- #1
paklin2
- 18
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- TL;DR Summary
- I’m curious if these matrices have been discussed elsewhere and if they have any significance. They may give some insight into how the different Spherical Harmonics interact because of the differential operators.
I’m New to the forum. I’m Interested if a certain set of matrices have any significance. To start out the unit vectors ##\vec i , \vec j, and ~\vec k ## are replaced with two dimensional matrices.
##\sigma r = \begin{pmatrix}1&0\\0&1\\\end{pmatrix}, ~\sigma z = \begin{pmatrix}1&0\\ 0&-1\\\end{pmatrix}, ~\sigma x = \begin{pmatrix}0&1\\ 1&0\\\end{pmatrix}, ~\sigma y = \begin{pmatrix}0&i\\ -i&0\\\end{pmatrix}##,
So ##r = x \cdot \sigma x + y \cdot \sigma y + z \cdot \sigma z## which is the same as ##\begin{pmatrix}z&x+yi\\ x-yi&-z\\\end{pmatrix}##
From the following differential operators ##L = \frac {1} {i} \vec r \times \nabla## with ##Lz = (x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x}),~Lx = (y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y}),~Ly = (z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})## a useful operator can be defined; ##Lxyz = Lx \cdot \sigma x + Ly \cdot \sigma y + Lz \cdot \sigma z## or ## Lxyz = \begin{pmatrix}
(x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})&(y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})+(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i\\
(y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})-(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i&-(x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})\\
\end{pmatrix}##
If the following matrix is multiplied by the differential operator Lxyz
##Rlm = \begin{pmatrix}
m \cdot Y^m_l(x,y,z) &\sqrt {(
l-m) \cdot (l+1+m)}\cdot Y^{m+1}_l(x,y,z) \\
\sqrt {(l+m) \cdot (l+1-m)} Y^{m-1}_l(x,y,z) &-m \cdot Y^m_l(x,y,z) \\
\end{pmatrix}\\
+l \cdot Y^m_l(x,y,z)##
then this matrix acts like an eigenvector with Eigen value l + 1.
The ## Y^m_l(x,y,z)## are spherical Harmomics.
See en.wikipedia.org/wiki/Table_of_spherical_harmonics
Also Quantum Mechanics, Albert Messiah, Volume 1, Appendix B, Article 10
Here are some comments about Rlm;:
The vector from Lxyz, ##(y \cdot \sigma x - x \cdot \sigma y )##, associated with ##\frac {\partial } {\partial z}## is perpendicular to the z axis. Similarly the other two vectors associated with ##\frac {\partial } {\partial x}## and ##\frac {\partial } {\partial y}## are perpendicular to the x vector and to the y vector. Taking the sum of the three vectors contained in Lxyz that are associated with partials but excluding the partials defines ##LDelta = (y \cdot \sigma x - x \cdot \sigma y ) + (z \cdot \sigma y - y \cdot \sigma z ) + (x \cdot \sigma z - z \cdot \sigma x ) ##. Taking the LDelta dot product with the x, y, z vector, ##\vec r##, gives ##(y \cdot \sigma x - x \cdot \sigma y )z\cdot (\sigma z)^2 +(z \cdot \sigma y - y \cdot \sigma z )x\cdot (\sigma x)^2 +(x \cdot \sigma z - z \cdot \sigma x )y\cdot (\sigma y)^2 ## which is zero so that LDelta is perpendicular to , ##\vec r##. LDelta is also perpendicular to the x = y = z vector. LDelta##\cdot \Delta \theta## with x, y, and z as specific functions of ##\theta## can generate R10(##\theta##) that’s similar to ##e^{i\theta}##. The LDelta path could define a circle around the x = y = z vector but maybe not for more than R10(##\theta##) which is similar to ##e^{i\theta}##
For certain Rlm in order for ##Lxyz\cdot Rlm – (l+1)\cdot Rlm## to be zero it must include the factor ##(r^2- z^2 –x^2 – y^2)## where the factor determines the difference is zero rather than zero itself determining it. A Legendre Polynomial is evident when ##(x^2 + y^2)## is replaced by ##(r^2- z^2 )##.
Also significant is that the square of ##(y \cdot \sigma x - x \cdot \sigma y )## turns out to be ##x^2 + y^2##. This is because the xy cross terms in ##x^2 + y^2##.cancel with this math.
