# Harmonic Oscillator Spectrum and Eigenstates

The one-dimensional harmonic oscillator consists of a particle moving under the influence of a harmonic oscillator potential, which has the form, $V(x)=\frac{1}{2}k x^2,$ where $k\!$ is the "spring constant".

We see that $V(x)\rightarrow \infty$ as $x\rightarrow \pm\infty.$ Therefore, all stationary states of this system are bound, and thus the energy spectrum is discrete and non-degenerate. Furthermore, because the potential is an even function, the parity operator commutes with Hamiltonian, and thus the wave functions will be either even or odd.

The energy spectrum and the energy eigenstates can be found by either an algebraic method using raising and lowering operators, which is described below, or by solving the Schrödinger equation for the system, as described in the next section.

## Solution of the Harmonic Oscillator by Operator Methods

The Hamiltonian of the one-dimensional harmonic oscillator is:

$H=\frac{\hat{p}^2}{2m}+\tfrac{1}{2}k\hat{x}^2,$

or, in terms of the natural frequency, $\omega=\sqrt{\frac{k}{m}},$

$H=\frac{\hat{p}^2}{2m}+\tfrac{1}{2}m\omega^2\hat{x}^2.$

With the aid of the operator identity,

$\hat{A}^2+\hat{B}^2=(\hat{A}-i\hat{B})(\hat{A}+i\hat{B})-i[\hat{A},\hat{B}],\!$

we may factorize the Hamiltonian as follows.

$H=\hbar\omega\left (\frac{m\omega}{2\hbar}\hat{x}^2+\frac{\hat{p}^2}{2m\hbar\omega}\right )=\hbar\omega\left (\sqrt{\frac{m\omega}{2\hbar}}\hat{x}-i\frac{\hat{p}}{\sqrt{2m\hbar\omega}}\right )\left (\sqrt{\frac{m\omega}{2\hbar}}\hat{x}+i\frac{\hat{p}}{\sqrt{2m\hbar\omega}}\right )+\tfrac{1}{2}\hbar\omega$

If we now define the operators,

$\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}+i\frac{\hat{p}}{\sqrt{2m\hbar\omega}}$

and

$\hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}\hat{x}-i\frac{\hat{p}}{\sqrt{2m\hbar\omega}},$

we may write the Hamiltonian as

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

One may easily show that the operators $\hat{a}\!$ and $\hat{a}^{\dagger}$ satisfy the commutation relation, $[\hat{a},\hat{a}^{\dagger}]=1.$

Let us now define the Hermitian operator, $\hat{n}=\hat{a}^{\dagger}\hat{a}.$ We denote the (normalized) eigenstate of $\hat{n}$ associated with the eigenvalue $n\!$ as $|n\rangle;$ i.e., $\hat{n}|n\rangle=n|n\rangle.$ Note that any eigenstate of $\hat{n}$ is also an eigenstate of the Hamiltonian, with eigenvalue

$E_n=\left (n+\tfrac{1}{2}\right )\hbar\omega.$

One may verify that the eigenvalue $n\geq 0\!$ by acting on the left of this definition with $\langle n|.$ There is therefore a lower bound of $\tfrac{1}{2}\hbar\omega$ on the energy of any state of the harmonic oscillator. This "zero-point energy" is a remarkable and significant feature peculiar to quantum mechanics. One may view this as a consequence of the Heisenberg uncertainty principle; because it is impossible to perfectly localize a particle in both position and momentum spaces, a particle in a harmonic oscillator potential will always possess a non-zero energy relative to the minimum of the potential.

We may now determine what the operators $\hat{a}$ and $\hat{a}^{\dagger}$ do to the eigenstates of $\hat{n}.$ Acting to the left on the definition, $\hat{n}|n\rangle=n|n\rangle,$ with each of these operators and employing the above commutation relation, we may show that

$\hat{a}|n\rangle=\sqrt{n}|n-1\rangle$

and

$\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle.$

We have thus shown that $\hat{a}$ is a "lowering operator", in the sense that, when applied to the eigenstate of $\hat{n}$ with eigenvalue $n,\!$ we obtain a result that is proportional to the eigenstate with eigenvalue $n-1.\!$ For a similar reason, $\hat{a}^{\dagger}$ is a "raising operator".

We may use the above results to further restrict the possible values of $n.\!$ We may show that it is quantized, and can only take non-negative integer values; i.e., $n=0,\,1,\,2,\,\ldots$ Let us suppose that an eigenstate of $\hat{n}$ with a positive non-integer eigenvalue $n\!$ exists. Without loss of generality, let us suppose that $0 since one can generate such a state from any other such eigenstate by repeated application of the lowering operator. If we act on this state with the lowering operator $\hat{a}$, then we will generate an eigenstate with a negative eigenvalue (note that the factor, $\sqrt{n},$ does not vanish in this case!), which cannot exist, as pointed out earlier.

