above is the PDF that is hosted on his website. The equation is on page 22 (pg 30 in the pdf). In chapter 2. It is the second equation of the subsection "derivation of the holonomy".

On equation 2.27 he splits up the integral from the exponential of a time evolution operator into finite steps (that are infinitely small). Because they are within an exponential the infinite series becomes an infinite product. I get all of that.

WHat I don't get is how the two Unitary matrices (the $U$ and the $U^{\dagger}$) come out of the exponential. This has been vexing me for a while. This is as he goes from the center series to the right-hand side product.

I'm wondering if this is a mistake or not. I'm not asking for a complete solution since I want to work it out but a hint would be nice.

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    $\begingroup$ Comment to the post (v2): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. $\endgroup$ – Qmechanic Jul 18 '14 at 13:57
  • $\begingroup$ Unitaries can always be taken out of exponentials, i.e. $\exp(UAU^\dagger)=U\exp(A)U^\dagger$. Is this what you are asking about? $\endgroup$ – Norbert Schuch Jun 22 '15 at 21:44

Edit: Sorry, my first answer was not the one expected. I let it below the correct answer I detail now:

It's always tricky to discuss exponentials of operators. The good thing is that it's always the same trick: you use $e^{A}\approx1+A$ in both directions, valid for small $A$. To get exact results you also use that $e^{At}=\prod_{i}\left(1+A\Delta t_{i}\right)$ sometimes, here we use both.

So here you start from $$U\left(0,T\right)=\hat{T}\lim_{N\rightarrow\infty}\exp\left[-\mathbf{i}\sum_{i=1}^{N}U_{i}H_{0}U_{i}^{\dagger}\Delta t\right]$$

and you use the trick

$$\exp\left[-\mathbf{i}\sum_{i=1}^{N}U_{i}H_{0}U_{i}^{\dagger}\Delta t\right]=\prod_{i=1}^{N}\left(1-\mathbf{i}U_{i}H_{0}U_{i}^{\dagger}\Delta t\right)=\prod_{i=1}^{N}U_{i}\left(1-\mathbf{i}H_{0}\Delta t\right)U_{i}^{\dagger}$$

because $U_{i}U_{i}^{\dagger}=1$. Then you use the trick $1-\mathbf{i}H_{0}\Delta t\approx e^{-\mathbf{i}H_{0}\Delta t}$ back, valid for small $\Delta t$ and you get

$$\exp\left[-\mathbf{i}\sum_{i=1}^{N}U_{i}H_{0}U_{i}^{\dagger}\Delta t\right]\approx\prod_{i=1}^{N}U_{i}e^{-\mathbf{i}H_{0}\Delta t}U_{i}^{\dagger}$$

which finally ends up with the desired expression

$$U=\hat{T}\lim_{N\rightarrow\infty}\prod_{i=1}^{N}U_{i}e^{-\mathbf{i}H_{0}\Delta t}U_{i}^{\dagger}$$

Nothing else ! After a few practising, it seems no longer tricky :-)

Is the time-ordering operator which perturbs you ?

I start from eq.(2.26) $$U\left(0,T\right)=\hat{T}\exp\left[-\mathbf{i}\int_{0}^{T}U\left(\lambda\left(t\right)\right)H_{0}U^{\dagger}\left(\lambda\left(t\right)\right)\right]$$

which I expand as usual $$U=\hat{T}\lim_{N\rightarrow\infty}\prod_{i=1}^{N}U_{i}e^{-\mathbf{i}H_{0}\Delta t}U_{i}^{\dagger}=\hat{T}\lim_{N}\left[U_{1}e^{-\mathbf{i}H_{0}\Delta t}U_{1}^{\dagger}U_{2}e^{-\mathbf{i}H_{0}\Delta t}U_{2}^{\dagger}U_{3}\cdots U_{N}^{\dagger}\right]$$

but thanks to the time-ordering, the first-in-time operator has to be on the right, so we have as well

$$U=\hat{T}\lim_{N}\left[U_{N}\cdots U_{3}^{\dagger}U_{2}e^{-\mathbf{i}H_{0}\Delta t}U_{2}^{\dagger}U_{1}e^{-\mathbf{i}H_{0}\Delta t}U_{1}^{\dagger}\right]$$

(I could in principle drop the time ordering operator at that step, since everything looks well ordered, but I should care about the prefactor then, so I keep it, it's fancy by the way !) Then expanding $U_{i}^{\dagger}U_{i}\approx1+A_{i}\Delta\lambda_{i}$ you get eq.(2.30) $$U=\hat{T}\lim_{N}\left[U_{N}\left(1-\mathbf{i}H_{0}N\Delta t+\sum_{i=1}^{N-1}A_{i}\Delta\lambda_{i}\right)U_{1}^{\dagger}\right]$$

and the adiabatic criterion and the fact that $H_{0}\left|\Psi_{0}\right\rangle =0$ to get

$$U\left(0,T\right)=\hat{P}e^{\oint Adl}$$

when $U_{N}=U_{1}=1$. Due to adiabaticity, there is a mapping between time-evolution and path-following, so $\hat{T}\rightarrow\hat{P}$ in the last expression.

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  • $\begingroup$ Except, in the text, Pachos starts with the sum. I understand approximating the integral as a function with a series for an exponent. However, I do not see how the U's come outside of the exponent when it becomes an infinite product. Forgive my lack of LaTeX skill: I do not see how: exp(sigma(UHU)) ----->pi( Ue*xp(H)*U) I hope that makes sense. $\endgroup$ – Elliott Miller Apr 9 '14 at 10:05
  • $\begingroup$ @Elliot Ah, ok, sorry then. This trick is an easy one: see the first part of the edited answer. To get correct latex, you can copy-paste the expressions from Lyx, available for free there lyx.org Have fun. $\endgroup$ – FraSchelle Apr 9 '14 at 12:15

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