Recent content by Pchemaaah

d/dz (z*Psi)=Psi(z)+zPsi'(z) Ahhhh ok I totally forgot about product rule and was stopping the derivative operator before psi. Thank you so much!

Yes it is the derivative with respect to z. So if I have a dummy function psi and I take the derivative of it with respect to z, then I get Psi'(z) (prime of z) or I was calling it Psi(dz) in the event that it had other variables (d/dz psi(x,y,z)=psi (x,y,dz)). If this is incorrect notation I...

Well I have a hard time typing it out, but it is just Psi(dz) instead of Psi(z) when d/dz is applied to it correct? And can you show me where I am going wrong or how much further I have to go to get to that result because that will essentially enable me to solve this problem since I keep...

Ok, so even if I do put a function in I don't see how you get your result: (zP_z-P_z z)psi(z)=(z*(-i*hbar*d/dz*psi(z))-(-i*hbar*d/dz*z*psi(z))=-i*z*hbar*Psi(dz)+ihbar*Psi(dz)=(-i*z*hbar+i*hbar)Psi(dz) Where is i/hbar coming from?

Hmm without using that linearity equation then I would look for something to expand or factor to see if I can get some sort of cancellation, but since we are working with operators that strategy doesn't work out. I think there might be some way to split the left side integral apart based on the...

Before I do that, can you show me how it ends up being i/hbar? I cannot for the life of me get to that answer which is probably the root of my difficulties haha. Ill show my steps I guess so you can point out where I am wrong: zP_z-P_z...

So does that mean that [tex]\int (A(f+g)-(Af+Ag)) \overline{(A(f+g)-(Af+Ag))} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex] becomes [tex]\int (A(f+g)-(Af+Ag)) \overline{(A(f+g)-(A(f+g)} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex] [tex]\int (A(f+g)-(Af+Ag)) \overline{(0))} = \int |A(f+g)-(Af+Ag)|^2 = 0[/itex]...

Im sorry but I don't understand what you did. It seems like you threw up an arbitrary operator S and used that to define the solution to (z*P_z -P_z *z) as itself. If it is not 0 or i(hbar), then what is its exact solution?

Like I said, we haven't addressed how this applies to space, only in paper/calculus form so I don't know any better. And I am pretty far removed from the realm of proper math expression (over 3 yrs since I took any math class) so forgive me for these technical errors.

I guess what I mean is that seems really arbitrary to pick what you did and I do not understand why/how you would come to pick that since my class has solely been working with integrals etc and hasnt even touched vectors/dot products. Edit--I am sorry but my level of understanding is really...

My class isn't using dot products or anything like that to do anything. We have just been mired in calculus with little to no instruction on how to use it. Why do you want to involve a scalar? Is it possible to show this without using dot products/other 3D tools?

Thank you for the response and sorry I did not explicity post the definitions. I understand that you defined A to be hermitian as demonstrated by your first set where <Af,g>=<f,Ag>, but I have no idea what the second part means. What is lambda? Why is the same thing on both sides and why did...

Is the solution to your calculation 0 or zP_z+i(hbar)? This is where my operator math gets fuzzy. And I extracted that equation from a larger problem that asks me to prove that the angular momentum operator (L) squared commutes with each component of the angular momentum. The book goes about...

Homework Statement I am basically trying to show that LxLy-LyLx=i(hbar)LzHomework Equations Lx=yPz-zPy Ly=zPx-xPzThe Attempt at a Solution I get to the end where I have i(hbar)Lz-z*y*PxPz+z*xPyPz. How do I get these last two terms to cancel out? I am not too strong in operator math (it hasnt...

Homework Statement Simply--Prove that any Hermitian operator is linearHomework Equations Hermitian operator defined by: int(f(x)*A*g(x)dx)=int(g(x)*A*f(x)dx) Linear operator defined by: A[f(x)+g(x)]=Af(x)+Ag(x) Where A is an operatorThe Attempt at a Solution I am at a complete loss of how to...