# QM- Proving the uncertainty relation - Commutators

• Apr 6th 2011, 12:52 PM
bugatti79
QM- Proving the uncertainty relation - Commutators
Folks,

I am stuck on the derivation of the uncertainty relation when using the commutator and anti commutator...

given$\displaystyle \triangle A=\hat A -\langle A \rangle$

$\displaystyle \displaystyle \triangle A \triangle B=\frac{1}{2}[\triangle A, \triangle B]+\frac{1}{2}\left(\triangle A, \triangle B\right)$

Above on the RHS is the commutator and anti-commutator respectively. I dont understand the next line for the commutator

$\displaystyle [\triangle A, \triangle B]_\pm=[\hat A -\langle A \rangle , \hat B -\langle B \rangle ]_\pm=[\hat A, \hat B -\langle B \rangle ]_\pm-[\langle A \rangle , \hat B-\langle B \rangle ]_\pm$

the last term in the above line is 0 because <A> is a c number

$\displaystyle =[\hat A, \hat B]-[\hat A, \langle B \rangle ]_\pm$

where the last term in the above line is also 0 for same reason.

Now I do know [A,B] is AB-BA but I dont know where this algebaic derivation is coming out of..

Thanks
• Apr 8th 2011, 10:12 AM
bugatti79
Quote:

Originally Posted by bugatti79
Above on the RHS is the commutator and anti-commutator respectively. I dont understand the next line for the commutator

$\displaystyle [\triangle A, \triangle B]_\pm=[\hat A -\langle A \rangle , \hat B -\langle B \rangle ]_\pm=[\hat A, \hat B -\langle B \rangle ]_\pm-[\langle A \rangle , \hat B-\langle B \rangle ]_\pm$

The last term above can be written as

$\displaystyle [<A>, \hat B] - [<A>,<B>]$ where <A> and <B> are numbers and $\displaystyle \hat B$ is an operator.

The commutator of [<A>,<B>] is <A><B>-<B><A> but we know that<A><B>=<B><A> implies [<A>,<B>]=0. Simlarly for $\displaystyle [<A>, \hat B]$

The procedure is the same for the second last term above and finally yields

$\displaystyle [\triangle A, \triangle B]_\pm = [ \hat A, \hat B]$