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I'm reading those lecture notes on atomic physics. Yesterday I posed a question on reducible tensors, and today I have a question on their relation to the density matrix.

If there's any information missing, please let me know so that I could provide it.

Please observe the following test from the script:

The density matrix ρ can be expanded in the basis of the irreducible tensor operators $T^{(K)}_Q$ according to $$\rho = \sum\limits_{K = > 0}^{2F} {\sum\limits_{Q = - K}^K {{m_{K,Q}}\,{T_Q}^{\left( K \right)}} }\quad \quad (8.29)$$

where $T^{(K)}_Q$ is given by the formula

$${T_Q}^{\left( K \right)} = \sum\limits_{M,M'} {{{\left( { - 1} \right)}^{F - M'}}\left\langle {F,M;F, - M'|K,Q} \right\rangle \left| {F,M} \right\rangle \left\langle {F,M'} \right|}$$

At a later part the writer shows how he derives the spin-1/2 system density matrix from the first equation as follows

Consider as a first example the decomposition of the density matrix of spin 1/2 particles. According to Eq. (8.29) it is given by $$\begin{array}{c} \rho = \sum\limits_{K = 0}^1 {\sum\limits_{Q = - K}^K {{m_{K,Q}}\,{T_Q}^{\left( K \right)}} } \\ = {m_{0,0}}T_0^{(0)} + \sum\limits_{Q = - K}^K {{m_{1,Q}}\,{T_Q}^{\left( 1 \right)}} \\ = \frac{1}{2} + \sum\limits_{Q = - K}^K {{m_{1,Q}}\,{T_Q}^{\left( 1 \right)}} \end{array}$$

Now I have a few questions:

1- Where did that 1/2 in the last line come from? How is it calculated?

2- How can you calculate the other terms systematically?

3- Can you provide the same calculation for spin-1 and spin-3/2 particles? I want to learn the generalized process.

4- Could you please put any of all this in the frame of Liouville's equation? I'm having a hard time imagining an equation of motion with spins that way.

Thank you for any efforts.

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Which lecture notes? –  Qmechanic Oct 12 '13 at 19:33
    
@Qmechanic Polarized light and Polarized Atoms - Prof. Antoine Weis - Université de Fribourg. –  The Quantum Physicist Oct 12 '13 at 20:13
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1 Answer 1

As to your first question: part of your difficulty is with the notation.

The $\frac{1}{2}$ is really $\frac{1}{2} \times 1_{2\times 2}$ where $1_{2\times 2}$ is the unit matrix: the $\frac{1}{2}$ arises because the $2\times 2$ unit matrix has trace=2.

More generally the coefficients $m_{1,Q}$ can be obtained by first writing

$\rho= m_{00}1_{2\times 2} + m_{1,Q}T^{(1)}_Q$

and tracing the density matrix with $(T^{(K)}_{Q})^\dagger$. Orthonormality under trace of the tensors as you have defined them, i.e. Tr[$(T^{(K)}_{Q})^\dagger T^{(K)}_{Q'}]=\delta_{QQ'}$ guarantees that Tr[$(T^{(K)}_{Q})^\dagger \rho]=m_{1,Q}$

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