# Proof of operator relation

1. Nov 16, 2007

### Ene Dene

I'm having a problem proving this operator relation:

$$exp(-i\phi\hat{j_{i}})exp(i\theta\hat{j_{k}})exp(i\phi\hat{j_{i}})=exp(i\theta(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}})$$ (1)

where

$$[\hat{j_{i}}, \hat{j_{k}}]=i\epsilon_{ikl}\hat{j_{l}}$$. (2)

I can prove this for:

$$exp(-i\phi\hat{j_{i}})\hat{j_{k}}exp(i\phi\hat{j_{i}})=cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}}$$ (3)

using Baker-Hausdorff lemma.

Now what I do when I'm trying to prove the first expresion, I expand the middle term in Taylor series, and then trying to use this lemma again, but problem arisses with higher powers of $$\hat{j_{k}}$$.

$$exp(-i\phi\hat{j_{i}})(1+i\theta\hat{j_{k}}+\frac{(i\theta\hat{j_{k}})^2}{2!}+\frac{(i\theta\hat{j_{k}})^3}{3!}+...)exp(i\phi\hat{j_{i}})$$

The first term:

$$exp(-i\phi\hat{j_{i}})exp(i\phi\hat{j_{i}})=1$$

Second term (what I was able to prove (3)):

$$i\theta(exp(-i\phi\hat{j_{i}}))\hat{j_{k}}exp(i\phi\hat{j_{i}})=i\theta(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}})$$

And now a problem arisses:

$$\frac{(i\theta)^nexp(-i\phi\hat{j_{i}})(\hat{j_{k}})^nexp(i\phi\hat{j_{i}})}{n!}$$

If (1) is true than it should be:

$$(i\theta)^nexp(-i\phi\hat{j_{i}})(\hat{j_{k}})^nexp(i\phi\hat{j_{i}})/n!=\frac{(i\theta(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}}))^n}{n!}$$

becoase

$$exp(i\theta(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}})=1+(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}})+\frac{(i\theta(cos(\phi)\hat{j_{k}}+sin(\phi)\hat{j_{l}}))^2}{2!}+...$$

but, I can't prove this. Using Baker-Hausdorff lemma for each term becomes too complicated and I get lose in all that mess.

2. Nov 16, 2007

### Hurkyl

Staff Emeritus
What happens when you square equation 3?

3. Nov 16, 2007

### Ene Dene

Nooo, it can't bee :).
I spent all night trying to solve this in most complicated ways and I didn't saw this...

Thank you very much!