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The generalized uncertainty principle can be derived and shown to be this which is fine and rigorous.

$\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle \geq \dfrac{1}{4} \vert \langle [ A,B ] \rangle \vert^{2} + \dfrac{1}{4} \vert \langle \{ \Delta A, \Delta B \} \rangle \vert^{2}$

On the right hand side, both quantities are real and positive. Sakurai argues that if you omit the anticommutator (as is typically done) the inequality is still true since the right hand side gets even smaller. However, this is bothersome since doesn't it mean that the square of the dispersion can approach even smaller values that allowed by the more rigorous uncertainty?

If I naively think about a number line, remove the anticommutator just lowers the floor for how much we can know about a quantum object doesn't it? Apologies for the bad drawing

enter image description here

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  • $\begingroup$ You do understand the entropic UP is a tighter bound, no? $\endgroup$ Oct 3 '20 at 17:07
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The anticommutator does not have immediate algebraic properties and no obvious physical interpretation beyond the “brute force” one of $\hat A\hat B+\hat B\hat A$.

For instance, whereas $[L_x,L_y]=i\hbar L_z$ the anticommutator is just the symmetrized product $L_xL_y+L_yL_x$ which isn’t anything special in the theory of angular momentum.

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  • $\begingroup$ But that doesn't really answer the question, even if you argue physically that it shouldn't be there, it's omission would increase the possible accuracy of a measurement wouldn't it? $\endgroup$ Oct 1 '20 at 22:45
  • $\begingroup$ I’m not saying it shouldn’t be there; I’m saying there not obvious interpretation to this anticommutator. The uncertainty relation is NOT meant to be a tight bound. $\endgroup$ Oct 1 '20 at 22:49
  • $\begingroup$ That might be my source of confusion then. Could you elaborate on why the uncertainty relation is not a tight bound? $\endgroup$ Oct 1 '20 at 22:51
  • $\begingroup$ well... usually a Lie algebra is defined via a commutator so the anticommutator takes you outside the algebra, and that means the anticommutator is always a bit weird to understand in terms of the original operators in the algebra. $\endgroup$ Oct 1 '20 at 22:55
  • $\begingroup$ so in this sense it was never meant to be a tight bound. Plus the commutator of two operators are much more common than anticommutators. $\endgroup$ Oct 1 '20 at 22:57

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