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

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The discussion revolves around differentiating exponentials with non-commuting operators, specifically in the context of Peskin's Quantum Field Theory (QFT) on page 84, formulas (4.17) and (4.18). Participants clarify that the expression for the time evolution operator U(t, t₀) cannot be simplified to e^(i(H₀-H)(t-t₀)) due to the non-commutativity of operators H and H₀. The Campbell-Hausdorff formula is highlighted as a necessary tool when dealing with such non-commuting operators. The key takeaway is to avoid moving operators through each other during differentiation unless they commute.

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gremezd
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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?
 
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gremezd said:
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
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.
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?
You just differentiate as usual, making sure that you never pass an operator "through" another operator that does not commute with it.
 
Thanks a lot! This has been tormenting me for ages!
 
Thank you, nrqed, for pointing out my mistake. I appreciate it :)
 
gremezd said:
Thank you, nrqed, for pointing out my mistake. I appreciate it :)

:smile: You are very very welcome.

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

Patrick
 

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