Recent content by Hennessy

yeah unfortunately my lecturer didn't give any reasoning , we asked why and he just said it was a "standard result" so i understood that as just " just take it as a fact" :/, not the best for learning but i've learnt to take some things on the chin on my degree when i don't understand them...

Hi Hill, Okay this makes more sense now and i see that we can pull out the 1/2 for sqr3/2 aswell so thank you for clarifying this and helping me be able to answer it appreciate it :).

Hello All, 2nd year undergrad taking my first course in modern physics. We have been given this question in a mock exam and at the bottom is the solution. When looking at a general cubit it seems the argument of both sin and cos functions should be (pi/2) not (pi/3). I have figured out a...

saved this now , thank you!

tried putting it into an online integral calculator and it came back saying erg functions was all. So perhaps i should try to use complex forms to try and make some of the integrals easier as in general exponentials are easier to work with ?

Yeah thats what ive learned so far on my course. Its more of a attempt the question yourself and when your really completely stuck look at someone else who has solved something similar and try to map it on to my problem and justify the reasoning. I appreciate being made aware of tools like this...

This is now the updated Answer. $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ satisfies a free particle in a infinite square well. For the ground state we have $$ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi n x}{L}\right)$$ where n = 1 for the ground state and so...

DUDE WHAT THIS IS HACK ON LIFE BROOOOOOOOO, thank you so much

so perhaps to simplify the integral after using a trig substitution is to use a change of variables to integrate it?

Hello so the correction i made was that i changed $$\frac{1}{L} \left[ x - \frac{L}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L}$$ and changed it to $$\frac{1}{L} \left[ x - \frac{1}{2\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_{0.4L}^{0.5L}$$ this didn't change the value...

apologies sin^2(x)=0.1 or sin^2(x)=0.19 , a very narrow slice, and 0.19 is a bigger width

I have recalculated the integrals and now i have that the probability of finding the particle at the said point is 0.19 ?

Apologies, i thought as it was introductory quantum mechanics i felt that was the most appropiate place to putt it, apologies.

yeah Sin^2(0.1x) looks very unreasonable, so now i need to figure out where i went wrong :(

what i was thinking , i had two combative thought, one was it makes sense as its based in the middle of the function which to me is towards the peak of the wavefunction and so its a thin slice around the peak of the wavefunction which made it a low probability. My other thought was although P...