# Differentiation of an exponential with operators (Peskin p.84)

1. Jul 23, 2008

### gremezd

Does anyone know how to differentiate an exponential, which has an operator in its power? I found it quite a trouble in Peskin's QFT (page 84, formulas (4.17), (4.18)).
Here we have these two formulas of Peskin:

$$U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }e^{-iH\left( t-t_{0}\right) }$$;
$$i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=e^{iH_{0}\left( t-t_{0}\right) }\left( H-H_{0}\right) e^{-iH\left( t-t_{0}\right) }$$.

I agree with this. However, if we write $$U\left( t,t_{0}\right)$$ as $$U\left( t,t_{0}\right)=e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }$$, then

$$i\frac{\partial}{\partial t}U\left( t,t_{0}\right)=\left( H-H_{0}\right)e^{i\left( H_{0}-H\right) \left( t-t_{0}\right) }$$

and we cannot transport $$e^{iH_{0}\left( t-t_{0}\right) }$$ to the left of $$\left( H-H_{0}\right)$$ so easily to obtain Peskin's result, since, according to my calculations, $$\left[ H,H_{0}\right]\neq0$$. Do we have a rule, which explains where to put the operators from the exponential after differentiation, when we have several noncummuting operators in the power of exponential?

2. Jul 23, 2008

### nrqed

But you can't write this since H and H_0 don't commute. $$e^A e^B = e^{A+B}$$ only when A and B commute. Otherwise you have to use the Campbell-Hausdorf formula.
You just differentiate as usual, making sure that you never pass an operator "through" another operator that does not commute with it.

3. Jul 23, 2008

### wasia

Thanks a lot! This has been tormenting me for ages!

4. Jul 23, 2008

### gremezd

Thank you, nrqed, for pointing out my mistake. I appreciate it :)

5. Jul 23, 2008

### nrqed

You are very very welcome.

And thank you for posting your question since this apparently helped Wasia too!!

Patrick