[Pol. p target] Modeling Microwave Unit Signal

February 15, 2011

GoLuckyRyan circuits, experimental, Lab Log , , Leave a comment

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.

Clebsch – Gordan Coefficient II

February 15, 2011

GoLuckyRyan Basic, operational, theory , Leave a comment

As last post discussed, finding to CG coefficient is not as straight forward as text book said by recursion.
However, there are another way around, which is by diagonalization of J^2

first we use the identity:

J^2 = J_1^2+J_2^2 + 2 J_{1z} J_{2_z} + J_{1+} J_{2-} + J_{1-} J_{2+}

when we “matrix-lize” the operator. we have 2 choice of basis. one is \left| j_1,m_1;j_2;m_2 \right> , which give you non-diagonal matrix by the J_{\pm} terms. another one is \left|j,m\right>, which give you a diagonal matrix.

Thus, we have 2 matrixs, and we can diagonalized the non-diagonal. and we have the Unitary transform P, from the 2-j basis to j basis, and that is our CG coefficient.

oh, don’t forget the normalized the Unitary matrix.

i found this one is much easy to compute.

Clebsch – Gordan Coefficient

February 15, 2011

GoLuckyRyan Basic, operational, theory, Working on Leave a comment

i am kind of stupid, so, for most text book with algebra example, i am easy to lost in the middle.

Thus, now, i am going to present a detail calculation based on Recursion Relations.

we just need equation and few observations to calculate all. i like to use the J- relation:

K(j,-m-1) C_{m_1 m_2}^{j m}= K(j_1,m_1) C_{m_1+1 m_2}^{j m+1}+ K(j_2,m_2) C_{m_1 m_2+1}^ {j m+1}

K(j,m) = \sqrt{ j(j+1) - m(m+1)}

C_{m_1 m_2}^{j m} is the coefficient.

Notice that the relation is only on fixed j, thus, we will have our m_1 m_2 plane with fixed j, so, we have many planes from j = j_1+j_2 down to j = |j_1-j_2| .

We have 2 observations:

  1. C_{j_1, j_2}^{j_1+j_2 , j_1+j_2} = 1 which is the maximum state. the minimum state also equal 1.
  2. For m \ne m_1+m_2 the coefficient is ZERO.

Thus, on the j = j_1 + j_2 plane. the right-upper corner is 1. then using the relation, we can have all element down and left. and then, we can have all element on the plane.

the problem comes when we consider j = j_1 + j_2 -1 plane. no relation is working! and no book tells us how to find it!

Lets take an example, a super easy one, j_1 = 1/2 , j_2 = 1/2 . possible j = 0, 1 , so we have 2 planes.

The j = 1 plane is no big deal.

but the j = 0 plane, there are only 2 coefficient. and we can just related them and know they are different only a sign. and we have to use the orthonormal condition to find out the value.

See? i really doubt is there somebody really do the actually calculation. J.J.Sakuarai just skip the j = l-1/2 case. he cheats!

when going to higher j1+j2 case, we have w=to use the J- relation to evaluate all coefficient. the way is start from the lower left corner, and use the J- relation to find out the relationship between each lower diagonal coefficients. then, since all lower diagonal coefficients have same m value, thus, the sum of them should be normalized. then, we have our base line and use the J+ to find the rest.

i will add graph and another example. say, j1 = 3, j2 = 1.