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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?

Solidification
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1 Answers1

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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 $.

Himanshu
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