I am reading a paper about QCNN and BP problem. And in the paper there is a part illustrating the relation between the trainability and variance of the cost. The cost function is written as,

$C(\theta)=Tr[V(\theta)\sigma V^\dagger (\theta)\tilde O]$

$\theta$ is the vector of the trainable parameters, $V(\theta)$ is the unitary that contains the gates in the convolutional and pooling layers plus the fully connected layer, $\tilde O$ is defined on the input Hilbert space such that its expectation value represents the measured value.

And we want to evaluate the partial derivative of $C(\theta)$ with respect to $\theta_\mu$, and the explicit form is

$\partial_\mu C = Tr[W_A V_L \sigma V_L^{\dagger} W_A^{\dagger} [H_\mu, W_B^\dagger V_R^\dagger\tilde O V_R W_B]]\space (*)$

where $V=V_RWV_L$, where $V_R$ and $V_L$ contain all gates in the QCNN except for W. And $W=W_BW_A, W_A=\prod_{\eta \leq \mu}e^{-i\theta_\eta H_\eta},W_A=\prod_{\eta > \mu}e^{-i\theta_\eta H_\eta}$.

I am quite confused about (*), since based on the direct derivation of $C(\theta)$, I get,

$\partial_\mu C = Tr[V_R W_B (\partial_\mu W_A) V_L \sigma V_L^\dagger W_A^\dagger W_B^\dagger V_R^\dagger \tilde O] + Tr[V_R W_B W_A V_L \sigma V_L^\dagger (\partial_\mu W_A^\dagger ) W_B^\dagger V_R^\dagger \tilde O]$

Why the relative positions of each operators are different from what I got above, and what is the specific definition of $H_\mu$ (to be more specific, how does $H_\mu$ relates to $\partial_\mu$) ?

For your reference, the the arxiv link of the original paper is https://arxiv.org/abs/2011.02966.

Any help would be appreciated!!


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