# Quantum Bayesian Networks

## February 15, 2019

### Derivative of matrix exponential wrt each element of Matrix

Filed under: Uncategorized — rrtucci @ 1:26 am

In the quantum neural net field, in order to do backpropagation, one often wishes to take the derivative of a unitary matrix with respect to a parameter it depends on. The wonderful software PennyLane by Xanadu evaluates such derivatives using a simple formula which gives an exact answer, albeit only in special cases. Here I will discuss a simple formula that is fully general, albeit only an approximation, although reputedly a very good approximation, probably due to its symmetric nature and the smoothness of exponential functions. The method is a simple symmetric finite difference approximation.

In a StackExchange question with exactly the same title as this post, somebody called Doug suggested what he calls Higham’s “Complex Step Approximation”, to wit:

If $A$ is a Real matrix and $E_{rs}$ is the matrix which is 1 at position $r,s$ and zero elsewhere, $\frac{d}{dA_{rs}}e^{A} \approx \frac{ e^{A + ihE_{rs}} - e^{A-ihE_{rs}} }{2ih} = \frac{{\rm Im}(e^{A + ihE_{rs}})}{h}$

But what if $A$ is a Hermitian matrix and we want the derivative of $exp(iA)$? Here is a simple adaptation of Higham’s formula to that case.

Let $E^\pm_{rs} = E_{rs} \pm E_{sr}$. Note that $(E^\pm_{rs})^\dagger = \pm E^\pm_{rs}$.

Define a matrix $M(A)$ by $M(A) = \left[ \begin{array}{cc} 0 &-e^{-iA}\\ e^{iA} & 0 \end{array}\right]$

Then $M(A+ih) = \left[ \begin{array}{cc} 0 &-e^{-iA+h}\\ e^{iA-h} & 0 \end{array}\right]$

so $M(A+ih)^\dagger = \left[ \begin{array}{cc} 0 &e^{-iA-h}\\ -e^{iA+h} & 0 \end{array}\right] =-M(A-ih)$

From this, one learns the following simple recipe: the effect of the dagger on $M(A)$ is to put a minus sign in front of the $M$ and to take the Hermitian of the argument too.

Therefore, $\frac{d}{d{\rm Re\;}A_{rs}}M(A) \approx \frac{ M(A + ihE^+_{rs}) - M(A-ihE^+_{rs}) }{2ih}= \frac{{\rm Re\;}[M(A + ihE^+_{rs})]}{ih}$

and $\frac{d}{d{\rm Im\;}A_{rs}}M(A) \approx \frac{ M(A + hE^-_{rs}) - M(A-hE^-_{rs}) }{2h}= \frac{{\rm Re\;}[M(A + hE^-_{rs})]}{h}$.

Since $\frac{d}{dx}M(A) = \left[ \begin{array}{cc} 0 &-\frac{d}{dx}e^{-iA}\\ \frac{d}{dx}e^{iA} & 0 \end{array}\right]$,

it follows that $\frac{d}{dx}e^{iA} = \left[\frac{d}{dx}M(A)\right]_{10}$

for $x = {\rm Re\;} A_{rs}, {\rm Im\;} A_{rs}$

Note:
When $A$ is Hermitian, $A_{rs}$ and $A_{sr} = A^*_{rs}$ are complex conjugates so they are not independent, but the real and imaginary parts of $A_{rs}$ are independent, so one can treat $A_{rs}$ and $A^*_{rs}$ as independent and do a change of variables from $(A_{rs}, A^*_{rs})$ to $({\rm Re}A_{rs}, {\rm Im}A_{rs})$.