Recent content by Manilzin

Hi, cerburos, welcome! I'm pretty new here myself, so it's not like I'm the proper representative of these forums, but anyway... if there's anything you want to chat about, I'll chat! I'm something equivalent to a physics major I guess, so it seems I'm on a different level of learning, but you...

Your solution seems perfectly reasonable to me, it only differs from the textbook one with a phase factor, it seems, and as we know, for real physical quantities such factors doesn't matter. So I'd say you can use your solution.

A spinor is an egienstate of the spin operator, which, in the case of spin 1/2-particles, can be represented as a two-component vector. You don't construct the spin operator from the spinors - you can construct a representation of the operator by finding its matrix elements, or by finding its...

I'm not sure how to actually measure J^2, but for your second comment: It is possible to measure both J^2 and the x or y (or z) component of J. It's not possible to measure all of them at once, though, since the x,y,z-components don't commute. So you can measure J^2 plus one of either Jx, Jy or...

I agree, this seems to be the effect we learn about in basic quantum mechanics. The comment by Griffiths, and the fact that this "antitunneling" was given as an exercise in his textbook, as stated in this text, indicates that this indeed is not a new effect. Perhaps the novelty of this is that...

Well, thanks, I guess that explains part of it. But as you said, what is the corresponding reasoning for spinor fields? Why does the solutions to the Dirac equations describe spin 1/2-particles?

Consider this quote from Mandl and Shaw, p. 237 ...this interaction coulpes the field W_{\alpha}(x) to the leptonic vector current. Hence it must be a vector field, and the W particles are vector bosons with spin 1. Could someone explain this for me? I do not understand the "hence"...

In your equation, \sigma_{\mu\nu} means the mu'th-nu'th component of the tensor (or matrix) sigma. When you have an expression like \sigma_{\mu\nu}F^{\mu\nu}, Einstein's summing convention is implied - that is, you should sum over repeated indices, in this case mu and nu, from...

Hm, yes, the law of association holds for matrices, so those two last expressions are equal.

Hm, could you post the exact argument given by this person? I agree, if it is a momentum eigenstate, measurement of momentum should give the exact eigenvalue...

I agree with all this, and the fact that the Hamiltonian generates time translation is easily seen from the Schrödinger equation itself. What I mean is that the Hamiltonian and the time derivative operator mentioned are not the same operators. They are related through the Schrödinger equation...

Ah ok, with time derivatives it makes more sense. However, I still disagree that this is the "energy operator". Nor is it the Hamiltonian, any more than an eigenvalue of H is the Hamiltonian. Rather, it's just what gives the time evolution of a system.

I'm not sure I understand you correctly, also, your second image didn't show. But I have never seen E being represented as an operator, it has always been the eigenvalue of the operator H. Also, the operator you wrote has the dimensions of momentum, not of energy, so I don't understand how this...

I have an idea, but I'm not sure whether it will be helpful.. I mean, if you rewrite sin(kr) as (exp(ikr)-exp(-ikr))/2i, then you have something on the form k*exp(f(k)). Then maybe you can use integration by parts to integrate? Maybe it won't work...

No, P(x) is the probability function, which is the wavefunction squared (actually, absolute value squared)... So P(x)dx gives the probability for finding the particle on a bit of length dx at position x.