Harmonic oscillator (quantum)

In quantum mechanics, the one-dimensional harmonic oscillator is one of the few systems that can be treated exactly. Classically, the prototype of a harmonic oscillator is a mass m vibrating back and forth around an equilibrium position. A large number of systems are governed (at least approximately) by the harmonic oscillator equation. Whenever one studies the behavior of a physical system in the neighborhood of a stable equilibrium position, one arrives at equations which, in the limit of small oscillations, are those of a harmonic oscillator. Two well-known examples are the vibrations of the atoms in a diatomic molecule about their equilibrium position and the oscillations of atoms or ions of a crystalline lattice. Also the energy of electromagnetic waves in a cavity can be looked upon as the energy of a large set of harmonic oscillators.

The four lowest harmonic oscillator eigenfunctions (also known as oscillator wave functions) are shown in the figure. They are shifted upward such that their energy eigenvalues coincide with the asympotic levels, with the zero levels of the wave functions at x = ±∞. As an example the energy of the n = 2 level (black) is plotted.

According to quantum mechanics, the wave function squared of a point mass, |ψ(x)|², is the probability of finding the mass in the point x. In the figure we see that the probability of finding the oscillating mass at points to the left or to the right of V(x) is non-zero and we notice that for these points the potential energy V(x) is larger than the energy eigenvalue. As an example two small vertical lines are plotted: to the left of the left line the potential energy is larger than $E_{2}={\tfrac {5}{2}}h\nu \equiv {\tfrac {5}{2}}\hbar \omega$, the energy of the n = 2 level. The same holds for the region to the right of the small black line on the right.

In classical mechanics the potential energy V can never surpass the total energy E, because V = E − T and the classical kinetic energy T is non-negative, so that V ≤ E. This is why the region, where the energy eigenvalue of an eigenfunction is smaller than the potential energy, is called a classically forbidden region. The fact that the probability of finding a mass in a classically forbidden is not zero, is often expressed by stating that the point mass can tunnel into the potential wall. This tunnel effect is one of the more intriguing aspects of quantum mechanics.

First four harmonic oscillator functions ψ_n. Function values are shifted upward by the corresponding energy values $(n+{\tfrac {1}{2}})h\nu .$ The quadratic potential V(x) is shown as reference. Small vertical (black) lines indicate boundaries of the classical forbidden regions of the n=2 level.

Schrödinger equation

The time-independent Schrödinger equation of the harmonic oscillator has the form

$\left[-{\frac {\hbar ^{2}}{2m}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}+{\frac {1}{2}}kx^{2}\right]\psi =E\psi$

The two terms between square brackets are the Hamiltonian (energy operator) of the system: the first term is the kinetic energy operator and the second the potential energy operator. The quantity $\hbar$ is Planck's reduced constant, m is the mass of the oscillator, and k is Hooke's spring constant. See the classical harmonic oscillator for further explanation of m and k.

The solutions of the Schrödinger equation are characterized by a vibration quantum number n = 0,1,2, .. and are of the form

$\psi _{n}(x)=\left({\frac {\beta ^{2}}{\pi }}\right)^{1/4}\;{\frac {1}{\sqrt {2^{n}\,n!}}}\;e^{-(\beta x)^{2}/2}\;H_{n}(\beta x)\quad {\hbox{with}}\quad E_{n}=(n+{\tfrac {1}{2}})\hbar \omega .$

Here

$\beta \equiv {\sqrt {\frac {m\omega }{\hbar }}}\quad {\hbox{and}}\quad \omega \equiv {\sqrt {\frac {k}{m}}}$

The functions H_n(x) are Hermite polynomials; the first few are:

$H_{0}(x)=1,\quad H_{1}(x)=2x,\quad H_{2}(x)=4x^{2}-2,\quad H_{3}(x)=8x^{3}-12x,\quad H_{4}(x)=16x^{4}-48x^{2}+12.$

The graphs of the first four eigenfunctions are shown in the figure. Note that the functions of even n are even, that is, $f_{2n}(-x)=f_{2n}(x)\,$, while those of odd n are antisymmetric $f_{2n+1}(-x)=-f_{2n+1}(x)\,.$

Solution of the Schrödinger equation

We rewrite the Schrödinger equation in order to hide the physical constants m, k, and h. As a first step we define the angular frequency ω ≡ √k/m, the same formula as for the classical harmonic oscillator:

${\frac {1}{2}}\left[-{\frac {\hbar ^{2}}{m}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}+m\omega ^{2}x^{2}\right]\psi =E\psi .$

