More QM trouble:(

Dathascome

Hi there,
I'm having a bit of trouble with something dealing with functions of operators and commutators. It's two different examples actually. For the first one I have that A and B are hermitian operators and their expected values with respect to a normalized state vector /S> are <A>=<S/A\S>
where [tex]
\Delta
[/tex]A=sqrt(A-<A>), and similarly for B.
Now here's the thing I'm having trouble with. They're trying to derive the uncertainty relation in this book, and for part of it they say that [[tex]
\Delta
[/tex]A,[tex]
\Delta
[/tex]B]=[A,B]
(these are the commutators)
I know it's probably something really obvious but for some reason I don't see it. I tried actually writing out the full expression for the commutator but I don't see why they would be equal.

As for the second problem I'm having, it has to do with functions of operators. Again I have 2 operators (this time not necessarily hermitian) that do not commute, [A,B] not equal to 0, which implies that [B,F(A)] (some function of A) is also not equal to 0. So here comes the parts I'm not sure of, they say that e^A*e^Bnot equal to e^(A+B), which I don't see why. How could I show this using a taylor expansion. I tried but didn't really get what I should have.
They then proceed to say that e^A*e^b = e^(A+B)*e^[A,B]/2

and also,
e^A*B*e^-A= B+ [A,B] + 1/2![A,[A,B]]+ 1/3![A,[A,[A,B]]]+...

neither of which I fully understand...I mean, I can see that the last one is a taylor series but I don't fully understand either of the last two things.

OlderDan

Check post #52 in this thread.

http://www.physicsforums.com/showthread.php?t=78949

Dathascome

Forgive me for not noticing that thread. Suprisingly that is the problem I was looking at. The thing is it didn't really answer my question. I'm sorry I know that there's probably something really stupid that I'm missing but I don't see this
[quote]the commutator
[tex][\Delta A,~ \Delta B] = [A - \langle A \rangle I,~B - \langle B \rangle I] = [A,B] ~~~-~(4)[/tex]
I do not get for some reason.
Writing out the commutator
[tex]
AB-A\langle B \rangle I - \langle A \rangle I B + \langle A \rangle I \langle B \rangle I -BA + B\langle A\rangle I +\langle B \rangle I A- \langle B \rangle I \langle A \rangle I
[/tex]

So I can pick out the AB-BA=[A,B], but what happens to all the rest of that stuff there?
OlderDan, thanks for trying to point me in the right direction, any further help would be great.

Gokul43201

Notice that [itex]\langle X \rangle[/itex] is just a scalar and [itex]IX = XI = X [/itex], so all the other terms cancel off nicely.

Dathascome

Damnit, I knew it was something really simple. Those are things that always get me, the obvious or somewhat obvious things

Is it something just as obvious that I missed concerning my second question with the exponentials?

Gokul43201

First, Taylor expand the exponentials :

[tex]a^A \cdot e^B = (I + A + \frac {A^2}{2!} + \frac {A^3}{3!} + ...)(I + B + \frac {B^2}{2!} + \frac {B^3}{3!} + ...)[/tex]
[tex]=I + A + B + AB + \frac {A^2 + B^2}{2!} + ...
= I + (A+B) + \frac {A^2 + B^2 + 2AB}{2!} + ... [/tex]

But notice that, for instance
[tex](A+B)^2 = A^2 + B^2+AB+BA = A^2 + B^2+2AB+BA-AB=(A+B)^2+[B,A] \neq (A+B)^2 [/tex]

Use this above and hopefully, the result will follow.