Second Matrices from Spherical Harmonics with Eigenvalue l+1

[paklin2]

TL;DR Summary

I’m curious if these matrices have been discussed elsewhere and if they have any significance. There's a question why it seems x, y, and z can be replaced by functions of a single variable Theta.

See the first post in the previous thread ‘Matrices from Spherical Harmonics with Eigenvalue l+1’ first.

Originally when I came across the Lxyz operator and the Rlm matrices I had a different question. If this had to do with something like the quantum Hydrogen atom then why did it appear that the Lxyz operator appears to generate a path from infinitesimals that are proportional to x, y, and z with x, y, and z being a function of a single variable, $\theta$, rather than the two independent variables in spherical harmonics.

To demonstrate this, as discussed previously, 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 and multiplied by Rlm; $$Lxyz \cdot Rlm= (Lx \cdot \sigma x+ Ly \cdot \sigma y + Lz \cdot \sigma z) \cdot Rlm = (y\frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})\cdot \sigma x+ (z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z}) \cdot \sigma y + (x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x}) \cdot \sigma z$$

or $$ Lxyz\cdot Rlm = \begin{pmatrix}

(x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})&(y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})+(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z}) (y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})-(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z})i&-(x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})\\

\end{pmatrix}$$

Rearranging terms gives $$ Lxyz\cdot Rlm = (z\cdot \sigma y - y \cdot \sigma z)\cdot \frac {\partial Rlm } {\partial x} + (x\cdot \sigma z - z \cdot \sigma x)\cdot \frac {\partial Rlm } {\partial y} + (y\cdot \sigma x - x \cdot \sigma y)\cdot \frac {\partial Rlm } {\partial z} $$

The vector $(z\cdot \sigma y - y \cdot \sigma z )$ is perpendicular to the x direction, the vector $(x\cdot \sigma z - z \cdot \sigma x)$ is perpendicular to the y direction, and the vector $(y\cdot \sigma x - x \cdot \sigma y)$ is perpendicular to the z direction. If x, y,z represent $\Delta x, \Delta y , and \Delta z$ then the perpendicular vectors represent $\Delta \theta$ for each axis . Adding these three vectors together gives $(y-z)\sigma x +(z-x)\sigma y+ (x-y)\sigma z$. This vector is perpendicular $x \cdot \sigma x+ y \cdot \sigma y + z \cdot \sigma z$ and the $x =y =z$ vector. Does this means there are circular paths around the x = y = z axis that can be generated by Lxyz for say an orbit? This is the same as x, y, and z being functions of a single variable $\theta$. Still for l larger than one in Rlm it doesn’t look like there’s something similar to $e^{I \theta}$ if you’re replacing a term with two different partials with a single derivative of $\theta$. The Legendre Polynomial seems to replace $e^{I \theta}$.

Moderator note: edited such that the formulas are in a scroll box.

 

[paklin2]

Not sure how this got in there
$Lxyz\cdot Rlm = \begin{pmatrix} (x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})&(y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})+(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z}) (y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})-(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z})i&-(x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})\\ \end{pmatrix}$

 

[fresh_42]

Double dollars do the job:

Not sure how this got in there
$$ Lxyz\cdot Rlm = \begin{pmatrix} (x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})&(y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})+(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z}) (y \frac {\partial Rlm } {\partial z}- z \frac {\partial Rlm } {\partial y})-(z \frac {\partial Rlm } {\partial x}- x \frac {\partial Rlm } {\partial z})i&-(x \frac {\partial Rlm } {\partial y}- y \frac {\partial Rlm } {\partial x})\\

\end{pmatrix}$$

 

[paklin2]

As far as I can determine the infinitesimals for the paths generated here by Lxyz are added to x, y, and z by the partials but are not influenced by the partials themselves. They appear to be circular paths. Except for R10 most of the Rlms if not all with l larger than 1 still generate circular paths but cannot be represented by the form $e^{I \theta}$ due to the second degree differential equation, the Legendre Polynomial equation, generated from $Lxyz \cdot Rlm = (l+1) Rlm$.

Is this just a curiosity. Otherwise are the Dirac central field equations still used.

