Recent content by Logan Rudd

Ah great. Thanks!

Homework Statement Consider the following state constructed out of products of eigenstates of two individual angular momenta with ##j_1 = \frac{3}{2}## and ##j_2 = 1##: $$ \begin{equation*} \sqrt{\frac{3}{5}}|{\tiny\frac{3}{2}, -\frac{1}{2}}\rangle |{\tiny 1,-1}\rangle +...

Sounds like a good plan. It looks like the IOIO will do everything I need it to. I'm looking forward to seeing what I can do with it. Thanks for the idea!

It sounds pretty ambitious given the amount of time I'll have to complete it and my lack of experience with microcontrollers and programming, but I think I could pull it off if I can get a good head start on things over winter break and find a good lab partner next semester. I might as well...

I actually really like that idea! Are you saying I could transmit data collected from the sensors by my Edison module via my cellphone service to my laptop by interfacing the Edison and cellphone to the IOIO-OTG? My only concern is not being familiar with Java. I'm finishing my first programming...

As an experimentalist, I am very excited to be taking my first upper division physics lab next semester! The course covers basic electronics (filters, diodes, transistors, op-amps, analog & digital circuits, D/A conversion, and LabView Programming, etc.) and measurement techniques with an...

After this I am trying to figure out what ##\Delta\left(\frac{d\psi}{dx}\right)## is. Integrating the potential part of SWE and taking the limit as ##\epsilon## approaches ##\pm a## I get: $$ \Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m}{\hbar ^2}\left[\alpha\psi(a)+\alpha\psi(-a)\right] $$...

1)So from my understanding, as long as ##E>0## you will have scattering states and these scattering states will always result in an imaginary ##\psi##, but bound states can also have an imaginary ##\psi##? Is this correct and or is there a better way of looking at this maybe more conceptually...

Thanks! That makes it much clearer. I figured it was a typo but was kind of confused to begin with.

I'm working on a problem out of Griffith's Intro to QM 2nd Ed. and it's asking to find the bound states for for the potential ##V(x)=-\alpha[\delta(x+a)+\delta(x-a)]## This is what I'm doing so far: $$ \mbox{for $x\lt-a$:}\hspace{1cm}\psi=Ae^{\kappa a}\\ \mbox{for $-a\lt x\lt...

Ahh, I see! But if the potential I am working with is the sum of two negative delta potentials then would there be two bound states? I'm trying to work it out in a similar fashion as the text works it out for a a single negative delta potential centered at 0 but since both of mine are centered...

Why is that, and why is it not the case for a negative delta-function potential?

I'm reading through Griffiths Intro to QM 2nd Ed. and when it comes to bound/scattering states (2.5) they say: ##E<0 \implies## bound state ##E>0 \implies## scattering state Why doesn't this change depending on whether you have a positive or negative delta-function potential?

I think I see my confusion. It looks as if ET is being applied to ## \psi(x) ## but I think its just because the time dependent part disappears when you multiply it by its complex conjugate. Thanks!

I know when the initial state (##\Psi(x,0)##) is given, ##\frac{d<x>}{dt} \not=<p>##. I thought you can only apply Ehrenfest's theorem when ##\Psi## is a function of x and t, however it seems like you can also apply it to the time-independent part (##\psi(x)##) by itself as well. Can someone...