First experiment of 6He with a polarized proton target

February 4, 2011

GoLuckyRyan experimental, papers reading , , , , , , , , , , , , , , Leave a comment

DOI : 10.1140/epjad/i2005-06-110-5

this paper reported a first spin polarized proton solid target under low magnetic field ( 0.08 T ) and hight temperature ( 100K )

the introduction overview the motivation of a solid target.

  • a polarized gas target is ready on many nuclear experiment.
  • on the radioactive beam ( IR beam ), the flux of a typical IR beam is small, since it is produced by 2nd scattering.
  • a solid target has highest density of solid.
    • most solid target can only be polarized on low temperature ( to avoid environmental interaction to reduced the polarization )
      • increase the experimental difficult, since a low temperature should be applied by a cold buffer gas.
    • high field ( the low gyromagnetic  ratio ).
      • high magnetic field make low energy scattered proton cannot get out from the magnetic field and not able to detect.
  • a solid target can be polarized at high temperature and low magnetic field is very useful

the material on use is a crystal of naphthalene doped with pentacene.

the procedure of polarizing the proton is :

  1. use optical pumping the polarize the electron of pentacene
    • the population of the energy states are independent of temperature and magnetic field.
  2. by Dynamic Nuclear Polarization (DNP) method  to transfer  the polarization of the electron to the proton.
    • if the polarization transfer is 100% and the relaxation time is very long. the expected polarization of proton will be 72.8%

The DNP method is archived under a constant microwave frequency with a sweeping magnetic field. when the magnetic field and  microwave frequency is coupled. the polarization transfer will take place.

the next paragraph talks about the apparatus’s size and dimension, in order to fit the scattering experiment requirements.

the polarization measurement is on a scattering experiment with 6He at 71 MeV per nucleons. By measuring the polarization asymmetry \epsilon , which is related to the yield. and it also equal to the polarization of the target P_t  times the analyzing power A_y .

\epsilon = P_t \times A_y

with a reasonable guess of the target polarization. the analyzing power of  6He was found.

the reason why the polarization-asymmetry is not equal to the analyzing power is that, the target is not 100% polarized, where the analyzing power is defined. when the polarization of the target is 100%, both are the same.

in the analysis part. it used optical model and Wood-Saxon central potential to simulate the result. And compare the result from 6He to 6Li at same energy. the root mean square of 6Li is larger then 6He. it suggest the d-α core of 6Li may responsible for that.

they cannot go further discussion due to the uncertainly on the polarization of the target.

a review on Hydrogen’s atomic structure

February 2, 2011

GoLuckyRyan advance, atomic, Basic, review, theory , , , , , , , , , , , , , , , , , , Leave a comment

I found that most of the book only talk part of it or present it separately. Now, I am going to treat it at 1 place. And I will give numerical value as well. the following context is on SI unit.

a very central idea when writing down the state quantum number is, is it a good quantum number? a good quantum number means that its operator commute with the Hamiltonian. and the eigenstate states are stationary or the invariant of motion. the prove on the commutation relation will be on some post later. i don’t want to make this post too long, and with hyperlink, it is more reader-friendly. since somebody may like to go deeper, down to the cornerstone.  but some may like to have a general review.

the Hamiltonian of a isolated hydrogen atom is given by fews terms, deceasing by their strength.

H = H_{Coul} + H_{K.E.} + H_{Rel} + H_{Darwin} + H_{s-0} + H_{i-j} + H_{lamb} + H_{vol} + O

the Hamiltonian can be separated into 3 classes.

__________________________________________________________

Bohr model

H_{Coul} = - \left(\frac {e^2}{4 \pi \epsilon_0} \right) \frac {1}{r}

is the Coulomb potential, which dominate the energy. recalled that the ground state energy is -13.6 eV. and it is equal to half of the Coulomb potential energy, thus, the energy is about 27.2 eV, for ground state.

