Recent content by hasan_researc

Hi, here are my credentials. 2008-2010: First two years of BSc physics at Imperial College London - 89% in my first year and 69% in the second year. 2010-2012: took gap years due to financial difficulty 2012-2013: Final year of BSc Physics at Imperial College London - expected to obtain...

What do you think?

So, this is what I've got to so far! \psi(x, t) = \frac{ \sqrt{2}u_{1}(x)e^{-iE_{1}t/hcross} + u_{2}(x)e^{-iE_{2}t/hcross} }{ \sqrt{3} } \phi_{1}(x) = \frac{\sqrt{2}u_{1}(x) + u_{2}(x)}{\sqrt{3}} \phi_{1}(x) = \frac{u_{1}(x) - \sqrt{2}u_{2}(x)}{\sqrt{3}} Solving the second and third...

Whow! This looks so easy! I'll post my solution in a minute.

OK, let's be clear. The probability is given by the square of the modulus of the relevant coefficient. And given what you have said right now, the coefficient is the exponential times the real number (associated with each u(x)) ?? Am I right? If I am, then the exponential cancels out...

Or e^{-iE_{2}t/hcross} for that matter?

Does e^{-iE_{1}t/hcross} have no effect whatsoever on my calculation?

I am sorry, Arya svit-kon, but would you allow me and vela get to through the end of the problem, please?

\lim_{z\rightarrow 0} \left(\frac{\sin z}{z}\right)=1, so \lim_{z\rightarrow 0} \left(\frac{2}{\pi z}\right)^{1/2} (\left \sin z \right) = \lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2}\left(\frac{\sin z}{z}\right) = \lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2}...

OK, so (b) possible outcomes of the second measurement = E_{1} and E_{2} and probability of E_{1} = 2/3 and E_{2} = 1/3 . What do you think?

I don't know how to solve for (b). Any ideas?

\psi(x, t) = \frac{ \sqrt{2}u_{1}(x)e^{-iE_{1}t/hcross} + u_{2}(x)e^{-iE_{2}t/hcross} }{ \sqrt{3} } What do you think?

I will have a think about this method while we move on to the next part of the question. 2. By means of the substitution z = kx, show that J_{1/2}(kx) and J_{-1/2}(kx) are solutions of the following equations: x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + \left(k^{2}x^{2} - p^{2}\right)y = 0...

Thanks for your help! I don't understand what 'order' means in this context. How do we compare the orders? "an estimate of the form J_{1/2}(z) = O(f(z)) as z \to 0 for an appropriately chosen bound f.": I don't understand!

Homework Statement This is the how the question begins. 1. Bessel's equation is z^{2}\frac{d^{2}y}{dz^{2}} + z\frac{dy}{dz} + \left(z^{2}- p^{2}\right)y = 0. For the case p^{2} = \frac{1}{4}, the equation has two series solutions which (unusually) may be expressed in terms of elementary...