## A reminder for the use of Dirac notation and Translation

The momentum operator $\hat{P}$ acts on a position ket or bar is tricky.

$\left

$B A \hat{P}\left|x\right> = B i\frac{d}{dx} ( A \left|x\right>)$

I added $A$ and $B$ to emphasize the tricky point, where $A$ or $B$ can be an operator, a ket, a bar, or a variable. We can see, after the first quantization, the position bar or ket is always waiting for something and the derivative is differentiate the whole thing, not only the bar or ket. Miss use will result incorrect result. for example, the non-commute property of the position and the momentum operator,

$\left= -i\frac{d}{dx} (\left =-i\frac{d}{dx} (\left$

$=-i x \frac{d}{dx} (\left - i = x \left - i = \left$

If you use it wrongly, you may have something like

$\left = -i\frac{d\left$

or

$\left = -i\frac{d}{dx} (\left) = -i\frac{d}{dx}(x) = -i$

The translation operator.

In many text book, the translation operator is stated as

$D(\epsilon) = 1 - i \hat{P} \epsilon$,

where $\hat{P}$ is momentum operator or the generator of displacement.

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Actually, there is a reason, notice that the translation operator is unitary operator,

$D(\epsilon)D^{\dagger}(\epsilon) = 1$

then , since any unitary operator can be written as

$D(\epsilon) = \exp(-i \epsilon \hat{Q}) \sim 1 - i \epsilon \hat{Q} +....$

where $\hat{Q}$ is a Hermitian operator.

But what is $\hat{Q}$ ? since $\hat{Q}$ is Hermitian, it has eigen ket and eigenvalue,

$\hat{Q} \left|q\right> = \left|q\right> q$.

The, we can check

$\left = \left = \left =\exp(-i \epsilon q) \left$

also

$\left = \left = \exp(-i \epsilon q) \left$

we set a wave function $\psi_q(x) = \left$

$\psi_q(x-\epsilon) = \exp(-i\epsilon q ) \psi_q(x)$

Then, we can have a derivative of the wave function

$\frac{d}{dx}\phi_q(x) =\lim_{\epsilon\to 0} \frac{\exp(i\epsilon q ) -1 }{\epsilon} \psi_q(x) = i q \psi_q(x)$

the solution is $\psi_q(x) = \exp(iqx)$, which is a plane wave with momentum q. Therefore, the operator $\hat{Q}$ is the momentum operator, and it is really a generator a translation on the wave function $\psi_q(x) \rightarrow \psi_q(x+\epsilon)$. To find the form of the momentum operator acting on position bar,

$\left = q \exp(-i q x) = -i \frac{d}{dx}( \exp(-iqx)) = -i\frac{d}{dx} (\left)$

$\left

*********************************

another mathematical way to do is, using Taylor series,

$D(\epsilon) \left|x\right> = \left|x+\epsilon\right> = \left|x\right> + \epsilon \frac{d}{dx}\left|x\right> + ...$

compare with the exponential, than

$\hat{Q}\left|x\right> = i\frac{d}{dx} ( \left|x\right>)$.

$\left = -i \frac{d}{dx}(\left) =\left q$.

The time propagation and angle rotation can also do in the same way. The rotation around z-axis is trivial, but a general rotation is not so easy in detail, but the idea can be generalized that the rotation generator is $\vec{\hat{J}}\cdot \vec{n}$.

Why the general translation operator is in exponential form? because it is a unitary operator. Why it is unitary? because it is symmetric that translate back and forth result no change. In general, if we have a quantity that can be “translate” and the translation is symmetry, than, the generator must take the same exponential form.

It is interesting that, the translation generator $\hat{Q}$ is a differential of the position $\alpha$, here $\alpha$ is a position of a general coordinate.  When I started QM, I always feel insecure that change of position like this way, because classically, the change of a position means it moves in time, and there is no time in the $\hat{P}$. However, look at the rotation generator. The same derivation (between the *****) is applicable on the rotation about z-axis that replace the $x \rightarrow \phi$, azimuth angle. Then, at the end, we will recognize the generator $\hat{Q} = -i\frac{d}{d\phi} = L_z$, which is an angular momentum operator on the z-axis, and classically in the sense that the $L_z = \frac{d}{d\phi}$ is known in EM in spherical coordinate (the wave equation). The key point is that the rotation on z-axis takes the form $\frac{d}{d\phi}$!

There is still some things need to clarify, like the time translation operator is unitary, but the time-reversal operator is anti-unitary. Also, the detail of the general rotation operator $\hat{D}_n(\epsilon) \left|\vec{r}\right> = \left|\vec{r'}\right>$. we know that the general rotation operator is a matrix form, How?

## Euler angle

with the help of the post changing frame, we are now good to use the Euler angle.

recall

$V_R = R_n ( - \theta ) V_S$

for the rotating frame axis is rotating positive with the static frame.

the Euler angle is performed on 3 steps

1. rotate on $Z_S$, the z-axis with $\alpha$, which is $R_{zS} ( - \alpha )$. the x-axis and the y-axis is now different, we notate this frame with a 1 .
2. rotate on $Y_1$, the y-axis in the 1- frame  by angle $\beta$, which is $R_{y1} ( - \beta )$. the new axis is notated by 2.
3. rotate on $Z_2$, the z-axis in the 2-frame by angle $\gamma$, which is $R_{z2} ( - \gamma )$. the new axis is notated by R.