H_{K.E.} = \frac {P^2}{ 2 m}

is the non-relativistic kinetic energy, it magnitude is half of the Coulomb potential, so, it is 13.6 eV, for ground state.

comment on this level

this 2 terms are consider in the Bohr model, the quantum number, which describe the state of the quantum state, are

n = principle number. the energy level.

l = orbital angular momentum. this give the degeneracy of each energy level.

m_l = magnetic angular momentum.

it is reasonable to have 3 parameters to describe a state of electron. each parameter gives 1 degree of freedom. and a electron in space have 3. thus, change of basis will not change the degree of freedom. The mathematic for these are good quantum number and the eigenstate \left| n, l, m_l \right> is invariant of motion, will be explain in later post. But it is very easy to understand why the angular momentum is invariant, since the electron is under a central force, no torque on it. and the magnetic angular momentum is an invariant can also been understood by there is no magnetic field.

the principle quantum number n is an invariance. because it is the eigenstate state of the principle Hamiltonian( the total Hamiltonian )!

the center of mass also introduced to make more correct result prediction on energy level. but it is just minor and not much new physics in it.

Fine structure

H_{Rel} = - \frac{1}{8} \frac{P^4}{m^3 c^2}

is the 1st order correction of the relativistic kinetic energy. from K.E. = E - mc^2 = \sqrt { p^2 c^2 + m^2c^4} - mc^2 , the zero-order term is the non-relativistic kinetic energy. the 1st order therm is the in here. the magnitude is about 1.8 \times 10^{-4} eV . ( the order has to be recalculate, i think i am wrong. )

H_{Darwin} = \frac{\hbar^{2}}{8m_{e}^{2}c^{2}}4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)

is the Darwin-term. this term is result from the zitterbewegung, or rapid quantum oscillations of the electron. it is interesting that this term only affect the S-orbit. To understand it require Quantization of electromagnetic field, which i don’t know. the magnitude of this term is about 10^{-3} eV

H_{s-o} = \left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\left(\frac{1}{2m_{e}^{2}c^{2}}\right)\frac{1}{r^3} L \cdot S

is the Spin-Orbital coupling term. this express the magnetic field generated by the proton while it orbiting around the electron when taking electron’s moving frame. the magnitude of this term is about 10^{-4} eV

comment on this level

this fine structure was explained by P.M.Dirac on the Dirac equation. The Dirac equation found that the spin was automatically come out due to special relativistic effect. the quantum number in this stage are

n = principle quantum number does not affected.

l = orbital angular momentum.

m_l = magnetic total angular momentum.

s = spin angular momentum. since s is always half for electron, we usually omit it. since it does not give any degree of freedom.

m_s = magnetic total angular momentum.

at this stage, the state can be stated by \left| n, l, m_l, m_s \right> , which shown all the degree of freedom an electron can possible have.

However, L_z is no longer a good quantum number. it does not commute with the Hamiltonian. so, m_l does not be the eigenstate anymore. the total angular momentum was introduced J = L + S . and J^2 and J_z commute with the Hamiltonian.  therefore,

j = total angular momentum.

m_j = magnetic total angular momentum.

an eigenstate can be stated as \left| n, l, s, j, m_j \right> . in spectroscopy, we denote it as ^{2 s+1} L _j , where L is the spectroscopy notation for l .

there are 5 degrees of freedom, but in fact, s always half, so, there are only 4 real degree of freedom, which is imposed by the spin ( can up and down).  the reason for stating the s in the eigenstate is for general discussion. when there are 2 electrons, s can be different and this is 1 degree of freedom.

Hyperfine Structure

H_{i-j} = \alpha I \cdot J

is the nuclear spin- electron total angular momentum coupling. the coefficient of this term, i don’t know. Sorry. the nuclear has spin, and this spin react with the magnetic field generate by the electron. the magnitude is 10^{-5}

H_{lamb}

is the lamb shift, which also only affect the S-orbit.the magnitude is 10^{-6}

comment on this level

the hyperfine structure always makes alot questions in my mind. the immediate question is why not separate the orbital angular momentum and the electron spin angular momentum? why they first combined together, then interact with the nuclear spin?

may be i open another post to talk about.