Multiplying the sum of squares by Lxyz gives zero suggesting ##\vec r ## is a constant
##\sigma r = \begin{pmatrix}1&0\\0&1\\\end{pmatrix}, ~\sigma z = \begin{pmatrix}1&0\\ 0&-1\\\end{pmatrix}, ~\sigma x = \begin{pmatrix}0&1\\ 1&0\\\end{pmatrix}, ~\sigma y = \begin{pmatrix}0&i\\ -i&0\\\end{pmatrix}##,
So ##r = x \cdot \sigma x + y \cdot \sigma y + z \cdot \sigma z## which is the same as ##\begin{pmatrix}z&x+yi\\ x-yi&-z\\\end{pmatrix}##
From the following differential operators ##L = \frac {1} {i} \vec r \times \nabla## with ##Lz = (x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x}),~Lx = (y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y}),~Ly = (z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})## a useful operator can be defined; ##Lxyz = Lx \cdot \sigma x + Ly \cdot \sigma y + Lz \cdot \sigma z## or ## Lxyz = \begin{pmatrix}
(x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})&(y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})+(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i\\
(y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})-(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i&-(x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})\\
\end{pmatrix}##
If the following matrix is multiplied by the differential operator Lxyz
##Rlm = \begin{pmatrix}
m \cdot Y^m_l(x,y,z) &\sqrt {(
l-m) \cdot (l+1+m)}\cdot Y^{m+1}_l(x,y,z) \\
\sqrt {(l+m) \cdot (l+1-m)} Y^{m-1}_l(x,y,z) &-m \cdot Y^m_l(x,y,z) \\
\end{pmatrix}\\
+l \cdot Y^m_l(x,y,z)##
then this matrix acts like an eigenvector with Eigen value l + 1.
The ## Y^m_l(x,y,z)## are spherical Harmomics.
See en.wikipedia.org/wiki/Table_of_spherical_harmonics
Also Quantum Mechanics, Albert Messiah, Volume 1, Appendix B, Article 10
Here are some comments about Rlm;:
The vector from Lxyz, ##(y \cdot \sigma x - x \cdot \sigma y )##, associated with ##\frac {\partial } {\partial z}## is perpendicular to the z axis. Similarly the other two vectors associated with ##\frac {\partial } {\partial x}## and ##\frac {\partial } {\partial y}## are perpendicular to the x vector and to the y vector. Taking the sum of the three vectors contained in Lxyz that are associated with partials but excluding the partials defines ##LDelta = (y \cdot \sigma x - x \cdot \sigma y ) + (z \cdot \sigma y - y \cdot \sigma z ) + (x \cdot \sigma z - z \cdot \sigma x ) ##. Taking the LDelta dot product with the x, y, z vector, ##\vec r##, gives ##(y \cdot \sigma x - x \cdot \sigma y )z\cdot (\sigma z)^2 +(z \cdot \sigma y - y \cdot \sigma z )x\cdot (\sigma x)^2 +(x \cdot \sigma z - z \cdot \sigma x )y\cdot (\sigma y)^2 ## which is zero so that LDelta is perpendicular to , ##\vec r##. LDelta is also perpendicular to the x = y = z vector. LDelta##\cdot \Delta \theta## with x, y, and z as specific functions of ##\theta## can generate R10(##\theta##) that’s similar to ##e^{i\theta}##. The LDelta path could define a circle around the x = y = z vector but maybe not for more than R10(##\theta##) which is similar to ##e^{i\theta}##
For certain Rlm in order for ##Lxyz\cdot Rlm – (l+1)\cdot Rlm## to be zero it must include the factor ##(r^2- z^2 –x^2 – y^2)## where the factor determines the difference is zero rather than zero itself determining it. A Legendre Polynomial is evident when ##(x^2 + y^2)## is replaced by ##(r^2- z^2 )##.
Also significant is that the square of ##(y \cdot \sigma x - x \cdot \sigma y )## turns out to be ##x^2 + y^2##. This is because the xy cross terms in ##x^2 + y^2##.cancel with this math.
Multiplying the sum of squares by Lxyz gives zero suggesting ##\vec r ## is a constant