Recent content by Thoros

The photon doesn't split in two, in fact the statement is meaningless. A photon is an excitation of the electromagnetic field which by definition cannot be split as it's already the fundamental quanta of the field. Instead it's wave function is scattered through the slits which adds up as an...

I was reading this paper http://venables.asu.edu/quant/Gas/gassupp16.pdf on the Aharanov-Bohm effect and became confused regarding the formula (16A-7) which reads e \hbar \vec \nabla \Lambda (\vec r) + e \vec A = 0. Now i understand that (16A-7) can only be true if ## \vec B = 0 ##. This is...

Thank you, Kurros, for the excellent reply. You seem to be correct, the indices L and R, identifying chirality, do not correspond to real handedness. Handedness is given by helicity. In the massless limit the two become identical. I quess the way to understand this is that in some way, L...

Am i correct when i say that the fermions get a mass and interaction term with the Higgs from the SU(2)_{L}\times U(1)_{Y} invariant Yukawa interaction -g_{y}\bar\psi_{L}\phi\psi_{R} - g_{y}\bar\psi_{R}\bar\phi\psi_{L} where \psi is the fermion field and \phi the Higgs field. My...

How did you get exp[iHt/h]exp[-iHt/h] on either side? The correct way should be: In the Heisenberg picture you have <0|x(t)x(to)|0> So using the evolution operator to switch the states to the Schrödinger picture you get: <0|exp[iHt/h] x(t) x(to) exp[-iHt/h]|0> = <0|U+ x(t) x(to) U |0>...

The photon is the quanta of the EM-field just as the electron is a quanta in the electron-field. These fields are fundamentally different and require distinct approaches. I think though that if you are really interested in exploring the possibility that you pointed out, i would search for papers...

Just check if your answer satisfies the inital condition and determine the constants A and B from normalization.

In the Heisenberg picture all time dependence is attributed to the operators, the states carry no time dependence. In the Schrödinger picture this is reversed, the operators are constants but the states change in time. The connection between the pictures is established by requiring physics to...

I would use the given current density to determine the vector potential \vec A which is related to the B and E fields by: \vec B = \nabla \times \vec A \vec E = - \frac{\partial \vec A}{\partial t}- \vec \nabla \phi After the gauge choice div A = 0 i got after omitting constants, like...

Yes, but beware. Since your state vector components are allowed to be complex. So a normal scalar product won't do, because it may become negative. This if fixed by the convention that you usually take the left column vector, transpose it to a row vector and take the complex conjugate of each...

The (1 0) and (0 1) are the eigenstates of spin. Also multiplication of states is ambiguous unless explicitly saying what do you mean under multiplication. You can basically multiply them in two ways. Either to get a scalar using a scalar product, or a bigger object using a tensor product our...

Good point, i think the word "iteratively" was used by my professor. But i discussed it with others and they were also confused about the perturbative/iterative part. As of now, i still have no real progress. Edit: I must thank you for brining this up. I ignored that part of the question and...

When you measure an observable of an arbitary system, for example it's spin, you will get a result which is an eigenvalue of this observable, that is to say it's operator representation. Thus the arbitary system collapses into it's appropriate eigenstate corresponding to the measured eigenvalue...

Homework Statement Given the gauge invariant Dirac equation (i\hbar \gamma^\mu D_{\mu} - mc)\psi(x, A) = 0 Show that the following holds: \psi(x, A - \frac{\hbar}{e} \partial\alpha) = e^{i\alpha}\psi(x, A) Homework Equations The covariant derivative is D_\mu = \partial_{\mu} +...

Alright so i reached the point where you get an interaction term in the lagrangian density leading to the inhomogenuous pair of Maxwell's equations. But to me the intrudiction of a covariant derivative is a little confusing. It seems perfectly reasonable to require that physics stay the same...