The quantum number are:

n = principle quantum number

l = orbital angular momentum

s = electron spin angular momentum.

j = spin-orbital angular momentum of electron.

i = nuclear spin. for hydrogen, it is half.

f = total angular momentum

m_f = total magnetic angular momentum

a quantum state is $\left| n, l, s, j,i, f , m_f \right>$. but since the s and i are always a half. so, the total degree of freedom will be 5. the nuclear spin added 1 on it.

Smaller Structure

H_{vol}

this term is for the volume shift. the magnitude is 10^{-10} .

in diagram:

NMR (nuclear magnetic resonance)

January 27, 2011

GoLuckyRyan Basic, no math , , , , , , , , , , Leave a comment

NMR is a technique to detect the state of nuclear spin. a similar technique on electron spin is call ESR ( electron spin resonance)

The principle of NMR is simple.

  1. apply a B-field, and the spin will align with it due to interaction with surrounding and precessing along the B-field with Larmor frequency, and go to Boltzmann equilibrium. the time for the spin align with the field is call T1, longitudinal relaxation time.
  2. Then, we send a pule perpendicular to the B-field, it usually a radio frequency pulse. the frequency is determined by the resonance frequency, which is same as the Larmor frequency. the function of this pulse is from the B-field of it and this perpendicular B-field with perturb the spin and flip it 90 degrees.
  3. when the spin are rotate at 90 degrees with the static B-field, it will generate a strong enough signal around the coil. ( which is the same coil to generate the pule ) and this signal is called NMR signal.
  4. since the spins will be affected by its environment, and experience a slightly different precession frequency. when the time goes, they will not aligned well, some precess faster, some slower. thus, the transverse magnetization will lost and look as if it decay. the time for this is called T2, transverse relaxation time.

by analyzing the T1 and T2 and also Larmor frequency, we can known the spin, the magnetization, the structure of the sample, the chemical element, the chemical formula, and alot many others thing by different kinds of techniques.

For nuclear physics, the use of NMR is for understand the nuclear spin. for example, the polarization of the spin.

 

 

Larmor Precession (quick)

January 23, 2011

GoLuckyRyan Spherical Stuffs, theory , , , , , , , , , , , , , , , 1 Comment

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.

Spin

January 22, 2011

GoLuckyRyan Basic, no math, Porperties, Spherical Stuffs, Working on , , , , , , , , , , , Leave a comment

( this is just a draft, not organized )

Spin is a intrinsics property of elementary particle, such as electron, proton, and even photon. Intrinsics means it is a built-in property, like mass, charge. Which extrinsic properties are speed, momentum.

Spin is a vector or tensor quality while charge and mass are scaler.

Spin can react with magnetic field, like charge reacts with electric field or mass react with force produce acceleration. Thus, spin is like a bar-magnet inside particle, counter part of charge.

The magnitude of spin is half integer or integer of reduced Planck’s constant \hbar . Particles with half integer of spin are classified as Fermion, and those with integer spin are Boson. they follow different statistic while interact together, thus, this creates different physics for different group.

we are not going to the mathematic description this time.

the effect of spin causes the magnetic moment, that’s why it react with magnetic field. the other thing that creates magnetic moment is angular momentum for charge particle, like electron orbiting around nucleus. So, both spin and angular momentum can be imagined as a little magnet, thus, they can interact, in physics, we call the interaction between spin and angular momentum is coupling. for example, spin-orbital coupling, spin-spin coupling, etc..

when the spin interact with external magnetic field, it will precess around the magnetic field with Larmor frequency. and the direction of the spin while undergoes procession can only be certain angle. for spin half, like electron or proton. there are only 2 directions, and we called it up and down.

Hall effect

January 20, 2011

GoLuckyRyan Basic, others, theory , Leave a comment

It is a short review. for more detail, wiki is a good place.

