Recent content by wood

I get a positive number so I thnk I might be an idiot and have had the correct answer all along...ie $$\left(\frac{k_R-k_L}{k_R+k_L}\right)^2=\left(\frac{k_L-k_R}{k_R+k_L}\right)^2$$

No, I keep getting kL-kR when I try to solve for A and B algebraically.

I am working through a past exam paper and this one has me stumped 1. Homework Statement Consider a particle of mass m with kinetic energy E incident from the left upon a step-up potential: $$U(x)=\begin{cases} 0 & \quad \text{for } x <0\\ V & \quad \text{for } x>0\\...

And that's where the constants come in?

Are you saying I just separate the ^3 terms into three sin(nπx/a)

Thanks. I'm not sure how to write my wave function as a linear combination of eigenfunctions, not sure if we have seen that. Is there another way perhaps? Ps nice quote in your sig.

Homework Statement For a particle of mass m in a one-dimensional infinite square well 0 < x < a, the normalised energy eigenfunctions ψn and eigenvalues En (integer n = 1, 2, 3, ...) are $$ \psi_n(x) =\sqrt{\frac{2}{a}} sin \left( \frac{n \pi x}{a} \right) \;$$ inside the well otherwise...

So it should read ##<x>=\int x\Psi^*(x)\Psi(x)dx## and I and doing the correct thing? (thanks for helping by the way...)

To calculate ##\langle x \rangle## I used should have been $$<x>=\int \langle x\rangle\Psi^*(x)\Psi(x)dx$$ And I didn't check if the cross terms canceled I just 'presumed' they did as they were normalised and therefore ##\int\psi_1^*(x)e^{+iE_1t/\hbar} \psi_2(x) e^{-iE_2t/\hbar} =0## I think...

From ##\psi_1^*\psi_1## I get ##-\psi_1^*(x)e^{+iE_1t/\hbar} \psi_1(x) e^{-iE_2t/\hbar}## which gives ##\psi_1^*(x)\psi_1(x) ## and as they are normalized to one particle ##\int\psi_1^*(x)\psi_1(x) = 1 ## leaving me with just the coefficients. Yes??

I thought I had by calculating <x> using $$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$

I think I am getting somewhere...(thanks) for a function to be normalized I need ##\int \psi^*(x)\psi(x)\,dx = 1## so it is normalised and not stationary for 3 Here $$\langle E\rangle = \int \Psi^*(x)E\Psi(x)\,dx$$ spits out an answer that doesn't depend on t Here i need ## \langle...

Thanks very much I think I am getting the hang of what is going on. I don't have to treat ##\langle x^2 \rangle## any different to how I dealt with ##x## i.e.I just use ##x^{2}## when workingout ##\langle x^2\rangle ##

Homework Statement Consider an infinite square well defined by the potential energy function U=0 for 0<x<a and U = ∞ otherwise Consider a superposed state represented by the wave function ## \Psi(x,t)## given at time t=0 by $$\Psi(x,0) = N \{(-\psi_1(x) + (1+ i)\psi_2(x)\}$$ 1. Assume that...

That is what the equation was meant to look like. My problem is do I square the function ##\psi## and multiply that by the mod squared of the function or multiply the function by x2. In short i am not sure what x2 is in the above formula thanks