Advanced atomic physx: From Liouville Equation to the Bloch equations

1. Jun 9, 2014

TheDestroyer

This question was on Stackexchange, and no one was able to solve it. So now I'm posting it here, and I wish someone could help.

I'm trying to derive the Bloch equations[1] from the Liouville equation[2]. This should be possible according to this paper[3], where it 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[5] 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[4]:

$$\left[F_{\pm1},T_{q}^{k}\right]=\hbar\sqrt{\left(k\mp q\right)\left(k\pm q+1\right)}T_{q\pm1}^{k}$$
$$\left[F_{0},T_{q}^{k}\right]=\hbar qT_{q}^{k}$$

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[1] 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[5], 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[1]! 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?

Thank you for any efforts.

[1]: http://en.wikipedia.org/wiki/Bloch_equations
[2]: http://en.wikipedia.org/wiki/Liouville's_theorem_(Hamiltonian)#Quantum_Liouville_equation
[3]: http://arxiv.org/pdf/physics/0605234.pdf
[4]: http://en.wikipedia.org/wiki/Tensor_operator#Spherical_tensor_operators
[5]: http://en.wikipedia.org/wiki/Spherical_basis

Last edited: Jun 9, 2014
2. Jun 9, 2014

TheDestroyer

In that line $k=1,q=0$, which makes the $\omega_0$ term vanish because it's muliplied by $q$.

Best regards.

3. Jun 9, 2014

Bill_K

One way to realize there's a sign error in the final equations is to calculate

$$m_x\frac{d}{dt}m_x + m_y\frac{d}{dt}m_y + m_z\frac{d}{dt}m_z$$
which should come out zero, and does not.

I think it's the second equation which has the wrong sign throughout, check it again. It should be
$$\frac{d}{dt}m_{y}=+\sqrt{2}m_{z}\left(i\omega_{y}\right)+m_{x}\omega_{z}$$

4. Jun 9, 2014

TheDestroyer

Thanks for your response. I checked it and I found no sign error; though maybe I'm wrong, but this is a secondary problem, since even if a sign error is there, this won't resolve the problem I'm facing where those equations have nothing to do with Bloch equations the way we know them.

Any other ideas?

5. Jun 9, 2014

Bill_K

Yes, that's not the only sign error! Sign errors are definitely your problem.

Write out the dm0/dt equation, converting it to Cartesian components. You'll have four terms on the RHS. Two of them cancel out, but the wrong two! You're supposed to be left with the cross product, mxωy - myωx, but with your signs those terms cancel out and you're left instead with the dot product, mxωx + myωy

EDIT: If you still haven't found the first sign error, better check it a third time. That one's pretty obvious, and if you can't find that one you probably won't find the others!

Last edited: Jun 9, 2014
6. Jun 10, 2014

TheDestroyer

Thank you for your response. As reasonable as what you're saying sounds, I verified the signs like 5 times in different times to make sure there's no mistake, and I can't find any. Let me give an example. Let's talk about the z equation.

The equation is:

$$i\frac{d}{dt}m_{1,0}=m_{1,1}\omega_{-1}\sqrt{2}+m_{1,-1}\omega_{+1}\sqrt{2}$$

We use the spherical bases and we get (ignore the sqrt(2) for now):

$$i\frac{d}{dt}m_{z}=-\left(m_{x}-im_{y}\right)\left(\omega_{x}+i\omega_{y}\right)-\left(m_{x}+im_{y}\right)\left(\omega_{x}-i\omega_{y}\right)$$

Do the brackets and we get:

$$i\frac{d}{dt}m_{z}=-m_{x}\omega_{x}-m_{x}i\omega_{y}+im_{y}\omega_{x}-im_{y}\omega_{y}-m_{x}\omega_{x}+m_{x}i\omega_{y}-im_{y}\omega_{x}-im_{y}\omega_{y}$$

And by simplifying:

$$i\frac{d}{dt}m_{z}=-\sqrt{2}m_{x}\omega_{x}-\sqrt{2}im_{y}\omega_{y}$$

This still isn't a cross product. Could you please verify this equation yourself?

Thank you.

7. Jun 12, 2014

TheDestroyer

Thanks to the people who've helped, though the problem was pointed to me by my Professor.

The problem was in the definition of raising and lowering operators with respect to spherical basis.

$$\vec{\omega}\cdot\vec{F}=\sum_{q=-k}^{k}\left(-1\right)^{q}\omega_{q}F_{-q}=-\omega_{-1}F_{1}+\omega_{0}F_{0}-\omega_{1}F_{-1}$$

Then $F_+$, the raising operator is not equivalent to $F_{+1}$ the spherical basis component. This I didn't know before, and now I learned. The right relation between those elements is

$$F_{\pm1}=\mp\frac{1}{\sqrt{2}}F_\pm$$

And the minus sign in this relation will solve the problem. And I eventually got the Bloch equations.