Measuring x and ##p_{y}## precisely

[gfd43tg]

Homework Statement

Homework Equations

The Attempt at a Solution

1.
$$ [\hat {x}, \hat {p}_{y}] = \hat {x}\hat {p}_{y} - \hat {p}_{y} \hat {x}$$
$$= \hat {x}(-i \hbar \frac {\partial f}{\partial y}) - (- i \hbar \frac {\partial (fx)}{\partial y}) $$
Since x is not a function of y, it can be taken out of the derivative, so the commutator is zero.

However, I have no clue how to answer the next two questions. I am very confused regarding measurements in quantum mechanics. Does it have something to do with Heisenberg Uncertainty principle?

 

[QuasiParticle]

I am very confused regarding measurements in quantum mechanics. Does it have something to do with Heisenberg Uncertainty principle?

In this context it indeed does. It can be shown that the product of the imprecisions in the measurements of two observables A and B follows the Heisenberg uncertainty relation:

$$\Delta A \Delta B \geq \frac{1}{2}\left | \left\langle \left[ \hat{A}, \hat{B} \right] \right\rangle \right |$$
With this you can answer the second question.

For the third one, do the exact same thing as you did for the first question: operate with the commutator on a test function f and see if you can rearrange the order of the derivatives so that the result is zero.

 

[gfd43tg]

Since the commutator is zero, does that mean that I can measure it precisely? Because $\sigma_{x} \sigma_{p_{y}} \ge 0$?

 

[QuasiParticle]

That's right, their measurements are not constrained by the uncertainty relation.