The Schrödinger Equation
The Quantum Harmonic Oscillator
One of the most important systems in all of quantum mechanics is the quantum harmonic oscillator. Beyond modelling an actual oscillating particle attached to a spring, it appears everywhere in physics: molecular vibrations, quantized electromagnetic fields, phonons in solids, and even quantum field theory. The remarkable feature is that the entire system can be solved exactly.
We begin with the quadratic potential
$$ V(x)=\frac{1}{2}m\omega^2x^2 $$Substituting this into the time-independent Schrödinger equation gives
$$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2x^2\psi = E\psi $$It is useful to first nondimensionalize the equation so that the essential structure becomes clearer. Since \( [\omega]= s^{-1} \), we have
$$ [m\omega^2x^2] = \mathrm{kg\,m^2\,s^{-2}} = \mathrm{J} $$and therefore the dimensions match those of energy. Also,
$$ [E] = \left[ \frac{\hbar^2}{ma^2} \right] $$where \( a \) is a length scale. Equating the characteristic energy scales,
$$ \frac{\hbar^2}{ma^2} \sim m\omega^2a^2 $$gives
$$ a^4 = \frac{\hbar^2}{m^2\omega^2} $$and therefore
$$ a^2 = \frac{\hbar}{m\omega} $$Multiply the eigenvalue equation by \( \displaystyle \frac{2}{\hbar\omega} \)
$$ -\frac{\hbar}{m\omega}\frac{d^2\psi}{dx^2} + \frac{m\omega}{\hbar}x^2\psi = \frac{2E}{\hbar\omega}\psi $$Define the dimensionless energy parameter
$$ \varepsilon = \frac{2E}{\hbar\omega} $$and introduce the dimensionless variable
$$ u = \frac{x}{a} $$Then
$$ \frac{d}{dx} = \frac{1}{a}\frac{d}{du} $$ $$ \frac{d^2}{dx^2} = \frac{1}{a^2}\frac{d^2}{du^2} $$Substituting into the Schrödinger equation gives
$$ -\frac{d^2\psi}{du^2} + u^2\psi = \varepsilon\psi $$To understand the asymptotic behaviour, consider the limit \( u\to\infty \). In this regime, the \( u^2\psi \) term dominates and we obtain approximately
$$ \psi'' = u^2\psi $$whose solutions behave like
$$ \psi(u) = c_1e^{-u^2/2} + c_2e^{u^2/2} $$The second term is not normalizable since it diverges exponentially as \( |u|\to\infty \). Therefore physically acceptable solutions must behave asymptotically like a Gaussian decay.
Since the integral of any polynomial multiplied by a Gaussian converges, we write
$$ \psi(u) = h(u)e^{-u^2/2} $$Differentiating,
$$ \frac{d\psi}{du} = \left( \frac{dh}{du} - uh \right)e^{-u^2/2} $$ $$ \frac{d^2\psi}{du^2} = \left( \frac{d^2h}{du^2} - 2u\frac{dh}{du} + (u^2-1)h \right)e^{-u^2/2} $$Substituting back into the differential equation yields
$$ \frac{d^2h}{du^2} - 2u\frac{dh}{du} + (\varepsilon-1)h = 0 $$We now seek power-series solutions of the form
$$ h(u) = \sum_{j=0}^{\infty} a_j u^j $$Then
$$ \frac{dh}{du} = \sum_{j=0}^{\infty}ja_ju^{j-1} $$ $$ -2u\frac{dh}{du} = -2\sum_{j=0}^{\infty}ja_ju^j $$and
$$ \frac{d^2h}{du^2} = \sum_{j=0}^{\infty}j(j-1)a_ju^{j-2} = \sum_{j=2}^{\infty}j(j-1)a_ju^{j-2} $$Reindexing with \( j=j'+2 \),
$$ \frac{d^2h}{du^2} = \sum_{j'=0}^{\infty}(j'+2)(j'+1)a_{j'+2}u^{j'} $$Substituting into the equation gives
$$ \sum_{j=0}^{\infty} \left( (j+2)(j+1)a_{j+2} - 2ja_j + (\varepsilon-1)a_j \right)u^j = 0 $$Therefore each coefficient must vanish independently, leading to the recursion relation
$$ a_{j+2} = \frac{2j+1-\varepsilon}{(j+2)(j+1)}a_j $$The solutions are fixed once \( a_0 \) and \( a_1 \) are specified. The coefficient \( a_0 \) generates the even solutions, while \( a_1 \) generates the odd solutions.
