state (and hence $COC$ to transform an operator) the result would differ only by a phase and hence physically be the same.
Nov
13
comment How many bits are encoded on the surface of the smallest black hole?
"Perhaps $r$ is defined so as to make this so" Yeah, $r$ is defined to be square root of surface area/$4\pi$. In a curved geometry it need not be equal to the distance from the center.
Nov
12
comment If space and time are equivalent, what's Spin in time dimension
Boost transformation correspond to rotation of time into space; so, in principle, their generators can be taken as analog of spin in time direction. However generators of boosts satisfy very different algebra from the spin algebra. This is because space and time aren't fully "equivalent" as can be seen from the signature of the metric (-1,1,1,1).
Oct
7
comment Definition of phase transitions in statistical mechanics
For a finite (quantum) system with #microstates <<Avogadro's number, partition function loses its thermodynamical interpretation. So even though the partition function of your system has discontinuity in limit of large $\lambda$, this discontinuity can not be interpreted to mean any "physical" phase transition in the system. Statistical mechanical formulation works well only for "large" systems.