# Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$?

In Griffiths' QM, he uses two inequalities (here numbered as $$(1)$$ and $$(2)$$) to prove the following general uncertainty principle: $$\sigma_A^2 \sigma_B^2\geq\left(\frac{1}{2i}\langle [\hat A ,\hat B]\rangle \right)^2$$

defining $$\lvert f\rangle=(\hat A-\langle A\rangle)\lvert \Psi\rangle$$ and $$\lvert g\rangle=(\hat B-\langle B\rangle)\lvert \Psi\rangle$$, he uses

• Schwarz inequality: $$\langle f\lvert f\rangle\langle g\lvert g\rangle\geq|\langle f\lvert g\rangle|^2\tag{1}$$ and with $$\sigma_B^2=\langle g\lvert g\rangle$$ and $$\sigma_A^2=\langle f\lvert f\rangle$$, he arrives at $$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$.

• the fact that for any complex number $$z$$ we have $$|z|^2\geq(\mathrm{Im}(z))^2=[\frac{1}{2i}(z-z^*)]^2\tag{2}$$ here $$z=\langle f\lvert g\rangle$$ and so $$z^*=\langle g\lvert f\rangle$$ , and we find that $$\langle f\lvert g\rangle=\langle\hat A \hat B\rangle-\langle A \rangle\langle B \rangle$$ and $$\langle g\lvert f\rangle=\langle\hat B \hat A\rangle-\langle A \rangle\langle B \rangle$$, so $$\langle f\lvert g\rangle-\langle g\lvert f\rangle=\langle [\hat A,\hat B]\rangle$$. Replacing this into $$(1)$$ gives the uncertainty principle.

Why he doesn't use $$|z|^2\geq(\mathrm{Re}(z))^2=[\frac{1}{2}(z+z^*)]^2$$ instead of $$(2)$$? This could give us a (correct though different) relation between the $$\sigma_A$$ and $$\sigma_B$$ too: $$\boxed{\sigma_A^2 \sigma_B^2\geq\frac{1}{4}\left(\langle \hat A \hat B\rangle +\langle \hat B \hat A\rangle -2 \langle \hat A\rangle\langle \hat B\rangle\right)^2}$$

• Because the fundamental equation of quantum mechanics is $[\hat A,\hat B]=i\hbar \hat I$, for conjugate variables $\hat A$ and $\hat B$. – David H Apr 17 '14 at 18:54

If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be:
$$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert f\rangle=(\langle A B\rangle-\langle A\rangle \langle B\rangle)(\langle B A\rangle-\langle A\rangle \langle B\rangle)$$ $$|\langle f\lvert g\rangle|^2=\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})\to$$ $$\boxed{\sigma_A^2 \sigma_B^2\geq\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})}$$ The above relation can be written better as $$|\langle f|g\rangle|^{2} = \bigg(\frac{\langle f|g\rangle+\langle g|f\rangle}{2}\bigg)^{2} + \bigg(\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\bigg)^{2}$$ $$\langle f|g\rangle-\langle g|f\rangle =\langle [{A},{B}]\rangle$$ $$\langle f|g\rangle+\langle g|f\rangle = \langle \{{A},{B}\}\rangle -2\langle {A}\rangle\langle {B}\rangle\to$$ $$\boxed{\sigma_A^2 \sigma_B^2\geq\Big(\frac{1}{2}\langle\{{A},{B}\}\rangle - \langle {A} \rangle\langle {B}\rangle\Big)^{2}+ \Big(\frac{1}{2i}\langle[{A},{B}]\rangle\Big)^{2}}$$ So, Griffiths ignores the first term in the above relation. It remains to tell why he does so. The answer is provided in the last paragraph of the Lubos Motl's answer to this question:
Well, in normal cases, the stronger version is not "terribly" useful because the anticommutator term is only nonzero if there is a "correlation" in the distributions of $A,B$ - i.e. if the distribution is "tilted" in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula.