The vacuum functional for the fermionic oscillator is given by

$$ Z[0] = N\int\mathcal{D}\overline{\psi}\mathcal{D}\psi \exp\left(i\int_0^Tdt\left(i\overline{\psi}\psi-w\overline{\psi}\psi \right)\right). \tag{5.80}$$

Using Weyl ordering and discretizing the time integral, it is claimed that this could be written as

$$ Z[0] = \lim_{\epsilon\rightarrow0}\lim_{N\rightarrow\infty}N\int d\overline{\psi}_1\dots d\overline{\psi}_{N-1}d\psi_1\dots d\psi_{N-1}$$ $$\times \exp\left(i\epsilon\sum_{n=1}^N\left(i\overline{\psi}_n\frac{\psi_n-\psi_{n-1}}{\epsilon}-w\overline{\psi}_n\frac{\psi_n+\psi_{n-1}}{2} \right)\right),\tag{5.81} $$

where the mid-point prescription of the Weyl ordering was used.

My question is pretty simple and naive. To write this expression, the author says on the bottom of p.92 that

$\overline{\psi}$ represents the momentum conjugate to $\psi$.

But if I make the computation I get (using left derivatives)

$$ \Pi_\psi = \frac{\partial L}{\partial \dot{\psi}}=-i\overline{\psi}.\tag{5.43} $$

Why is it valid to ignore the $-i$ factor and just to consider $\overline{\psi}$ as the conjugate momentum?

  • $\begingroup$ ${}$ Which page? $\endgroup$ – Qmechanic Jun 25 at 4:47
  • $\begingroup$ p92, second edition. Thanks for looking it up! $\endgroup$ – user2820579 Jun 25 at 17:29
  • $\begingroup$ Does it make you equation for $Z[0]$ invalid? $\endgroup$ – Oбжорoв Jun 25 at 18:06

Ashok Das is not ignoring any factor of $-i$ in formulas. The word "represent" in the sentence $\overline{\psi}$ represents the conjugate to $\psi$ is here used semantically in a weaker sense than "is equal", e.g. "is equal up to a multiplicative constant".

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  • $\begingroup$ General question. What is then the rule in order to apply the mid-point prescription? It seems to me that this is a loosely rule, and I'm worried that different cases yield different answers. Does this only happens for fermions? $\endgroup$ – user2820579 Jun 26 at 18:19

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