Undergrad Is the Heisenberg-Robertson Uncertainty Relation Always Consistent?

[DuckAmuck]

TL;DR

Something about this is not clear.

The general uncertainty principle is derived to be:
\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2} \langle \{A,B\} \rangle -\langle A \rangle \langle B \rangle \right)^2 + \left(\frac{1}{2i} \langle [A,B] \rangle \right)^2
Then it is often "simplified" to be:
\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2i} \langle [A,B] \rangle \right)^2
But this simplification only works if:
\left(\frac{1}{2} \langle \{A,B\} \rangle -\langle A \rangle \langle B \rangle \right)^2 \geq 0
However is this condition always met? *

 

[PeroK]

The anticommutator of two Hermitian operators is Hermitian, hence has a real expectation value. Therefore, the term you are discarding is always positive.

If $M$ is supposed to be a Hermitian operator, then $B = iM$ certainly isn't.

 

[vanhees71]

Note that also
$$\hat{C}=\frac{1}{\mathrm{i}} [\hat{A},\hat{B}]$$
is self-adjoint if $\hat{A}$ and $\hat{B}$ are self-adjoint, and thus also the somewhat weaker Heisenberg-Robertson uncertainty relation is consistent, i.e., the left-hand side of the inequality is positive semidefinite. The original stronger uncertainty relation is due to Schrödinger (if needed, I can look for the citation).