## Q-value and Biding energy

I already talked on binding energy.

And the Q-value is the mass different between nuclear reaction. In same case binding energy is same as the Q-value.

Q-value is :

$Q= \sum{m_i} - \sum{m_f}$

By energy conservation, it can be rewritten by kinetic energy.

$Q= \sum{T_f} - \sum{T_i}$

But using energy is bit troublesome due to the frame transform. Using mass is simple, since it is a Lorentzian invariance.

However, during scattering experiment. The kinetic energy is much more easy to measure. Thus the kinetic form is used more frequently in experiment.

I will give the expression of K.E. in lab frame later.

## [Pol. p target] Modeling Microwave Unit Signal

a period is 23ms. with in this period, the modulation signal voltage is

$V_{ms}(b,s) = b + ( t - 12.5) s [V]$

$Max(V_{ms}) = 29.2 [V]$   $Min(V_{ms} = 0 [V]$

on the Gunn Oscillator, there is a mechanical switch, which can adjust the base frequency but changing the length.

$f_{base} (l) = 0.61 + 6.433 [GHz]$

this data is provided by 3 data point in the manual. the output frequency of the microwave is

$f_{out} = f_{base} + F_m ( V_{ms} )$

where $F_m$ is the modulation function, that we have to find out. linear?quadratic? at least get a good approximation for it.

the resonance frequency and its FWHM should depend only on the microwave cavity. an a absorption signal can be formulated by a Lorentzian distribution. and this signal will be converted to voltage by a linear conversion factor. ( the green words is an assumption )

$L( f_{res} , f_{out} , FWHM_{res} ) = 1/ ( 1 + (\frac { f_{res} - f_{out} } {FWHM_{res}} )^2 )$

From the relation between the length and voltage at peak. we can find out the modulation function. since the output frequency is equal to the peak frequency. thus, the output frequency is fixed

$f_{out} = f_{res} = 0.6 l + 6.433 + F_m (V_{ms})$

if we measure l and V_{ms} we can find out F.

by further measurement,  the modulation is non-linear. That’s also explained the FWHM on the CRO change with frequency. since the FWHM of the microwave cavity should be same and the change of the FWHM in CRO reflected that the gradient of the frequency output. for a linear frequency output, the FWHM should be the same. but if the gradient change with due to the modulation signal, the FWHM will change.

## 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.