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Maybe stupid questions! But:

  1. Why do we need fermionic HOs? And what do they describe?

  2. And how do we come up with the Hamiltonian $$ H=\frac{1}{2} [c^\dagger,c]~? $$

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  • $\begingroup$ Fermions (such as the electron) exist in nature. A fermionic QHO is a first step towards understanding the physics of electrons, etc. so they are certainly important from that perspective. $\endgroup$ – Prahar Nov 17 '17 at 16:45
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A harmonic oscillator, whether it's bosonic or fermionic, is a single-particle state that can be occupied by noninteracting particles. A fermionic state can be occupied by one particle at most, while a bosonic state can be occupied by an unlimited number of particles. Both

$$H_{\mathrm{BHO}}=\frac{\hbar\omega}{2}\{b^\dagger,b\}=\hbar\omega\left(b^\dagger b+\frac{1}{2}\right)$$

and

$$H_{\mathrm{FHO}}=\frac{\hbar\omega}{2}[c^\dagger,c]=\hbar\omega\left(c^\dagger c+\frac{1}{2}\right)$$

are in fact of the same form. We can derive the Hamiltonian form of the harmonic oscillator from that of one particle, be it bosonic or fermionic. The difference between $[b,b^\dagger]=1$ and $\{c,c^\dagger\}=1$ does not arise in the Hamiltonian form, but in the restriction of occupancy. The zero-point energy $\hbar\omega/2$ has the same meaning in both situations.

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Try to have a look to this answer...

Harmonic oscillator is the fundamental beginning of quantum field theory! Every elementary particle moreover can be seen as a quantum fluctuation of the quantum field and (almost) every perturbation among a equilibrium state can be well approximate by the harmonic oscillator-like behaviour.

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As for the calculative part of your question, the Hamiltonian follows immidiately from the analogy to the bosonic harmonic oscillator

$$\hat{H}_{\rm bosonic}=\hbar\omega\left(\hat{N}+\frac{1}{2}\right)=\frac{1}{2}\hbar\omega\left\{\hat{a}^{\dagger},\hat{a}\right\}$$

$$\hat{H}_{\rm fermionic}=\hbar\omega\left(\hat{N}-\frac{1}{2}\right)=\frac{1}{2}\hbar\omega\left[\hat{c}^{\dagger},\hat{c}\right]$$

We just take the same Hamiltonian, but now with anti-commuting ladder operators. This acounts for the replacement $\left[\cdot,\cdot\right]\leftrightarrow\left\{\cdot,\cdot\right\}$ (notice that the Hamiltonians differ by a constant which can be discarded).

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    $\begingroup$ Yes, but not every analogy gives rise to the correct description. $\endgroup$ – Immanuel Nov 17 '17 at 17:02
  • $\begingroup$ You are correct. See the edited text. $\endgroup$ – eranreches Nov 17 '17 at 17:03
  • $\begingroup$ You didn't answer my question. My question is why are we allowed to build the Hamiltonian in an analogous manner. My guess is: to allow the zero energy state for fermions. $\endgroup$ – Immanuel Nov 17 '17 at 17:21
  • $\begingroup$ Why not? Lets go in reverse: we built a system with two levels and an Hamiltonian matrix $\frac{1}{2}\hbar\omega\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ and if one defines ladder operators this turns out to be analogous to the QHO case. $\endgroup$ – eranreches Nov 17 '17 at 17:25
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Fermionic harmonic oscillators are useful due to the fact that in the same way the usual harmonic oscillator arises from quantising a bosonic field, the former arises from quantising a Dirac fermion. In particular, take the Dirac field as,

$$\psi(\vec x) = \sum_s \int \frac{\mathrm d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vec p}}} \left\{ b^s_{\vec p} u^s(\vec p)e^{-i\vec p \cdot \vec x} + c^s_{\vec p} \mathscr{v}^s(\vec p)^\dagger e^{i\vec p\cdot \vec x} \right\}.$$

Upon using the anti-commutation relations such as, $\{b^r_{\vec p}, {b^s_{\vec q}}^\dagger\} = (2\pi)^3 \delta^{rs}\delta^{(3)}(\vec p -\vec q)$ and the same for $c^s_{\vec p}$ with all others vanishing, one can compute the Hamiltonian,

$$H = \int \frac{\mathrm d^3 p}{(2\pi)^3} E_{\vec p} \left\{{b^s_{\vec p}}^\dagger b^s_{\vec p} +{c^s_{\vec p}}^\dagger c^s_{\vec p}\right\}$$

after normal ordering, which is the analogue of a Hamiltonian for an infinite number of harmonic oscillator, written in terms of anti-commuting operators, describing fermionic statistics.

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  • $\begingroup$ Why do we take this form? Is this because of the fact that there is no zero energy state for fermions? $\endgroup$ – Immanuel Nov 17 '17 at 17:07
  • $\begingroup$ @Immanuel The bosonic harmonic oscillator doesn't have zero energy... $\hat H |0\rangle = \frac12 \hbar \omega |0\rangle$. $\endgroup$ – JamalS Nov 17 '17 at 17:11
  • $\begingroup$ Sorry, I meant there is zero energy state for fermions. $\endgroup$ – Immanuel Nov 17 '17 at 17:20
  • $\begingroup$ I'm a little confused. Am I right? $\endgroup$ – Immanuel Nov 17 '17 at 17:37
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One way to understand it is using quanta. The bosonic oscillator has energy levels $E_{n} = \omega (n + 1/2)$. One says that at energy level $E_{n}$, there are $n$ quanta present. This borrows terminology from field theory, and is literally the basis of the particle interpretation. We have bosonic particle states as excitations of these harmonic oscillators, due to the fact that we can have an arbitrary number of quanta present. In contrast, in the fermionic oscillator, the Hilbert space is only 2 dimensional. In terms of quanta, this says that the quanta obey the Pauli exclusion principle. Hope this helped.

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protected by Qmechanic Nov 18 '17 at 1:21

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