# Tag Info

## New answers tagged grassmann-numbers

2

It seems that OP is pondering about the notion of supernumbers, and the generalization of Fubini and Tonnelli's theorems for integration over superdomains and supermanifolds. See e.g. this Phys.SE post and the references listed therein for details. Example: Consider the integral $$\tag{1} ... 0 I don't think there are miracles here. With an interaction term like \lambda \bar \psi \psi \phi, ou may always write something like : Z(j,\eta, \bar \eta) \sim e^{\large i\int d^4x ~\lambda~ {\frac{\delta}{\delta \eta(x)}\frac{\delta}{\delta \bar \eta(x)}\frac{\delta}{\delta J(x)}}} \int \mathcal{D\phi}\mathcal{D\psi}\mathcal{D \bar \psi} e^{i \int ... 2 The idea is correct (it's called the Hubbard-Stratonovich transformation), but I can't say more without the details of the action. It is discussed in any good textbook on quantum field theory for condensed matter. Concerning you questions (if you're okay with the physicist's approach to Grassman numbers) : the product of two Grassmann numbers commutes with ... 2 EDIT The method you want to use is ok, and gives a quick result. Here it is:$$I=\int\prod d\theta^{*}d\theta\theta_{k}^{*}\theta_{l}exp(\theta^{*}B\theta+\eta^{*}\theta+\theta^{*}\eta)=\left(\frac{\partial}{\partial\eta_{k}^{*}}\right)\left(\frac{\partial}{\partial\eta_{l}}\right)\int\prod d\theta^{*}d\theta ...

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Although these equations are correct, it is not quite correct to treat them as eigenvalue equations. In quantum theory, both bosonic and fermionic operators act on a Hilbert space, thus have ordinary numerical matrix elements and eigenvalues. One of the possible realizations of the Hilbert space in the case of bosonic operators is as a Hilbert space of ...

2

First discretize the spacetime, assign a fermion pair $\bar{\psi}(i)$ and $\psi(i)$ at each point i. Then assuming operator $\hat{A}$ is symmetric, hence which can be diagonalized by a unitary operator whose determinant is one, the path integral can be written in the following way: \int \Pi_{i} d\psi(i)d\bar{\psi}(i) e^{i \delta \sum ...

1

Let's say $f$ admits a taylor series $f(\bar\theta \theta) = A + B \bar\theta \theta + C\bar\theta \theta \bar\theta \theta + \dots$. Now, $\bar\theta \theta \bar\theta \theta = -\bar\theta^2 \theta^2 = 0$, etc., so our function terminates at the linear term. Furthermore, the integral of $f$ over $d \bar\theta\, d\theta$, by the rules of Berezin integration, ...

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