The rotating frame is related with the static frame by:

$V_R = R_{z2} ( - \gamma ) R_{y1} ( - \beta ) R_{zS} ( - \alpha ) V_S$

or

$R_R ( \alpha, \beta, \gamma ) =$$R_{zS} ( - \gamma )$ $R_{y1} ( - \beta )$ $R_{zS} ( - \alpha )$

for each rotation is on a new frame, the computation will be ugly, since, after each rotation, we have to use the rotation matrix in new coordinate.

There is another representation, notice that:

$R_{y1} ( -\beta ) =$ $R_{zS} ( - \alpha )$ $R_{yS} ( - \beta )$ $R_{zS} ( \alpha)$

which mean, the rotating on y1 -axis by $\beta$ is equal to rotate it back to $Y_S$  on zS -axis and rotated it by $\beta$ on yS – axis, then rotate back the $Y_S$ to $Y_1$ on zS – axis.

i use a and b for the axis between the transform.

and we have it for the z2-axis.

$R_{z2} ( -\gamma ) = R_{y1} ( - \beta ) R_{z1} ( - \gamma ) R_{y1} ( \beta )$

by using these 2 equation and notice that the z1-axis is equal to zS-axis.

$R_R ( \alpha , \beta, \gamma ) = R_{zS} ( - \alpha ) R_{yS} (- \beta ) R_{zS} ( - \gamma )$

which act only on the the same frame.

## on angular momentum adding & rotation operator

the angular momentum has 2 kinds – orbital angular momentum $L$, which is caused by a charged particle executing orbital motion, since there are 3 dimension space. and spin $S$, which is an internal degree of freedom to let particle “orbiting” at there.

thus, a general quantum state for a particle should not just for the spatial part and the time part. but also the spin, since a complete state should contains all degree of freedom.

$\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>$

when we “add” the orbital angular momentum and the spin together, actually, we are doing:

$J = L \bigotimes 1 + 1 \bigotimes S$

where the 1 with L is the identity of the spin-space and the 1 with S is the identity of the 3-D space.

the above was discussed on J.J. Sakurai’s book.

the mathematics of $L$ and $S$ are completely the same at rotation operator.

$R_J (\theta) = Exp( - \frac {i}{\hbar} \theta J)$

where $J$ can be either $L$ or $S$.

the $L$ can only have effect on spatial state while $S$ can only have effect on the spin-state. i.e:

$R_L(\theta) \left| s \right> = \left| s\right>$

$R_S(\theta) \left| x \right> = \left| x\right>$

the $L_z$ can only have integral value but $S_z$ can be both half-integral and integral. the half-integral value of $Sz$ makes the spin-state have to rotate 2 cycles in order to be the same again.

thus, if the different of $L$ and $S$ is just man-made. The degree of freedom in the spin-space is actually by some real geometry on higher dimension. and actually, the orbital angular momentum can change the spin state:

$L \left| s \right> = \left | s' \right > = c \left| s \right>$

but the effect is so small and

$R_L (\theta) \left| s\right > = Exp( - \frac {i}{\hbar} \theta c )\left| s \right>$

but the c is very small, but if we can rotate the state for a very large angle, the effect of it can be seen by compare to the rotation by spin.

$\left < R_L(\omega t) + R_S(\omega t) \right> = 2 ( 1+ cos ( \omega ( c -1 ) t)$

the experiment can be done as follow. we apply a rotating magnetic field at the same frequency as the Larmor frequency. at a very low temperature, the spin was isolated and $T_1$ and $T_2$ is equal to $\infty$. the different in the c will come up at very long time measurement and it exhibit a interference pattern.

if $c$ is a complex number, it will cause a decay, and it will be reflected in the interference pattern.

if we find out this c, then we can reveal the other spacial dimension!

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the problem is. How can we act the orbital angular momentum on the spin with out the effect of spin angular momentum? since L and S always coupled.

one possibility is make the S zero. in the system of electron and positron. the total spin is zero.

another possibility is act the S on the spatial part. and this will change the energy level.

__________________________________

an more fundamental problem is, why L and S commute? the possible of writing this

$\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>$

is due to the operators are commute to each other. by why?

if we break down the L in to position operator x and momentum operator p, the question becomes, why x and S commute or p and S commute?

$[x,S]=0 ?$

$[p,S]=0 ?$

$[p_x, S_y] \ne 0 ?$

i will prove it later.

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another problem is, how to evaluate the Poisson bracket? since L and S is not same dimension. may be we can write the eigenket in vector form:

$\begin {pmatrix} \left|x, t \right> \\ \left|s\right> \end {pmatrix}$

i am not sure.

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For any vector operator, it must satisfy following equation, due to rotation symmetry.

