For two compatible observables A and B i.e. if $[A, B]=0$, the uncertainty principle says that $$(\Delta A)_\psi(\Delta B)_\psi\geq 0$$ in any state $|\psi\rangle$ where $(\Delta A)_\psi=(\langle \psi|A^2|\psi\rangle-(\langle \psi|A|\psi\rangle)^2)^{1/2}$ . I know that these uncertainties have nothing to do with the precision of measurement. It is however not clear to me when will we get equality and when inequality?
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asked Mar 27, 2021 at 17:55
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4$\begingroup$ If you look at the derivation of the general uncertainty principle, you should be able to tell which terms are neglected there to obtain this inequality from a more complicated equality, and so this will be an equality when these neglected terms are zero. Do you have some difficulty with doing this on your own? $\endgroup$– ACuriousMind ♦Mar 27, 2021 at 18:02
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1$\begingroup$ Consider $A=B$, when would $(\Delta A)_\psi$ be non-zero? $\endgroup$– ACatMar 27, 2021 at 18:04
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$\begingroup$ I think, there exist some states $\psi$ for which either $(\Delta A)_\psi$ or $(\Delta B)_\psi$ or both are zero. In those states, the uncertainty product is zero. Am I right? $\endgroup$– SolidificationMar 27, 2021 at 18:11
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$\begingroup$ @DvijD.C. When $\psi$ is not an eigenstate of $A$. Also, see my comment above. $\endgroup$– SolidificationMar 27, 2021 at 18:16
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1 Answer
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The Heisenberg uncertainty relation looks like $$(\Delta A)^2(\Delta B)^2\geq \frac{1}{4}\langle \psi|[\hat{A},\hat{B}]_+|\psi\rangle^2+\frac{\hbar^2}{4}$$ The above inequality becomes (if you look for the whole derivation) equality only if
- $\hat{A}|\psi\rangle =c\hat{B}|\psi\rangle $
- $\langle \psi|[\hat{A},\hat{B}]_+|\psi\rangle =0$
where $\hat{A}=A-\langle A\rangle $ and $\hat{B}=B-\langle B\rangle $.
answered Mar 27, 2021 at 18:11