Recent content by hnicholls

Got it! Thanks! Did not correctly apply l = 1 as the subspace and the m values (1,0,-1) as the basis states.

Applying the general formula $$|\hat{L}_+ |l,m \rangle=\sqrt{(l-m)(l+m+1)} |l,m+1 \rangle.$$ $$|\hat{L}_+ |l,m \rangle=\sqrt{(0-(-1))(0+(-1)+1)} |l,m+1 \rangle.$$ $$|\hat{L}_+ |l,m \rangle=\sqrt{(+1)(0)} |0,0 \rangle.$$ $$|\hat{L}_+ |l,m \rangle=\sqrt{0} |0,0 \rangle.$$ I don't see how...

L+ with l = 1 and m = -1, 0, 1 L+|0,-1> = √[0(0+1)-(-1)((-1)+1)]|0,-1+1> = 0ħ|0,0>. This would be a 0 value not √2ħ. Not sure why this wrong.

L+ with l = 1 and m = -1, 0, 1 L+|0,-1> = (0(0+1)-(-1)((-1)+1))^1/2|0,-1+1> = 0ħ|0,0>. This would be a 0 value not √2ħ. Not sure why this wrong.

In calculating the matrix elements for the raising operator L(+) with l = 1 and m = -1, 0, 1 each of my elements conforms to a diagonal shifted over one column with values [(2)^1/2]hbar on that diagonal, except for the element, L(+)|0,-1>, where I have a problem. This should be value...

Sorry part of Ψ did not load Ψ = [ βxe −3iωt/2 + []e −5iωt/2]

I have the following non-stationary state of the QSHO: Ψ = [ e −3iωt/2 + []e −5iωt/2] where β = mω/ħ in calculating <E2> The answer I see in the textbook is 6.17 ħ2ω2. This answer suggests that in calculating <E2> = ∫ Ψ*Ĥ2Ψ dx where Ψ is composed of the above two terms Ψ1 and Ψ2 (a linear...

(n/2)!/[(n+2)/2]! (n/2)! = (n/2)(n/2 - 2/2) = (n/2)[(n-2)/2] [(n+2)/2]! = [(n+2)/2][(n+2)/2 - 2/2][(n+2)/2 - 4/2] = [(n+2)/2][(n/2][(n- 2)/2] cancel out the (n/2)[(n-2)/2] and you are left with 1/[(n+2)/2] or 2/(n+2) Thanks DrClaude. The last point was a BIG help. Henry

(k+1)k(k-1)(k-2) etc.

I'm not clear how a substitution of k = (n/2)! would help.

I am trying to work through a simplication of this factorial with variables: (n/2)!/[(n+2)/2]! I get, 2[n(n-1)]/2[(n+2)(n+1)n(n-1)] cancelling the 2[n(n-1)] leaves me with 1/[(n+2)(n+1)] However, Wolfram Alpha tells me this can be simplified as 2/(n+2) and I don't see that. Thanks

I have. Are you subtly referring to ψ(x) = Aeipx/ħ oscillating between +1 and -1 as | x | → ∞, i.e. that the solution to the TISE for a free particle is not normalizable?

Thanks to everyone for all your help. I finally was able to calculate full reflection and zero transmission qualitatively for E = V in a manner analogous to the reflection and transmission coefficients for E < V and E > V. At the risk of irritating Simon further, I still had one issue on which...

I've studied vector calculus and linear algebra. By the way, you stated previously, "In your original wavefunction, the probabilities blow up as x gets large ... which is nonsense and a good clue you got it wrong. I couldn't answer your question until you got that. An easy check would have...

I have a mediocre understanding of undergraduate calculus and differential equations.