Double variation of Schwinger action principle

Question

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"

This post imported from StackExchange Physics at 2014-04-20 15:03 (UCT), posted by SE-user Natanael

Answer 1

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:

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

This post imported from StackExchange Physics at 2014-04-20 15:03 (UCT), posted by SE-user Qmechanic

Comments

What would cause us to get off-diagonal contributions?

This post imported from StackExchange Physics at 2014-04-20 15:03 (UCT), posted by SE-user Natanael

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I updated the answer. The generic contribution is off-diagonal.

This post imported from StackExchange Physics at 2014-04-20 15:03 (UCT), posted by SE-user Qmechanic