Physical normalizability requires the power series to terminate. To terminate the series, we require
$$ 2j+1-\varepsilon=0 $$for some integer \( j=n \), so that
$$ a_{n+2}=0 \quad \, n\in\mathbb{Z}^+ $$Hence
$$ h(u) = a_nu^n+a_{n-2}u^{n-2}+\cdots $$and the allowed energies satisfy
$$ \varepsilon = 2n+1 = \frac{2E}{\hbar\omega} $$Therefore
$$ E_n = \left( n+\frac{1}{2} \right)\hbar\omega $$The even states arise from \( a_1=0 \), while the odd states arise from \( a_0=0 \).
The resulting polynomials are precisely the Hermite polynomials:
$$ h_n(u)=H_n(u) $$whose leading behaviour is
$$ H_n(u) = 2^nu^n+O(u^{n-2}) $$Hermite polynomials can be generated efficiently from
$$ e^{-z^2+2zu} = \sum_{n=0}^{\infty} \frac{z^n}{n!}H_n(u) $$The first few are
$$ H_0=1 $$ $$ H_1=2u $$ $$ H_2=4u^2-2 $$ $$ H_3=8u^3-12u $$ $$ H_4=16u^4-48u^2+12 $$The normalized wavefunctions are therefore
$$ \psi_n(x) = \frac{1}{\sqrt{2^nn!}} \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} H_n(u) \, e^{-u^2/2} $$Hermite polynomials satisfy the differential equation
$$ \frac{d^2H_n}{du^2} - 2u\frac{dH_n}{du} + 2nH_n = 0 $$Operator Method
The harmonic oscillator also admits a remarkably elegant algebraic solution. Begin with the Hamiltonian operator
$$ \hat{E} = \frac{\hat{p}^2}{2m} + \frac{m\omega^2}{2}\hat{x}^2 $$Dimensional analysis gives
$$ [\hbar]=[px] $$ $$ [m]=[m] $$ $$ [\omega]=[t^{-1}] $$Hence
$$ [\hbar m\omega]=[p]^2 $$ $$ \left[ \frac{\hbar}{m\omega} \right] = [x]^2 $$ $$ [\hbar\omega]=[E] $$Define the characteristic scales
$$ x_0 = \sqrt{\frac{2\hbar}{m\omega}} $$ $$ p_0 = \sqrt{2\hbar m\omega} $$so that
$$ x_0p_0=2\hbar $$Then the Hamiltonian becomes
$$ \hat{E} = \hbar\omega \left( \frac{\hat{p}^2}{p_0^2} + \frac{\hat{x}^2}{x_0^2} \right) $$Using the identity
$$ c^2+d^2=(c-id)(c+id) $$we compute
$$ \left( \frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} \right) \left( \frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} \right) = \frac{\hat{x}^2}{x_0^2} + \frac{\hat{p}^2}{p_0^2} + \frac{i}{2 \hbar} [\hat{x},\hat{p}] $$Since \( [\hat{x},\hat{p}]=i\hbar \), this becomes
$$ \frac{\hat{x}^2}{x_0^2} + \frac{\hat{p}^2}{p_0^2} - \frac{1}{2} $$Therefore
$$ \hat{E} = \hbar\omega \left( \left( \frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} \right) \left( \frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} \right) + \frac{1}{2} \right) $$Define the annihilation and creation operators
$$ \hat{a} = \frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} $$ $$ \hat{a}^{\dagger} = \frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} $$Then
$$ \hat{E} = \hbar\omega \left( \hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right) $$Hermitian Operators
Given a linear operator \( \hat{O} \), its adjoint \( \hat{O}^{\dagger} \) is defined through
$$ \int dx\,f^*(\hat{O}^{\dagger}g) = \int dx\,(\hat{O}f)^*g $$For a complex number \( c\in\mathbb{C} \),
