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Fermions $f_i,\,f_j$ with respective momenta $\pi_i,\,\pi_j$ satisfy the equal-time canonical anticommutation relations $$\left\{\ f_i,\,f_j \right\} = \left\{\ \pi_i,\,\pi_j \right\} = 0,\,\left\{\ f_i\left(t,\,\mathbf{x}\right),\,\pi_j \left(t,\,\mathbf{x'}\right)\right\} = i\hbar \delta_{ij} \delta \left(\mathbf{x},\,\mathbf{x'}\right),$$where the second ...


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Fermionic creation and annihilation operators always satisfy commutation relations with bosonic (or more generally even) operators and anticommutation relations with othe fermionic creation and annihilation operators (or more generally odd operators). In particular, the creation operators for distinct, orthogonal modes always anticommute.


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Fleshing out @Peter Morgan's answer a bit, to the effect that $x$ and $p$ are not bounded operators, so their commutator need not be bounded. First note that $$[x^n,p] = i\hbar nx^{n-1} ~,$$ hence the operator norms of both sides satisfy $$ 2\| p\| ~ \|x\|^n \geq n \hbar \|x\|^{n-1}$$ so that, for any $n$, $$2\|x\|~\|p\|\geq n \hbar~. $$ Since $n$ ...


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@ACuriousMind thank you; I'll read the rules as soon as possible. I was wrong. Look at this: If $\hbar=1$, then \begin{align}\hat{P}_{j}&=-i\frac{\partial}{\partial X_{j}}\\ \left[ \frac{r_i}{r},\hat{P}_j \right]f&=\frac{r_{i}}{r}\left(-i\frac{\partial}{\partial X_{j}}\right)f+i\frac{\partial}{\partial X_{j}}\left( \frac{r_{i}}{r} \right ...



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