# More QM trouble

1. Jul 29, 2005

### Dathascome

More QM trouble:(

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 $$\Delta$$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 [$$\Delta$$A,$$\Delta$$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.

Last edited: Jul 29, 2005
2. Jul 29, 2005

### OlderDan

3. Jul 29, 2005

### 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

Last edited: Jul 29, 2005
4. Jul 29, 2005

### Gokul43201

Staff Emeritus
Notice that $\langle X \rangle$ is just a scalar and $IX = XI = X$, so all the other terms cancel off nicely.

Last edited: Jul 29, 2005
5. Jul 29, 2005

### 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?

6. Jul 29, 2005

### Gokul43201

Staff Emeritus
First, Taylor expand the exponentials :

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

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

Use this above and hopefully, the result will follow.