Partial decoherence from interaction between two qubits $\renewcommand{\ket}[1]{\left\lvert #1 \right \rangle}$
If a quantum system $A$ becomes entangled with another quantum system $B$, then $A$ can no longer be described by a pure quantum state.
For example, given a Bell state
$$
    \ket{00} + \ket{11} \, ,
$$
the state of either qubit by itself is a classical mixture of $\ket{0}$ and $\ket{1}$.
In this case where we have complete entanglement, the state of each qubit uniquely determines the state of the other and the quantum coherence of each individual qubit is completely lost.
However, if the two qubits interact only weakly, the effect on each individual qubit state should be equivalent to a weak measurement and result in only partial loss of coherence.
Starting with an initial product state of two qubits
$$
  \ket{\psi(0)} = \frac{1}{2} ( \underbrace{\ket{0} + \ket{1}}_A ) \otimes (\underbrace{\ket{0} + \ket{1}}_B) \, ,
$$
can we show that physical interaction through an interaction Hamiltonian
$$
  H/\hbar = \frac{g}{2} \left( \sigma_x \otimes \sigma_x + \sigma_y \otimes \sigma_y \right)
    = g \left(
        \begin{array}{cccc}
          0 & 0 & 0 & 0 \\
          0 & 0 & 1 & 0 \\
          0 & 1 & 0 & 0 \\
          0 & 0 & 0 & 0
        \end{array}
      \right)
$$
can result in partial decoherence of qubit $A$?
We use the basis state ordering $\{ \ket{00}, \ket{01}, \ket{10}, \ket{11} \}$ implicit in the definition of the Kronecker product.
Note, therefore, that qubit $A$ is the most significant bit in the two-qubit kets.
Furthermore, in the chosen basis state ordering, $\ket{\psi(0)}$ is represented as
$$
    \frac{1}{2}
    \left(
        \begin{array}{c} 1 \\ 1 \\ 1 \\ 1
        \end{array}
    \right) \, .
$$
 A: $\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$
The propagator for this Hamiltonian is
$$
U(t) = \exp(-iHt/\hbar) = 
  \left(
  \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \cos(gt) & -i \sin(gt) & 0 \\
    0 & -i \sin(gt) & \cos(gt) & 0 \\
    0 & 0 & 0 & 1
  \end{array}
  \right)
  \, .
$$
Applying $U(t)$ to the initial state $\ket{\Psi(0)}$, we get
$$
  \ket{\psi(t)} =
  U \ket{\psi(0)} = 
  \frac{1}{2}
  \left(
    \begin{array}{c}
      1 \\ e^{-i g t} \\ e^{-i g t} \\ 1
    \end{array}
  \right) \, .
$$
The density matrix is therefore
$$
  \rho(t) = \frac{1}{4}
  \left(
  \begin{array}{cccc}
    1         & e^{igt}  & e^{igt}  & 1        \\
    e^{-igt}  & 1        & 1        & e^{-igt} \\
    e^{-igt}  & 1        & 1        & e^{-igt} \\
    1         & e^{igt}  & e^{igt}  & 1
  \end{array}
  \right) \, .
$$
Now take the partial trace over qubit $B$ (the least significant bit).
Doing this leaves the density matrix of qubit $A$ (the most significant bit) as
$$
  \rho_A(t) =
  \frac{1}{2}
  \left(
    \begin{array}{cc}
      1 & \cos(gt) \\
      \cos(gt) & 1
    \end{array}
  \right)
  \, .
$$
After a time $t = \pi/(2g)$, the coherence of qubit $A$ is completely gone.
This time is precisely the time it takes for the two qubits to "swap" their excitations, i.e. for the process
$$\ket{1} \otimes \ket{0} \rightarrow \ket{0} \otimes \ket{1} \, .$$
Note, however, that because of the initial state where both qubits are in a superposition of excited and not excited, we observe in $\rho_A(t)$ no change in population of the two levels.
Instead, we see only an oscillation in the qubit's coherences.

Thanks to Emilio Pisanty for help with this question and answer.
