13. Example #3: The Harmonic Oscillator (Analytic Method)

Let us return to the Schrödinger equation for harmonic (quantum) oscillator from the previous section:

$-\frac{\hbar }{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\psi =E\psi $                                          (13.1)

In this section, we are going to solve this equation analytically. By introducing a dimensionless variable: $\xi =\sqrt{{}^{m\omega }/{}_{\hbar }}x$ into this equation, the Schrödinger equation (13.1) then reads:

$\frac{{{d}^{2}}\psi }{d{{\xi }^{2}}}=({{\xi }^{2}}-K)\psi $                                                 (13.2)

Here, $K={}^{2E}/{}_{\hbar \omega }$ is the energy, per units of $\left( {}^{1}/{}_{2} \right)\hbar \omega $.

For large ξ (large x), the Schrödinger equation (13.2) can be expressed:

$\frac{{{d}^{2}}\psi }{d{{\xi }^{2}}}-{{\xi }^{2}}\psi \approx 0$                                              (13.3)

And we are looking for a solution of this equation with ‘non-asymptotic behavior’, i.e. normalizable and no ‘blow up’ terms as |x| → ∞:

$\psi (\xi )\sim {{e}^{-{}^{{{\xi }^{2}}}/{}_{2}}}$                                               (13.4)

We assume a solution as follows:

$\psi (\xi )=h(\xi ){{e}^{-{}^{{{\xi }^{2}}}/{}_{2}}}$                                            (13.5)

Substituting this solution to the Schrödinger equation (13.2) and the equation becomes:

$\frac{{{d}^{2}}h}{d{{\xi }^{2}}}-2\xi \frac{dh}{d\xi }+(K-1)h=0$                                     (13.6)

This equation is the Hermite polynomials differential equation. According to Taylor’s theorem, any reasonably well-behaved function can be expressed as a power series. So, the proposed solution to equation (13.6) is in the form of a power series in ξ:

$h(\xi )={{a}_{0}}+{{a}_{1}}\xi +{{a}_{2}}{{\xi }^{2}}+...=\sum\limits_{j=0}^{\infty }{{{a}_{j}}{{\xi }^{j}}}$                               (13.7)

Differentiating equation (13.7), we have:

$\frac{dh}{d\xi }={{a}_{1}}+2{{a}_{2}}\xi +3{{a}_{3}}{{\xi }^{2}}+...=\sum\limits_{j=0}^{\infty }{j{{a}_{j}}{{\xi }^{j-1}}}$                         (13.8)

And the second order derivative of equation (13.7) is:

$\frac{{{d}^{2}}h}{d{{\xi }^{2}}}=2{{a}_{2}}+6{{a}_{3}}\xi +...=\sum\limits_{j=0}^{\infty }{(j+1)(j+2){{a}_{j+2}}{{\xi }^{j}}}$                               (13.9)

Substituting equations (13.7), (13.8) and (13.9) in (13.6), we find:

$\sum\limits_{j=0}^{\infty }{\left[ (j+1)(j+2){{a}_{j+2}}-2j{{a}_{j}}+(K-1){{a}_{j}} \right]}{{\xi }^{j}}=0$                             (13.10)

And hence, that:

${{a}_{j+2}}=\frac{2j-K+1}{(j+1)(j+2)}{{a}_{j}}$                                          (13.11)

Equation (13.11) is the recursion formula, this equation enables us to generate a₂, a₄, a₆ … when a₀ is given and it generates a₃, a₅, a₇ … when a₁ is given. For normalizable solutions, the power series must terminate at some ‘highest’ j (say, at n), such that a_n+2 = 0. Therefore, we must set equation (13.11) to zero for a certain n:

$2n+1-K=0$                                              (13.12)

Or,

$K=2n+1$                                                 (13.13)

And the allowed energy E_n, must be of the form:

  for n = 0,1,2,3,...                               (13.14)

For the allowed values of K, the recursion formula (13.11) becomes:

${{a}_{j+2}}=\frac{-2(n-j)}{(j+1)(j+2)}{{a}_{j}}$                                          (13.15)

In general, h_n (ξ) will be a polynomial of degree n in ξ, involving even powers only if n is an even integer and odd powers only if n is an odd integer. For instance, by applying (13.15), if n = 0: h₀ (ξ) = a₀ and hence, ${{\psi }_{0}}(\xi )={{a}_{0}}{{e}^{-{{{\xi }^{2}}}/{2}\;}}$ . If n = 1: ${{h}_{1}}(\xi )={{a}_{1}}\xi $ and ${{\psi }_{1}}(\xi )={{a}_{1}}\xi {{e}^{{-{{\xi }^{2}}}/{2}\;}}$ . And so on.

Then, the normalized stationary states for the harmonic oscillator are:

${{\psi }_{n}}(x)={{\left( \frac{m\omega }{\pi \hbar } \right)}^{{1}/{4}\;}}\frac{1}{\sqrt{{{2}^{n}}n!}}{{H}_{n}}(\xi ){{e}^{-{{{\xi }^{2}}}/{2}\;}}$                                 (13.16)

Here, H_n (ξ) is the Hermite polynomials, which is generally formulated as follows:

${{H}_{n}}(\xi )={{(-1)}^{n}}{{e}^{{{\xi }^{2}}}}\left( \frac{{{d}^{n}}}{d{{\xi }^{n}}}{{e}^{-{{\xi }^{2}}}} \right)$ for n = 0, 1, 2, 3, …                           (13.17)

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.