1
$\begingroup$

When computing a generating functional, $Z[J]$, in terms of the generating functional of Green functions, $Z[0]$, in my lecturer's notes we reach the following terms:

$$Z[J]= \mathcal{N} \int Dq \hspace{1mm}\text{exp} \left( \frac{-i}{\hbar} \int dt V(q(t))\right)\text{exp}\left(\frac{i}{\hbar} \int ds \hspace{1mm} L_0 (q(t), \dot{q}(t)) + J(s) q(s)\right)\tag{1}$$

$$Z[J]= \mathcal{N} \int Dq \hspace{1mm}\text{exp} \left( \frac{-i}{\hbar} \int dt V\left(-i\hbar\frac{\delta}{\delta J(t) }\right)\right)\text{exp}(...)\tag{2}$$

Can anyone explain to me the identity used for the substitution, $q(t) = -i\hbar\frac{\delta}{\delta J(t)}$ , as this makes no sense to me.

I believe the $-i\hbar$ is just a convention but I don't understand why does $q(t) \to -i\hbar \frac{\delta}{\delta J(t)}$?

$\endgroup$
1
  • 2
    $\begingroup$ Expand the dots in $\exp(\ldots)$ so that the dependence on $J$ is explicit... $\endgroup$
    – MannyC
    May 17, 2020 at 11:15

1 Answer 1

2
$\begingroup$

Since $$q(t)e^{i\int ds J(s)q(s)} = - i\frac{\delta}{\delta J(t)}e^{i\int ds J(s)q(s)},$$ we can write for an arbitrary functional of $q(t)$ $$F[q(t)]e^{i\int ds J(s)q(s)} = \sum_{n=0}^{+\infty}\int dt_1...dt_n\frac{\delta^n F[q(t)]}{\delta q(t_1)...\delta q(t_n)}|_{q(t)=0}q(t_1)...q(t_n)e^{i\int ds J(s)q(s)}=\\ \sum_{n=0}^{+\infty}\int dt_1...dt_n\frac{\delta^n F[q(t)]}{\delta q(t_1)...\delta q(t_n)}|_{q(t)=0} (- i)^n\frac{\delta}{\delta J(t_1)}...\frac{\delta}{\delta J(t_n)} e^{i\int ds J(s)q(s)}=\\ F\left[- i\frac{\delta}{\delta J(t)}\right] e^{i\int ds J(s)q(s)}.$$ In general, such relations are easier to derive first in terms of ordinary derivatives, before switching to functional ones, where the time arguments and integrals clutter the essence. (Derivation rules are identical.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.