# Relation between the bosonic harmonic oscillator and the 'ordinary' harmonic oscillator

What exactly is the connection between the Harmonic oscillator that we study in Quantum Mechanics I, and the bosonic (and fermionic) harmonic oscillator ?

In some sources, it is claimed that my 'ordinary' harmonic oscillator is the same as the bosonic one.

In the normal harmonic oscillator :

$$E_n=\left(n+\frac12\right)\hbar\omega$$

In this case, $$n$$ represents a particular energy level. So, the harmonic oscillator has energy levels of energies $$\frac{1}{2}\hbar\omega,\frac{3}{2}\hbar\omega,\frac{5}{2}\hbar\omega$$.. and so on. This is similar to the energy levels of the hydrogen atom for example.

These levels can be filled with particles. For example, in case of bosons, we can fill up each state with as many bosons as we want. In case of fermions, each of these energy levels can have $$(2s+1)$$ fermions, where $$s$$ is the spin.

In case of the bosonic oscillator, we have the same expression,

$$E_n=\left(n+\frac12\right)\hbar\omega$$

However, now, $$n$$ no longer represents the energy levels, and is interpreted the number of bosons. Then the total energy of the system of $$n$$ bosons must be $$E_n$$.

Why does all the bosons have the same energy $$\hbar\omega$$ in this representation. This new oscillator doesn't have energy levels anymore ?

In the previous case, $$|n\rangle$$ would represent a single particle in the $$n$$th energy level of a quantum oscillator. For representing multiple particles, we would use symmetric and antisymmetric wavefunctions. In the second representation, $$|n\rangle$$ seems to represent an oscillator with $$n$$ bosons occupying the same energy level $$\hbar\omega$$.

So what exactly is the similarity and the difference between the two representations ?

Moreover, suppose, I have an ordinary harmonic oscillator, with $$N$$ non-interacting bosons in the ground state. This can be broken into $$N$$ individual and independent harmonic oscillators. The total energy would be $$E=N(0+\frac{1}{2})\hbar\omega=\frac{1}{2}N\hbar\omega$$.

However, according to the second representation, the total energy of $$N$$ bosons would be equal to $$(N+\frac{1}{2})\hbar\omega$$.

These seem to be clearly different. However, I'm unable to wrap my head around the two.

Are they called similar just because of how similar the expressions look ? Or is there some deeper meaning here ?

What is the potential in case of the bosonic harmonic oscillator ? In the ordinary oscillator, there is a quadratic potential, and different energy levels, that a particle can occupy. The bosonic interpretation seems to be quite different.

It's the same thing. Think of a vibrating string with modes of frequency $$\omega_n$$. Each of those modes is a harmonic oscillator, so when we quantize the string each mode can have energy $$(N+1/2)\hbar \omega_n$$. We then say that the string has $$N$$ phonons in mode $$n$$. The same for photons: If we have a cavity with standing wave modes of frequency $$\omega_n$$, we may have energy $$E= \hbar \omega (N+1/2)$$. We then then have $$N$$ photons in mode $$n$$. Any linear system of bosonic fields works the same way. If we have an infinite translationaly invariant system with dispersion equation $$\omega(k)= \sqrt{c^2 k^2+M^2c^4}$$ then we can assign momentum $$p=\hbar k$$ to the mode and get mass $$M$$ bosons with $$E^2=c^2 p^2+ M^2c^4$$.
• As you have mentioned, the bosonic oscillator represents a bosonic field, which can be treated as a harmonic oscillator. Each natural mode, can be occupied by any $N$ number of bosons, and the total energy is given by the formula in the answer. In this case, we are treating each mode as a separate oscillator. However, in the first case, we had an oscillator which had several energy levels(modes) that the particle could be in. Jun 13 at 5:20
• So, intuitively these seem different from me, even if they are mathematically identical. For example, a single boson would have energy $\frac{3}{2}\hbar\omega_n$ according to the bosonic oscillator. However, according to the ordinary oscillator, it would have energy $(n+\frac{1}{2})\hbar\omega$ depending on which energy level it occupies in the ordinary SHO. Jun 13 at 5:23