Hamiltonian of quantum harmonic oscillator - how it affects the dynamics of a system?

Question

The quantum harmonic oscillator can be described with the creation and annihilation operators of its eigen states:

$$H=\hbar \omega\left(a^+a+\frac{1}{2}\right) \, .$$

Which possesses the following qualities regarding the eigenstate $|n\rangle$ $$ a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle $$ $$ a|n\rangle=\sqrt{n}|n-1\rangle. $$

My question is, can anyone give me some intuition regarding this peculiar Hamiltonian? For example, in QM, according to the time-dependent Schrodinger equation, we know that the Hamiltonian is an operator that describes the dynamics of the system. So from what I get from this particular Hamiltonian is that a harmonic oscillator oscillates between eigenstates (because of the creation and annihilation operators).

Any explanation or intuition is welcome.

Answer 1

It is actually misleading to say that harmonic oscillator is oscillating between eigenstates. Well, it depends on in which eigenstate we are. If we measured the position, the oscillator will be in an position eigenstate, but not in an energy eigenstate. Yes in that case the position eigenstate will be a superposition of energy eigenstates, one could in a simplifying way say, the harmonic oscillator "is oscillating" between energy eigenstates. However, after an energy measurement we are in an energy eigenstate, the harmonic oscillator "is oscillating" in this energy eigenstate and not between energy eigenstates. In order to demonstrate this in the formalism of creation and annihilation operators we just apply the Hamilton operator on a state $|n\rangle$.

$$H |n\rangle = \hbar\omega \left(a^\dagger a|n\rangle + \frac{1}{2}|n\rangle \right) = \hbar\omega \left(a^\dagger |n-1\rangle \sqrt{n} + \frac{1}{2}|n\rangle \right) = \hbar\omega \left(\sqrt{n-1+1}\sqrt{n}|n\rangle +\frac{1}{2}|n\rangle \right) = \hbar\omega\left( n +\frac{1}{2}\right) |n\rangle $$

so the Hamilton operator applied on this state provides the energy value of this state. Or in other words, $|n\rangle $ is an eigenstate of the Hamilton operator as expected. And the harmonic oscillator will oscillate in that state after an energy measurement.

Answer 2

$H=\hbar \omega\left(a^+a+\frac{1}{2}\right)$ comes from the following classical-looking Hamiltonian: $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2 = \frac{1}{2m}[p^2+(m\omega x)^2]$$ by defining $a^{\pm}=\frac{1}{2\hbar m\omega}(\mp ip+m\omega x)$. The fact that $a^\pm$ allows one to oscillate between eigenstates follows from $$H(a^\pm\psi)=\cdots=(E\pm\hbar\omega)(a^\pm\psi)$$ where $\psi$ is an eigenstate with energy $E$. Then $a^\pm\psi$ is still an eigenstate—provided that $\psi$ is not the lowest-energy eigenstate. (see e.g. Griffiths, Section 2.3 for the calculation.)

Answer 3

What you mention about the creation and annihilation operators is true. I'm not sure if you just didn't mention it or if this is the intuition you are looking for: note that if you combine creation and annihilation operator like in the Hamiltonian it becomes $$ a^{\dagger}a|n\rangle = a^{\dagger} \sqrt{n}|n-1\rangle = \sqrt{(n-1)+1}\sqrt{n}|n\rangle = n |n\rangle $$ so the Hamiltonian just counts how many particles are in each eigenstate and gives you the corresponding energy.

Answer 4

This is the so called operator approach of the Harmonic oscillator(HO). In the real space representation, the Hamiltonian has the familiar form $$\hat{H}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2$$. By solving the time-independent Schrodinger equation, we can obtain the eigenstates and eigenenergies $E_n=\hbar \omega(n+\frac{1}{2})$. That is the HO has different states with corresponding energy level.

In the algebraic approach, the Hamiltonian takes the form $$\hat{H}=\hbar \omega\left(\hat{a}^\dagger \hat{a}+\frac{1}{2}\right)$$. The physical interpretation is that now there is only one state for the HO and we can put bosons in this state. For example, the ground state corresponds to the vacuum state(no bosons). The first excited state corresponds to putting one boson in that state and the second excited state corresponds to putting two bosons on that state and so on. So we can use the occupation number representation |n> as the eigenstate. For example, |0> means the ground state with 0 boson. |4> means there are 4 bosons in the state and |n> means there are n bosons. As you can find in any textbook, the term $\hat{a}^\dagger \hat{a}$ is the so-called number operator that counts how many number of bosons in the state |n>. i.e. $$\hat{a}^\dagger \hat{a}|n>=\hat{N}|n>=n|n>$$ Thus, the Hamiltonian now becomes $$\hat{H}=\hbar \omega\left(\hat{N}+\frac{1}{2}\right)$$ and the energy is just the same as before $E_n=\hbar \omega(n+\frac{1}{2})$. Now you can interpret the Hamiltonian as the total energy of bosons with the energy of each boson being $\hbar \omega$ which is just the energy difference between each consecutive eigenstates in the real space representation. The reason we can interpret it in this way is because the energy difference between each two neighbouring eigenstates(in real space representation) is the same(i.e. $\hbar \omega$).

Lastly, you can see that in the above Hamiltonian we only need to replace $\hat{N}$ with its eigenvalue $n$ to obtain the eigenenegies. This is because $\hat{N}$ commutes with the Hamiltonian and they share common eigenstates so $n$ is a good quantum number to describe the system.