Consider the following hamiltonian, describing a system of independent bosons: $$ \hat H = \sum_{q \neq 0} \hbar c_q |q|\hat \beta^\dagger_q \hat \beta_q \tag{1}$$ where $\hat \beta_q$ and $\hat \beta^\dagger_q$ are the destruction and creation operators of a bosonic mode, $q$ is the momentum carried by a boson of the mode, and $c_q$ is a sort of "sound speed" depending on $q$.

As a side note, this hamiltonian emerges in the context of the Luttinger Model of the interacting 1D fermionic liquid, where the aforementioned bosons represent the collective modes of the fermionic liquid. I'm studying it on the book Quantum Theory of the Electron Liquid by Giuliani and Vignale, Chapter 9.

What confuse me are the following "momentum" and "position" operators, defined in Exercise 9.4 of Giuliani and Vignale's book: $$ \hat P_q = \frac{\hat \beta^\dagger_{-q} + \hat \beta_q}{\sqrt{2}}, \qquad \hat X_q = \frac{\hat \beta^\dagger_{-q} - \hat \beta_q}{i\sqrt{2}} \tag{2} $$

First of all, why are $\hat P_q$ and $\hat X_q$ defined mixing the $q$ and $-q$ modes? From the theory of the harmonic oscillator, I would have defined them as $\hat P_q = \frac{\hat \beta^\dagger_q + \hat \beta_q}{\sqrt{2}}$ and $\hat X_q = \frac{\hat \beta^\dagger_q - \hat \beta_q}{i\sqrt{2}}$. With definition $(2)$, hamiltonian $(1)$ becomes: $$\hat H = \frac{1}{2} \sum_{q \neq 0} \hbar c_q |q| \left( \hat P_{-q} \hat P_q + \hat X_{-q} \hat X_q \right) \tag{3}$$ while with "my" definition it would have become the much more familiar-looking $\hat H = \frac{1}{2} \sum_{q \neq 0} \hbar c_q |q| \left( \hat P_q^2 + \hat X_q^2 \right)$.

Moreover, after that Giuliani and Vignale consider the ground state wavefunction in the "momentum representation" given by the operators in $(2)$. But they are not even hermitian, since $\hat P^\dagger_q = \hat P_{-q}$ and $\hat X^\dagger_q = \hat X_{-q}$, so how can I treat them as regular momentum and position operators? Can I consider the $P_q$ variables of this "momentum representation" as real variables just like I do in the regular momentum representation? Or rather, since $\hat P_q$ isn't hermitian, they're complex variables?

  • $\begingroup$ Tip: you need to look at how the field is represented in terms of plane waves, and particularly how it behaves under complex/hermitian conjugation. I don't remember the details in this book, but these expressions do not strike me as particularly odd - it often happens for other wave fields as well, e.g., phonons. $\endgroup$
    – Roger V.
    Apr 10, 2020 at 10:05

1 Answer 1


The advantage of the $P_q\propto \beta^\dagger_q-\beta_{-q}$ definition is that $P_q$ has a well defined momentum: It creates an excitation of momentum $+q$ and destroys one of momentum $-q$, in each case changing the momentum of the system by $+q$. This means that $P_q$ and $X_q$ have simple commutators with the total momentum operator (generator of translations) for the system. $$ [{\mathcal P}_{tot}, X_q]= qX_q, \quad etc. $$ If you did not have the $\beta_{-q}$ this would not be true.

  • $\begingroup$ I see... It makes sense, thank you. Could you also help me with the second part of my question? That is, making sense of the "momentum representation" of the wavefunction, even if our momentum isn't an observable since it is not hermitian. $\endgroup$
    – DoeJohn
    Apr 10, 2020 at 13:55
  • $\begingroup$ The actual displacement of the boson at site $n$ corresponding to $X$ would be $X(n) = \sum_q (e^{iqn}X_q+e^{-iqn} X_{-q})$, and this is hermitian. $\endgroup$
    – mike stone
    Apr 10, 2020 at 15:38

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