Double variation of Schwinger action principle The Schwinger action principle is given by
$$\delta_{1}\big\langle b\big|a\big\rangle= i\int_{t_{a}}^{t_{b}}\text{d}t\,\sum_{c,d}\big\langle b\big|c\big\rangle\big\langle c\big|\delta_{1}L(t)\big|d\big\rangle\big\langle d\big|a\big\rangle$$
where the state $|c\big\rangle$ is in time $t_c$ and so on.

Now we perform another variation $\delta_{2}$ which is independent of the first variation $\delta_{1}$
$$\delta_{2}\delta_{1}\big\langle b\big|a\big\rangle= i\int_{t_{a}}^{t_{b}}\text{d}t\,\sum_{c,d}\bigg[\bigg(\delta_{2}\big\langle b\big|c\big\rangle\bigg)\big\langle c\big|\delta_{1}L(t)\big|d\big\rangle\big\langle d\big|a\big\rangle
 +\big\langle b\big|c\big\rangle\big\langle c\big|\delta_{1}L(t)\big|d\big\rangle\bigg(\delta_{2}\big\langle d\big|a\big\rangle\bigg)\bigg]$$.

Tom D.J. writes (in the book "The Schwinger Action Principle and Effective Action" page: 345.)
"Note that since the second variation in the structure of the Lagrangian
is independent of the first, there is no term like $\delta_2\big\langle c\big|\delta_{1}L(t)\big|d\big\rangle$ in the above equation."

Could someone elaborate on this and maybe show with an example why this is true?

The closest I could think of was something of the lines "$\delta_{2}\big\langle c\big|\delta_{1}L(t)\big|d\big\rangle=0$
  since if $\delta_{2}$
  and $\delta_{1}$
  are with respect to different functions this term will be zero. If they are variations of the same variables this will be of second order and will be ignored"
 A: Well, it becomes a bit clearer when we see the final formulas of Ref. 1:
$$\delta \langle  a_f , t_f |a_i , t_i \rangle 
~=~ \frac{i}{\hbar} \int_{t_i}^{t_f} \! dt 
\langle  a_f , t_f |  \delta L(t)  |a_i , t_i \rangle \tag{7.126}
$$
$$ \delta^{\prime} \delta \langle  a_f , t_f |a_i , t_i \rangle ~=~\frac{1}{2}\left(\frac{i}{\hbar}\right)^2 
\int_{t_i}^{t_f} \! dt \int_{t_i}^{t_f} \!dt^{\prime} $$
$$\times \langle  a_f , t_f |T[ \delta L(t)~\delta^{\prime}L(t^{\prime}) ] |a_i , t_i \rangle. \tag{7.131}$$
Recall that the Schwinger action principle can be described via a time integral $\int_{t_i}^{t_f}\!dt~ \delta L(t)$ of an operator $\delta L(t)$. Imagine that the time interval 
$$[t_i,t_f]~=~\cup_{n=1}^N I_n, \qquad I_n~:=~[t_{n-1},t_n],$$ 
is divided into a sufficiently fine discretization $t_i=t_0<t_1, \ldots t_{N-1} < t_N=t_f $, where the integer $N$ is sufficiently large.
By inserting many completeness identities $\sum_b  |b , t_n \rangle\langle  b , t_n |={\bf 1}$, we can split a total variation (7.126) into many small contribution labelled by the time intervals $I_n$, $n\in\{1,2, \ldots, N\}$.
Similarly, when performing a double variation (7.131), we will get $N$ diagonal and $N(N-1)$ off-diagonal contributions labelled by two time intervals $I_n$ and $I_m$, where $n,m\in\{1,2, \ldots, N\}$. (A diagonal contribution $n=m$ refers to the same time interval $I_n=I_m$.) If in the limit $N\to\infty$, the $N(N-1)$ off-diagonal contributions dominate over the $N$ diagonal contributions, the two variations becomes effectively independent. 
However in hindsight, it seems that Ref. 1 is assuming that the two variation $\delta$ and $\delta^{\prime}$ are manifestly independent and not just effectively independent. Manifest independence here means that the $\delta^{\prime}$ variation simply doesn't act on the $\delta L(t)$ operator, and vice-versa.
References:


*

*D.J. Toms, The Schwinger Action Principle and Effective Action, 1997, Section 7.6.

