How do we prove that two conjugate operators $X$ and $Y$ induce $\sigma_x$ and $\sigma_y$ driving terms when restricted to a two level subspace? Suppose I have a Hamiltonian for a particle moving in a one dimensional potential
$$H = H(X,Y) \qquad [X,Y] = i$$
where $X$ is the dimensionless position, $Y$ is the dimensionless momentum, and $\epsilon_0$ is an energy scale.
Both $X$ and $Y$ are Hermitian.
Now suppose we add a driving Hamiltonian
$$H_x = \epsilon_x \, f_x(t) \, X$$
where $f_x(t)$ is a dimensionless function of time and $\epsilon_x$ is another energy scale.
If $f_x(t)$ is on resonance with a transition between two energy eigenstates of $H$, we expect it to produce transitions between those two states.
Restricting considerations to just those two levels, we can write $H_x$ as a 2x2 matrix:
$$H_x \sim \epsilon_x f_x(t)\left(
\begin{array}{cc}
0 &
\langle 0 | X | 1 \rangle \\
\langle 1 | X | 0 \rangle &
0
\end{array}
\right)
= \epsilon_x \langle 1 | X | 0 \rangle f_x(t) \sigma_x \, .
$$
where $\sim$ means "is represented by" and we've assumed $\langle 0 | X | 0 \rangle$ and $\langle 1 | X | 1 \rangle$ are both zero.$^{[a]}$
Now suppose we have a different driving Hamiltonian
$$H_y = \epsilon_y f_y(t) Y \, .$$
By similar reasoning to what we did above, we could reason that
$$H_y \sim \epsilon_y f_y(t) \left(
\begin{array}{cc}
0 & \langle 0 | Y | 1 \rangle \\
\langle 1 | Y | 0 \rangle & 0
\end{array}
\right)
= \epsilon_y \langle 1 | Y | 0 \rangle f_y(t) \sigma_x \, .$$
This, however, is not correct.
By looking at examples such as the harmonic oscillator we find that the driving Hamiltonian which couples to $Y$ is proportional to $\sigma_y$, not $\sigma_x$.
How, in general, can we argue that two conjugate operators, when restricted to a certain two state subspace, must be proportional to $\sigma_x$ and $\sigma_y$?
$[a]$: Or at least, we can ignore the diagonal elements of $H_x$. In general the diagonal part can be written as a sum of $\sigma_z$ and the identity. The identity part can be completely ignored. The $\sigma_z$ part can be dropped if $f(t)$ is oscillatory because is just induces a net-zero fluctuation in the level splitting between $|0\rangle$ and $|1\rangle$. I think.
 A: Consider a Hermitian operator $X$, and denote by $x$ its projection to the two-dimensional subspace. Then, $x$ is Hermitian as well.  If you assume that the diagonal is zero (e.g. because you shift the energy + choose a rotating frame accordingly), then $x$ is of the form $x=\alpha_x\sigma_x+\beta_x\sigma_y$ with real $\alpha$ and $\beta$. The same is true for the projection of $Y$, $y=\alpha_y\sigma_x+\beta_y\sigma_y$.
Note that we cannot say that $x=\sigma_x$, as can be seen by changing to a different canonical basis such as $(X\pm Y)/\sqrt{2}$ (which gives $x,y\propto\sigma_x\pm\sigma_y$).
In particular, note that $[x,y]$ is a sum of commutators of $\sigma_x$ and $\sigma_y$, i.e., it is proportional to $i\sigma_z$ (with a real prefactor).
Note that we have not used that $X$ and $Y$ are a conjugate pair. An open question is whether this would imply that $x$ and $y$ are orthogonal (i.e., $\alpha_x\alpha_y+\beta_x\beta_y=0$).
A: Good question,
hopefully this won't backfire since I didn't do any math to check this... but try the commutator of $[H_{x},H_{y}]$ before your 'represented as' notation (that is, in matrix form), and you should see they don't commute, hopefully in a similar way to $\sigma_{x}$ and $\sigma_{y}$.  This should be a result of your given info about $[X,Y]$.
Tell me how it goes.
