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.

9. Quantum Mechanics: Further Interpretation of Wave Function

Born’s statistical interpretation of wave function says that: ${{\left| \Psi (x,t) \right|}^{2}}dx$ is the probability that the particle will be found at a coordinate between x and x+dx at time t. Consequently, the total probability of finding the particle somewhere should be equal to 1. Referring to the total probability for continuous distributions: $\int\limits_{-\infty }^{\infty }{\rho (x)dx}=1$ (please see Equation (8.10) in my previous post), the integral of ${{\left| \Psi (x,t) \right|}^{2}}dx$ must be equal to 1:

$\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx=1}$                                                    (9.1)

The wave function $\Psi (x,t)$ is a solution of Schrödinger equation:

$i\hbar \frac{\partial \Psi }{\partial t}=-\frac{\hbar }{2m}\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}+V\Psi$                                             (9.2)

And also, $A\Psi (x,t)$ is a solution of equation (9.2), where A is any (complex) constant. Since the condition in equation (9.1) has to be satisfied, we have to normalize the wave function to find this multiplicative constant. Please note that the wave function $\Psi (x,t)$  must go to zero as |x|→∞, this physically realizable states correspond to the ‘square-integrable’ solutions to Schrödinger’s equation.

Now, we are going to proof that a normalized wave function stays normalized for all future time, i.e. the normalization of the wave function is preserved.

Straightforwardly, by taking the first order derivative of the term on the left hand side of equation (9.1) with respect to time t, we obtain:

$\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx}=\int\limits_{-\infty }^{\infty }{\frac{\partial }{\partial t}{{\left| \Psi (x,t) \right|}^{2}}dx}$                                       (9.3)

The wave function $\Psi (x,t)$ is a function of place x as well as time t, therefore we take the partial derivative of the integrand on the right hand side of equation (9.3). Working out further the integrand on the right hand side of equation (9.3) with the help of product rule, we obtain:

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{\partial }{\partial t}$(Ψ^*Ψ) = Ψ^*$\frac{\partial \Psi }{\partial t}$ + $\frac{\partial {{\Psi }^{*}}}{\partial t}$Ψ                                         (9.4)

Taking the complex conjugate of Schrödinger equation (9.2):

$\frac{\partial {{\Psi }^{*}}}{\partial t}=-\frac{i\hbar }{2m}\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}+\frac{i}{\hbar }V{{\Psi }^{*}}$                                          (9.5)

Substituting the Schrödinger equation and its complex conjugate into equation (9.4):                                

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{i\hbar }{2m}$(Ψ^*$\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}$ - $\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}$Ψ)                                          (9.6)

Resolving this equation further, we get:

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{\partial }{\partial x}$[$\frac{i\hbar }{2m}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ)]                                      (9.7)

                                    

By substituting equation (9.7) into (9.3),  the integral can then be analyzed explicitly: 

$\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{}}$|Ψ(x,t)|²dx = $\frac{i\hbar }{2m}$(Ψ^*$\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}$ - $\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}$Ψ)|$_{-\infty }^{\infty }$                                      (9.8)

However, the wave function $\Psi (x,t)$ must go to zero as x goes to infinity, therefore we have: $\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx=0}$  hence that the integral is constant, i.e. independence of time. Thus the wave function stays normalized for all future time. QED

Consider the expectation value of x for a particle in state Ψ given by:

$\left\langle x \right\rangle =\int\limits_{-\infty }^{\infty }{x}$|Ψ(x,t)|²dx                                                          (9.9)

Equation (9.9) gives the average of measurements performed on all particles in the state Ψ. Thus, the expectation value does not give the average of repeated measurements on one identical system. Rather it is the average of repeated measurements on a group of identical systems.

Due to the time dependence of Ψ, the expectation value of x will change and it is interesting in knowing how fast $\left\langle x \right\rangle $ will change. By taking the derivative of equation (9.9) with respect to time, we get:

$\frac{d\left\langle x \right\rangle }{dt}=\int{x\frac{\partial }{\partial t}}$|Ψ|² dx                                                       (9.10)

By replacing the term $\frac{\partial }{\partial t}$|Ψ|² for $\frac{\partial }{\partial x}$[$\frac{i\hbar }{2m}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ)] (referring to equation (9.7)), equation (9.10) becomes:

$\frac{d\left\langle x \right\rangle }{dt}=\frac{i\hbar }{2m}$∫ x$\frac{\partial }{\partial x}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ) dx                                   (9.11)

Using integration by parts: $\int\limits_{a}^{b}{f\frac{dg}{dx}dx}=fg$|$_{a}^{b}-\int\limits_{a}^{b}{\frac{df}{dx}gdx}$; and under the condition that the ∫ |Ψ|² dx  must be finite: Ψ and its derivative ${}^{\partial \Psi }/{}_{\partial x}$ must go to zero at ± ∞, equation (9.11) can be simplified:

$\frac{d\left\langle x \right\rangle }{dt}=-\frac{i\hbar }{2m}$ ∫ (Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ) dx                                   (9.12)

 

Carrying out another integration by parts, we conclude that:

$\frac{d\left\langle x \right\rangle }{dt}=-\frac{i\hbar }{m}\int{{{\Psi }^{*}}\frac{\partial \Psi }{\partial x}dx}$                                                 (9.13)

Equation (9.13) is the velocity of the expectation value of expectation value of x which is not the same thing as the velocity of the particle. Since the particle doesn’t even have a determinate position prior to the measurement, it is not even clear what velocity means in quantum mechanics. It is just the probability of getting a particular value of velocity. However, equation (9.13) tells us how to calculate the expectation value of velocity straightforwardly from Ψ and for our present purposes, we adequately could postulate that the expectation value of the velocity <v> is equal to the time derivative of the expectation value of position <x>:

$\left\langle v \right\rangle =\frac{d\left\langle x \right\rangle }{dt}$                                                           (9.14)

 Figure 9.1. Working out of expectation value of velocity v

In quantum mechanics it is more convenient to work with momentum (p=mv) rather that velocity:

$\left\langle p \right\rangle =m\frac{d\left\langle x \right\rangle }{dt}=-i\hbar \int{\left( {{\Psi }^{*}}\frac{\partial \Psi }{\partial x} \right)dx}$                                    (9.15)

To summarize, we rewrite the expressions for <x> and <p>:

$\left\langle x \right\rangle =\int{{{\Psi }^{*}}(x)\Psi dx}$                                                    (9.16)

$\left\langle p \right\rangle =\int{{{\Psi }^{*}}(\frac{h}{i}\frac{\partial }{\partial x})\Psi dx}$                                                 (9.17)

The term x that represents position and the term $\left( \frac{\hbar }{i}\frac{\partial }{\partial x} \right)$ that represents momentum respectively in equation (9.16) and (9.17) are the operators in quantum mechanics.

All dynamical variables can be expressed in terms of position and momentum, for instance, kinetic energy: $T={}^{1}/{}_{2}m{{v}^{2}}={}^{{{p}^{2}}}/{}_{2m}$ . Generally, to calculate the expectation value of a quantity, the following equation can be applied:

$\left\langle Q(x,p) \right\rangle =\int{{{\Psi }^{*}}Q(x,\frac{\hbar }{i}\frac{\partial }{\partial x})\Psi dx}$                                   (9.18)

And for kinetic energy:

$\left\langle T \right\rangle =\frac{-{{\hbar }^{2}}}{2m}\int{{{\Psi }^{*}}\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}dx}$                                               (9.19)