I know that Heisenberg Uncertainty principles states that the momentum and position of a quantum object can not be determined at the same time. This is very strange to me. I want the basic reason behind it, i.e why is this so.


The basic reason is that the position operator X and the momentum operator P cannot be simultaneously diagonalized because $[X,P]=i\hbar I\ne{0}$. This means there is no state which is simultaneously an eigenstate of both X and P.

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    $\begingroup$ I'm sure you are trying to make it as simple as possible, but please only make it as simple as possible. Don't cross the line into over simplifying. And you did. You make it sound like a nonzero commutator implies there is no common eigenvector. But that isn't true. The reason no state is eigen to both is because that particular commutator has a nonzero expectation value for every state. In general the product of the standard deviations is bounded below by a multiple of the expectation value of the commutator. $\endgroup$ – Timaeus Mar 6 '16 at 8:29
  • $\begingroup$ @Timaeus Good point. I tried to clarify by adding the previously implied identity matrix on the RHS of the commutator. Now the expectation of I for all the state vectors is more clearly zero, where the state vectors are not zero and belong to the carrier space of the matrices X,P and I. Do I still have a mathematical flaw? $\endgroup$ – Gary Godfrey Mar 6 '16 at 9:45
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    $\begingroup$ The expectation of $\mathbb 1$ being nonzero for every state is what is important. Your answer makes it sound like the commutator being nonzero is what disallows common eigenvectors. And your comment says the expectation of $\mathbb 1$ is zero for some reason. $\endgroup$ – Timaeus Mar 6 '16 at 9:51
  • $\begingroup$ Whoops, typo. Should read "Now the expectation of I for all the state vectors is more clearly non-zero ...". $\endgroup$ – Gary Godfrey Mar 6 '16 at 10:08

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