The quantum harmonic oscillator is the quantum-mechanical analog of the classical harmonic oscillator. Because an arbitrary smooth potential can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point, it is one of the most important model systems in quantum mechanics. Furthermore, it is one of the few quantum-mechanical systems for which an exact, analytical solution is known.
One-dimensional harmonic oscillator
Hamiltonian and energy eigenstates
The Hamiltonian of the particle is:
H ^ = p ^ 2 2 m + 1 2 k x ^ 2 = p ^ 2 2 m + 1 2 m ω 2 x ^ 2 , {\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}k{\hat {x}}^{2}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}m\omega ^{2}{\hat {x}}^{2}\,,}
where m is the particle's mass, k is the force constant, ω = k / m {\textstyle \omega ={\sqrt {k/m}}} is the angular frequency of the oscillator, x ^ {\displaystyle {\hat {x}}} is the position operator (given by x in the coordinate basis), and p ^ {\displaystyle {\hat {p}}} is the momentum operator (given by p ^ = − i ℏ ∂ / ∂ x {\displaystyle {\hat {p}}=-i\hbar \,\partial /\partial x} in the coordinate basis). The first term in the Hamiltonian represents the kinetic energy of the particle, and the second term represents its potential energy, as in Hooke's law. The time-independent Schrödinger equation (TISE) is,
H ^ | ψ ⟩ = E | ψ ⟩ , {\displaystyle {\hat {H}}\left|\psi \right\rangle =E\left|\psi \right\rangle ~,}
where E {\displaystyle E} denotes a real number (which needs to be determined) that will specify a time-independent energy level, or eigenvalue, and the solution | ψ ⟩ {\displaystyle |\psi \rangle } denotes that level's energy eigenstate. Then solve the differential equation representing this eigenvalue problem in the coordinate basis, for the wave function ⟨ x | ψ ⟩ = ψ ( x ) {\displaystyle \langle x|\psi \rangle =\psi (x)} , using a spectral method. It turns out that there is a family of solutions. In this basis, they amount to Hermite functions,
ψ n ( x ) = 1 2 n n ! ( m ω π ℏ ) 1 / 4 e − m ω x 2 2 ℏ H n ( m ω ℏ x ) , n = 0 , 1 , 2 , … . {\displaystyle \psi _{n}(x)={\frac {1}{\sqrt {2^{n}\,n!}}}\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}e^{-{\frac {m\omega x^{2}}{2\hbar }}}H_{n}{\left({\sqrt {\frac {m\omega }{\hbar }}}x\right)},\qquad n=0,1,2,\ldots .}
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