network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 2 years |
| seen | Feb 8 at 14:47 | |
| stats | profile views | 32 |
|
Feb 13 |
awarded | Tumbleweed |
|
Feb 6 |
asked | What is the difference between Mean Field Theory and Effective Medium Theory? |
|
Dec 13 |
awarded | Popular Question |
|
May 18 |
awarded | Yearling |
|
Mar 14 |
comment |
General procedure for Clebsch-Gordan expansions Thank you,I need to find the analogous equation that you gave for the spherical harmonics above, for harmonic functions on $S^3$; do you know of a book/paper where I can find the derivation you mentioned? |
|
Mar 14 |
accepted | General procedure for Clebsch-Gordan expansions |
|
Mar 14 |
asked | General procedure for Clebsch-Gordan expansions |
|
Jan 24 |
comment |
Coupling Coefficients in SO(4) I have noticed one thing that might help, the author of eq. 1 states that "The canonical basis states of the symmetric (class-one) irreducible representation $l=l_{(n)}$ for the chain $SO(n)\supset SO(n-1)\supset...\supset SO(3)\supset SO(2)$ are labelled by the $(n-2)$-tuple $M=(l_{(n-1)},N)=(l_{(n-1)},...,l_{(3)},m_{(2)})$ of integers $l_{(n)}\geq l_{(n-1)}\geq ...\geq l_{(3)}\geq |m_{(2)}|$". Also in his notation here I believe "SO(n) irreducible representation $l_{(n)}=l$ [has] SO(n-1) irrep labels $l_{(n-1)}=l'$". |
|
Jan 24 |
asked | Coupling Coefficients in SO(4) |
|
Jan 21 |
awarded | Commentator |
|
Jan 21 |
comment |
What is an isoscalar factor? Its in eqs. 4.6-4.7 from "Coupling coefficients of SO(n)and integrals involving Jacobi and Gegenbauer polynomials" by Sigitas Alisauskas, you can find it at iopscience.iop.org/0305-4470/35/34/307 |
|
Jan 21 |
asked | What is an isoscalar factor? |
|
Jun 29 |
comment |
Clebsch-Gordan Identity @Dan: The sums are over all valid values of the arguments, specifically $-a\leq\alpha\leq a, -b\leq\beta\leq b, -d\leq\delta\leq d$ |
|
Jun 28 |
asked | Clebsch-Gordan Identity |
|
May 26 |
comment |
Why are two eigen-state-kets with different eigenvalues orthogonal? yes, I was making this mistake. Thank you for clarifying, I think part of the confusion was the idea that the indices can take non-positive values (i.e. for $j_1=1$, $m_1$ can take any of $\{-1,0,1\}$ so $\vec{e}_m$ can be any of $\{ \vec{e}_{-1}, \vec{e}_0, \vec{e}_1 \}$. All of which are mutually orthogonal. Am I understanding this correctly now? |
|
May 26 |
comment |
Why are two eigen-state-kets with different eigenvalues orthogonal? @Peter: so I assume the proof would be something like: define $\left|j_1 j_2 m_1 m_2\right> = \sum_j \sum_m \left<j_1 j_2 m_1 m_2 \mid j_1 j_2 j m\right> \left|j_1 j_2 j m\right>$, where $\left<j_1 j_2 m_1 m_2 \mid j_1 j_2 j m\right>$ are the Clebsch-Gordan coefficients. Do the same thing for a different ket, say, $\left|j_1' j_2 m_1' m_2\right>$, then their inner-product is the sum of the products of their coefficients, which should be zero (I assume there will be some kronecker deltas in there somewhere too). |
|
May 26 |
comment |
Why are two eigen-state-kets with different eigenvalues orthogonal? @Peter: thank you for your explanation that was very helpful. So if I'm understanding you correctly the dual bra to a given eigenket (in my case a simultaneous eigenket) is defined to give 0 as their inner-product. If this is correct, then in practical terms, given some numerical values for $j_1, j_2, m_1, m_2, j, m$, how do I compute the inner product? (i.e. the definition we've imposed tells me what the inner-product of two orthogonal kets is, but it tells me nothing about what the inner-product of two general kets is so that I can check if they are orthogonal). |
|
May 26 |
comment |
Why are two eigen-state-kets with different eigenvalues orthogonal? I admit I'm a beginner and probably am making A LOT of beginner's mistakes. I know the derivation is incorrect, I want to know what I did wrong so that I can correctly understand. |
|
May 26 |
asked | Why are two eigen-state-kets with different eigenvalues orthogonal? |
|
May 25 |
accepted | state vector notation |