 

[paklin2]

I have to make a correction here. Circular paths are formed from $cos(\theta)$ and sin($\theta$) in three dimensions here, but as far as I know there’s also paths formed from cos(n$\theta$) and sin(n$\theta$) here using Lxyz that are not circular but there’s still only one variable, $\theta$ rather than two, $\theta$ and $\phi$.

 

[paklin2]

Here’s an argument that x, y, and z may not be a function of a single variable, $\theta$. Often $Lxyz \cdot Rlm - (l+1) Rlm = f(z) \cdot (x^2 +y^2 +z^2 – r^2)$. Lxyz doesn’t determine that this equation is zero by itself since $( x^2+ y^2 +z^2 – r^2)$ is determined to be zero by other means then direct application of Lxyz to Rlm. Lxyz is still involved since $Lxyz \cdot ( x^2+ y^2 +z^2)= 0$ suggests $x^2+ y^2 +z^2$ is a constant. The second constant $\phi$ appears in $x+iy = r^{t\phi} \cdot sin( \theta)$ in spherical harmonics.

 

[paklin2]

It appears that most likely the paths determined by Lxyz, at least those from Rlm with $m =0$, are circles. With $Lxyz \cdot Rlm - (l+1) Rlm = f(z) \cdot (x^2 +y^2 +z^2 – r^2)$ often, $x^2+ y^2 +z^2 - r^2$ is required to be zero and this establishes that the path that Lxyz determines is a circle. This means there’s a single variable rather than two independent variables.

The origin of the circles considered here are on the $x = y = z$ axis, The r here is not the origin of the circle that’s on this axis but gives the actual value of $r^2$ for $x^2+ y^2 +z^2$ when x, y, z originate at the origin of the xyz vector space.

Lxyz was suggested by possible factoring of the Shroedinger Equation but the one variable result suggests classical physics.

 

[paklin2]

Here is a summary.

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 $Lxyz = \begin{pmatrix} (x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})&(y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})+(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i\\ (y \frac {\partial } {\partial z}- z \frac {\partial } {\partial y})-(z \frac {\partial } {\partial x}- x \frac {\partial } {\partial z})i&-(x \frac {\partial } {\partial y}- y \frac {\partial } {\partial x})\\ \end{pmatrix}$. Lxyz was suggested by the possibility of factoring the Schoedinger Equation but this shouldn’t imply that it generates only quantum expressions or only classical physics expressions.

Lxyz is equivalent to $(y \cdot \sigma x - x \cdot \sigma y ) \cdot \frac {\partial } {\partial z} + (z \cdot \sigma y - y \cdot \sigma z ) \cdot \frac {\partial } {\partial x} + (x \cdot \sigma z - z \cdot \sigma x ) \cdot \frac {\partial } {\partial y}$. The partials are multiplied by terms in front of them and if the x, y,and z in these terms are replaced by $\Delta x, \Delta y, and \Delta z$ then the expression can be used as increments added to x, y , and z in the overall expression for the partials and Lxyz can be used to generate a path. For example $\frac {\partial f(x)} {\partial x} x$ will be changed to $\frac {\partial f(x)} {\partial x} \Delta x = f(x+ \Delta x) – f(x) = \frac {\partial f(x)} {\partial x} y \cdot \Delta \theta$ where $\frac {\partial x} {\partial \theta} = y$ by adding $\Delta x$ to x in $f(x+ \Delta x)$.

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. Therefore $Lxyz \cdot Rlm = (l+1) \cdot Rlm$.

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
Example of Spherical Harmonics

$Y^{-1}_1(x,y,z) = \frac {1} {2} \sqrt {\frac {3} {2 \pi}} \frac {x -iy} {r}$

$Y^0_1(x,y,z) = \frac {1} {2} \sqrt {\frac {3} {\pi}} \frac {z} {r}$

$Y^1_1(x,y,z) = -\frac {1} {2} \sqrt {\frac {3} {2 \pi}} \frac {x+iy} {r}$
$Y^{-2}_2(x,y,z) = \frac {1} {4} \sqrt {\frac {15} {2 \pi}} {\frac {(x -iy)^2} {r^2}}$

$Y^{-1}_2(x,y,z) = \frac {1} {2} \sqrt {\frac {15} {2 \pi}} {\frac {(x-iy)z} {r}}$

$Y^0_1(x,y,z) = \frac {1} {2} \sqrt {\frac {5} {\pi}} \frac {(2x^2-x^2-y^2)} {r}$

