Heisenberg's uncertainty principle and measurements

Question

While studying quantum key distribution, I came across this postulate which states that given two entangled particles A and B at Alice and Bob sites respectively, if Alice measures A to be $\uparrow$ polarized (along the x axis), then Bob can only measure B to be $\downarrow$ (along the x axis) but not the horizontal polarization $\leftarrow$ or $\rightarrow$ of B as well (along the y axis). This phenomena is interpreted by Heisenberg's uncertainty principle. That being said, I have the following question: From the articles that I have read about Heisenberg uncertainty principle, we cannot measure both the position and the momentum of a given particle with 100% accuracy. How this result is related to the interpretation mentioned above? I just cannot spot the relation.

Answer 1

The Heisenberg uncertainty principle is just a re-statement in quantum mechanics of what is known as the Cauchy-Schwartz inequality in vector spaces (pure maths).

I will quote the result of what you can find a proof of here.

The general definition of the uncertainty principle is that:

$$\hat{\sigma}_A^2 \cdot \hat{\sigma}_A^2 \geq \left |\frac{1}{2}\langle \{\hat{A}, \hat{B} \} \rangle - \langle \hat{A} \rangle \langle \hat{B} \rangle \right|^2 + \left |\frac{1}{2}\langle [\hat{A}, \hat{B} ] \rangle \right |^2, $$

where $\hat{A}$ and $\hat{B}$ are any two operators in quantum mechanics, $\hat{\sigma}$ the operator corresponding to their deviation from the expectation value, e.g. $\hat{\sigma}_A = \hat{A} - \langle \hat{A} \rangle$. The $[ \dots, \dots]$ is their commutator, and $\{ \dots, \dots\}$ the anticommutator.

So, you can now apply this to any pair of operators you want.

In your case, $\hat{A} = S_z$ and $\hat{B} = S_y$. You know the (anti)commutation relationships between the Pauli matrices so you work from there.