12. Example #2: The Harmonic Oscillator (Raising and Lowering Operators Method)

A particle with mass m is attached to a spring of force constant k. Under the assumption that the motion of the particle is frictionless, the Hooke’s law for the classical harmonic oscillator says:

$-kx=m\frac{{{d}^{2}}x}{d{{t}^{2}}}$                                                        (12.1)

The solution of this linear differential equation is given by:

$x(t)=A\sin (\omega t)+B\cos (\omega t)$                                           (12.2)

Here, $\omega \equiv \sqrt{{}^{k}/{}_{m}}$ is the angular frequency of oscillation. And the potential energy is given by:

$V(x)=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}$                                                   (12.3)

In this section, we are going to solve the time-independent Schrödinger equation for the potential given by (12.3). For this particular case, the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\psi =E\psi $                                             (12.4)

Rewriting equation (12.4) in term of momentum operator:

$\frac{1}{2m}\left[ {{\left( \frac{\hbar }{i}\frac{d}{dx} \right)}^{2}}+{{(m\omega x)}^{2}} \right]\psi =E\psi $                                       (12.5)

We are going to solve this equation first by means of raising a₊ and lowering a_- operators:

${{a}_{\pm }}=\frac{1}{\sqrt{2m}}\left( \frac{\hbar }{i}\frac{d}{dx}\pm im\omega x \right)$                                            (12.6)

We have to work with this operators carefully, since generally they do not commute. For instance, it can be shown that: (a₊a_-) f (x) ≠ (a_-a₊) f (x). We invoke here a test function f (x) in our analysis (for more details, please take a look at the attached calculation below), and we conclude that:

${{a}_{-}}{{a}_{+}}=\frac{1}{2m}\left[ {{\left( \frac{\hbar }{i}\frac{d}{dx} \right)}^{2}}+{{(m\omega x)}^{2}} \right]+\frac{\hbar \omega }{2}$                                  (12.7)

${{a}_{+}}{{a}_{-}}=\frac{1}{2m}\left[ {{\left( \frac{\hbar }{i}\frac{d}{dx} \right)}^{2}}+{{(m\omega x)}^{2}} \right]-\frac{\hbar \omega }{2}$                                  (12.8)

And  from these equations:

${{a}_{-}}{{a}_{+}}-{{a}_{+}}{{a}_{-}}=\hbar \omega $                                                     (12.9)

Therefore, in terms of raising and lowering operators, the Schrödinger equation can be expressed:

$({{a}_{-}}{{a}_{+}}-\frac{1}{2}\hbar \omega )\psi =E\psi $                                        (12.10)

$({{a}_{+}}{{a}_{-}}+\frac{1}{2}\hbar \omega )\psi =E\psi$                                       (12.11)

From these equations, it can be shown that: if ψ satisfies the Schrödinger equation, with energy E, then a₊ψ satisfies the Schrödinger equation with energy: (E+ħω). Also, a_-ψ is the solution of Schrödinger equation with energy (E-ħω).

When we apply the lowering operator repeatedly, we will reach a state with the lowest rung of energy ψ_o such that:

${{a}_{-}}{{\psi }_{0}}=0$                                                  (12.12)

Working out the operator on ψ_o further:

$\frac{1}{\sqrt{2m}}\left( \frac{\hbar }{i}\frac{d{{\psi }_{0}}}{dx}-im\omega x{{\psi }_{0}} \right)=0$                                  (12.13)

This follows:

$\frac{d{{\psi }_{0}}}{dx}=-\frac{m\omega }{\hbar }x{{\psi }_{0}}$                                            (12.14)

Afther solving this differential equation, we obtain:

${{\psi }_{0}}(x)={{A}_{0}}{{e}^{-\frac{m\omega }{2\hbar }{{x}^{2}}}}$                                           (12.15)

Filling ψ₀ into equation (12.11), and by exploting the fact that a_-ψ₀ = 0, we determine the energy of the ground state:

${{E}_{0}}=\frac{1}{2}\hbar \omega $                                                   (12.16)

And the excited states with their corresponding energies is given by:

${{\psi }_{n}}(x)={{A}_{n}}{{({{a}_{+}})}^{n}}{{e}^{-\frac{m\omega }{2\hbar }{{x}^{2}}}}$                                        (12.17)

${{E}_{n}}=(n+\frac{1}{2})\hbar \omega $                                                     (12.18)

The normalization constant A_n is given by:

${{A}_{n}}={{\left( \frac{m\omega }{\pi \hbar } \right)}^{{}^{1}/{}_{4}}}\frac{{{(-i)}^{n}}}{\sqrt{n!{{(\hbar \omega )}^{n}}}}$                                      (12.19)

Equation (12.19) can be derived by exploiting equation:

${{a}_{+}}{{\psi }_{n}}=i\sqrt{(n+1)\hbar \omega }{{\psi }_{n+1}}$                                              (12.20)

This equation can be derived from normalization condition and the Schrödinger equation.