# Why the partial derivative of cost function is written in this form?

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!!