- #1
Dathascome
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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 [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.
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.
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