Is the Heisenberg uncertainty principle limited to measurements in the same direction?

Question

Heisenberg's Uncertainty Principle follows from the commutational relationship between the position and momentum operators, namely: $[\hat x_i,\hat p_j]=i\hbar\delta_{ij}$. Of course, in one dimension, this is simply $[\hat x, \hat p]=i\hbar$; but what about measuring two different directions, say $\hat x_1$ and $\hat p_2$? In this case, the operators actually commute.

I got the following result using David Tong's reasoning for the Uncertainty Principle, found in his QM lectures section 3.4. I simply changed the fact that, now, the operators commute, and therefore:

$$ (\Delta_{\psi}x_1)(\Delta_{\psi}p_2)\geq 0$$

This doesn't necessarily imply that both uncertainties have to be zero, but it shows that they can be. Is this correct? Does the Heisenberg Uncertainty Principle only work for position and momentum measurements along the same axis?

EDIT: Link to the lectures, if anyone wants to follow the reasoning.

Answer 1

You are correct. There is a generalized Heisenberg uncertainty principle for two observables $A$ and $B$ is given by

$$\Delta A \Delta B \geq \frac{1}{2} |\langle \psi |[A,B]|\psi\rangle |$$

where $[A,B]=AB-BA$ the commutator. If you use position and momentum in the same direction $[x_j,p_j]=\mathrm i \hbar$ ($j\in\{1,2,3\}$), you recover the usual Heisenberg uncertainty relation.

Now all you have to ask is what is $[x_i,p_j]$ for $i\neq j$ (position/momentum operators in two different Cartesian axes). The answer is $[x_i,p_j]=0$. That means that you can find a common basis for the position $x_i$ and momentum $p_j$ ($i\neq j$) where the basis states have defined values of $x_i$ and $p_i$ at the same time.

Similarly, you have $[x_i,x_j]=0$ and $[p_i,p_j]=0$.

For other operators this relation is even more complicated. For example for angular momentum operators you have

$$\Delta L_x \Delta L_y \geq \frac{1}{2}|\langle\psi| L_z|\psi\rangle|$$.

Answer 2

Per your example, consider:

$$\psi(x, y, z, t) = \frac 1 N e^{i(p/\hbar -\omega t)}\delta(x)\delta(z) $$

In the $y$ coordinate, it's an eigenstate of momentum:

$$\hat p_y\psi = p\psi $$

while in the other coordinates, it has maximal momentum uncertainty and definite position (if the delta functions bother you, feel free to use a Gaussian limit).

Since $\sigma_{p_y} = 0$, you will always have:

$$ \sigma_{p_y}\sigma_x = 0 $$

no matter what the $x$ of $z$ dependence is.