Vacuum persistance amplitude E. Fradkin's Field Theories in Condensed Matter Physics formulas 3.57 and 3.58:
I feel really sad about it, but all my tries of getting from formula 
$$
Z = \operatorname{tr} \hat{T} \prod_{j=1}^{N_t}e^{i\Delta_t H(t_j)}
$$
to 
$$
Z = \sum_{\{\alpha_j\}} \prod_{j=1}^{N_t} <\alpha_j|e^{-i\Delta_t H(t_j)} |\alpha_{j+1}>
$$
led to nothing. Can anyone show more precise transformation or give some clues please.
 A: It looks like the sum appears because of the completeness relation
$$\mathbf{1} = \sum_{\alpha} \left| \alpha \right> \left< \alpha \right|$$
taken over all the $\{ \alpha_j \}$ to give
$$\mathbf{1} = \sum_{\{ \alpha_j \}} \left| \alpha_j \right> \left< \alpha_j \right|$$
Starting with the first expression, which I believe you corrected in your comment to be the path integral
$$\mathrm{tr} ~\hat T ~e^{i \int dt H(t)}$$
if you make the segments finite length $\Delta t$ you can write it, for three segments, as
$$\hat T\sum_{\beta} \left< \beta \mid \exp{(\{i(H(t_1)\Delta t + H(t_2)\Delta t + H(t_3)\Delta t)\}} \mid \beta \right>$$
Inserting the completeness relation, you have
$$Z = \hat T\sum_{\beta}~\sum_{\alpha_1, \alpha_2, \alpha_3}\langle \beta \mid \alpha_1 \rangle \langle \alpha_1 \mid e^{i H_1 \Delta t} \mid \alpha_2 \rangle \langle \alpha_2 \mid e^{i H_2 \Delta t}\mid \alpha_3 \rangle \langle \alpha_3 \mid e^{i H_3 \Delta t} \mid \beta \rangle$$
The rest should be pretty straightforward. Generalize to $N_t$ segments and simplify. I'll leave that to you.
