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So I'm rading Shankaar's book and got stuck in this place.

$$ (\Delta \Omega)^{2}(\Delta \Lambda)^{2} \geq \frac{1}{4}\left\langle\psi\left|[\hat{\Omega}, \widehat{\Lambda}]_{+}\right| \psi\right\rangle^{2}+\frac{\hbar^{2}}{4}. $$

We say the first term is positive definitive. But why does that mean we can neglect the term? The way I understand it, it means it will always be a positive term, or am I completely not thinking in the right direction?

Edit: I'm really tired and I missed the forest for the trees.

Qmechanic
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user220348
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3 Answers3

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If $T \geq 5 + x^2$, we can neglect $5$ to say $T \geq x^2$ because $5$ is positive definite.

lAPPYc
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  • I mean, yeah I kinda deduced that myself. Why though? – user220348 Feb 27 '19 at 21:42
  • @user220348 because the commutator not being zero sometimes brings about non-compatibility between two operators, and so is more important to questions of uncertainty. If the commutator is zero, there is no uncertainty relation, which has just made me realize that the relation you've written down is wrong. It is not an anti-commutator of $\Omega$ and $\Lambda$, but of their uncertainties. – lAPPYc Feb 27 '19 at 21:54
  • It's not wrong, check Shankaar page 237 and Zettili page 96. Identical answers.

    It comes from.

    $ \begin{aligned} \hat{\Omega} \hat{\Lambda} &=\frac{\hat{\Omega} \hat{\Lambda}+\hat{\Lambda} \hat{\Omega}}{2}+\frac{\hat{\Omega} \hat{\Lambda}-\hat{\Lambda} \hat{\Omega}}{2} \ &=\frac{1}{2}[\hat{\Omega}, \hat{\Lambda}]_{+}+\frac{1}{2}[\widehat{\Omega}, \hat{\Lambda}] \end{aligned} $

     – user220348 Feb 27 '19 at 22:00
  • I also don't get what you mean, of course the commutator between these 2 operators is not zero, we wouldn't get the uncertainty principle in the first place if they were zero. This seems obvious. We have defined it as $ [\Omega, \Lambda]=i \Gamma $ . I don't see how your comment give any kind of answer to my questions, seems a bit like cyclical thoughts. First you just say because it is like that and when I ask why, you make a random statement the commutator that, to me, has nothing to do with my question. – user220348 Feb 27 '19 at 22:04
  • @IAPPYc tagging you cause I forgot – user220348 Feb 27 '19 at 22:34
  • I think what @IAPPYc is trying to say in his answer is that since $5$ is positive definite and $T$ is greater than the rhs, the second inequality follows because it is less than the rhs of the first. –  Feb 27 '19 at 22:47
  • Yes I was being stupid. – user220348 Feb 27 '19 at 23:46
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if

$$ A \ge B + C $$

and $B\ge 0$ then

$$ A \ge C $$

We are not neglecting $B$ in writing this final statement we are using a very basic fact about mathematical inequality relations.

Jagerber48
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Adding another answer because the comment box got full. And didn't edit the old answer because wasn't sure if you'd get notified of it.

I do not have Shankaar or Zettili, but you've written the expression for $\Omega \Lambda$ (I don't know how to put hats on them, sry), while the uncertainty relation deals with $\Delta \Omega \Delta \Lambda$. The statement with the anti-commutator is:

$4(\Delta \Omega)^2 (\Delta \Lambda)^2 \geq \left|\left<\{\Delta\Omega,\Delta\Lambda\} \right>\right|^2 + \left|\left<[\Omega,\Lambda] \right>\right|^2 $

Do the calculations, you will find it. If your relation was correct, and the commutator was zero as well, then $\Omega\Lambda = \Lambda\Omega \Rightarrow \{\Omega,\Lambda\} = 2\Omega\Lambda$, which would result in your relation saying:

$4(\Delta \Omega)^2 (\Delta \Lambda)^2 \geq 4 \left|\left<\Omega\Lambda\right> \right|^2\geq 0$

contrary to what you said about there not being any uncertainty relation at all. This is happening because your relation is wrong.

As for your second comment: First I thought you wanted to know "why does it mean we can do this", and I gave the answer to why we can do this. Then you said you knew that, so I figured you were asking why we choose to ignore the anticommutator and not the commutator; because we could make a similar argument saying the mod square of the commutator is always positive, and so just write

$4(\Delta \Omega)^2 (\Delta \Lambda)^2 \geq \left|\left<\{\Delta\Omega,\Delta\Lambda\} \right>\right|^2 + \left|\left<[\Omega,\Lambda] \right>\right|^2 \geq \left|\left<\{\Delta\Omega,\Delta\Lambda\} \right>\right|^2 $

which would of course be right. But, if the commutator is zero, this relation gives us

$4(\Delta \Omega)^2 (\Delta \Lambda)^2 \geq 4(\Delta \Omega)^2 (\Delta \Lambda)^2$

which tells us exactly nothing about the codependency of the uncertainties of the operators.

However, if we neglect anticommutator and it is zero, the uncertainty relation does not become trivial. This is what I meant, and also that the uncertainty relation between incompatible operators results because of the commutator and not the anticommutator, when I said that the commutator is more important to the uncertainty relation.

All this, of course, if your question was "why not keep both", to which I can only say, the equation is more concise this way. Keeping both terms however, definitely strengthens the inequality.

lAPPYc
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