# More QM trouble

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.

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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
the commutator
$$[\Delta A,~ \Delta B] = [A - \langle A \rangle I,~B - \langle B \rangle I] = [A,B] ~~~-~(4)$$
I do not get for some reason.
Writing out the commutator
$$AB-A\langle B \rangle I - \langle A \rangle I B + \langle A \rangle I \langle B \rangle I -BA + B\langle A\rangle I +\langle B \rangle I A- \langle B \rangle I \langle A \rangle I$$

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.

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Gokul43201
Staff Emeritus
Gold Member
Dathascome said:
Writing out the commutator
$$AB-A\langle B \rangle I - \langle A \rangle I B + \langle A \rangle I \langle B \rangle I -BA + B\langle A\rangle I +\langle B \rangle I A- \langle B \rangle I \langle A \rangle I$$

So I can pick out the AB-BA=[A,B], but what happens to all the rest of that stuff there?
Notice that $\langle X \rangle$ is just a scalar and $IX = XI = X$, so all the other terms cancel off nicely.

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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
Staff Emeritus
Gold Member
Dathascome said:
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.
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. 