Laplacian in spherical coordinate

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the Momentum operator in spherical coordinate

\nabla^2 = \frac {1}{r^2}\frac {\partial } { \partial r} \left ( r^2 \frac {\partial} {\partial r} \right ) - \frac {1}{r^2} L^2

where L is the Reduced angular momentum operator. the minus sign is very important for giving a correct sign. the original angular momentum operator J is related by:

J=\hbar^2 L

by compare the Laplacian in spherical coordinate, the L is

L^2 = - \frac {1}{sin(\theta)} \frac {\partial}{\partial \theta} \left( sin(\theta) \frac {\partial}{\partial \theta} \right ) - \frac {1}{sin(\theta)} \frac{\partial^2} {\partial \phi ^2}

But this complicated form is rather useless, expect you are mathematic madman.

we can start from classical mechanic

\vec{L} = \vec {r} \times \vec{p}

L_x = y \frac {\partial} {\partial z} - z \frac {\partial}{\partial y }

L_y = z \frac {\partial} {\partial x} - x \frac {\partial}{\partial z }

L_z = x \frac {\partial} {\partial y} - y \frac {\partial}{\partial x }

with the change of coordinate

\begin {pmatrix} x \\ y \\ z \end{pmatrix} = \begin {pmatrix} r sin(\theta) cos(\phi) \\ r sin(\theta) sin(\phi) \\ r cos(\theta) \end{pmatrix}

and the Jacobian Matrix M_J , which is used for related the derivatives.

since

\frac {\partial}{\partial x} = \frac {\partial r}{\partial x} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial x} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial x} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial y} = \frac {\partial r}{\partial y} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial y} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial y} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial z} = \frac {\partial r}{\partial z} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial z} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial z} \frac {\partial} {\partial \phi}

which can be simplify

\nabla_{(x,y,z)} = M_J^T \nabla_{(r, \theta, \phi )}

M_J = \frac {\partial ( r, \theta, \phi) }{\partial (x,y,z)}

M_J^{\mu\nu} = \frac {\partial \mu}{\partial \nu}

then, we have

L_x = i sin(\phi) \frac {\partial }{\partial \theta} +i cot(\theta) cos(\phi) \frac { \partial }{\partial \phi}

L_y =-i cos(\phi) \frac {\partial }{\partial \theta} + i cot(\theta) sin(\phi) \frac { \partial }{\partial \phi}

L_z = - i \frac {\partial }{\partial \phi}

However, even we have the functional form, it is still not good.  we need the ladder operator

L_+ = L_x + i L_y = Exp(i \phi) \left( \frac {\partial }{\partial \theta} + i cot(\theta) \frac { \partial }{\partial \phi} \right)

L_- = L_x - i L_y = Exp(-i \phi) \left( \frac {\partial }{\partial \theta} - i cot(\theta) \frac { \partial }{\partial \phi} \right)

notice that

L_+^\dagger = L_-

so, just replacing i \rightarrow -i .

when we looking for the Maximum state of the spherical Harmonic Y_{max}(\theta, \phi)

L_+ Y_{max}(\theta,\phi) = 0 *)

use the separable variable assumption.

Y_{max}(\theta, \phi) = \Theta \Phi

L_+ \Theta \Phi = 0 = - Exp(i \phi) \left( \frac {d\Theta}{d \theta} \Phi + i cot(\theta) \frac { d\Phi}{d\phi} \right) \Theta

\frac {tan(\theta)}{\Theta} \frac { d \Theta} {d \Theta } = - \frac {i}{\Phi} \frac {d \Phi} {d \phi} = m

the solution is

Y_{max}(\theta,\phi) = sin^m(\theta) Exp(i m \phi )

L^2 Y_{max}(\theta, \phi) = m(m+1) Y_{max}(\theta,\phi)

an application on Hydrogen wave function is here.

angular momentum operator in matrix form

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in order to fine out the matrix elements, we just need to know 2 equations

J_z \left| j,m\right> = m \hbar \left|j,m\right>

J_\pm \left| j, m \right>=\hbar \sqrt{ j(j+1) -m ( m\pm 1) } \left |j,m \right >

The calculation is straight forward, but be careful of the sign.

i define the coefficient :

K_j(m) = \frac{\hbar}{2} \sqrt {j(j+1) -m(m+1) }

and the matrix coefficients are:

J_z ^{\mu\nu} (j) = \hbar (j -\mu +1) \delta_{\mu \nu}

J_x ^{\mu\nu} (j) = K_j(j-\mu) \delta_{\mu (\nu-1)}+K_j(j-\nu)\delta_{(\mu-1)\nu}

J_y ^{\mu\nu} (j) = -i K_j(j-\mu) \delta_{\mu (\nu-1)}+ i K_j (j-\nu)\delta_{(\mu-1)\nu}

where

\mu , \nu = 1,2,...,2j+1

the Kronecker Delta indicated that only 1st upper and lower diagonal elements are non-zero.

