The simple harmonic oscillator is a classical problem in classical physics. In this post, we are going to analyze this familiar problem in the context of quantum mechanics. Our approach is typical of quantum mechanics and quite different from what we are used to doing in classical mechanics, and we are going to obtain many surprising but insightful conclusions.

Establishing the Basics

After the previous post, you must be familiar with using Hamiltonian in analyzing problems in QM. Writing out the Hamiltonian and utilizing the Schrodinger equation is a common approach in QM, just like using Newton’s Laws in classical mechanics.

As a first steps, let’s try to write out the Hamiltonian for a harmonic oscillator. There is nothing special in the kinetic energy part, and for the potential energy part, we use a result that’s easily derived using classical mechanics:

Our goal is to somehow use this Hamiltonian to find the energy eigenstates (energy levels) and eigenvalues of the quantum harmonic oscillator. Notice that there are squared quantities in both terms on the RHS. Neglecting the operator symbols for a moment, we can easily see that the above expression can be written as the square of a complex number:

where “a” is the name we give to this new quantity. It is deceptively easy to go back to operators, we just need to add back the operator signs for a, x, and p. However, we need to remember that operators are not necessary commutative. In fact, we know for sure that $\hat x$ and $\hat p$ are not commutative. So we need to careful when translating the above expression to an expression for operators. Let’s see what happens: (remember to switch the star for complex conjugate to a dagger for operators)

Notice that we get back the Hamiltonian for the quantum harmonic oscillator in the parenthesis. This tells us that:

Of course we could have done things the other way around to compute $\hat a \hat a^\dagger$ and get:

This gives us an important commutator relationship:

Raising and Lowering Operator and the Spectrum of States

Now we seem to be stuck at this point. Let’s make some reasonable assumptions and see where they can take us.

First, define the number operator: (you will understand the reason for this name soon)

Given the relationship between this number operator and the Hamiltonian, we know that they must commute, and the energy eigenstates must be eigenstates of the number operator as well. Call an energy eigenstate $|n\rangle$, then

where n is the eigenvalue of that state.

At the same time, define the action of the operator $\hat a^\dagger$ on this state to be:

where $|\psi\rangle$ is an arbitrary state. Now consider:

Apply the commutator relationship we derived above:

This tells us that $|\psi\rangle$ is also an eigenstate of the number operator, and its eigenvalue is greater than the eigenvalue of $|n\rangle$ by 1. This result is very suggestive, it tells us that the function of $\hat a^\dagger$ is to raise the energy eigenstates higher by one level, and the number operator, when acted on the eigenstates, gives us eigenvalues that correspond to the # of energy level of that state, hence the name “number operator”.

Using a similar logic, we can also deduce the function of $\hat a$:

This tells us that $\hat a^\dagger$ and $\hat a$ forms a pair of raising and lowering operator for the energy eigenstates of the QHO hamiltonian.

This all seems well, and we have a spectrum of states labelled from 0 to arbitrarily high numbers. However, there are still a couple of things to clear up here. First, just like with the raising and lowering operator for the spin states, we need to be careful with the “beginning” and the “end” of the spectrum of states. We are dealing with energy here, so there isn’t really an “end”. At the “beginning”, it is logical to define:

Second, we need to normalize all the states. Assuming the state $|n\rangle$ is normalized, we have:

So after normalization, we have:

Using this result and assuming the ground state is normalized, we have:

Energy Eigenvalues and the Zero-point Energy

We’ve been talking about the spectrum of states a lot now, but we only looked at its behavior under $\hat N$. It’s time to look at its eigenvalues under $\hat H$. We can easily know that:

The most astonishing implication of this result is manifest when n=0, which is the lowest energy state of the quantum harmonic oscillator. Classically when we think of the lowest possible energy of a harmonic oscillator, we think of the system being at rest and having 0 energy. However, this is clearly not the case in QM. The ground state energy is $\frac {1}{2} \hbar \omega$ instead of 0! This lowest possible energy is known as the zero-point energy.

An example makes this result even more counter-intuitive. If our harmonic oscillator is a pendulum, QM tells us that it is impossible to place the pendulum at rest. It never stays still at the vertical position and oscillates at an energy of $\frac {1}{2} \hbar \omega$ even at the lowest energy level.

Deriving the Explicit Form of the Energy Eigenstates

We’ve been referring to the energy eigenstates as $|n\rangle$, but what are their algebraic forms?

To derive that, we start from the ground state and utilize the relationship:

Using the form of $\hat a$ we already know, we have:

Where $x_0 = \sqrt{\frac {\hbar}{m\omega}}$.

To transform from ket states to position space wave functions, we multiply all terms by $\langle x|$ to get:

This is a relatively simple differential equation. Solving it gives us:

I did not write out the pre-factor in the above expression because we need to obtain it using the normalization criteria. To do that, you need to be familiar with Gaussian integrals. Appendix D.7 of reference[1] provides detailed information on that. Carrying out all the calculation, we get:

Using the relationship between $|0\rangle$ and $|n\rangle$ we have above, we can easily obtain an expression for the nth energy eigenstate:

Where $H_n(x)$ is the Hermite polynomial.

Calculating the Expectation Value of Position and Momentum

One of the most frequent calculation we do in QM is the calculation of expectation values. In this problem of the harmonic oscillators, we are usually most interested in the expectation value of position and momentum. So how do we calculate them.

An inevitable step is to apply the position and momentum operators on the energy eigenstates. However, we don’t know the effect of these operators on the eigenstates up to this point. We’ve been dealing with the raising and lowering operators all the time. Lucky, we can express the position and momentum operators in terms of the raising and lowering operators. Remember the definitions:

Using these expression, we can solve for $\hat x$ and $\hat p$ as if we are solving a system of two equations with two unknowns. We get:

Using these expressions and the known effects of $\hat a$ and $\hat a^\dagger$ on the energy eigenstates, we can easily obtain quantities like $\langle x\rangle$ and $\langle p\rangle$.

References

[1] Zajc, William. PHYS GU4021-4022 Quantum Mechanics I-II, Department of Physics, Columbia University. 2019-2020.

[2] Townsend, John. A Modern Approach to Quantum Mechanics. 2nd Edition.