Is the Uncertainty Principle valid for information about the past?

The uncertainty principle should be understood as follows: The position and momentum of a particle are not well-defined at the same time. Quantum mechanically, this is expressed through the fact that the position and momentum operators don't commute: $[x,p]=ihbar$.

The most intuitive explanation, for me, is to think about it in terms of wave-particle duality. De Broglie introduced the idea that every particle also exhibits the properties of a wave. The wavelength then determines the momentum through $$p=frac{h}{lambda}$$ where $lambda$ is the De Broglie wavelength associated with the particle. However, when one thinks about a wave, it is clear that the object described by it will not be easy to ascribe a position to. In fact, one needs a specific superposition of waves to create a wave that is essentially zero everywhere except at some position $x$. However, if one creates such a wave packet, one loses information about the exact wavelength (since a wave with a single, well-defined wavelength will simply extend throughout space). So, there is an inherent limitation to knowing the wavelength (i.e. momentum) and position of a particle. On a more technical level, one could say that the uncertainty principle is simply a consequence of wave-particle duality combined with properties of the Fourier transform. The uncertainty is made precise by the famous Heisenberg uncertainty principle, $$sigma_xsigma_pgeq frac{hbar}{2}$$ More generally, for two non-comuting observables $A$ and $B$ (represented by hermitian operators), the generalized uncertainty principle reads $$sigma_A^2sigma_B^2geq left(frac{1}{2i}langle [A,B] rangleright)^2 implies sigma_Asigma_B geq frac{|langle [A,B]rangle| }{2}$$ Here, $sigma$ denotes the standard deviation and $langledotsrangle$ the expectation value. This holds at any time. Therefore, the measurement occurring right now, having occurred in the past or occurring in the future has nothing to do with it: The uncertainty principle always holds.