Recent content by Jesssa

Oh I see, Thanks for your answer! However I am still having trouble figuring out how the negative sign of the coefficients depends on j1 and m1 in this example. (-1)j1 + m1 From application of either of the ladder operators you get Cm = -Cm+1 But from this it's not clear...

Hello, I was wondering, is there a way to derive the expression for these coefficients without the use of the general CGC formula? For example, the J=0 case (this is taken from wikipedia) The 1/sqrt term is clear but the (-1)^.. term not so much, is there a way to find this...

Thanks for replying HallsofIvy, I understand what you have posted, but do you know how you get the (-1)^{j_1- m_1} term in the expression? I know, if you consider the formula it comes from the Wigner 3-j but I'm interested in finding out where it comes from (intuitively, if it is possible)...

Hello, I'm not sure whether I should have posted this in main Quantum mechanics thread because it's not really regarding homework, but I posted it in here just to be safe. I was wondering, is there a way to derive the expression for these coefficients without the use of the general CGC...

Ah okay cool, Thanks! So \sum_i R_{ip} R_{iq} = \sum_i R^T_{pi} R_{iq} = \delta_{pq} Then \sum_i \sum_p \sum_q R_{ip} J_{1p} R_{iq} J_{2q}= \sum_p \sum_q \delta_{pq} J_{1p} J_{2q} = \sum_p J_{1p} J_{2p} = J_{1x}J_{2x} + J_{1y}J_{2y} + J_{1z}J_{2z} Which is what I needed...

Do you mean that the R's are orthogonal (since they're rotation matrices) so R^T_{pi}R_{iq} = \hat{1} Is that correct? It seems like the indices would make it incorrect. Haha, I don't know why but the indices are giving me the most confusing, How come it is okay to write...

I am having a bit of trouble seeing what the trick is, Does R_{ip} J_{1p} R_{iq} J_{2q}=J_{1p} J_{2q} Because the whole system is being rotated by the same rotation? Then p and q both run from 1-3 right, (x y and z)? But I'm really not sure, Or is it something like R_{iq}=R^T_{ip} ...

Cool thanks alot! So this reduces D_{1}^{\dagger }{{J}_{1,x}}{{D}_{1}}D_{2}^{\dagger }{{J}_{2,x}}{{D}_{2}}+\,\,D_{1}^{\dagger }{{J}_{1,y}}{{D}_{1}}D_{2}^{\dagger }{{J}_{2,y}}{{D}_{2}}+D_{1}^{\dagger }{{J}_{1,z}}{{D}_{1}}D_{2}^{\dagger }{{J}_{2,z}}{{D}_{2}} To...

Hey Oxvillan, Thanks a lot for your reply, Sorry what I was trying to ask was if the tensor product between D_1 and D_2 is equal to the multiplication of the two rotation matrices, D_1 D_2. I have worked out why this is true now. Sorry I just realized I was inconsistent with my indexes and...

another way to think of it, Sy|a> = <a|Sy Sy|a> = matrix times column vector = column vector <a|Sy = row vector times matrix = row vector it is not possible that Sy|a> = <a|Sy

Do you mean that, Sy|a> = a |a> where a is real? <a| Sy = row vector times matrix = row vector

They just added them together |I> + |II> = 2/sqrt(2) |1> = sqrt(2) |1> So dividing by sqrt(2) gives what you have posted.

Hey, I have a question regarding the invariance of a 'mixed' Casimir operator under rotation, By 'mixed' Casimir operator I refer to: \vec{J}_1\cdot \vec{J}_2 Where J1 and J2 are two independent angular momenta. I want to show that this 'mixed' Casimir operator is invariant under...

thanks for replying! sorry if i am miss understanding something, but it seems like you are describing the steps of part a of the question. i'm having trouble seeing how the sign of r(a) and r(b) effects P(x) = p'(x) + kp(x) (part b) since here the zero of P(x) is in the interval (a,b)...

hey i'm trying to figure out how to approach part b of this problem, http://imageshack.us/a/img850/6059/asdasdno.jpg so i can see that you can apply the mean value theorem to p'(x) so there exists some c between a and b such that f'(c) = [f(b) - f(a)] / (b-a)=0 so p'(x)...