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From what I understood from wikipedia, as well as some other resources, each phonon corresponds to a normal mode oscillation, and the creation operator to create a phonon of wavevector $k$ is: $$ a^{\dagger}_k $$

However, from this post, a [sic] general single-phonon is described by a superposition of normal modes, and so the operator appears as: $$ \sum_{k} f(k)\,a^{\dagger}_{k} $$

My question is which operator describes a phonon, and if it is the latter, what does it mean to be a "general phonon" and it would also be wonderful if some resources that would help me understand this whole link between phonon and normal modes be included as well.

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This is actually a general principle not specific to phonons.

Given a collection of creation operators $a^\dagger(k)$ creating 1-particle momentum eigenstates of momentum $k$, these states are not actually viable states - they do not lie inside the Hilbert space, since their "inner product" with each other is not a finite number, but usually a delta function: $\langle k' \vert k \rangle = \delta(k - k')$.

An actual particle state is now, generally, given by "smearing out" the creation operators with some sufficently smooth "profile" $f(k)$, e.g. a Gaussian, so a general 1-particle state is created by $$ \int f(k)a^\dagger(k)\mathrm{d}k$$

Nevertheless, one often uses the states created by $a^\dagger(k)$ alone because they are easier to work with than the smeared out states. Since they occur for the (formal) choice $f(k) = \delta(k-k')$, you might imagine them as the limit of a series of smearing functions that are increasingly more sharply peaked around $k'$ - choose any of the usual approximations of a $\delta$-function by smooth function, again, for example, Gaussians.

Since both $a^\dagger$ and the smeared out version create 1-particle states, it's not possible to say that one describes a particle but the other not - they both do, but in different states.

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  • $\begingroup$ How do I interpret the profile $f(k)$? Can I say it is associated with the probability that the particle is in state $k$? Can I also say that both describe single particle states because the choice of initial basis is arbitrary (if we ignore smoothing)? $\endgroup$
    – NaOH
    Jul 30, 2015 at 15:03
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    $\begingroup$ @NaOH: Yes! If it is normalized ($\int f(k) = 1$), then it is precisely the probability density for that. $\endgroup$
    – ACuriousMind
    Jul 30, 2015 at 15:07

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