Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-18T00:47:22Z https://physics.stackexchange.com/feeds/question/108812 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/108812 13 Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$? user215721 https://physics.stackexchange.com/users/24459 2014-04-17T14:32:41Z 2019-06-10T19:31:03Z <p>In Griffiths' QM, he uses two inequalities (here numbered as <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>) to prove the following general <a href="http://en.wikipedia.org/wiki/Uncertainty_principle" rel="nofollow noreferrer">uncertainty principle</a>: <span class="math-container">$$\sigma_A^2 \sigma_B^2\geq\left(\frac{1}{2i}\langle [\hat A ,\hat B]\rangle \right)^2$$</span></p> <p>defining <span class="math-container">$\lvert f\rangle=(\hat A-\langle A\rangle)\lvert \Psi\rangle$</span> and <span class="math-container">$\lvert g\rangle=(\hat B-\langle B\rangle)\lvert \Psi\rangle$</span>, he uses </p> <ul> <li><p>Schwarz inequality: <span class="math-container">$$\langle f\lvert f\rangle\langle g\lvert g\rangle\geq|\langle f\lvert g\rangle|^2\tag{1}$$</span> and with <span class="math-container">$\sigma_B^2=\langle g\lvert g\rangle$</span> and <span class="math-container">$\sigma_A^2=\langle f\lvert f\rangle$</span>, he arrives at <span class="math-container">$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$</span>.</p></li> <li><p>the fact that for any complex number <span class="math-container">$z$</span> we have <span class="math-container">$$|z|^2\geq(\mathrm{Im}(z))^2=[\frac{1}{2i}(z-z^*)]^2\tag{2}$$</span> here <span class="math-container">$z=\langle f\lvert g\rangle$</span> and so <span class="math-container">$z^*=\langle g\lvert f\rangle$</span> , and we find that <span class="math-container">$\langle f\lvert g\rangle=\langle\hat A \hat B\rangle-\langle A \rangle\langle B \rangle$</span> and <span class="math-container">$\langle g\lvert f\rangle=\langle\hat B \hat A\rangle-\langle A \rangle\langle B \rangle$</span>, so <span class="math-container">$\langle f\lvert g\rangle-\langle g\lvert f\rangle=\langle [\hat A,\hat B]\rangle$</span>. Replacing this into <span class="math-container">$(1)$</span> gives the uncertainty principle.</p></li> </ul> <p>Why he doesn't use <span class="math-container">$$|z|^2\geq(\mathrm{Re}(z))^2=[\frac{1}{2}(z+z^*)]^2$$</span> instead of <span class="math-container">$(2)$</span>? This could give us a (correct though different) relation between the <span class="math-container">$\sigma_A$</span> and <span class="math-container">$\sigma_B$</span> too: <span class="math-container">$$\boxed{\sigma_A^2 \sigma_B^2\geq\frac{1}{4}\left(\langle \hat A \hat B\rangle +\langle \hat B \hat A\rangle -2 \langle \hat A\rangle\langle \hat B\rangle\right)^2}$$</span></p> https://physics.stackexchange.com/questions/108812/why-is-the-uncertainty-principle-not-sigma-a2-sigma-b2-geq-langle-a-b-rang/108846#108846 3 Answer by Mo_ for Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$? Mo_ https://physics.stackexchange.com/users/20898 2014-04-17T18:03:57Z 2014-04-18T10:37:43Z <p>The most correct relation is the <a href="http://arxiv.org/pdf/quant-ph/0011115v3.pdf" rel="nofollow noreferrer">following general relation</a>, that actually contains both terms.</p> <p>If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be:<br> $$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert f\rangle=(\langle A B\rangle-\langle A\rangle \langle B\rangle)(\langle B A\rangle-\langle A\rangle \langle B\rangle)$$ $$|\langle f\lvert g\rangle|^2=\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})\to$$ $$\boxed{\sigma_A^2 \sigma_B^2\geq\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})}$$ 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 <a href="https://physics.stackexchange.com/questions/4049/meaning-of-the-anti-commutator-term-in-the-uncertainty-principle/4055#4055">this question</a>:</p> <blockquote> <p>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.</p> </blockquote> https://physics.stackexchange.com/questions/108812/why-is-the-uncertainty-principle-not-sigma-a2-sigma-b2-geq-langle-a-b-rang/108849#108849 5 Answer by rob for Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$? rob https://physics.stackexchange.com/users/44126 2014-04-17T18:29:03Z 2014-04-17T18:29:03Z <p>Griffiths' formulation makes it explicit that operators which <i>commute</i> are <i>not</i> restricted by the uncertainty principle. Your boxed expression obscures this physical and mathematical insight.</p>