# Derivatives for a density operator

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• Haorong Wu
Haorong Wu
TL;DR Summary
How to properly calculate the derivatives for a density operator?
Hi. Suppose I have a state ##\left | \psi (0)\right >=\sum_m C_m \left | m\right >## evolving as $$\left | \psi (0+dz)\right>=\left | \psi (0)\right >+dz \sum_iD_i\left | i\right >=\sum_m C_m \left | m\right >+dz \sum_iD_i\left | i\right >=\sum_m( C_m+dz D_m)\left |m\right >.$$
Then the density operator at ##0+dz## is \begin{align}\rho(0+dz)&=\left | \psi (0+dz)\right>\left< \psi (0+dz)\right|=\sum_{mn}( C_m+dz D_m)( C^*_n+dz D^*_n)\left |m\right >\left< n\right|\nonumber \\&=\sum_{mn}(C_mC^*_n+dz(D_mC^*_n+C_mD^*_n)+dz^2D_mD^*_n)\left |m\right >\left< n\right|.\end{align}

I have seen in a paper, Roux F S. Infinitesimal-propagation equation for decoherence of an orbital-angular-momentum-entangled biphoton state in atmospheric turbulence[J]. Physical Review A, 2011, 83(5): 053822, that the author take the derivative of the density matrix as $$\partial_z \rho(z)=\sum_{mn}(D_mC^*_n+C_mD^*_n)\left |m\right >\left< n\right|,$$ i.e., terms with ##dz##.

Then when the author tries to recover the density matrix at some point ##z##, the result is given by just integrating the above derivative.

My question is from ##\rho(0+dz)=\left | \psi (0+dz)\right>\left< \psi (0+dz)\right|##, clearly, its a pure state, but if we calculate it from the derivative ##\rho(0+dz)=\rho(0)+dz \partial_z \rho(z)##, then it is not pure since terms with ##dz^2## is lost. Why there is a conflict? Should we discard terms with ##dz^2## or not?

Thanks!

Gold Member
then it is not pure since terms with ##dz^2## is lost. Why there is a conflict?
A more correct way to do it is to avoid dealing with infinitesimal numbers ##dz##. The truncated Taylor expansion
$$\rho(z) = \rho(0)+\left.\frac{\partial\rho(z)}{\partial z}\right|_{z=0} z+{\cal O}(z^2)$$
is not necessarily pure. Only the full Taylor expansion is guaranteed to be pure. But if you are doing approximation for small (but finite!) ##z##, then a truncated expansion gives you approximate purity. The nonpurity is ##{\cal O}(z^2)##, which is consistent with expansion up to the terms linear in ##z##.

An apparent conflict in your computation arises from the fact that ##\rho(0+z)=|\psi(0+z)\rangle\langle\psi(0+z)|## is exact equality, while ##\rho(0+z)=\rho(0)+z\rho'(0)## is only an approximation. But to see that, you must work with finite ##z## (not with infinitesimal ##dz##, which, as a number, is not a well defined object).

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Haorong Wu