Results 1 to 2 of 2

Math Help - QM- Proving the uncertainty relation - Commutators

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    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 \triangle A=\hat A -\langle A \rangle

     <br />
\displaystyle \triangle A \triangle B=\frac{1}{2}[\triangle A, \triangle B]+\frac{1}{2}\left(\triangle A, \triangle B\right)<br />

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

     <br />
[\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

     <br />
=[\hat A, \hat B]-[\hat A, \langle B \rangle ]_\pm<br />

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by bugatti79 View Post
    Above on the RHS is the commutator and anti-commutator respectively. I dont understand the next line for the commutator

     <br />
[\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

     <br />
[<A>, \hat B] - [<A>,<B>] where <A> and <B> are numbers and \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 [<A>, \hat B]

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

     <br />
[\triangle A, \triangle B]_\pm = [ \hat A, \hat B]<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 27th 2010, 08:57 AM
  2. Proving a subgroup relation
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 18th 2009, 03:42 PM
  3. Proving a subgroup relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 15th 2009, 08:56 PM
  4. Finding Absolute Uncertainty From Relative Uncertainty
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 16th 2009, 10:16 AM
  5. commutators
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: September 14th 2008, 12:14 AM

/mathhelpforum @mathhelpforum