The generalized uncertainty principle can be written as (where A and B are observables): $$ \sigma_A\sigma_B \geq \left| \frac{1}{2i}\langle [A,B]\rangle_\Psi \right| $$ But the average value of the commutator depends on $\Psi$, the state of the particle in question. Hence is it right to say that for some states the observables A and B will have to respect the uncertainty relation and in some other cases it won't? I guess it depends on the shape of the commutator so when can we be sure that the average value will be null even if the commutator itself is non-null? Or when can we be sure that the average value won't be null?

I thought that one case where this principle won't depend on the state is when the commutator $[A,B] = \alpha I$ where $I$ is the identity matrix as it commutes with any state. And thus, $\langle [A,B]\rangle = \langle\Psi \mid [A,B]\mid\Psi\rangle = \alpha \langle \Psi\mid\Psi\rangle= \alpha. $

Are there any other cases where the uncertainty principle is independent of the state?


1 Answer 1


In general the uncertainty inequality does depend on the state $\Psi$, i.e . the bound on the right hand side changes with $\Psi$. However, for any $\Psi$, the inequality is always respected so the product of uncertainty of observables always satisfies the bound, provided the observables are self-adjoint and the state $\Psi$ is normalizable. These last two conditions are more or less implicit (sometimes more than less) in the derivation.

In general, the bound will not depend on $\Psi$ when the commutator is diagonal and thus proportional to the unit operator, as you suggest.

  • $\begingroup$ Is there any other cases where the lower limit of the Heisenberg Inequality won't depend on the state? $\endgroup$
    – E. Morell
    Nov 10, 2017 at 1:20
  • $\begingroup$ @E.Morell If the commutator is $0$ but this is a trivial case. I don’t know of any other but I would not be surprised to learn there are lesser known examples. $\endgroup$ Nov 10, 2017 at 1:46

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