Recent content by alexgs

Thanks. That makes a lot of sense. So just define P with a + or - so as to give the usual operator from regular quantum mechanics (P^i = -i d/dx^i, for i=1,2,3). The unitary operator that shifts the field for that choice of P is exp(+i a_\mu P^\mu) (using the +,-,-,- metric).

Some basic questions: When you write (T.\phi)(x) is this the same as [\exp(-i a_\mu P^\mu)\phi](x)? And if so, would T^{-1}.x = x-a or x+a? If you can write T.\phi in terms of P^\mu, is that enough to get the form of P^\mu using samalkhaiat's Eq.1? If so, can someone write it out...

Not sure what you mean by "compatible" and "consistent". If you are saying that the equality with the substitutions you mention is consistent with the third equation of my post I'm pretty sure that's not true. The reason I suggested the substitutions I did was so that the left hand sides would...

I am still confused because it looks like the operator P^\mu is transforming the coordinate, not the function, in Polyrhythmic's last equation. Forgive me for making things painfully explicit but perhaps someone can spot the exact source of my confusion if I write everything out as clearly as...

Hi Polyrythmic. Can you explain why what I did is flawed? I agree that the line you quoted is where my derivation differs from Maggiore's by a minus sign, which leads to the difference in my result for P^\mu . I do not understand Maggiore's Eq. 2.108, which I will copy here: \phi'(x'-a)...

Hi, I am working through Maggiore's QFT book and have a small problem that is really bothering me. It involves finding the representation of the Poincare algebra generators on fields. I always end up with a minus sign for my representation of a translation on fields compared to Maggiore...

As in the last two posts, the most elegant and general way to see that <p> comes from the fact that p is a hermitian operator. But if you only know about p in position space (i.e. p = -ih d/dx) then you can see that <p> is real by showing that <p> = <p>* (is this statement obvious?). Do...

You are right, when V>E the wave function is a dying exponential. This situation is never happens classically but in quantum mechanics this usually manifests itself as the particle being able to tunnel into a region where V>E. The dying exponential indicates that you are less and less likely to...

Even if you "scale down" deltaX I believe the proof still breaks down: As long as deltaX is finite (.1 or .01 or even .0000000001) the proof doesn't work because sin(deltaX) does not exactly equal deltaX and cos(deltaX) does not exactly equal 1. But your manipulations are still valuable...

Hi, First, you have to convert your second equation into a an equation with bx, by, bz. For your third equation how about one saying that the length of the new vector B is the same as the length as the original B. Another way you might think about the problem is to find the direction...

Your formula does work for n=1/2. (If you square 2^(1/2-1) [1,1;1,1] you get back your original matrix.) In fact, the formula works for n=1/b for any positive integer b. Since you know it works for positive integers it also works for n=a/b for positive integers a/b. So it works for positive...

I think there's something called Cramer's rule that tells you how to construct the inverse of a matrix in terms of determinant of the matrix and the cofactor of the matrix. Since both of these objects are polynomials the entries of inverse matrix are just polynomial functions of the entries of...

It might be easier just to try to figure out how a basis of k-forms transforms. I.e. what is the dual of e_1 ^ e_2 ^ ... ^ e_k? Then you can show that that the Hodge dual is linear and write your form A in terms of the basis.

What exactly do you mean by isomorphic?

Hi, I've got two related questions. You can decompose a (semisimple) lie algebra into root spaces, each of which are 1-dimensional. If X has root a and Y has root b then [X,Y] has root a+b. If the root space of a+b is not zero (i.e. there is a root a+b) then is it possible for [X,Y] to still...