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This is a question on the mathematics of path integration.

If we take the action density $S[\phi](t) = \frac{1}{2}\dot{\phi}\dot{\phi}$ and we take the path integral

$$K_T(A,B) = \left. \int e^{-\int_0^T S[\phi](t)} D\phi\right|_{\phi(0)=A}^{\phi(T)=B} = \frac{1}{\sqrt{2\pi T}} e^{-(A-B)^2/(2T)}\tag{1}$$

Where the path integral is over all fields $\phi(t)$ with $0<t<T$ with boundary conditions set. (I don't if there is a common way of writing boundary conditions for function integrals). One can prove this by breaking the functional integral into an infinite set of integrals over each of $\phi(t)$. It satisfies:

$$\int\limits_{-\infty}^\infty K_{t_1}(A,B)K_{t_2}(B,C) dB= K_{t_1+t_2}(A,C).\tag{2}$$

So I wanted to do the same for the simplest Fermion type action $$S[\psi](t)=\dot{\psi}\psi - \psi\dot{\psi}.\tag{3}$$ i.e. where $\psi$ is Grassmann-valued and hence $$\psi(t_1)\psi(t_2) = -\psi(t_2)\psi(t_1)\tag{4}$$ for any $t_1$ and $t_2$

$$K_T(a,b) = \left. \int e^{-\int_0^T S[\psi](t)} D\psi\right|_{\psi(0)=a}^{\psi(T)=b} = \text{undefined?}\tag{5}$$

Is there a way to make sense of such an integral or is it ill defined? It seems to have two different solutions depending on if we divide the time interval into an even or an odd number of small pieces!

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If we consider the coherent state path integrals, they still enjoy the semigroup property (2), and we don't have to worry about functional determinants.

  1. The bosonic coherent state path integral is $$\begin{align} \langle z_f^{\ast}, t_f | z_i, t_i \rangle ~=~& \int_{z(t_i)=z_i}^{\bar{z}(t_f)=z^{\ast}_f} \! {\cal D}\bar{z}~{\cal D}z ~e^{iS[z,\bar{z}]}, \cr {\cal D}\bar{z}~{\cal D}z~:=~& \prod_{n=1}^N \frac{d\bar{z}_n~dz_n}{2 \pi i}, \end{align}\tag{A}$$ $$\begin{align} iS[z,z^{\ast}]~:=&~ (1-\lambda)z^{\ast}(t_f)~z(t_f) + \lambda z^{\ast}(t_i) z(t_i) \cr +& \int_{t_i}^{t_f}\! dt \left[\lambda \dot{z}^{\ast} z -(1-\lambda) z^{\ast} \dot{z}- iH_N(z^{\ast},z) \right],\end{align}\tag{B}$$ $$\begin{align} | z, t \rangle~:=~& e^{\hat{a}^{\dagger}(t)z}| 0 \rangle,\cr \int_{\mathbb{C}} \frac{d\bar{z}~dz}{2 \pi i} e^{-\bar{z}z} |z \rangle\langle \bar{z} |~=~&\mathbb{1}. \end{align}\tag{C}$$ where $\lambda\in \mathbb{R}$ is a real constant which the action (B) does not actually depend on, due to the fundamental theorem of calculus. For more details, see e.g. this related Phys.SE post.

  2. The fermionic coherent state path integral is similarly $$\begin{align} \langle \eta_f^{\ast}, t_f | \eta_i, t_i \rangle ~=~& \int_{\eta(t_i)=\eta_i}^{\bar{\eta}(t_f)=\eta^{\ast}_f} \! {\cal D}\bar{\eta}~{\cal D}\eta ~e^{iS[\eta,\bar{\eta}]}, \cr {\cal D}\bar{\eta}~{\cal D}\eta~:=~& \prod_{n=1}^N d\bar{\eta}_n~d\eta_n, \end{align}\tag{D}$$ $$\begin{align} iS[\eta,\eta^{\ast}]~:=&~ (1-\lambda)\eta^{\ast}(t_f)~\eta(t_f) + \lambda \eta^{\ast}(t_i) \eta(t_i) \cr +& \int_{t_i}^{t_f}\! dt \left[\lambda \dot{\eta}^{\ast} \eta -(1-\lambda) \eta^{\ast} \dot{\eta}- iH_N(\eta^{\ast},\eta) \right],\end{align}\tag{E}$$ $$\begin{align} | \eta, t \rangle~:=~& e^{\hat{c}^{\dagger}(t)\eta}| 0 \rangle,\cr \int_{\mathbb{C}^{0|1}} d\bar{\eta}~d\eta ~e^{-\bar{\eta}\eta} |\eta \rangle\langle \bar{\eta} |~=~&\mathbb{1}. \end{align}\tag{F}$$

The complex Grassmann-odd variable $\eta\in \mathbb{C}^{0|1}$ can be decomposed into two real Grassmann-odd variables $\in \mathbb{R}^{0|1}$. There is a similar construction for a single real Grassmann-odd variable, which is OP's question.

References:

  1. A. Altland & B. Simons, Condensed matter field theory, 2nd ed., 2010; p. 160-164.

  2. T. Lancaster & S.J. Blundell, QFT for the Gifted Amateur, 2014; section 28.2.

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  • $\begingroup$ No doubt this is correct. But I'm not sure how to use this to work out the integral in the question in which I wrote "undefined" which is related to the specific action stated? $\endgroup$
    – user84158
    Commented Jun 1, 2022 at 20:23
  • $\begingroup$ Notes to self: Given a Hermitian fermionic mode $\hat{c}^2=\hbar\hat{\bf 1}$, how does the corresponding coherent states work out? $\hat{c}|0\rangle=|1\rangle$ and $\hat{c}|1\rangle=|0\rangle$. $\endgroup$
    – Qmechanic
    Commented Jun 2, 2022 at 7:31
  • $\begingroup$ Notes to self: Def. (F) agrees with Altland & Simons (4.17). $\endgroup$
    – Qmechanic
    Commented Sep 27, 2022 at 12:45

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