The generalized uncertainty principle says,

$$\sigma_A^2\sigma_B^2~ \ge ~\frac14\langle i[A, B]\rangle^2.$$

But the complex field is not ordered, i.e, inequalities like $\le$, $\ge$, etc are absurd. For instance, is $i>1$? Doesn't $i$ in there makes the whole inequation meaningless?

  • 2
    $\begingroup$ The commutator has its own factor of $i$ $\endgroup$ – Yuzuriha Inori Mar 10 '18 at 6:51
  • 1
    $\begingroup$ The more familiar, equivalent, formula is $\sigma_A^2\sigma_B^2~ \ge ~\frac14|\langle i[A, B]\rangle|^2.$. The absolute value arises form Cauchy-Schwartz' inequality...The factor $i$ can be therefore omitted as it shows up in the absolute value. Alternatively (J.Murray's answer below), as $i[A,B]$ is real you can keep it, dropping the absolute value. $\endgroup$ – Valter Moretti Mar 10 '18 at 10:33

To elaborate on Yuzuriha's comment, if $A$ and $B$ are self-adjoint operators, then

$$\langle [A,B]\rangle = \langle\psi,AB\psi\rangle - \langle\psi,BA\psi\rangle = \langle A \psi,B\psi\rangle - \langle B\psi,A\psi\rangle$$


$$\overline{\langle [A,B]\rangle} = \overline{\langle A \psi,B\psi\rangle} - \overline{\langle B\psi,A\psi\rangle} = \langle B\psi,A\psi\rangle -\langle A \psi,B\psi\rangle = -\langle [A,B]\rangle $$

which implies that $\langle [A,B]\rangle$ is purely imaginary. The explicit factor of $i$ makes the whole thing make sense.

Of course, if you're in the mood for mindless pedantry, then complex numbers $z$ with $Im(z)=0$ are still complex numbers and therefore are members of an unordered set; in that case, you can define the obvious total order on the subset $R:=\{z\in\mathbb{C} | Im(z)=0\}$ and note that the objects in question belong to $R$.

  • $\begingroup$ The step $\langle\psi,AB\psi\rangle - \langle\psi,BA\psi\rangle = \langle A \psi,B\psi\rangle - \langle B\psi,A\psi\rangle$ is true only for Hermitian operators. What about non-Hermitian observables? $\endgroup$ – Ayatana Mar 10 '18 at 7:15
  • 4
    $\begingroup$ All observables are Hermitian. $\endgroup$ – J. Murray Mar 10 '18 at 7:16
  • 4
    $\begingroup$ In fact, all observables are self-adjoint, which is a subtly stronger requirement. $\endgroup$ – J. Murray Mar 10 '18 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.