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I was reading these lecture notes and they show how if you start with two operators $p$ and $q$ such that $[q,p]=i$, you can define $a:=\frac{1}{\sqrt{2}}(q+ip)$ and $a^\dagger:=\frac{1}{\sqrt{2}}(q-ip)$, such that $[a,a^\dagger]=1$, and that this in turn implies that these are bosonic ladder operators. This left me with a few questions.

  1. The creation/annihilation commutation relations are different for fermions and bosons. Does that mean that the moment/position commutation relations also differ? And if not, can we then construct bosonic creation/annihiliation ladder operators for all kinds of particles?

  2. Suppose I have a wavefunction $\vert \psi\rangle$ with position and momentum operators $Q$ and $P$ such that $Q\vert x\rangle=x\vert x\rangle$ for a wavefunction representing a particle precisely at position $x$, and similarly $P\vert p\rangle= p\vert p\rangle$ for a particle with momentum precisely $p$. Then I can define creation and annihilation operators $a=\frac{1}{\sqrt{2}}(Q+iP)$ and $a^\dagger = \frac{1}{\sqrt{2}}(Q - iP)$. These satisfy $[a,a^\dagger]=1$ (up to normalization), so they are ladder operators. Hence, it has a bunch of eigenstates of the form $\vert 0\rangle, \vert 1\rangle, \dots, \vert n\rangle,\dots$. What do these eigenstates represent?

  3. When I see descriptions of systems like (2), they always seem to use creation/annihilation operators indexed by position or momentum, like $a_p^\dagger$ and $a_p$. This would imply a set of operators $q_p:= \frac{1}{\sqrt{2}}(a_p^\dagger + a_p)$ and $p_p:=\frac{i}{\sqrt{2}}(a_p^\dagger - a_p)$. What do the position/momentum operators $q_p$ and $p_p$ represent?

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    $\begingroup$ Q2) They refer to energy eigenstates, $a_+a_- \psi =( E - 1/2 \hbar \omega)\psi$ where $E_n= (n+1/2)\hbar \omega$ $\endgroup$
    – Tim Crosby
    Jul 20, 2020 at 18:16
  • $\begingroup$ I can see why they refer to energy eigenstates if we have a simple harmonic oscillator Hamiltonian $H= Q^2+P^2$ (omitting constants). Does that still hold if there is a different Hamiltonian? $\endgroup$
    – Sam Jaques
    Jul 21, 2020 at 14:03

1 Answer 1

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The creation/annihilation commutation relations are different for fermions and bosons. Does that mean that the moment/position commutation relations also differ?

No. The position and momentum operator have the usual commutation relations for fermions.

However, to create a spin-1/2 fermion, we don't just need to specify it's position (or momentum), but also whether it has spin component $\pm \hbar/2$ along the $z$ axis. This spin degree of freedom is an extra part of the creation/annhilation operator and leads to the anti-commutation relation. Note that since the vacuum has no spin, to create a fermion state we must apply a creation operator with an associated spin.

If you write down the anticommutation relations carefully, you should get something like

\begin{equation} \{b_\sigma,b^\dagger_{\sigma'}\} = \delta_{\sigma,\sigma'} \end{equation} where $\sigma=\pm 1$ represents the orbital angular momentum quantum number of the fermion.

And if not, can we then construct bosonic creation/annihiliation ladder operators for all kinds of particles?

Mathematically you can construct whatever you want, but bosonic creation/annhilation operators will not help you represent fermionic degrees of freedom.

Suppose I have a wavefunction |𝜓⟩ with position and momentum operators 𝑄 and 𝑃 such that 𝑄|𝑥⟩=𝑥|𝑥⟩ for a wavefunction representing a particle precisely at position 𝑥, and similarly 𝑃|𝑝⟩=𝑝|𝑝⟩ for a particle with momentum precisely 𝑝. Then I can define creation and annihilation operators 𝑎=12√(𝑄+𝑖𝑃) and 𝑎†=12√(𝑄−𝑖𝑃). These satisfy [𝑎,𝑎†]=1 (up to normalization), so they are ladder operators. Hence, it has a bunch of eigenstates of the form |0⟩,|1⟩,…,|𝑛⟩,…. What do these eigenstates represent?

You can mathematically construct these states, but they don't have any relation to eigenstates of the Hamiltonian and do not carry an obvious physical meaning (at least not that I know of). In particular, none of these states would have spin 1/2.

When I see descriptions of systems like (2), they always seem to use creation/annihilation operators indexed by position or momentum, like 𝑎†𝑝 and 𝑎𝑝. This would imply a set of operators 𝑞𝑝:=12√(𝑎†𝑝+𝑎𝑝) and 𝑝𝑝:=𝑖2√(𝑎†𝑝−𝑎𝑝). What do the position/momentum operators 𝑞𝑝 and 𝑝𝑝 represent?

In quantum field theory, we think of each Fourier mode of the field as a harmonic oscillator. So each Fourier mode as an associated creation and annhilation operator. The "position" and "momentum" of the Fourier mode are essentially the (complex-valued) amplitude of the Fourier mode and time derivative of the Fourier mode.

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  • $\begingroup$ If I define $q_\sigma = \frac{1}{\sqrt{2}}(b_\sigma^\dagger + b_\sigma)$ and $p_\sigma = \frac{1}{\sqrt{2}}(b_\sigma^\dagger - b_\sigma)$, then $[q_\sigma,p_\sigma] = i(\frac{1}{2} - b_\sigma^\dagger b_\sigma)$. Which seems to suggest $q$ and $p$ are not position/momentum operators. What I'm getting from all your answers is that the creation/annihilation and position/momentum correspondence from those notes isn't as general as I hoped. So as a follow-up: any ideas on why there is such a nice correspondence for bosons and only bosons? $\endgroup$
    – Sam Jaques
    Jul 21, 2020 at 14:10
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    $\begingroup$ Caveat: it's possible I am not using standard terminology, I'm a bit more familiar with how this works in field theory. However, the way I would think of it, is that position and momentum are a "bosonic" degrees of freedom -- classically they commute. It's really the internal spin degree of freedom that's the fermionic part. The $q_\sigma$ and $p_\sigma$ operators that are constructed out of the Hamiltonian don't just encode the position and momentum, they also include the internal spin information. The correspondence you are used to is because in basic quantum mechanics... $\endgroup$
    – Andrew
    Jul 21, 2020 at 15:57
  • $\begingroup$ ...you solve the Schrodinger equation for particles without spin, and so the only degree of freedom is position or momentum. If you dealt with a spin-1 boson, the creation and annhilation operators should properly have indices telling you what polarization you are creating, and the corresponding "p" and "q" operators would also contain spin information and would not purely be "position and momentum" of the particle. $\endgroup$
    – Andrew
    Jul 21, 2020 at 15:58
  • $\begingroup$ YMMV on whether the following comment is helpful but: in field theory the harmonic oscillator arises because each Fourier mode of a field obeys a harmonic oscillator equation. Then you have creation and annihilation operators for each mode. The "p" and "q" operators, built out of the creation and annihilation operators, then don't represent position and momentum of anything, they describe the (complex valued) amplitude of the Fourier mode and its time derivative. Then the spin index maybe makes more sense -- each polarization state has an independent Fourier mode for a given frequency. $\endgroup$
    – Andrew
    Jul 21, 2020 at 16:01

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