## high pass filter, low pass filter and OpAmp

a resistor and capacitor in series can create a filter to let either high frequency AC or low frequency AC to pass.

the high pass filter is connect to the resistor.

the low pass filter is connected to the capacitor.

the reason can be understand when consider the charging and dis-charging process of the capacitor. the charge rate is equal to $1 / RC$. For a low frequency, the time vary slowly, and the capacitor can charged up and equal to to supply voltage when $t >= RC$ or $\omega <= 1/ RC$. thus, the voltage across the capacitor can always equal to the supply and the voltage across the resistor is always zero. thus, low pass filter connected to capacitor.

when high frequency, time vary fast, and the capacitor cannot charge up and oppose the supply voltage. thus, current always flow in an out and make the resistor voltage follow the supply voltage. thus, for frequency $\omega >= 1/RC$, only high  frequency can “live” in the resistor and therefore the high pass filter is connected to the resistor.

in mathematic, one can use impedance for calculation, that i don’t show in here.

to calculate the output of an OpAmp, just need to remember 3 things:

1. the open loop amplification is very large.  $G \rightarrow \infty$
2. the impedance of the input is very large, no current flow between $V_+$ and $V_-$
3. the output is the different of the input times amplification.  $V_{out} = G (V_+ - V_-)$

and OpAmp can be modified and only let high or low frequency to be amplified.

under the Inverting amplifing, such that V+ in grounded.

add a parallel capacitor with the Rf give you a low pass filter.

add a series capacitor with the Rin give you a high pass filter.

## Transmission Line

a Transmission Line is any thing used to transmit a electric signal.

in a AC circuit, the voltage does not only variate on time but also on space. For low frequency, the wavelength is long and this can be neglected. But at hight frequency, the variation is significant.

since the voltage is changing from different location. we cut the transmission line in a small sector, and each sector is analog to some circuit elements, no matter the shape of the line. this gives us a easy understanding of what is going on for the voltage and current. But this analogy neglected the effect of temperature, material non-linearity and magnetic hysteresis effect.

the equation of voltage across a section is

$V(x) = ( R + i \omega L )\Delta x I(x)+ V(x+\Delta x )$

since the resistance and inductance have unit per length. rearrange and take limit of x.

$- \frac {d V(x)}{dx} = ( R + i \omega L ) I(x)$

the equation of current is

$I(x) = ( Q + i \omega C )\Delta x V(x+ \Delta x) + I(x+ \Delta x )$

The Q is conductance, which is NOT an invert of resistance in case of AC. both Q and C will draw some current away in AC circuit. take limit gives

$- \frac { d I(x) }{dx} = (Q + i \omega C ) V(x)$

now we have 2 coupled equations. If we de-couple them, we will have

$\frac {d^2 V(x)}{dx^2} = k^2 V(x)$

$k = \sqrt{ (R+i \omega L ) ( Q + i \omega C ) }$

and the current share the same equation. notice that k is a complex number.

the solution are:

$V(x) = V_f Exp( - k x ) + V_b Exp( k x)$

$I(x) = I_f Exp( - k x) + I_b Exp( k x )$

where the subscript means forward and backward. from the coupled equation of Current, we can related the voltage and current and find out the impedance.

$V(x) = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C } } (I_f Exp(-kx) - I_b Exp(k x)$

Thus, we define the Characteristic Impedance for forward wave.

$Z_0 = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C }}$

the Characteristic Impedance for backward wave is a minus sign. there fore, we can rewrite the current in term of voltage.

