# Work principle of Heisenberg Uncertainty Principle

Let us assume that one quantum particle is moving in some space (phase space). I appoint two observers to observe that particle. Now at any instant of time I ask the $$1^{st}$$ observer to measure the x-position and the y-component of linear momentum ; and I ask the $$2^{nd}$$ observer to measure the particle's y-position and x-component of linear momentum.

As $$[x,p_y] = 0$$ and $$[y,p_x] = 0$$ so H.U.P will not come into effect between these two pairs. And after they make their measurements if I combine their data and I will get the precise x-position , x-momentum , y-position , y-momentum of that particle at that moment. Seems like H.U.P failed. But pretty surely this is not the scenario. What is the discrepancy in this thought situation/experiment ?

• You can ignore the y-direction and just consider the x-direction. Then you find that the second observer will have an uncertainty due to the measurement of the first observer that satisfies HUP. Commented Mar 17, 2023 at 8:04
• Thank you for your answer. I got it now. Commented Mar 17, 2023 at 8:06

What has to be kept in mind in the experiment described in the OP, that the initial wave function collapses after the first measurement (into a state with definite values of $$x$$ and $$p_y$$) and the second measurement is done on this collapsed wave function, not on the initial state. SO one cannot really combine the data.
Your formulation "$$=0$$" doesn't hold, as it's an inequality "you can't know better than ...". You can see easily if you note down $$|\vec r \cdot \vec p| \le \frac{h}{2 \pi}$$, so the first observer already spoiled it.