The only way to guarantee that no states with negative values of $n\!$ are generated is to restrict $n\!$ to be a non-negative integer. This is guaranteed because, by repeated application of the lowering operator, we will eventually obtain the state, $|0\rangle,$ and $\hat{a}|0\rangle=0.$ Therefore, we have shown that $|0\rangle$ is the ground state of the harmonic oscillator.

So, starting from any energy eigenstate, we can construct all other energy eigenstates by applying $\hat{a}\!$ or $\hat{a}^{\dagger}\!$ repeatedly. In particular, by repeated application of the raising operator, we may generate all of the eigenstates of the harmonic oscillator from its ground state:

$|n\rangle=\frac{(\hat{a}^{\dagger})^n}{\sqrt{n!}}|0\rangle$

## The Ground State Wave Function

We may use the above results to easily determine the ground state of the harmonic oscillator in position space. Starting from the fact that $\hat{a}|0\rangle=0,\!$ we may write, remembering that $\hat{p}\rightarrow-i\hbar\frac{d}{dx}$ in the position basis,

$\left( \sqrt{\frac{m\omega}{2\hbar}}x+\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right )\psi_0(x)=0\!$

This is a first-order ordinary differential equation, which can easily be solved; the solution is

$\psi_0(x)=A e^{-m\omega x^2/2\hbar},$

where $A\!$ is a normalization constant. Upon normalization, we find that the ground state wave function is

$\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^2/2\hbar}.$

We may obtain the ground state wave function in momentum space as well. Remembering that, in this case, $\hat{x}\rightarrow i\hbar\frac{d}{dp},$ the differential equation satisfied by the momentum-space wave function is

$i\left (\sqrt{\frac{m\hbar\omega}{2}}\frac{d}{dp}+\frac{p}{\sqrt{2m\hbar\omega}}\right )\phi_0(p)=0,$

and its normalized solution is

$\phi_0(p)=\frac{1}{(\pi m\hbar\omega)^{1/4}}e^{-p^2/2m\hbar\omega}.$

## The Excited State Wave Functions

Illustration of the harmonic oscillator potential along with its eigenstates and its energy spectrum.

Given the ground state wave function, we may obtain the excited state wave functions by repeated application of $\hat{a}^{\dagger},$ as described earlier. In position space,

$\psi_n(x)=\frac{1}{\sqrt{n!}}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}\left( \sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right )^ne^{-m\omega x^2/2\hbar}.$

We may show that

$\psi_n(x)=\frac{1}{\sqrt{2^nn!}}\left (\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-m\omega x^2/2\hbar}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right ),$

where $H_n(\xi)=(-1)^n e^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2}$ is a Hermite polynomial.

In the momentum representation, the excited states may be written as

$\phi_n(p)=\frac{i^n}{\sqrt{n!}}\frac{1}{(\pi m\hbar\omega)^{1/4}}\left (\sqrt{\frac{m\omega \hbar}{2}}\frac{d}{dp}-\frac{p}{\sqrt{2m \omega\hbar}}\right )^ne^{-p^2/2m\hbar\omega},$

or

$\phi_n(p)=\frac{(-i)^n}{\sqrt{2^n n!}}\frac{1}{(\pi m\hbar\omega)^{1/4}}H_n\left (\frac{p}{\sqrt{m\omega\hbar }}\right )e^{-p^2/2m\hbar\omega}.$

Note the appearance of the imaginary unit $i\!$ that is not present in the position representation of these states. This solution can also be obtained via a Fourier transformation of the position representation wave functions.

Notice that there are two factors in the wave functions for the excited states, a Gaussian function and a Hermite polynomial. The former causes the wave function to decrease exponentially as $x\rightarrow\pm\infty,$ as required of any bound state, while the later accounts for the behavior of the wave function at short distances and the number of nodes of the wave function.

The quantum harmonic oscillator is of particular interest as a problem due to the fact that it can be used to (at least approximately) describe many different systems. A few examples include molecular vibrations, quantum LC circuits, and phonons in solids.

## Problems

(1) Calculate the expectation value of the position $\hat{x}\!$ in an eigenstate of the harmonic oscillator.

(2) Calculate the expectation value of the momentum $\hat{p}\!$ in an eigenstate of the harmonic oscillator.

(3) Show that the average kinetic energy, $\langle\hat{T}\rangle,$ is equal to the average potential energy, $\langle\hat{V}\rangle.$ This is a special case of the virial theorem, which we will discuss in a later section.

(See Liboff, Richard Introductory Quantum Mechanics, 4th Edition, Problem 7.10 for reference.)