The hall probe is perpendicular to the B field( pointing up) and have a current I passing through ( going forward ).

Due to the Lorentz force. The positron is moving to right and accumulate. The accumulating charge creates a electric force to the left to against further positron accumulate. The magnetic force will be balanced by the electric force. Due to the electric force, there is associated voltage across the hall probe. This voltage is called hall voltage.

F_B = e v B = V_H d

Where e is positron charge, v is speed of positron, B is the magnetic field, V_H is the Hall voltage and d is the distance across the hall probe.

The current I is

I = A n e v

Where A is the cross section area of the hall probe, n is density of the positron carrier, v is the positron velocity.

Thus,

V_H = \frac { B}{ V n e } I

Where V is the volume of the hall probe. But the V n is equal to the total number of positron N.

V_H = \left ( \frac { B}{N e} \right ) I

Which is to say, the hall voltage is proportional to the magnetic field.

Type of accelerator (Ring type)

December 18, 2010

GoLuckyRyan Basic, no math , , , , Leave a comment

Ring type accelerator solved the difficulties by linear type.

  1. the particle circulate inside the accelerator, so, it can be accelerate infinite time in principle.
  2. The space require is smaller compare with similar energy output Linac.

However, there is a draw back is, for charged particle running in a circular path, it will radiate energy by EM wave due to the centripetal acceleration, thus, even it is just running in constant speed, it will radiate and energy lost. this is called synchrotron radiation.

There are mainly 2 types of ring accelerators, 1) cyclotron, 2) synchrotron.

Cyclotron


cyclotron is the simplest type, it has 2 D shape cavities and the 2 D shape formed a circle. the 2 D shape cavities is under a magnetic field to blend the particle. and the 2 Ds have different electric potential. when a particle pass from 1 D to the others, due to the potential different, it will be accelerate.

as you can imagine, the potential of the Ds has to be oscillating so that the particle is accelerated when passing each gap between the 2Ds. That frequency is called cyclotron frequency. and it also reflects the particle circulating frequency. surprisingly, the cyclotron frequency only depends on the magnetic field strength, the charge and the mass of the particle.

\omega = 2\pi f = \frac { B q} {m}

which means, no matter the particle position, it moves in same frequency. Thus, the outer particle move faster then the inner one.

now a day, cyclotron may not just contains 2 D cavities but any \frac { 2 \pi }{n} cavities. where n is number of cavities. thus, 2Ds is also called π – cavity.

The typical speed it can reach is about 10% of speed of light.

The only draw back is, the energy it can reach is limited, if using fixed B field or E field, due to relativistic effect. (i.e. the cyclotron frequency also depend on the speed ) the particle cannot match the frequency and accelerated, after it goes to relativistic speed.

another factor is the B field strength is limited, even using super conductive magnetic. and the limit of B field, limited the max output.

particle is released at the center of the B-field and go outward as it acquire speed. Thus, the limitation of radius also limited the max speed. and also, a large radius means a large B field area, which raise a problem on uniform on the B-field.

So, there are another type of cyclotron, which changing the B field or E field to cope with the changing frequency. such cyclotron is called Synchrocyclotron. but due to the velocity dependent of the frequency, only certain speed of particle can be accelerate, thus, the intensity of the beam is smaller then cyclotron.

Synchrotron

synchrotron can reach a great energy and accelerate particle very close to speed of light.

it uses a lot beam focusing devices and accelerating devices to accelerate the beam in a very large radius. each device is well tuned, and all devices are well synchronized for different particle. Thus it is a very delicates and sophisticated machine.

  1. particle can have every high energy
  2. high intensity of beam
  3. it can have some section only for linear motion with accelerate.
  4. it is not limited by the B field. since the narrow of the beam, a higher forcing B field can be applied.

The only factor reduced the power output is the synchrotron radiation. that’s why they build a bigger and bigger one, since the large radius can reduce the radiation lost.