I'm trying to derive the Bloch equations from the Liouville equation. This should be possible according to this paper, which discusses higher order Bloch equations (second order spherical tensors). I'm trying to derive the same for the simple vector Bloch equations.
The way I'm doing this is the Following.
The Lioville equation is this: $$\frac{d}{dt}\rho=\frac{1}{i\hbar}\left[H,\rho\right]$$
The Hamiltonian for an interaction with the magnetic field can be written as
$$H=-\vec{\mu}\cdot\vec{B}=-\frac{g\mu_{B}}{\hbar}\vec{B}\cdot\vec{F}=-\vec{\omega}\cdot\vec{F},$$ where $$\left\{ \omega_{x},\omega_{y},\omega_{z}\right\} =\frac{g\mu_{B}}{\hbar}\left\{ B_{x},B_{y},B_{z}\right\} $$
And I expand the dot product using spherical basis dot product: $$\vec{\omega}\cdot\vec{F}=\sum_{q=-k}^{k}\left(-1\right)^{q}\omega_{q}F_{-q}=-\omega_{-1}F_{1}+\omega_{0}F_{0}-\omega_{1}F_{-1}$$ The density matrix can be expanded in terms of spherical operators as: $$\rho=\sum_{k=0}^{2F}\sum_{q=-k}^{k}m_{k,q}T_{q}^{\left(k\right)},$$
Now I try to compute the commutator using all the information I calculated:
$$\left[H,\rho\right]=\sum_{q=-k}^{k}\left[\omega_{-1}F_{1}-\omega_{0}F_{0}+\omega_{1}F_{-1},m_{k,q}T_{q}^{\left(k\right)}\right]$$
$$=\sum_{q=-k}^{k}m_{k,q}\left[\omega_{-1}F_{1}-\omega_{0}F_{0}+\omega_{1}F_{-1},T_{q}^{\left(k\right)}\right]$$
$$=\sum_{q=-k}^{k}m_{k,q}\omega_{-1}\left[F_{1},T_{q}^{\left(k\right)}\right]-\sum_{q=-k}^{k}m_{k2,q}\omega_{0}\left[F_{0},T_{q}^{\left(k\right)}\right]+\sum_{q=-k}^{k}m_{k,q}\omega_{1}\left[F_{-1},T_{q}^{\left(k\right)}\right]$$
Now we use the commutation relations with tensor operators:
$$\left[F_{\pm1},T_{q}^{k}\right]=\hbar\sqrt{\left(k\mp q\right)\left(k\pm q+1\right)}T_{q\pm1}^{k}$$ $$\left[F_{0},T_{q}^{k}\right]=\hbar qT_{q}^{k}$$
And we get:
$$\left[H,\rho\right]=\sum_{q=-k}^{k}m_{k,q}\omega_{-1}\hbar\sqrt{\left(k-q\right)\left(k+q+1\right)}T_{q+1}^{k}-\sum_{q=-k}^{k}m_{k,q}\hbar q\omega_{0}T_{q}^{k}+\sum_{q=-k}^{k}m_{k,q}\omega_{+1}\hbar\sqrt{\left(k+q\right)\left(k-q+1\right)}T_{q-1}^{k}$$
Now we substitute this in the Liouville equation
$$i\sum_{q}\frac{d}{dt}m_{k,q}T_{q}^{\left(k\right)}=\sum_{q=-k}^{k}m_{k,q}\omega_{-1}\sqrt{\left(k-q\right)\left(k+q+1\right)}T_{q+1}^{k}-\sum_{q=-k}^{k}m_{k,q}q\omega_{0}T_{q}^{k}+\sum_{q=-k}^{k}m_{k,q}\omega_{+1}\sqrt{\left(k+q\right)\left(k-q+1\right)}T_{q-1}^{k}$$
Then we multiply this equation from the left by $T_{q^{\prime}}^{\left(k^{\prime}\right)\dagger}$ and use the orthogonality relation $\text{Tr}\left(T_{q}^{\left(k\right)\dagger}T_{q^{\prime}}^{\left(k^{\prime}\right)}\right)=\delta_{qq^{\prime}}\delta_{kk^{\prime}}$ and we finally get:
$$i\frac{d}{dt}m_{k,q}=m_{k,q+1}\omega_{-1}\sqrt{\left(k-q\right)\left(k+q+1\right)}-m_{k,q}q\omega_{0}+m_{k,q-1}\omega_{+1}\sqrt{\left(k+q\right)\left(k-q+1\right)},$$
My question starts here: How do we get the standard Bloch equations from this? I'm gonna try to do this. So use the vector version of this by saying that $k=1$ and $q={-1,0,1}$, which gives me the set of equations:
$$i\frac{d}{dt}m_{1,-1}=m_{1,0}\omega_{-1}\sqrt{2}+m_{1,-1}\omega_{0}$$ $$i\frac{d}{dt}m_{1,0}=m_{1,1}\omega_{-1}\sqrt{2}+m_{1,-1}\omega_{+1}\sqrt{2}$$ $$i\frac{d}{dt}m_{1,1}=-m_{1,1}\omega_{0}+m_{1,0}\omega_{+1}\sqrt{2}$$
Now to go to Cartesian coordinates I use the conversion of spherical basis, which are:
$$m_{1,+1}=\frac{m_{x}-im_{y}}{\sqrt{2}}$$ $$m_{1,0}=m_{z}$$ $$m_{1,-1}=-\frac{m_{x}+im_{y}}{\sqrt{2}}$$
But this doesn't give the Bloch equations! When I substitute those in the last result I get (by also adding and subtracting the first and last equations to separate mx and my equations)
$$\frac{d}{dt}m_{x}=-\sqrt{2}m_{z}\left(i\omega_{x}\right)-m_{y}\omega_{z}$$ $$\frac{d}{dt}m_{y}=-\sqrt{2}m_{z}\left(i\omega_{y}\right)-m_{x}\omega_{z}$$ $$\frac{d}{dt}m_{z}=\sqrt{2}m_{x}\left(i\omega_{x}\right)-\sqrt{2}m_{y}\left(i\omega_{y}\right)$$
Which are weird and I don't understand... could someone please explain where I did a mistake? Why am I not getting the Bloch equations?