$I(x) = \frac {1}{Z_0} ( V_f Exp(-kx) - V_b Exp(kx)$

_________________________________________

## impedance matching

for a load at the end of the transmission line, the wave will get reflected. to see this, we have to consider the load, which imposed another equation. the voltage across the load is:

$V(L) = V_f Exp( - k L ) + V_b Exp( k L)$

and the current input to the load is:

$I(L) = \frac {1}{Z_0} ( V_f Exp(-kL) - V_b Exp(kL)$

the current and the voltage is related by:

$V(L) = Z_L I(L)$

solve it, and find the ratio of :

$\frac { V_b }{V_f} = Exp( - 2 k L ) \frac { Z_L - Z_0 }{ Z_L + Z_0 }$

since we can do nothing on the exponential, thus, we define a reflection coefficient:

$\Gamma_0 = \frac { Z_L - Z_0} {Z_L - Z_0}$

for no reflected wave, the impedance of the load and the transmission line should be equal and it is called impedance matching.

the power of the load is:

$P = V(L)I*(L) = \frac { 1}{Z_0} ( V_f^2 Exp( - 2 k L ) - V_b^2 Exp( 2k L) + V_f V_b* -V_f* V_b )$

when the impedance is matched, the power is maximum.

## on Relaxation in NMR

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.

## NMR (nuclear magnetic resonance)

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)

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

( 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.

## decay time constant and line width

the spectrum of energy always has a peak and a line width.

the reason for the line width is, this is decay.

i give 2 explanations, once is from classical point of view and i skipped the explanation for the imaginary part. so, i am not fully understand. the 2nd explanation is look better, but it is from QM. however, there is one hide question for that explanation is, why the imaginary energy is negative?

the simplest understanding of the relation is using fourier transform. (i think)

Fourier transform is changing the time-frame into the frequency frame. i.e, i have a wave, propagating with frequency w. we can see a wave shape when plot with time. and we only see a line, when we plot with frequency, since there is only 1 single frequency. however, for a general wave, it is composite of many different frequencies, using fourier transform can tell us which frequency are involved. And energy is proportional to frequency.

when the particle or state under decay. the function is like

$f(t) = Exp(-R t) Exp ( i \omega_0 t)$

where the R is decay constant, and ω0 is the wave frequency.

after fourier transform, assume there is nothing for t < 0

$F(t) = \frac {1} { R + i ( \omega_0 - \omega )}$

the real part is

$Re(F(t)) = \frac {R} { R^2 + ( \omega_0 - \omega )^2}$

which is a Lorentzian shape and have Full-Width-Half-Maximum (FWHM) is 2R. it comes from the cosine part of the fourier transform. thus, the real part.

and the imaginary part is

$Im(F(t)) = \frac {\omega_0 - \omega}{R^2 + ( \omega_0 -\omega )^2 }$

the imaginary part is corresponding to the since part, so, we can neglect it. (how exactly why we can neglect it? )

Thus, we can see, if there is no decay, R → 0, thus, there is no line width.

therefore, we can see the line width in atomic transition, say, 2p to 1s. but there are many other mechanism to the line width, like Doppler broadening, or power broadening. So, Decay will product line width, but not every line width is from decay.

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

another view of this relation is from the quantum mechanics.

the solution of Schroedinger equation is

$\Psi (x,t) = \phi(x) Exp \left( - i \frac {E}{\hbar} t \right)$

so, the probability conserved with time, i.e.:

$|\Psi(x,t)|^2 = |\Psi (x,0)|^2$

if we assume the energy has small imaginary part

$E = E_0 - \frac {i} {2} R \hbar$

( why the imaginary energy is nagative?)

$|\Psi(x,t)|^2 = |\Psi (x,0)|^2 Exp ( - R t)$

that make the wavefunction be :

$\Psi (x,t) = \phi (x) Exp( - i \frac {E}{\hbar} t ) Exp( - \frac {R}{2} t )$

what is the meaning of the imaginary energy?

the wave function is on time-domain, but what is “physical”, or observable is in Energy -domain. so, we want Psi[x,E] rather then Psi[x,t], the way to do the transform is by fourier transform.

and after the transform, the probability of finding particle at energy E is given by

$|\Psi(x,E)|^2 = \frac {Const.}{R^2 +(\omega_0 - \omega )^2}$

which give out the line width in energy.

and the relation between the FWHM(line width) and the decay time is

mean life time ≥ hbar / FWHM

which once again verify the uncertainty principle.