Exponential operator multiplication

[Dracovich]

1. I have a fairly straight forward problem (or basically, i need help to just get started on my problem), i have forgotten all my QM and am in a "bit" over my head here. Basically i am having problems remembering how one would go about treating operators that are inside exponentials.



2. So i have $$e^{\hat{U^{\dagger}}}\hat{A}e^{\hat{U}}$$ and basically was just wondering how one combines the A and U operator in some way



3. I honestly didn't get very far, looking through some of the basic coursebooks for QM i didn't find it at least in an obvious place (as in where they explain the basic properties of operators) and looking around/googling etc the closest i got was to represent it as a power series, i tried writing that out and see if it got me somewhere but i couldn't see how it was suppose to help me (although I'm sure it should).

 

[malawi_glenn]

Use the Taylor Expansion of exp with the U-operator as argument.

$$\mathrm{e}^{x} = \sum^{\infin}_{n=0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

 

[Dracovich]

Thanks, i had kind of gotten that far (as noted in 3, perhaps it's not called power series, sorry about that). Guess i got it right but I'm just not good enough to go from there lol, well at least i know that's definitely what i should be working with, cheers!

 

[genneth]

Is there a particular A and U that you have in mind?

 

[Dracovich]

Yes, didn't mention it since i figured i'd try to just get a boost here and finish on my own, but the actual thing looks like this:

$$e^{-i\theta \hat{J}^{\dagger}}\big(\frac{\hat{a_{0}}}{\hat{a_{1}]} \big) e^{i\theta \hat{J}}$$

I can't find the command for vector notation so i had to use \frac, but that's not a fraction, it's a vector

$$\hat{J}=(\hat{a_{0}}^{\dagger}\hat{a_{1}]+\hat{a_{1}}^{\dagger}\hat{a_{0}})/2$$

where $$a_0$$ and $$a_1$$ (and the daggers that is) are the creation and annihilation operators.

I tried writing it out and seeing what happens when the operators J and a get multiplied, but I'm honestly horrible at this, it was probably a mistake taking this course seeing as how it's been over 2 years since i last even looked in a quantum book and wasn't that great to begin with hah, but there you go, might as well stick it out as i can't opt out anymore.

 

[malawi_glenn]

\vec{your vector}

Why don't you show us a bit of your attempt to solution?

It is required to get help you know..

 

[Dracovich]

I would if i had much to show :) I'm afraid I'm fairly clueless, like i mentioned i had only gotten so far as writing the exponential operator out in a power series, and also writing out explicitly the a operators multiplying with a single J operator in hopes that it would give me some insight, but no luck. But don't worry about it, the class is today so it's too late anyway :) Thanks though, i really do appreciate the help given so far.

 

[a19grey]

There's an exponential identity that you can use instead of expanding it all out.

$$e^{\hat{-\hat{B}}}\hat{A}e^{\hat{B}} = \hat{A} + [A,B] + 1/2![A,[A,B]] + 1/3![A,[A,[A,B]]] + ...$$

Because often times, the dagger of a unitary operator is just the negative of it. And the brackets denote the commutator. In many problems, the commutator [A,B] = scalar so only the first two terms are non-zero since [A,(any constant)] = 0.