$$ \int dx\,f^*(c^{\dagger}g) = \int dx\,(cf)^*g $$ $$ = c^* \int dx\,f^*g $$ $$ = \int dx\,f^*(c^*g) $$Since this must be true for all \( f \) and \( g \), we have
$$ c^{\dagger}=c^* $$Next consider the derivative operator acting on the space of normalizable functions:
$$ \int dx\,f^*(\partial_x^{\dagger}g) = \int dx\,(\partial_xf)^*g $$Integrating by parts,
$$ \int uv' = uv - \int vu' $$ $$ u = g, \quad u' = \partial_x g $$ $$ v = f^*, \quad v' = \partial_x f^* $$ $$ \int dx \, \partial_x f^* g = f^*g\Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty}dx\,f^*\partial_xg $$Since normalizable wavefunctions vanish at infinity, the boundary term disappears
$$ = \int dx\,f^*(- \partial_x g) $$and so
$$ \partial_x^{\dagger} = -\partial_x $$For the position operator,
$$ \hat{x}f(x)=xf(x) $$and
$$ \int dx\,f^*(\hat{x}^{\dagger}g) = \int dx\,(\hat{x}f)^*g $$ $$ = \int dx\,xf^*g $$ $$ = \int dx\,f^*(\hat{x}g) $$Therefore
$$ \hat{x}^{\dagger}=\hat{x} $$Operators satisfying
$$ \hat{O}^{\dagger}=\hat{O} $$are Hermitian operators and possess real eigenvalues (see proof). Since physical observables must be real, operators corresponding to observables must be Hermitian.
$$ \hat{p}^{\dagger} = \hat{p} \qquad \left( \frac{\hbar}{i} \partial_x \right)^{\dagger} = \frac{\hbar}{i} \partial_x $$The annihilation operator \( \hat{a} \) is not Hermitian, and therefore does not correspond directly to an observable. However,
$$ \hat{E} = \hbar\omega \left( \hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right) $$is Hermitian.
The commutator of the ladder operators is
$$ [\hat{a},\hat{a}^{\dagger}] = - \frac{i}{2 \hbar} [\hat{x},\hat{p}] + \frac{i}{2 \hbar} [\hat{p},\hat{x}] $$ $$ = \frac{1}{2} + \frac{1}{2} = 1 $$ $$ (\hat{A}\hat{B}^{\dagger})^{\dagger} = \hat{B}^{\dagger} \hat{A}^{\dagger} $$Using this,
$$ [\hat{E},\hat{a}] = \hbar\omega [\hat{a}^{\dagger} \hat{a},\hat{a}] = \hbar\omega (\hat{a}^{\dagger} \hat{a} \hat{a} - \hat{a} \hat{a}^{\dagger} \hat{a}) $$ $$ = \hbar\omega (\hat{a}^{\dagger} \hat{a} - \hat{a} \hat{a}^{\dagger}) \hat{a} = \hbar\omega [\hat{a}^{\dagger},\hat{a}] \hat{a} $$ $$ \therefore [\hat{E},\hat{a}] = -\hbar\omega\hat{a} $$ $$ [\hat{E},\hat{a}^{\dagger}] = \hbar\omega [\hat{a}^{\dagger} \hat{a},\hat{a}^{\dagger}] = \hbar\omega (\hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a}^{\dagger} \hat{a}) $$ $$ = \hbar\omega \hat{a}^{\dagger} (\hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a}) = \hbar\omega \hat{a}^{\dagger} [\hat{a},\hat{a}^{\dagger}] $$ $$ \therefore [\hat{E},\hat{a}^{\dagger}] = \hbar\omega\hat{a}^{\dagger} $$Consider the energy eignenvalue equation
$$ \hat{E}\phi_E=E\phi_E $$and define
$$ \psi=\hat{a}\phi_E $$We want to find the energy eigenvalue of \( \psi \). Acting on \( \psi \) with \( \hat{E} \), we get
$$ \hat{E}\psi = \hat{E}\hat{a}\phi_E $$Using
