Recent content by Zero1010

Thanks for the reply. Sorry that's my typo it should be: ##0 \lt r \leq b## and the second one should be ##r \gt b## Also in the question it says to suppose there is a deviation from Coulombs law at very small distances (even though there is no evidence for it)

Hi, I just need someone to check if I am on the right track here Below is a mutual Coulomb potential energy between the electron and proton in a hydrogen atom which is the perturbed system: ##V(x) = \begin{cases} - \frac{e^{2}}{4\pi\epsilon_{0}}\frac{b}{r^{2}} \text{for } 0<r \leq 0 \\ -...

That all makes sense and good proof of the answer. I have looked at the approach in #13 and #23 and I can now see where that factor came from. thanks

Couldn't resist. I understand the approach here and it is so much clearer than what I was doing before, the only thing I'm not clear on is the ##\frac{2}{L}## factor you have in the approach above. When I worked out the spatial wave function at the start I multiplied the ##\frac{\sqrt2}{\sqrt...

Thanks for all your help and advice with this. Its gotten me so much closer to an answer than I would have been. Time is becoming a bit of an issue as this is due in on Monday and I have one more question to look at. So I'm going to leave this for now and hopefully come back to it over the...

No but the factor I am working with from the probability density is ##\frac{1}{L}## not ##\frac{2}{L}## in which is in post 13. Should I be using ##\frac{2}{L}##?

Of course. Any hints where?

I've had a bit of time this morning to look at this and I have had another go. Here is my attempt: ##\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{\pi x_{1}}{L})dx_{1} = \frac{1}{L} (\frac{L}{4}) = \frac{1}{4}## ##\frac{1}{L}\int_0^\frac{L}{2} sin^{2}(\frac{2\pi x_{2}}{L})dx_{2} = \frac{1}{L}...

Thanks for the tip, makes it all a lot clearer and easier to set out. Thanks for all your help so far. I'm going to have to leave it for now and come back to it tomorrow but I'm happy I'm much closer to a solution now.

Ok thanks. I'll take another look at it.

Sorry, missed the last bit: ##\frac{L^{4}}{18\pi^{2}}##

Ok, fingers crossed: For the quadratic expansion I now get (fixing the factor and 2 in front of the second term): ##\frac{1}{L}[sin^{2}(\frac{\pi x_{1}}{L})sin^{2}(\frac{2\pi x_{2}}{L})+2sin(\frac{\pi x_{1}}{L})sin(\frac{\pi x_{2}}{L})sin(\frac{2\pi x_{1}}{L})sin(\frac{2\pi...

You're a life saver. Its amazing how it sometimes only takes one or two points to open a question up. I think I have it now. Thanks again

Ok if possible it would be great to check over my work here and maybe some advice on how to proceed as I've hit a brick wall with some integration. In the next part of the question I have two indistignishable particles in an infinite square well with walls at x=0 and x=L. Firstly, I need to...

Cool thanks. I'll have another look at the second part of the question tomorrow.