Derivatives for a density operator

[Haorong Wu]

TL;DR

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!

 

[Demystifier]

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).