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However, I also found out that apparently you can still apply spin operators to eigenstates that don't correspond to that particular spin operator.

Example: For the spin operator σ on the x-axis:

σ

_{11}and σ

_{22}= 0

σ

_{12}and σ

_{21}= 1

Now if you apply this operator to the spin up eigenstate <1 , 0> (which corresponds to the z-axis), then the product of your matrix vector multiplication comes out to be:

<0 , 1> (which is the spin down state on the z-axis)

Now what exactly does this mean physically? I initially thought that the fact that you can apply the x-axis spin operator to a z-axis eigenstate and simply get a flipped z-axis eigenstate meant that spin up on the z-axis would turn into spin left on the x-axis (since spin down and spin left are both -1 eigenvalues) if you were to redo a measurement of spin on the x-axis this time. However, wouldn't that interpretation violate the uncertainty principle since I would know that spin up on the z-axis = spin left on the x-axis? If this interpretation is incorrect, then what exactly is the implication of the result of applying a spin operator to an eigenstate for another axis?