Exponent of tensor product of operators (context: weak measurement)

Question

I'm stuck on some identity for an exponentiated tensor product of operators, $e^{\hat A\otimes\hat B}$.

I'm learning weak measurement, reading the review by Kofman et al (2012, DOI: 10.1016/j.physrep.2012.07.001).

They study a discrete system S and a continuous meter M (In their Section 2.2). S and M get entangled in a pre-measurement via a brief interaction Hamiltonian: $$ \hat{H}=\hat{A}\otimes\hat{p} $$

where $\hat{A}$ is a discrete operator on S (with eigenstates $|{a_i}\rangle$), and $\hat{p}=-i\partial_q$ is a continuous operator M (on the variable q). During pre-measurement, states in $S\otimes M$ then evolve per the unitary operator $$ \hat{U}=e^{-i\hat{H}}= e^{-i\hat{A}\otimes\hat{p}} $$

If S starts in $|a_0\rangle, $ (an eigenstate of $\hat{A}$ with eigenvalue $a_0$) and M in $|m^{(0)}\rangle = \psi_0(q)$, the initial product state is $|a_0\rangle\otimes|m^{(0)}\rangle \equiv |a_om^{(0)}\rangle $. By unitary evolution we get (Kofman's equation 2.13): $$ |s_om^{(0)}\rangle \rightarrow e^{-i\hat{A}\otimes\hat{p}} |a_0\rangle\psi_0(q) = |a_0\rangle e^{-ia_0\hat{p}}\psi_0(q) $$

Now, that last equality has me confused.

So I'll spell it out with $\hat{A} = a_0\hat\sigma_z = a_0 \left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$, and $|a_0\rangle = \left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$, and $\hat\sigma_z\otimes-i\partial_q = \left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)(-i\partial_q) $, for the left-hand side:

$$ e^{-a_0\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)\partial_q} (\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)\psi_0(q)) $$

Now, I believe that the exponent can be viewed as three commuting operators, so that we could write it $$ e^{a_0}e^{-i\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)}e^{-i\partial_q} (\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)\psi_0(q)) $$

[Edit: Per Oleg's answer, the above belief is wrong.]

Also, I believe that $\hat{A}$ only operates on $|a\rangle,$ and $\hat{p}$ only on $\psi_0(q)$. Then:

$$ \rightarrow e^{a_0} (\left(\begin{smallmatrix}e^{-i}&0\\0&e^{i}\end{smallmatrix}\right)\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)) (e^{-\partial_q}\psi_0(q)) = e^{a_0}\left(\begin{smallmatrix}e^{-i}\\0 \end{smallmatrix}\right)e^{-\partial_q}\psi_0(q) $$

$$ = e^{a_0+i}|a_0\rangle e^{-\partial_q}\psi_0(q) = |a_0\rangle e^{a_0-i-i\partial_q}\psi_0(q)$$

Which is not at all like the right-hand side in Kofman's equation 2.13(above)! They get a multiplication by $a_0$; I get an addition by $a_0-i$.

Answer 1

Here how it comes out: just Taylor expand the exponent, apply it, then sum it back $$e^{A\otimes B}\vert a_0\rangle\vert \psi_B\rangle= \sum_{n=0,1...}\frac{1}{n!} A^n\otimes B^n\vert a_0\rangle\vert \psi_B\rangle =\sum_{n=0,1...}\frac{1}{n!} a_0^nI\otimes B^n \vert a_0\rangle\vert \psi_B\rangle =e^{a_0I\otimes B}\vert a_0\rangle\vert \psi_B\rangle$$

In your calculation you said

Now, I believe that the exponent can be viewed as three commuting operators

followed by calculation that states something like

$$e^{A\otimes B}=e^{A\otimes I}e^{I\otimes B}$$

which is not true even when $A$ and $B$ are scalars, let alone operators.