$$ [\hat{E},\hat{a}] = \hat{E}\hat{a} - \hat{a}\hat{E} = -\hbar\omega\hat{a} $$we have
$$ \hat{E}\hat{a} = \hat{a}\hat{E} - \hbar\omega\hat{a} $$Therefore
$$ \hat{E}\psi = (\hat{a}\hat{E} - \hbar\omega\hat{a}) \phi_E = (\hat{a} E - \hbar\omega\hat{a}) \phi_E $$ $$ = ( E - \hbar\omega) \hat{a} \phi_E = (E-\hbar\omega)\psi $$so \( \hat{a} \) lowers the energy eigenvalue by one quantum. Likewise, if
$$ \psi=\hat{a}^{\dagger}\phi_E $$Then
$$ \hat{E}\psi = \hat{E}\hat{a}^{\dagger} \phi_E $$Using
$$ [\hat{E},\hat{a}^{\dagger}] = \hat{E}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{E} = \hbar\omega\hat{a}^{\dagger} $$we have
$$ \hat{E}\hat{a}^{\dagger} = \hat{a}^{\dagger}\hat{E} + \hbar\omega\hat{a}^{\dagger} $$Therefore
$$ \hat{E}\psi = (\hat{a}^{\dagger}\hat{E} + \hbar\omega\hat{a}^{\dagger}) \phi_E = (\hat{a}^{\dagger} E + \hbar\omega\hat{a}^{\dagger}) \phi_E $$ $$ = (E + \hbar\omega) \hat{a}^{\dagger} \phi_E = (E+\hbar\omega)\psi $$Thus \( \hat{a}^{\dagger} \) raises the energy by one quantum.
Therefore
$$ \hat{a} \; \text{is the lowering operator} $$ $$ \phi_E \xrightarrow{\hat{a}} \phi_{E-\hbar\omega} $$ $$ \hat{a}^{\dagger} \; \text{is the raising operator} $$ $$ \phi_E \xrightarrow{\hat{a}^{\dagger}} \phi_{E+\hbar\omega} $$Since the energy cannot decrease indefinitely, there must exist a minimum-energy state \( \phi_0 \) satisfying
$$ \hat{a}\phi_0=0 $$This is the ground state. Applying the Hamiltonian,
$$ \hat{E}\phi_0 = \hbar\omega \left( \hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right)\phi_0 $$ $$ = \frac{1}{2}\hbar\omega\phi_0 $$so the ground-state energy is
$$ E_0 = \frac{1}{2}\hbar\omega $$To determine the explicit ground-state wavefunction, use
$$ \hat{a} = \frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} = \frac{x}{x_0} + \frac{\hbar}{p_0} \partial_x $$Then
$$ \hat{a}\phi_0=0 \quad\Rightarrow\quad \left( \frac{p_0}{\hbar x_0}x+\partial_x \right)\phi_0=0 $$Since
$$ x_0p_0=2\hbar $$we obtain
$$ \left( \partial_x + \frac{2}{x_0^2}x \right)\phi_0 = 0 $$Therefore
$$ \phi_0'(x) = -\frac{2}{x_0^2}x\phi_0(x) $$Integrating,
$$ \phi_0(x) = ce^{-x^2/x_0^2} $$Equivalently,
$$ \phi_0(x) = ce^{-m\omega x^2/2\hbar} $$Normalization requires
$$ |c|^2 \int_{-\infty}^{\infty} e^{-m\omega x^2/\hbar}\,dx = 1 $$Using the Gaussian integral,
$$ |c|^2 \sqrt{\frac{\pi\hbar}{m\omega}} = 1 $$Hence
$$ \phi_0(x) = \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} e^{-x^2/x_0^2} $$All excited states can be generated algebraically through repeated application of the raising operator:
$$ \phi_n(x) = A_n(\hat{a}^{\dagger})^n\phi_0(x) $$Choosing the normalization convention,
$$ \phi_n(x) = \frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n\phi_0(x) $$with energies
$$ E_n = \left( n+\frac{1}{2} \right)\hbar\omega \qquad n=0,1,2,\ldots $$The eigenstates satisfy the orthonormality condition
$$ \int_{-\infty}^{\infty} \phi_m^*\phi_n\,dx = \delta_{mn} $$