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.


1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.