How to picture the Quantum Harmonic Oscillator with particle creation interpretation?

Question

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?

Answer 1

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. :)

Answer 2

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, $$ H= \sum_i \epsilon (n_i+ \frac{1}{2}). $$

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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.