Entangled spins in a uniform magnetic field

I have been trying to calculate the time derivative of the expectation value of the operator representing the spin components of a a singlet state (e.g. sigma(z)tau(z)). I formed the tensor product of the two spins using the regular Pauli matrices and chose a Hamiltonian representing a uniform magnetic field in the sigma(z)tau(z) direction. Using the commutation relations for the time derivative of an expectation value I found that all three time derivatives for the expectation values of the three composite spin components end up being zero, which is disappointing since I was expecting to see something that looked liked Larmour precession. Something even wierder is that I found that sigma(z)tau(z) and sigma(x)tau(x) commute, which I know is ridiculous.

My question is how do I properly treat this problem because I don't think I'm doing this right.

Also, when I write "sigma(z)tau(z)" I really mean the operator that represents the tensor product of two spin components in a composite system.

• If sigma(z)tau(z) is a tensor product of two spin components in a composite system then it is not a direction so how does the magnetic field point in that direction? Is this a classical external magnetic field or are they each responding to the magnetic field from the magnetic dipole moment of the other one? Commented Oct 30, 2015 at 0:43
• Well I thought the Hamiltonian for two spins could be written as H=(hw/2)sigma(z)tau(z); where h is h-bar and w is "omega" where you would bury everything about the magnetic field (strength, frequency etc). Commented Oct 30, 2015 at 18:39

I tried to ask you to clarify the problem so I'll assume a strong external classical field.

The Hamiltonian for a magnetic moment $\vec \mu=\gamma \vec S$ (where $\gamma$ is the gyromagnetic ratio) in an external magnetic field is

$$H=(P_x^2+P_y^2+P_z^2)/2m-\vec \mu \cdot \vec B$$

Which assumes no other interactions, such as electric charge or electric fields. For two particles we can get

$$H=(P_{1x}^2+P_{1y}^2+P_{1z}^2)/2m_1+(P_{2x}^2+P_{2y}^2+P_{2z}^2)/2m_2-\vec \mu_1 \cdot \vec B-\vec \mu_2 \cdot \vec B$$

Which ignores any electric fields, and even any interaction between the magnetic fields of the dipoles on each other. Ignoring the dipole dipole interactions can be a small effect if the external magnetic field is much larger (e.g. when the particles are far apart from each other).

We can relate this to the spin by using the gyromagnetic ratio. $$H=(P_{1x}^2+P_{1y}^2+P_{1z}^2)/2m_1+(P_{2x}^2+P_{2y}^2+P_{2z}^2)/2m_2-\gamma_1\vec S_1 \cdot \vec B-\gamma_2\vec S_2 \cdot \vec B$$

And if we have an external magnetic field all in the z direction and switch to operators we get:

$$\hat H=(\hat P_{1x}^2+\hat P_{1y}^2+\hat P_{1z}^2)/2m_1+(\hat P_{2x}^2+\hat P_{2y}^2+\hat P_{2z}^2)/2m_2-\gamma_1B\hat\sigma_{z1} -\gamma_2B\hat\sigma_{z2}$$

For such a Hamiltonian (a sum of two single particle Hamiltonians) you can literally take as an eigenstate, a product of single particle eigenstates. So for instance you can take a free particle spatial part $\Psi_1(\vec r_1),$ and $\Psi_2(\vec r_2)$ and a spin state like $\left[\begin{matrix}1\\0\end{matrix}\right]_1\otimes\left[\begin{matrix}1\\0\end{matrix}\right]_2$ and for an eigenvalue you get the sum of the eigenvalues of the four states on their own.

In your case you want a singlet, so your spin state is $\frac{1}{\sqrt 2}\left[\begin{matrix}1\\0\end{matrix}\right]_1\otimes\left[\begin{matrix}0\\1\end{matrix}\right]_2-\frac{1}{\sqrt 2}\left[\begin{matrix}0\\1\end{matrix}\right]_1\otimes\left[\begin{matrix}1\\0\end{matrix}\right]_2$ and so your state is $$\Psi(\vec r_1,\vec r_2)=\frac{\Psi_1(\vec r_1)\Psi_2(\vec r_2)}{\sqrt 2}\left(\left[\begin{matrix}1\\0\end{matrix}\right]_1\otimes\left[\begin{matrix}0\\1\end{matrix}\right]_2-\left[\begin{matrix}0\\1\end{matrix}\right]_1\otimes\left[\begin{matrix}1\\0\end{matrix}\right]_2\right).$$

And it is a linear combination of energy eigenstates such as $\Psi_1(\vec r_1)\Psi_2(\vec r_2)\left[\begin{matrix}1\\0\end{matrix}\right]_1\otimes\left[\begin{matrix}0\\1\end{matrix}\right]_2$ and $\Psi_1(\vec r_1)\Psi_2(\vec r_2)\left[\begin{matrix}0\\1\end{matrix}\right]_1\otimes\left[\begin{matrix}1\\0\end{matrix}\right]_2,$ which have eigenvalues $T_1+T_2+B(\gamma_2-\gamma_1)\hbar/2$ and $T_1+T_2+B(\gamma_1-\gamma_2)\hbar/2$ respectively. So each eigenstate just has a nice linear phase evolution over time. And by linearity you know how the whole singlet state evolves.

• Sorry for not being clear sooner. Commented Oct 31, 2015 at 3:34
• On one note, I would like to know what exactly you mean by T1 and T2 at the very end. Commented Oct 31, 2015 at 19:13
• @user2216571 Kinetic energies of particle one and particle two respectively. Commented Oct 31, 2015 at 19:14
• Ah! I knew I was going to something obvious... Thanks. Commented Oct 31, 2015 at 19:29