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The uncertainty principle between the position $x$ and the momentum $p$ is given by: $$ \sigma_x \sigma_p \geq \hbar/2,$$ whereas for the $x$ and $y$ components of the angular momentum is given by: $$ \sigma_{L_x} \sigma_{L_y} \geq \frac{\hbar}{2}\langle L_z\rangle .$$

What is the physical meaning of the Right Hand Side being just a number or an expectation value?

EDIT: I realise the expectation value itself is just a number, but it can take several different values, as opposed to a constant

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    $\begingroup$ An expectation value is just a number. $\endgroup$ – ACuriousMind Mar 10 '15 at 14:48
  • $\begingroup$ Yeah I but it can vary, I meant as opposed to just a constant $\endgroup$ – SuperCiocia Mar 10 '15 at 14:52
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    $\begingroup$ I still don't understand what the question is, then. The thing on the right in your first equation is also just an expectation value - that of a multiple of the identity. $\endgroup$ – ACuriousMind Mar 10 '15 at 14:55
  • $\begingroup$ @SuperCiocia it can vary only if you pass from one state to another state. The quantities $\sigma$ on the LHS don't vary from state to state? The same with the RHS. I your first inequality you have on the RHS a number. But in the 2nd inequality you have a variable, which is state-dependent. $\endgroup$ – Sofia Mar 11 '15 at 12:51
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The Heisenberg uncertainty principle in the most general form $$\Delta_\omega(A)\Delta_\omega(B)\geq\frac12|\omega([A,B])|$$ depends on the state $\omega$ on which it is evaluated. In the special case of the canonical commutation relations $[q,p]= i\hbar I$, $\omega(I)=1$ for any state and therefore the RHS reduces to a constant. For more general commutators however this won't be the case and a dependence on the state $\omega$ will remain.

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  • $\begingroup$ Is there a physical meaning if the uncertainty does not depend on the state? $\endgroup$ – SuperCiocia Mar 10 '15 at 15:31
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The uncertainty product is bounded from below by the expectation value of the commutator of the relevant observables. If $A$ and $B$ are any two observables, then the generalized Heisenberg uncertainty relation reads as $$ \sigma_A\sigma_B \geq \frac{1}{2}\vert \langle[A,B]\rangle\vert .$$ For the case of position - linear momentum pair, the commutator is $[x,p]=i\hbar$ and so the right-hand side of the inequality above becomes independent of the state.

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  • $\begingroup$ Is there a physical meaning if the uncertainty does not depend on the state? $\endgroup$ – SuperCiocia Mar 10 '15 at 15:31

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