这个图片生成的markdown格式，里面的latex公式包裹符号不对，应该把[和]改为$$
whereas
[
i\left\langle \bar{z}_n \left| \hat{A}n \right| z{n-1} \right\rangle \epsilon = i \left\langle \bar{z}n \left| \bar{z}n A_n z{n-1} \right| z{n-1} \right\rangle \epsilon
= i \epsilon \bar{z}n A_n z{n-1} \left\langle \bar{z}n | z{n-1} \right\rangle.
]
Thus
[
\left\langle \bar{z}_n \left| \left( 1 + i \epsilon \hat{A}n \right) \right| z{n-1} \right\rangle = e^{\bar{z}n z{n-1}} + i \epsilon \bar{z}n A_n z{n-1} e^{\bar{z}n z{n-1}}
= e^{\bar{z}n z{n-1}} \left[ 1 + i \epsilon \bar{z}n A_n z{n-1} + O(\epsilon^2) \right]
= e^{\bar{z}n z{n-1} - 1 + i \epsilon \bar{z}n A_n z{n-1} + O(\epsilon^2)}.
]
Combining this expression with the exponent in (\boxed{29}) we obtain
[
e^{-\bar{z}_n z_n} e^{\bar{z}n z{n-1} + i \epsilon \bar{z}_n A_n z_n + O(\epsilon^2)}
= e^{-\bar{z}n (z_n - z{n-1}) + i \epsilon \bar{z}_n A_n z_n + O(\epsilon^2)}
= \exp \left[ (-\bar{z}_n \dot{z}_n + \epsilon \bar{z}_n A_n z_n) \epsilon + O(\epsilon^2) \right].
]
Finally,
[
U(z'', z'; t'', t') = \left\langle z'' \left| P \exp \left[ -i \int_{t'}^{t''} \hat{H}(t) dt \right] \right| z' \right\rangle
\equiv \left\langle z'' \left| P \exp \left[ i \int_{t'}^{t''} \hat{A}(t) dt \right] \right| z' \right\rangle
= \int [D^2 z] \exp \left[ \bar{z}(t'') \dot{z}(t'') + i \int_{t'}^{t''} \left( i \bar{z}(t) \dot{z}(t) + \bar{z}(t) A(t) z(t) \right) dt \right],
]
[
\equiv \int [D^2 z] \exp \left[ \bar{z}(t'') \dot{z}(t'') + i \int_{t'}^{t''} L dt \right],
]
where
[
[D^2 z] = \prod_{t' < t < t''} \frac{d \bar{z}(t) d z(t)}{2 \pi i},
]
and (L) is of "classical" form (\boxed{25}).
Let us confine our attention to the one-particle subspace of the Fock space. As the number operator (\hat{N}) is conserved by virtue of Eq.(\boxed{28}), if we start from the one-particle subspace of the Fock space, we shall remain in this subspace during all the evolution. The transition amplitude (U_{kl}(t'', t')) between the one-particle
这个图片生成的markdown格式，里面的latex公式包裹符号不对，应该把[和]改为$$
whereas
[
i\left\langle \bar{z}_n \left| \hat{A}n \right| z{n-1} \right\rangle \epsilon = i \left\langle \bar{z}n \left| \bar{z}n A_n z{n-1} \right| z{n-1} \right\rangle \epsilon
= i \epsilon \bar{z}n A_n z{n-1} \left\langle \bar{z}n | z{n-1} \right\rangle.
]
Thus
[
\left\langle \bar{z}_n \left| \left( 1 + i \epsilon \hat{A}n \right) \right| z{n-1} \right\rangle = e^{\bar{z}n z{n-1}} + i \epsilon \bar{z}n A_n z{n-1} e^{\bar{z}n z{n-1}}
= e^{\bar{z}n z{n-1}} \left[ 1 + i \epsilon \bar{z}n A_n z{n-1} + O(\epsilon^2) \right]
= e^{\bar{z}n z{n-1} - 1 + i \epsilon \bar{z}n A_n z{n-1} + O(\epsilon^2)}.
]
Combining this expression with the exponent in (\boxed{29}) we obtain
[
e^{-\bar{z}_n z_n} e^{\bar{z}n z{n-1} + i \epsilon \bar{z}_n A_n z_n + O(\epsilon^2)}
= e^{-\bar{z}n (z_n - z{n-1}) + i \epsilon \bar{z}_n A_n z_n + O(\epsilon^2)}
= \exp \left[ (-\bar{z}_n \dot{z}_n + \epsilon \bar{z}_n A_n z_n) \epsilon + O(\epsilon^2) \right].
]
Finally,
[
U(z'', z'; t'', t') = \left\langle z'' \left| P \exp \left[ -i \int_{t'}^{t''} \hat{H}(t) dt \right] \right| z' \right\rangle
\equiv \left\langle z'' \left| P \exp \left[ i \int_{t'}^{t''} \hat{A}(t) dt \right] \right| z' \right\rangle
= \int [D^2 z] \exp \left[ \bar{z}(t'') \dot{z}(t'') + i \int_{t'}^{t''} \left( i \bar{z}(t) \dot{z}(t) + \bar{z}(t) A(t) z(t) \right) dt \right],
]
[
\equiv \int [D^2 z] \exp \left[ \bar{z}(t'') \dot{z}(t'') + i \int_{t'}^{t''} L dt \right],
]
where
[
[D^2 z] = \prod_{t' < t < t''} \frac{d \bar{z}(t) d z(t)}{2 \pi i},
]
and (L) is of "classical" form (\boxed{25}).
Let us confine our attention to the one-particle subspace of the Fock space. As the number operator (\hat{N}) is conserved by virtue of Eq.(\boxed{28}), if we start from the one-particle subspace of the Fock space, we shall remain in this subspace during all the evolution. The transition amplitude (U_{kl}(t'', t')) between the one-particle