Uncertainty Principle in 3 dimensions

Question

I'm trying to understand how to write Heisenberg uncertainty principle in 3 dimensions. What I mean by that is to prove something of the form $f(\Delta p_x,\Delta p_y,\Delta p_z,\Delta x,\Delta y,\Delta z) \geq A$.

This is what I got: The unknown volume that a single particle can be in is $\Delta V=\Delta x \Delta y \Delta z$. The uncertainty in the size of the momentum is $\Delta p = \sqrt{\Delta p_x^2 + \Delta p_y^2 + \Delta p_z^2}$

Now this is where I get stuck. In my textbook, for the 1d case, they used De-Broglie equation for connecting the uncertainty of the particle wavelength and its momentum along the $x$-axis. But does De-Broglie equation is correct per axis or for the size of the vectors?

Answer 1

If you choose any direction in space, the 1D uncertainty principle applies in that direction. So, if you know the component of a particle's momentum in that direction well, you cannot know its projected position in that direction well. The directions of good and poor localization need not be aligned with your coordinate axes.

Answer 2

There is no need for a 3D uncertainty principle.

The operators commute between dimensions:

$$[\hat{x}^i, \hat{p}^j] = i \hbar \delta^{ij}$$

Momentum in one dimension and position in another can be measured with arbitrary precision simultaneously (at least there is no restriction from QM).

The closest thing I can think of for what you're asking is:

$$ \sum_{i=1}^{3} \Delta x^i \Delta p^i \geq \frac{3}{2}\hbar $$

which is a weaker inequality than (and not as useful as) $\Delta x^i \Delta p^i \geq \frac{1}{2}\hbar $.

Answer 3

The uncertainty relation holds for each direction in space-time, so $\Delta x_i \Delta p_i \geq \hbar/2$ for $i=0,1,2,3$ ($t,x,y,z$).