Harmonic oscillator (quantum)

First four harmonic oscillator functions ψ_n. Potential V(x) is shown as reference. Function values are shifted upward by the corresponding energy values $(n+{\tfrac {1}{2}})\hbar \omega .$

In quantum mechanics, the one-dimensional harmonic oscillator is one of the few systems that can be treated exactly. Its time-independent Schrödinger equation 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 so as 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 out of sight 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 with eigenvalue

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

This encourages us to try 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

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