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.

Solved Example #1: The Infinite Square Well

Solved Example #1: The Infinite Square Well is attached.

11. Example #1 (Application of Schrödinger equation): The Infinite Square Well

Consider a particle with mass m, subjected in a potential:

V (x) = 0 for 0 ≤ x ≤ a, and V (x) → ∞ elsewhere.

The particle is completely free, except at the two ends x = 0 and x = a where an infinite force prevents the particle from escaping. In this case, we would like to find the general solution to the Schrödinger equation for this particular potential.

 Figure 11.1. Infinite square well potential.

The probability of finding the particle outside the well is zero, i.e. ψ (x) = 0. While inside the well, the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                    (11.1)

Or:

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $, with $k=\frac{\sqrt{2mE}}{\hbar }$                                        (11.2)

Equation (11.2) is the simple harmonic oscillator equation, and the general solution is given by:

$\psi (x)=A\sin kx+B\cos kx$                                           (11.3)

With A and B are arbitrary constants. These constants are fixed by boundary conditions of the problem. ψ is continuous, when the potential V (x) goes to infinity, i.e. ψ (a) =  ψ (0) = 0.

Substituting the boundary condition ψ (0) = 0 into equation (11.3), we obtain: B = 0, and hence:

$\psi (x)=A\sin kx$                                                (11.4)

The boundary condition ψ (a) = A sin ka = 0, this condition can be satisfied for A = 0, then ψ (x) = 0. However, if A = 0, we get a trivial, “non-normalizable” solution. Therefore, sin ka = 0 which means that:

ka = 0, ±π, ±2π, ±3π, …                                           (11.5)

The negative solution gives nothing new and the minus sign can be absorbed into A. So the distinct solution is given by:

${{k}_{n}}=\frac{n\pi }{a}$  with n = 1, 2, 3, …                                         (11.6)

With the help of boundary condition at x = a, the possible allowed values of energy E_n can be determined:

${{E}_{n}}=\frac{{{\hbar }^{2}}k_{n}^{2}}{2m}=\frac{{{n}^{2}}{{\pi }^{2}}{{\hbar }^{2}}}{2m{{a}^{2}}}$                                                   (11.7)

In order to fix the constant A, we have to normalize ψ:

$\int\limits_{0}^{a}{{}}$|A|² sin² (kx) dx = 1, so that: |A|² = ${}^{2}/{}_{a}$                                (11.8)

The simplest is to pick the positive real root for constant A: A = $\sqrt{{2}/{a}\;}$then, the solutions inside the potential well are:

${{\psi }_{n}}(x)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right)$                                              (11.9)

ψ₁ is the state with the lowest energy, the so-called ground state, and the other states with higher energies are the excited states. Notice that the energy increases in proportion to n² and each successive state has one more node as we go up in energy (see figure 11.2 below where first three initial states are depicted). And this is a general property, that applies for any shape of the potential.

 Figure 11.2. The first three states of infinite square well.

As can be seen in figure 11.2, the states are alternately even and odd, with respect to the center of the potential well. This property applies whenever the potential is an even function.

Another important property of the states ψ_n is that, they are mutually orthogonal. And this property can be shown:

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx}=0$                                        (11.10)

For m ≠ n. And for m = n, the integral in equation (11.10) is equal to 1. We can also expressed equation (11.10) in terms of Kronecker delta function which is defined by: δ_mn = 0 if m ≠ n and if m = n, δ_mn = 1. Therefore,

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx={{\delta }_{mn}}}$                                   (11.11)

We say that the ψ’s are orthonormal.

Please note that any other function f (x), can be expressed as linear combination of state in equation (11.9), this is called Dirichlet’s theorem:

$f(x)=\sqrt{\frac{2}{a}}\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sin \left( \frac{n\pi }{a}x \right)}$                                       (11.12)

This expression for f (x), is in fact the Fourier expansion. The expansion coefficients c_n can be evaluated for a given f (x) and by applying the orthonormality of ψ_n. Thus, the m-th coefficient in the expansion of f(x) is given by:

${{c}_{m}}=\int{{{\psi }_{m}}}{{(x)}^{*}}f(x)dx$                                      (11.13)

The stationary states for the infinite square well are:

${{\Psi }_{n}}(x,t)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}$                              (11.14)

And the most general solution to the time-dependent Schrödinger equation is a linear combination of stationary states:

$\Psi (x,t)=\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}}$                           (11.15)