Secondly, we divide through by $\hbar \omega$, a factor that has dimension energy,

${\frac {1}{2}}\left[-{\frac {\hbar }{m\omega }}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}+{\frac {m\omega }{\hbar }}x^{2}\right]\psi ={\frac {E}{\hbar \omega }}\psi .$

Write

$\beta \equiv {\sqrt {\frac {m\omega }{\hbar }}},\quad y\equiv \beta x,\quad {\mathcal {E}}\equiv {\frac {E}{\hbar \omega }}$

and the Schrödinger equation becomes

${\frac {1}{2}}\left[-{\frac {1}{\beta ^{2}}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}+\beta ^{2}x^{2}\right]\psi ={\mathcal {E}}\psi \quad \Longrightarrow \quad {\frac {1}{2}}\left[-{\frac {\mathrm {d} ^{2}}{\mathrm {d} y^{2}}}+y^{2}\right]\psi ={\mathcal {E}}\psi .$

Note that all constants are hidden in the equation on the right.

A particular solution of this equation is

$\psi _{0}(y)=e^{-y^{2}/2}.$

We verify this

$-{\frac {\mathrm {d} e^{-y^{2}/2}}{\mathrm {d} y}}=ye^{-y^{2}/2}\quad {\hbox{and}}\quad {\frac {\mathrm {d} (ye^{-y^{2}/2})}{\mathrm {d} y}}=e^{-y^{2}/2}\left(1-y^{2}\right),$

so that

${\frac {1}{2}}\left[-{\frac {\mathrm {d} ^{2}}{\mathrm {d} y^{2}}}+y^{2}\right]e^{-y^{2}/2}={\frac {1}{2}}\left[e^{-y^{2}/2}\left(1-y^{2}\right)+y^{2}e^{-y^{2}/2}\right]={\frac {1}{2}}e^{-y^{2}/2}.$

We conclude that ψ₀(y) ≡ exp(-y²/2) is an eigenfunction

${\frac {1}{2}}\left[-{\frac {\mathrm {d} ^{2}}{\mathrm {d} y^{2}}}+y^{2}\right]\psi _{0}={\frac {1}{2}}\psi _{0},$

with eigenvalue

${\mathcal {E}}_{0}={\frac {1}{2}}\quad \Longrightarrow \quad E_{0}={\tfrac {1}{2}}\hbar \omega .$

This encourages us to try to find a total solution ψ(y) by a function of the form

$\psi (y)=e^{-y^{2}/2}f(y)=\psi _{0}(y)f(y).$

Use the Leibniz formula

${\frac {d^{2}(\psi _{0}f)}{dy^{2}}}={\frac {d^{2}\psi _{0}}{dy^{2}}}f+2{\frac {d\psi _{0}}{dy}}{\frac {df}{dy}}+{\frac {d^{2}f}{dy^{2}}}\psi _{0}$

and

${\frac {1}{2}}\left[-{\frac {d^{2}\psi _{0}}{dy^{2}}}+y^{2}\psi _{0}\right]f={\frac {1}{2}}\psi _{0}f\quad {\hbox{and}}\quad {\frac {d\psi _{0}}{dy}}=-y\psi _{0}$

then we see

${\frac {1}{2}}\left[-{\frac {d^{2}(\psi _{0}f)}{dy^{2}}}+y^{2}\psi _{0}f\right]={\frac {1}{2}}\psi _{0}f+y\psi _{0}{\frac {df}{dy}}-{\frac {1}{2}}{\frac {d^{2}f}{dy^{2}}}\psi _{0}={\mathcal {E}}\psi _{0}f.$

Divide through by −ψ₀(y)/2 and we find the equation for f

${\frac {d^{2}f}{dy^{2}}}-2y{\frac {df}{dy}}-f=-2{\mathcal {E}}f.$

The differential equation of Hermite with polynomial solutions H_n(y) is

${\frac {d^{2}H_{n}}{dy^{2}}}-2y\,{\frac {dH_{n}}{dy}}+2nH_{n}=0.$

Due to the appearance of the integer coefficient 2n in the last term the solutions are polynomials. We see that if we put

$2{\mathcal {E}}-1=2n\quad \Longrightarrow \quad {\mathcal {E}}=n+{\tfrac {1}{2}}\quad \Longrightarrow \quad E=\hbar \omega (n+{\tfrac {1}{2}})$

that we have solved the Schrödinger equation for the harmonic oscillator. The unnormalized solutions are

$\psi (y)=e^{-y^{2}/2}\;H_{n}(y),\quad n=0,1,2,\ldots$