A trace formula of two noncommutative operators In many cases of quantum many-body problems, the Hamiltonian $H$ can always be divided into two parts, i.e. $H_0$ and $H'$. In this occasion, one can systemically calculate the partition function throught the formula
$$
Z ~= Tre^{-\beta(H_0+H')} ~=~ Tre^{-\beta H_0}Te^{-\int_0^{\beta}H'(\tau)d\tau}
$$
where 
$$H'(\tau)=e^{\tau H_0}H'e^{-\tau H_0},$$ 
and $T$ is the time ordering operator. Here, I want to ask that how can I prove the above formula? 
 A: Hints:


*

*OP's identity follows from standard manipulations in the interaction picture, cf. e.g. Ref. 1. 

*Start with the evolution operator
$$\tag{1} U(t_f,t_i)~:=~\exp\left(-\frac{i}{\hbar}H (t_f-t_i) \right), \qquad H~=~H_0+V, $$
which satisfies the Schrödinger equation
$$\tag{2} i\hbar\frac{\partial}{\partial t_f}U(t_f,t_i)~=~HU(t_f,t_i).$$

*Define the evolution operator in the interaction picture
$$\tag{3} U_I(t_f,t_i)~:=~\exp\left(\frac{i}{\hbar}H_0 (t_f-t_i) \right)U(t_f,t_i) ,$$ 
and define the interaction Hamiltonian 
$$\tag{4}  H_I(t_f) ~:=~\exp\left(\frac{i}{\hbar}H_0 (t_f-t_i) \right)V\exp\left(-\frac{i}{\hbar}H_0 (t_f-t_i) \right).  $$

*Show that the evolution operator (3) satisfies the 1st order ODE
$$\tag{5} -i\hbar\frac{\partial}{\partial t_f}U_I(t_f,t_i)~=~H_I(t_f)U_I(t_f,t_i).$$

*Deduce from (5) that evolution operator $U_I(t_f,t_i)$ in the interaction picture can be written as a time-ordered exponential 
$$\tag{6} U_I(t_f,t_i)~=~T\exp\left(-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~H_I(t)  \right).$$ 

*Finally, deduce that
$$\tag{7} {\rm Tr}\left[U(t_f,t_i)\right]
~\stackrel{(3)}{=}~{\rm Tr}\left[\exp\left(-\frac{i}{\hbar}H_0 (t_f-t_i)\right) U_I(t_f,t_i) \right], $$
and wick-rotate in order to obtain OP's identity.
References:


*

*M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 4.2.

