The Heisenberg Principle states that for each direction, $\Delta x\cdot \Delta p_x \ge \hbar , \Delta y\cdot \Delta p_y \ge \hbar$ and $\Delta z\cdot \Delta p_z \ge \hbar$.

But, can anything be said about $\Delta x\cdot \Delta p_y$? Can I measure the position in one direction and the momentum in an orthogonal direction in any required precision?


1 Answer 1


The uncertainty principle arises from non-commuting observables. More information can be seen in a discussion and derivation at wikipedia: here The result is the general uncertainty principle for two observables A,B

$$\sigma_A\sigma_B \ge \frac{1}{2} \left|\left\langle\left[{A},{B}\right]\right\rangle\right|$$

Since $p_y$ and $x$ commute, in principle you can specify both to arbitrary precision.

  • 2
    $\begingroup$ ...if you only measure them but neither $y$ nor $p_x$ $\endgroup$ Apr 21, 2011 at 9:58
  • $\begingroup$ In addition, you can construct a complete orthonormal eigenbasis with the $| \psi \rangle$ parametrized by the $y$-position and $x$-momentum. This would amount to taking the spatial representation of the wavefunction and Fourier-transforming only wrt to $x$. $\endgroup$
    – Lagerbaer
    Apr 21, 2011 at 15:02

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