# Measuring x and $p_{y}$ precisely

1. Mar 26, 2015

### Maylis

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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?

2. Mar 26, 2015

### QuasiParticle

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.

3. Mar 30, 2015

### Maylis

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

4. Mar 30, 2015

### QuasiParticle

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