I have studied the annihilation and creation operators and number operator $N$ in relation with the simple harmonic oscillator that is governed by: $\ H = \hbar\omega(N+ \frac{1}{2})$.
I don't understand the relation between the harmonic oscillator and, for example, this Hamiltonian $\ H = \hbar\omega_0a^{\dagger}a+\hbar\omega_1a^{\dagger}a^{\dagger}aa $ that I have found in an example in a lecture notes. They calculate the energies of this system.
They use the annihilation operator that is defined from the simple harmonic oscillator to solve that system. What is physically this system? Why can I use the SHO to calculate the energies? I feel that I am confused with the a operator. I thought that it was defined from the Hamiltonian of the simple harmonic oscillator, isn't it?
Consider one dimensional motion of a particle having mass $ m $. The Hamiltonian is given by$$\mathcal{H}=\hbar \omega_{0} a^{\dagger} a+\hbar \omega_{1} a^{\dagger} a^{\dagger} a a\tag{5.88}$$where$$a=\sqrt{\frac{m \omega_{0}}{2 \hbar}}\left(x+\frac{i p}{m \omega_{0}}\right)\tag{5.89}$$is the annihilation operator, $ x $ is the coordinate and $ p $ is its canonical conjugate momentum. The frequencies $ \omega_{0} $ and $ \omega_{1} $ are both positive.
a) Calculate the eigenenergies of the system.
b) Let $ |0\rangle $ be the ground state of the system. Calculate
i. $ \langle 0|x| 0\rangle $
ii. $ \langle 0|p| 0\rangle $
iii. $ \left\langle 0\left|(\Delta x)^{2}\right| 0\right\rangle $iv. $ \left\langle 0\left|(\Delta p)^{2}\right| 0\right\rangle $
Using the commutation relation$$\left[a, a^{\dagger}\right]=1\tag{5.300}$$one finds$$\mathcal{H}=\hbar \omega_{0} N+\hbar \omega_{1}\left(N^{2}-N\right)\tag{5.301}$$where$$N=a^{\dagger} a\tag{5.302}$$is the number operator.
a) The eigenvectors of $ N $$$N|n\rangle=n|n\rangle\tag{5.303}$$(where $ n=0,1, \cdots) $ are also eigenvectors of $ \mathcal{H} $ and the following holds$$\mathcal{H}|n\rangle=E_{n}|n\rangle\tag{5.304}$$where$$E_{n}=\hbar\left[\omega_{0} n+\omega_{1}\left(n^{2}-n\right)\right]\tag{5.305}$$
Best Answer
Let's quickly review the quantum harmonic oscillator. We have a single particle moving in one dimension, so the Hilbert space is $L^2(\mathbb{R})$: the set of square-integrable complex functions on $\mathbb{R}$. The harmonic oscillator Hamiltonian is given by
$$H= \frac{P^2}{2m} + \frac{m\omega^2}{2}X^2$$
where $X$ and $P$ are the usual position and momentum operators: acting on a wavefunction $\psi(x)$ they are $X \psi(x) = x\psi(x)$ and $P \psi(x) = -i\hbar\ \partial \psi / \partial x$. Of course, we can also think of them as acting on an abstract vector $|\psi\rangle$.
By letting $P \to -i\hbar\ \partial/\partial x$ we could solve the time independent Schrödinger equation $H \psi = E \psi$, but this is a bit of a drag. So instead we define operators $a$ and $a^\dagger$ as in your post. Notice that the definition of $a$ and $a^\dagger$ has nothing whatsoever to do with our Hamiltonian. It just so happen that these definitions are convenient because the Hamiltonian turns out to be $\hbar \omega (a^\dagger a + 1/2)$.
For convenience we define the number operator $N = a^\dagger a$; at this stage number is just a name with no physical interpretation. Using the commutation relation $[a,a^\dagger] = 1$ and some algebra we notice that $N$ has a nondegenerate spectrum given by the natural numbers. In other words, the eigenvalues of $N$ are $\{0,1,2,\dots\}$, and to each eigenvalue $n$ there corresponds a single state $|n\rangle$ with $N|n\rangle = n |n\rangle$. Notice that, again, $N$ is independent of our Hamiltonian. However, because the Hamiltonian turns out to be $\hbar \omega (N+1/2)$ we immediately know that the states $|n\rangle$ are its eigenvectors, with energies $\hbar \omega (n + 1/2)$.
Now you are given a different Hamiltonian. The Hilbert space is still exactly the same, and so are $a$, $a^\dagger$ and $N$, because their definition had nothing to do with the original Hamiltonian. You can still use their properties to find energies, eigenvectors, and so on. The states $|n\rangle$ are still the eigenstates of $N$, though a priori they might not be eigenstates of the new $H$ (exercise 31 asks you to prove that they in fact are eigenstates of the new $H$). The important point here is that operators are (usually) defined independently of the Hamiltonian. They characterize the physical system. After all, you know that there are operators $X$ and $P$, and you have no qualms about using them with different Hamiltonians. The Hamiltonian gives the energy and the time evolution, but the observables and related operators are independent of your choice of Hamiltonian.
About the physical interpretation... exercise 31 asks you to prove that $H=\hbar\omega_0 N + \hbar \omega_1 (N^2-N)$; notice that we have gotten rid of $\hbar\omega_0 /2$ since it is just a constant. I would usually expect $\omega_1$ to be smaller than $\omega_0$ so this is a small perturbation (for small $n$ at least), but we don't really care about that right now. You can see that $|n\rangle$ are still the eigenstates of the Hamiltonian; all we did is shift the energies by an amount $\hbar \omega_1 (N^2-N)$.