Recent content by heardie

Benny for 2) as well simply compare coefficents. The product of omegeas is (-1)^n, the right hand side is -1

2002 VCE Specialist Maths exam Benny?

Good to see some sensible, rational discussion in here, instead of childish name calling. Much better :)

Can Marcus even read? Or is he just ignorant!

At a guess its the Stern-Gerlach experiment being discussed. See http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html There is a discussion of Stern-Gerlach about 1/3 of the way down the page. Additionally you could Google it

The homogenous equation: \frac{d^2y}{dx^2}+xy=0 is a negative sign off the Airy equation: \frac{d^2y}{dx^2}-xy=0 Therefore the solution of the original DE \frac{d^2y}{dx^2}+xy=x^2 is given by y = CAiryAi(-x) + DAiryBi(-x)+x where AiryAi and AiryBi, are independent solutions of the...

the force is mediated by (virtual?) W and Z particles I believe

The verificiation of the De Broglie formula (I am lead to believe) comes from considering a de Broglie wave: $\psi ({\bf{r}},t) = Ae^{i({\bf{k}}.{\bf{r}} - \omega t)} $ If you assume the relationshup E = \hbar \omega holds for material particles you then write E = \hbar \omega =...

I'm not too sure with 1, but the first order perbutation correction is given by the mean of the perubtation, so if the state is symmetrical about 0, the mean (and therefore the permutation) will be zero.

the math used is hardcore - looks like a lot of tensor calculus to me

I can't see why the Hamiltonian can't be a matrix. We could still solve the eigenvalue problem (most people would be more used to solve matricies with eigenvalues) And I guess it would have to be self-adjoint, to ensure it has real e'values, so what we measure (the e'value) is real.

Well I have the 'official' answersr now: 2.89eV and 1.45 eV In both cases the opening of the enegry gap due to the lattic interaction or Bragg reflection, is well away from the Cu Fermi energy of 7eV. The condunction band of Cu is always partially filled, and Cu maintains its metallica nature.

In fact $\frac{{2\pi }} {\lambda } = k$ when k is the wavenumber, so the k^2 substution can be made even without the knowledge of how it helps solve the DE.

If electrons are propagating in the [100] direction they are traveling perpendicular to a plane though the x-axis. Thus when they interact with that plane, they are reflected at 90 deg. What about all other planes? this is bothering me!

Dont worry. Completly missed something here. All makes sesnse now. Can I delete this thread somehow?