# Almost Done with Derivation of the Uncertainty Principle, but stuck!

• May 16th 2011, 06:14 PM
Arturo_026
Almost Done with Derivation of the Uncertainty Principle, but stuck!
I would tremendously apreciate if anyone could please shed some light on a step in The uncertainty principle, in which I have gotten stuck.
HOW DID THEY GO FROM Eq. (4.93) to (4.94) [Page #51] ... I can't seem to find how this happened? Thanks.

http://farside.ph.utexas.edu/teaching/qmech/qmech.pdf
• May 16th 2011, 08:12 PM
Arturo_026
I think I have some idea on how this was obtained. Just iluminate me onto something. If I have B<A> where B and A are operators and <A> refers to the mean value of A. If I do <B<A>> can this equal <B><A> also, Can <<A><B>> = <A><B> ... ? Thank you

I know I am being very vague with this question, so forgive me if I commited any obvious mathematical flaw.
• May 17th 2011, 02:11 AM
Ackbeet
You are correct.

$\displaystyle \langle B\langle A\rangle\rangle=\int_{-\infty}^{\infty}\psi^{*}B\langle A\rangle\psi\,dx=\langle A\rangle\int_{-\infty}^{\infty}\psi^{*}B\psi\,dx=\langle A\rangle\,\langle B\rangle.$

So what you have is that, since $\displaystyle \langle A\rangle$ is just a number, it comes out of the integral.

In Equation 4.93, on the RHS, if you expand out the product in-between the wave functions, you get

$\displaystyle AB-\langle A\rangle B-A\langle B\rangle+\langle A\rangle\,\langle B\rangle.$

When you take the expectation of that whole expression, the right three terms end up being like terms. So you just get

$\displaystyle -\langle A\rangle\,\langle B\rangle.$

That is, the positive term cancels out exactly one of the negative terms.

Make sense?

Thanks for posting a QM question, by the way. In my opinion, we don't get nearly enough of those around here!