δμ(ν-1) means the 1st upper diagonal elements. since ν = μ+1 to make it non-zero.

For example:

J_x (1) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {2} & 0 \\ \sqrt{2} & 0 &\sqrt{2} \\ 0 & \sqrt{2} & 0\end{pmatrix}

J_x (\frac {3}{2}) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix}

To compute J_y , we just need to multiply the upper diagonal with i and the lower diagonal with - i .

The coefficient K_j(j-\mu)  is every interesting. if we only look at the first upper diagonal. and take the square of each element.

J_x(\frac {1}{2}) = 1

J_x(1) = \begin{pmatrix} 2 & 2 \end{pmatrix}

J_x(\frac{3}{2}) = \begin{pmatrix} 3 & 4 & 3 \end{pmatrix}

and we use this to form a bigger matrix

\begin {pmatrix} ... & 5 & 4 & 3 & 2 & 1 \\ ... & 10 & 8 & 6 & 4 & 2 \\ ... & ... & 12 & 9 & 6 & 3 \\ ... & ... & ...& 12 & 8 & 4 \\ ... & ... & ... & ... & 10 & 5 \end {pmatrix}

if we read from the top right hand corner, and take the diagonal element. we can see that they fit the 1st upper diagonal element of J_x ( j) , if we take square root of each one.

and the pattern are just the multiplication table! How nice~

so, i don’t have to compute for j = 5/2.

J_x ( \frac{5}{2} ) = \frac {\hbar}{2} \begin {pmatrix} 0 & \sqrt{5} & 0 & 0& 0 & 0 \\ \sqrt{5} & 0 & \sqrt{8} & 0 & 0 & 0 \\ 0 & \sqrt{8} & 0 & \sqrt{9} & 0 & 0 \\ 0 & 0 & \sqrt{9} & 0 & \sqrt{8} & 0 \\ 0& 0 &0 & \sqrt{8} &0 & \sqrt{5} \\ 0 & 0 & 0 & 0 & \sqrt{5} & 0 \end {pmatrix}

but the physical reason for this trick , i don’t know. for the Pascal triangle, we understand the reason for each element – it is the multiplicity.

In general:

\displaystyle J_x(L) = \frac{\hbar}{2} \begin{pmatrix} 0 & \sqrt{2L} & 0 & ... & ... & ... \\ \sqrt{2L} & 0 & \sqrt{2(2L-1)} & ... & ... & ... \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & \sqrt{r (2L-r+1)} & ... \\  .. & ... & ... & ... & ...  \end{pmatrix}

or the ij element of the matrix is

\displaystyle [J_x(L)]_{i, i>j} = \frac{\hbar}{2} \sqrt{i(2L-i+1)} \delta_{i+1,j}

on Relaxation in NMR

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If we only switch on the transverse magnetic field for some time \tau . after the field is off, the system will go back to the thermal equilibrium. it is due to the system is not completely isolated.

instead of consider a single spin, we have to consider the ensemble. and an ensemble is describe by the density matrix.

the reason for not consider a single spin state is, we don’t know what is going on for individual spin. in fact, in the previous section, the magnetization is a Marco effect. a single spin cannot have so many states, it can only have 2 states – up or down. if we insist the above calculation is on one spin, thus, it only give the chance for having that direction of polarization. which, is from many measurements.

so, for a single spin, the spin can only have 2 states. and if the transverse B field frequency is not equal to the Larmor frequency , and the pule is not a π-pulse, the spin has chance to go to the other state, which probability is given by a formula. and when it goes to relax back to the minimum energy state, it will emit a photon. but when it happen, we don’t know, it is a complete random process.

However, an ensemble, a collection of spins, we can have some statistic on it. for example, the relaxation time, T1 and T2.