When solving the harmonic oscillator we introduce the creation/annihilation operator (ignoring constants) $$ a = \frac{1}{\sqrt{2}} \left(q + i p\right), \, a^\dagger = \frac{1}{\sqrt{2}} \left(q + i p\right) \implies [a, a^\dagger] = 1. $$ This also gives the relationship $$ q = \frac{1}{\sqrt{2}} \left(a + a^\dagger\right), \, p = \frac{-i}{\sqrt{2}} \left( a - a^\dagger \right). $$ In many body QM, we also use creation operators. Here, we choose a set of quantum numbers $\lambda$, and use creation/annihilation operators $a_\lambda, \, {a_\lambda}^\dagger$ to create/destroy a particle in the state $\lambda$. Does the relationship to the position/momentum operator still hold? Defining $$ q_\lambda = \frac{1}{\sqrt{2}} \left(a_\lambda + a_\lambda^\dagger\right), \, p_\lambda = \frac{-i}{\sqrt{2}} \left( a_\lambda - a_\lambda^\dagger \right), $$ it will obviously still give the canonical commutation relation $[q_\lambda, p_\lambda] = i$, but what does the operator mean physically? Is it he position operator for the particle created by ${a_\lambda}^\dagger$? And, if we choose to change which quantum numbers we use to describe our system, $\lambda \rightarrow \nu$, does that mean that we have to change the position operator we use, $q_\lambda \rightarrow q_\nu$?
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$\begingroup$ No, the relationship to the position/momentum operator does not still hold: you may define such operators $q_\lambda,p_\lambda$, but they act in an abstract, notional space, and represent specific Fourier modes of quantum fields. Worse yet, x of these fields has nothing to do with these q s. $\endgroup$– Cosmas ZachosCommented Feb 13, 2021 at 21:37
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$\begingroup$ When you have a man body system, for example a lattice with electrons, one uses the operators $x_i$ for the position of particle $i$. These are then not related to $a_i$? Do the meaning of the position operator change if you describe the system with a different set of quantum numbers? $\endgroup$– Martin JohnsrudCommented Feb 13, 2021 at 21:45
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1$\begingroup$ Well, the displacements of the $x_i$ s are oscillator variables hopelessly coupled to their neighbors. Once you disentangle them by Fourier transforming $i\to k$ each such oscillator is related to the $a_k$s. Your $q_k$s then have little to do with the location variables i. See here. $\endgroup$– Cosmas ZachosCommented Feb 13, 2021 at 22:00
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The only situation where $\hat{q}$ and $\hat{p}$ can be interpreted to represent the physical position and momentum is the physical harmonic oscillator. In other physical scenarios, one can still define such quadrature operators in terms of the ladder operators, but they are only analogues to the position and momentum operator of the harmonic oscillator. Quadrature operators are useful in all scenarios, because they lead to an abstract phase space representation of the states.
answered Feb 14, 2021 at 3:37