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For the simple quantum harmonic oscillator we can solve Schrodinger's equation and derive the analytic form of the eigenstates of e.g. a non relativistic electron in a harmonic potential. We may then go on to define ladder operators which enable us to move between eigenstates of the potential. One interpretation of these ladder operators is that they 'create' or 'destroy' a photon of energy hw. One can go on to define the Number operator which enables us to determine how many times the oscillator has been excited or equivalently how many 'photons' there are in the system.

My question is: is it correct to say the wavefunction we derived for the different energy levels still just describes the single non-relativistic electron and doesn't describe the newly created/destroyed photons.

If this is correct then where does the interpretation of 'creation' of particles even come from because it seems like we are not describing the produced photons with any wavefunction we are merely postulating that they have been created? What is a good intuitive explanation for the interpretation in terms of photon creation?

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    $\begingroup$ The interpretation of the bosons comes from the algebra that satisfy the newly defined creation and annihilation operators. Check for instance:en.wikipedia.org/wiki/Creation_and_annihilation_operators From there, you can study other properties of these new particles and so on. $\endgroup$ Feb 9 at 9:41
  • $\begingroup$ Following this interpretation should I think of the QHO as a single particle state with energy h$\omega$ that is occupied by n bosons in the nth excited state? It's a strange way to describe a system that physically amounts to for example an electron oscillating in a harmonic potential trap. $\endgroup$ Feb 9 at 9:51
  • $\begingroup$ Not quite, let me write a full answer $\endgroup$ Feb 9 at 10:04
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In my opinion, the answer to your question is no. So we can write our problem as $\hat{H}\psi = E\psi$, and we can derive (check David Griffiths, "Introduction to Quantum Mechanics" for example) that $\psi^{n}\left( x\right) = A_n \left( a^{\dagger}\right)^n \psi_{0}\left( x\right)$, where $A_n$ is a normalization constant, and $\psi_{0}\left(x \right)$ is the ground-level wave function.

So the creation operator $a^{\dagger}$ does enter in the wavefunctions. :)

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  • $\begingroup$ I agree the mathematical formalism can be written in terms of creation and annihilation operators but my question is more where this interpretation of 'creating' and 'destroying' particles comes from- it seems a little counter-intuitive to me that we would describe a single particle system that is oscillating in terms of other created bosonic particles. $\endgroup$ Feb 9 at 9:49
  • $\begingroup$ Okay, I see, I am afraid I can only argue that it simply comes out of the math, maybe sb else has a more physical explanation. :) $\endgroup$
    – user248824
    Feb 9 at 9:50
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To me, the boson interpretation comes only clear when one considers a many particle situation.

When we deal with one single harmonic oscillator, one has one creation $a^{\dagger}$ and one annihilation operator $a$. Each time you act with $a^{\dagger}$ you are essential raising the total energy of the oscillator by one quanta of energy. One way to see this is that $H= \epsilon (n+ \frac{1}{2})$, i.e. your total energy only depends on how many times you acted with $a^{\dagger}$ as $H$ essentially counts how many times this happens and multiplies it by some energy gap $\epsilon$. The interpretation is that you have one single particle which can be excited upwards or downwards with the ladder ops. Note I didn't mention bosons yet as it wouldn't make sense as explained below.

Now consider a chain of independent harmonic oscillators. We are dealing with the same story as before as the oscillators are completely unaware of each other. Now we label each oscillator with an index $i$ to keep track of it. We also label their associated creation $a^{\dagger}_i$ and annihilation operators $a_i$. If you play a bit with the algebra, you can show that they satisfy the canonical commutation relationships, $$ [a_i, a^{\dagger}_j] =\delta_{ij} \quad [a_i,a_j]=[a_i^{\dagger},a_j^{\dagger}]=0 $$ which is a defining property for bosonic particles (whether or not they are fundamental is a different story, they can, and are, quasiparticles). The interpretation now is clear, at each site you can create/destroy a boson with its corresponding creation and annihilation operators. The energy of the system is simply the sum of the energies at each site, or the number of excitations/bosons we have at each site, $$ H= \sum_i \epsilon (n_i+ \frac{1}{2}). $$


To finish off, yes, the wave function that solves the Schrödinger equation for a particle in a harmonic potential will always be a valid wave function for such situation. Now, you can alternatively interpret it as bosonic quasiparticles exactly as explained above.

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    $\begingroup$ Thank you for taking the time to answer- this clears up some of my confusion. One minor edit I would make is to change 'at each site you have a boson' to 'at each site you have the possibility of creating one or more bosonic quasiparticles' as we can create multiple or not have any bosons at each site. Am I correct to say that we can interpret the different energy levels of the QHO to be described in terms of 'quasiparticles' that satisfy the bosonic commutation relations? The connection between this picture and an oscillating single non-relativistic particle is still not clear to me. $\endgroup$ Feb 9 at 11:07
  • $\begingroup$ Uh yeah I edited that. That assertion is vague: a boson is a (quasi-)particle that satisfies canonical commutation relationships (CCRs). A single QHO can't satisfy the CCRs as you need other sites (or in QFT position/momentum) to compare with (i.e. $[a_i,a_j]=0$ has no meaning with a single QHO). Thus the connection would be unclear cause it is ill-defined, a single QHO cant satisfy CCRs and thus can't have a notion of bosons, this is why I wrote all of that above. However you would be right provided you have many QHOs. $\endgroup$ Feb 9 at 14:27
  • $\begingroup$ Great that makes more sense now- I am currently studying QFT so wanted to make sure I understood simple QHO particle interpretation but I see now it is best defined when we are looking at a string of QHOs or in the continuum limit a QFT. Thanks again $\endgroup$ Feb 9 at 14:34
  • $\begingroup$ That is good practice! One typically sees this for the first time in precisely that context :). Make sure to select one of the posted answers by ticking it so that the post is closed if you have no further questions. Otherwise let me know if I can help. $\endgroup$ Feb 9 at 14:41

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