$[V_i, J_j] = i \hbar V_k$   run in cyclic

Thus,

where J is rotation operator. but i am not sure is it restricted to real space rotation. any way, spin is a vector operator, thus

$latex [S_x, L_y] = i \hbar S_z = – [S_y, L_x]$

so, L, S is not commute.

## Laplacian in spherical coordinate

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.

## Larmor Precession (quick)

Magnetic moment ($\mu$) :

this is a magnet by angular momentum of charge or spin. its value is:

$\mu = \gamma J$

where $J$ is angular momentum, and $\gamma$ is the gyromagnetic rato

$\gamma = g \mu_B$

Notice that we are using natural unit.

the g is the g-factor is a dimensionless number, which reflect the environment of the spin, for orbital angular momentum, g = 1.

$\mu_B$ is Bohr magneton, which is equal to

$\mu_B = \frac {e} {2 m}$ for positron

since different particle has different mass, their Bohr magneton value are different. electron is the lightest particle, so, it has largest value on Bohr magneton.

Larmor frequency:

When applied a magnetic field on a magnetic moment, the field will cause the moment precess around the axis of the field. the precession frequency is called Larmor frequency.

the precession can be understood in classical way or QM way.

Classical way:

the change of angular momentum is equal to the applied torque. and the torque is equal to the magnetic moment  cross product with the magnetic field. when in classical frame, the angular momentum, magnetic moment, and magnetic field are ordinary vector.

$\vec {\Gamma}= \frac { d \vec{J}}{dt} = \vec{\mu} \times \vec{B} = \gamma \vec {J} \times \vec{B}$

solving gives the procession frequency is :

$\omega = - \gamma B$

the minus sign is very important, it indicated that the J is precessing by right hand rule when $\omega >0$.

QM way:

The Tim dependent Schrödinger equation (TDSE) is :

$i \frac {d}{d t} \left| \Psi\right> = H \left|\Psi\right>$

H is the Hamiltonian, for the magnetic field is pointing along the z-axis.

$H = -\mu \cdot B = - \gamma J\cdot B = -gamma B J_z = \omega J_z$

the solution is

$\left|\Psi(t) \right> = Exp( - i \omega t J_z) \left| \Psi(0) \right>$

Thus, in QM point of view, the state does not “rotate” but only a phase change.

However, the rotation operator on z-axis is

$R_z ( \theta ) = Exp( - i \frac {\theta}{\hbar} J_z )$

Thus, the solution can be rewritten as:

$\left|\Psi (t)\right> = R_z( \omega t) \left|\Psi(0)\right>$

That makes great analogy on rotation on a real vector.

## Hydrogen Atom (Bohr Model)

OK, here is a little off track. But that is what i were learning and learned. like to share in here. and understand the concept of hydrogen is very helpful to understand the nuclear, because many ideas in nuclear physics are borrow from it, like “shell”.

The interesting thing is about the energy level of Hydrogen atom. the most simple atomic system. it only contains a proton at the center, um.. almost center, and an electron moving around. well, this is the “picture”. the fact is, there is no “trajectory” or locus for the electron, so technically, it is hard to say it is moving!

why i suddenly do that is because, many text books said it is easy to calculate the energy level and spectrum for it. Moreover, many famous physicists said it is easy. like Feynman, Dirac, Landau, Pauli, etc… OK, lets check how easy it is.

anyway, we follow the usual say in every text book. we put the Coulomb potential in the Schrödinger equation, change the coordinate to spherical. that is better and easy for calculation because the coulomb potential is spherical symmetric. by that mean, the momentum operator (any one don’t know what is OPERATOR, the simplest explanation is : it is a function of function.) automatically separated into 2 parts : radial and angular part. The angular part can be so simple that it is the Spherical harmonic.

Thus the solution of the “wave function” of the electron, which is also the probability distribution of  the electron location, contains 2 parts as well. the radial part is not so trivial, but the angular part is so easy. and it is just $Y(l,m)$.

if we denote the angular momentum as L, and the z component of it is Lz, thus we have,

$L^2 Y(l,m) = l(l+1) \hbar^2 Y(l,m)$

$L_z Y(l,m) = m \hbar Y(l,m)$

as every quadratic operator, there are “ladder” operator for “up” and “down”.

$L_\pm Y(l,m) =\hbar \sqrt{l(l+1) - m(m\pm 1)} Y(l,m \pm 1)$

which means, the UP operator is increase the z-component by 1, the constant there does not brother us.

it is truly easy to find out the exact form of the $Y(l,m)$ by using the ladder operator. as we know, The z component of the a VECTOR must have some maximum. so, there exist an $Y(l,m)$ such that

$L_+ Y(l,m) =0$

since there is no more higher z-component.

by solve this equation, we can find out the exact form of $Y(l,m)$ and sub this in to L2, we can know$Max(m) = l$. and apply the DOWN operator, we can fins out all $Y(l,m)$, and the normalization constant is easy to find by the normalization condition in spherical coordinate, the normalization factor is $sin(\theta)$, instead of 1 in rectangular coordinate.

$\int_0^\pi \int_0^{2 \pi} Y^*(l',m') Y(l,m) sin(\theta) d\theta d \psi = \delta_{l' l} \delta_{m' m}$

more on here