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2

The reason your logic fails is because $\psi$ is not simply a Grassmann variable; it is a four-component vector of complex Grassmann numbers (in four dimensions): $$\psi=\left(\begin{array}{c} \theta_1 \\ \theta_2 \\ \theta_3 \\ \theta_4 \end{array}\right)$$ With this knowledge, try computing $\bar{\psi}\psi\bar{\psi}\psi$ and ...

3

The argument is false in four dimensional space. The error is the assumption that you get one Grassman number per spinor. In fact, you get one Grassman number per spinor component! In 4d, spinors have multiple components. (Both Weyl spinors have 2 components, and Dirac spinors have 4.) In 1d space, this is a correct argument. In 2d, it is correct for ...

5

The terminology of a mode of a free quantum field $\phi(x)$ comes from writing it as a Fourier transform, often also called mode expansion: $$\phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$ where for a ...

2

The term mode is used to define a particular state of a system and may refer for instance to its spin, wavevector, polarisation, charge etc. If we wanted to create a boson at position $x$ with an up-spin and with wavevector $k$, we may use the field operator $\hat{a}^\dagger(x, k, \uparrow)$ on the vacuum state $\vert0\rangle$. The most clear distinction ...

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