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Sep
30
comment How to derive the form of the parity operator acting on Lorentz spinors?
@Stephen I am not sure if it would help but look at Osborne's lecture notes on SM (in particular sections 2 and 3.5)
Sep
29
comment How to derive the form of the parity operator acting on Lorentz spinors?
I guess Stephen wants to ask why the matrices are $iI$ when, according to his calculations, they could be of more general form.
Sep
25
comment AdS/CFT seminal papers?
Also see the answer by Genneth to this question: physics.stackexchange.com/q/8162
Sep
24
comment Renormalizability of the Polyakov Action
Yes, 1 means $\eta_{\mu\nu}$. Similarly $aX$ means $af^{(1)}_{\mu\nu\rho}X^{\rho}$ for some 'tensor' $f^{(1)}$, $(aX)^2$ means $a^2 f^{(2)}_{\mu\nu\rho\sigma}X^{\rho}X^{\sigma}$ for some 'tensor' $f^{(2)}$ and so on. Sign of kinetic term for $X^0$ would be wrong. This gives rise to negative normed (unphysical) states in the spectrum which are then eliminated by applying constraints (which come from the equation of motion of the worldsheet metric h ).
Sep
21
revised From quantization under external classical gauge field to a fully quantized theory
added 326 characters in body
Sep
20
comment When are Eigenfunctions/Wavefunctions real?
Sorry, now i understand that the question is not that trivial since you are asking about reality of wavefunctions (i.e. position representation of eigenstates) and not the eigenstates.
Sep
20
comment When are Eigenfunctions/Wavefunctions real?
The answer to 1 actually depends upon how you define reality structure on the complex space of states. One can always choose a basis of eigenstates of the Hamiltonian and call them real.
Sep
18
comment Motivation to introduce von Neumann algebras in addition to $C^*$algebras?
Just a personal view. In matters of relevance to physics one is usually concerned with studying a particular observable ( like momentum, energy etc) or at most a finite algebra of observables (like spin algebra). I personally haven't seen any physics question whose answer is to be found in studying the set of all operators at once. For a physicist the statement "an observable should be Hermitian" is more useful than the statement "set of all operators form Von Neumann algebra (or whatever)"
Sep
16
comment Scale-invariant differential operator
There is no need to make any operator or expression dimensionless unless you have to (say) take its exponential or log etc. $\nabla^2$ is dimensionful and to make it dimensionless you'll have to multiply it with some constant having dimensions of length^2. Mathematically a constant of course won't make any difference. Anyway, I am telling you only very rough and personal view. I hope someone would give a better answer:)
Sep
16
comment Scale-invariant differential operator
Yes, dimensionless quantities whose definition doesn't involve choice of any scale are scale invariant; e.g. (length of stick A)/ (length of stick B). { reply to comment2: Mass=inverse length in natural units ($\hbar=c=1$)}
Sep
16
comment Scale-invariant differential operator
Scale invariance (roughly) means invariance of a quantity independent of the scale used to measure it (e.g. meter or cm scale used for measuring length). The given operator has dimension of mass^2 and hence is not scale invariant. In general dimensionful quantities can never be scale invariant (e.g. lenth. A 1 meter long stick is 100 cm long on cm scale). There are some dimensionless quantities too whose definition itself depends upon choice of some scale and hence which are not scale invariant. One example is charge of an electron. It depends upon from how close you look at it.
Sep
16
comment Is spacetime all that exists?
@DImension10 Isn't it correct from string theory point of view where "spacetime fields" $X^{\mu}$ (along with fermions $\Psi^{\mu}$ which too can be thought of as anticommuting extension of spacetime) are the basic constituent of matter?
Sep
14
comment Renormalizability of the Polyakov Action
Not sure but consider following argument. Suppose G be of the form 1+aX+(aX)^2+ .. where a has units of mass. Now define new variable Y=\sqrt(T) X. Lagrangian will now have a free part, and interacting part. Coefficient of the first interaction term will be g=a/sqrt(T), of second interaction term g^2 and so on. Now note that g has mass dimension (1-p)/2. Now for renormalizability we should require that it be non negative and hence that p<=1
Sep
6
comment Central charge in a $d=2$ CFT
Poincare algebra doesn't have a central extension. This i think is proven in Weinberg Vol 1. Those symmetry algebras which admit nontrivial central extension are prone to develop anomaly upon quantization. I am not sure why quantization results into only central extension type deformation of the algebra but i think the reason is that definition of quantum operators involves normal ordering (or more generally renormalization) which makes the definition ambiguous by some constant term e.g. in case of Virasoro algebra quantum definition of L0 is ambiguous by addition of a constant.
Aug
29
comment Spinor irreducible reps of the Lorentz group and their algebra
sorry i didn't read it carefully. T(g) is usual vector representation and you are talking about its splitting into two su(2)'s right? By Dirac representation i meant 4d complex spinor representation which is direct sum of right and left Weyl.
Aug
29
comment Spinor irreducible reps of the Lorentz group and their algebra
Is representation T(g) Dirac representation ?
Aug
24
comment superposition of two states with different number of particles
@richard In absence of any mechanism for production or annihilation of particles, time evolution will not be able to mix up the states with different particles numbers and hence you can study them separately too.
Aug
22
comment A curious issue about Dyson-Schwinger equation(DSE): why does it work so well?
@Jia Yiyang I am not sure but i think any derivation of (1) making use of this assumption can't be correct. For a free field we can anyway derive (1) by actually first computing the time ordered expression (using mode expansion of fields) and then verifying that it satisfies (1)
Aug
21
comment A curious issue about Dyson-Schwinger equation(DSE): why does it work so well?
I think correlation function of any product of fields in path integral can be written equal to the expectation value of the corresponding time ordered expression of quantum fields. So in particular equation (4) is correct. But RHS of (4) is still not zero. I guess the rule is that whenever there is an integral or differential operator expression inside time ordering then one should use discretized version of the expression appearing inside time ordering. So the issue is to correctly interpret the time ordering expression, and not the path integral
Aug
20
comment Introduction to spinors in physics, and their relation to representations
There are two different notions of a "vector". First is the definition used by mathematicians according to which any element of a vector space is a vector. Another definition is what is often used by physicists according to which a vector is that which behaves as a vector under rotations. Spinors are of course vectors in mathematical sense however they are not vectors in physics sense.