Recent content by maverick_76

Yes I am looking for the bound state of a particle. Awesome! Thank you!

Okay so I am trying to solve a delta potential well with an infinite potential wall on one side a distance a away from the well. The other side is open so I am confused about how to set up the problem. Here is a picture of my work so far and if anyone has an insight into this I'd appreciate some...

Thank you, I was talking with my professor this morning and he said the same thing that tackling the problem with exponentials is much easier than using trig functions. The problem also is much clearer to understand this way too.

How is that possible? If k=n, n=1,2,3... then it has to equal 1/sqrt(pi) right? A^2 + B^2 = 1/sqrt(pi) is what I mean, how can it equal 0?

Okay so solving the integral of modulus psi squared, I find A^2 + B^2 = (1/pi) How exactly do I find A & B values with this, If I can't assume one is zero? I know: psi(0) = B = psi(2pi) & psi'(0) = Ak = psi'(2pi) but I am rusty here, can I use this info to solve for A & B?

So I am working on the problem of the particle bound to a ring of radius R. I am trying to solve it two ways, as a standing wave and as a running wave. I'm stuck right now solving for the standing wave. So far I have: ψ(x)=Asin(kx) + Bcos(kx) I know that it is periodic from 0 to 2π so if I...

My thoughts are g(P) unpacks as such: -iħ∇(p) (the partial derivative with respect to P) so g(P)r would be: -iħdg/dP I mean using the logic from the previous solution this allows for the second expression to hold true. Not sure if I'm following the math rules though...

Ah I see now. So now how would I go about proving this expression: [g(P), r] = -iħ dg/dP I'm not sure how the del operator works inside a function.

Maybe I'm missing something here, because my limited knowledge of commutators says that the left hand side unpacked is { Pg(r)-g(r)P } Ψ(r) I'm not sure how to show that g(r)iħ dΨ/dr = 0 in order for the expression above to hold true.

so I have an expression here: [P,g(r)]= -ih dg/dr P is the momentum operator working on a function g(r). Is this true because when you expand the left hand side the expression g(r)P is zero because the del operator has nothing to work on?

Okay that makes sense. Thanks!

Here is the original integral and resulting error function

Okay so I was integrating an expression and ended up getting an imaginary error function in the answer. I'm not sure where to go from there, I plugged it into wolfram and the root it gave me looks nice but is that worth anything to me? The integral was being evaluated from -∞ to ∞, would I need...

okay gotcha. Thanks!

okay so the limit would be lim_k->∞ sin(kx) xHow would I go about evaluating it? Is there a substitution trick?