$Y^1_2(x,y,z) = -\frac {1} {2} \sqrt {\frac {15} {2 \pi}} {\frac {(x+iy)z} {r}}$

$Y^2_2(x,y,z) = \frac {1} {4} \sqrt {\frac {15} {2 \pi}} {\frac {(x+iy)^2} {r^2}}$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 $LDelta = (y \cdot \Delta \theta \cdot \sigma x - x \cdot \Delta \theta \cdot \sigma y ) + ( z \cdot \Delta \theta \cdot \sigma y - y \cdot \Delta \theta \cdot \sigma z ) + (x \cdot \Delta \theta \cdot \sigma z - z \cdot \Delta \theta \cdot \sigma x )$. Rearranging the sigmas in Lxyz and taking the LDelta dot product with the x, y, z vector, $\vec r$, gives $(y \cdot \Delta \theta - z \cdot \Delta \theta)\cdot \sigma x^2 \cdot x +(z \cdot \Delta \theta - x \cdot \Delta \theta) \cdot \sigma y^2 \cdot y + (x \cdot \Delta \theta - y \cdot \Delta \theta) \cdot \sigma z^2 \cdot z$ which is zero. Therefore LDelta is perpendicular to , $\vec r$. LDelta is equivalent to $\Delta \theta$ acting as a vector along the circumference and is perpendicular to the axis $a \cdot (\sigma x + \sigma x +\sigma x)$. The path generated should be a circle as discussed in the last paragraph..

The origin of the circles considered here are on the $a \cdot (\sigma x + \sigma x +\sigma x)$ axis for some value of a. The radii for these circles are not r. The $\vec r = x\cdot \sigma x + y\cdot \sigma y +z\cdot \sigma z$ originates at the origin of the xyz vector space. This r would have to do with a cone not a circle.Let $Rlm = Xlm \cdot \sigma x + Ylm \cdot \sigma y + Zlm \cdot \sigma z$. LDelta corresponds to $\Delta \theta$ or momentum around an orbit and originates from $(Lx \cdot \sigma x + Ly \cdot \sigma y) \cdot (Zlm \cdot \sigma z)$. A similar delta originates from $(Lx \cdot \sigma x + Ly \cdot \sigma y) \cdot (Xlm \cdot \sigma x + Ylm \cdot \sigma x)$. This second delta is $LDelta2 = ((y \cdot \Delta \theta - z \cdot \Delta \theta ) - (z \cdot \Delta \theta - x \cdot \Delta \theta)) \cdot \sigma x + ((z \cdot \Delta \theta - x \cdot \Delta \theta ) - (x \cdot \Delta \theta - y \cdot \Delta \theta )) \cdot \sigma y + ((x \cdot \Delta \theta - y \cdot \Delta \theta ) - (y \cdot \Delta \theta - z \cdot \Delta \theta )) \cdot \sigma z$. LDelta2 is perpendicular to $\Delta \theta$ acting as a vector along the circumference and is also perpendicular to the axis $a \cdot (\sigma x + \sigma x +\sigma x)$. LDelta2 seems to be a vector on the tip of radius that points along the radius to the center of the circle. It generates the second degree differential Legendre equation for polynomials and could represent curvature or electrostatic force since force is consistent with a second degree differential equation. The Rlm’s seem to suggest the hydrogen atom with an orbit and with different energy states from a classical physics point of view.
If vector increments of a fixed length are perpendicular to two non parallel vectors then the increments can be joined together with certain procedures to form a path. Making the increments infinitesimal converges them to a unique path..

Since ths direction of the vector increments is perpendicular to the plane of the two perpendiculars only the position of the increments needs to be determined here. Both the first vector LDelta increment and vector LDelta2 increment are attached to to the tip of $\vec r$. The second vector LDelta increment is attached to a new perpendicular vector, $\vec r$ + first vector LDelta increment, and by continuing along the circumference and making it converge with infinitesimals a circle is formed. The second vector LDelta2 increment is attached under the first vector LDelta2 increment along the radius. This is continued and made to converge as before.

 

[paklin2]

A slight correction needed. For something like $\Delta x = y \cdot \Delta \theta$, x and y will change moving around the circumference in a single move but $\Delta \theta$ should stay constant for a single move around the circumference..