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 $q(t) \to -i\hbar \frac{\delta}{\delta J(t)}$?

  • 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


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.)


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