Module 2: Quantizing the Oscillator
From Classical Springs to Quantum Operators
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1 The Recipe: Canonical Quantization
In Module 1, we saw that the classical harmonic oscillator has Hamiltonian:
\[ H_{\text{classical}} = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 \]
where \(x\) is position and \(p = m\dot{x}\) is momentum.
The procedure of canonical quantization is the recipe for turning a classical system into a quantum one:
- Promote classical variables to operators: \(x \to \hat{x}\), \(p \to \hat{p}\)
- Impose the canonical commutation relation: \[[\hat{x},\, \hat{p}] = i\hbar\]
- The Hamiltonian becomes an operator: \(\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2\)
- The state of the system is a wavefunction \(\psi(x,t)\) obeying the Schrödinger equation: \[i\hbar\frac{\partial\psi}{\partial t} = \hat{H}\psi\]
This is the postulate of quantum mechanics — we cannot derive it from classical mechanics. But it works spectacularly well for every physical system we’ve ever tested.
Textbook reference: Shankar, Chapter 4; Griffiths Section 2.3; Cohen-Tannoudji Vol. 1, Chapter II.
1.1 The Commutation Relation: What Does It Mean?
The classical Poisson bracket \(\{x, p\}_{\text{PB}} = 1\) gets promoted to \([\hat{x}, \hat{p}] = i\hbar\).
In the position representation (the most common choice), we represent: \[ \hat{x} \to x \quad \text{(multiply by } x\text{)}, \qquad \hat{p} \to -i\hbar\frac{d}{dx} \]
Let’s verify the commutator:
Verification of \([\hat{x}, \hat{p}] = i\hbar\):
Apply \([\hat{x}, \hat{p}]\) to an arbitrary function \(f(x)\):
\[ [\hat{x}, \hat{p}]f = \hat{x}(\hat{p}f) - \hat{p}(\hat{x}f) = x\left(-i\hbar\frac{df}{dx}\right) - \left(-i\hbar\right)\frac{d(xf)}{dx} \]
\[ = -i\hbar x f' + i\hbar\left(f + xf'\right) = i\hbar f \]
Since this holds for any \(f(x)\), we have \([\hat{x}, \hat{p}] = i\hbar\). ✓
2 The Quantum Hamiltonian and Schrödinger Equation
The quantum Hamiltonian for the harmonic oscillator is:
\[ \hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2 x^2 \]
For stationary states (energy eigenstates), we seek solutions \(\psi_n(x)\) such that:
\[ \hat{H}\psi_n = E_n \psi_n \]
This gives the time-independent Schrödinger equation (TISE):
\[ \boxed{-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 x^2\psi = E\psi} \]
Our job: find all \(E\) values for which this equation has normalizable solutions.
3 Solving the TISE: The Analytic Method
3.1 Dimensional Analysis: Natural Units
Before solving, let’s introduce natural dimensionless variables. The problem has two natural scales:
- Length scale: \(x_0 = \sqrt{\dfrac{\hbar}{m\omega}}\) (called the oscillator length or zero-point length)
- Energy scale: \(\hbar\omega\)
Define the dimensionless variable: \[ \xi = \frac{x}{x_0} = x\sqrt{\frac{m\omega}{\hbar}} \]
And the dimensionless energy: \[ \varepsilon = \frac{2E}{\hbar\omega} \]
The TISE becomes: \[ \frac{d^2\psi}{d\xi^2} + (\varepsilon - \xi^2)\psi = 0 \]
The oscillator length \(x_0 = \sqrt{\hbar/m\omega}\) sets the quantum scale of fluctuations. For a typical molecule like \(\text{N}_2\) with \(\omega \sim 10^{14}\) rad/s, \(x_0 \sim 10^{-11}\) m — this is the scale of quantum vibrations around the bond length.
3.2 Asymptotic Behavior for Large \(\xi\)
For \(|\xi| \to \infty\), the equation becomes \(\frac{d^2\psi}{d\xi^2} \approx \xi^2\psi\).
The two independent solutions are \(\psi \sim e^{+\xi^2/2}\) (diverges, unphysical) and \(\psi \sim e^{-\xi^2/2}\) (decays, physical).
So we write: \[ \psi(\xi) = h(\xi)\, e^{-\xi^2/2} \]
where \(h(\xi)\) is a function that grows more slowly than \(e^{\xi^2/2}\).
3.3 The Hermite Differential Equation
Substituting \(\psi = h e^{-\xi^2/2}\) into the TISE:
Substitution:
\[\frac{d\psi}{d\xi} = h'e^{-\xi^2/2} - \xi h e^{-\xi^2/2}\]
\[\frac{d^2\psi}{d\xi^2} = h''e^{-\xi^2/2} - 2\xi h'e^{-\xi^2/2} + (\xi^2 - 1)he^{-\xi^2/2}\]
Plugging into \(\psi'' + (\varepsilon - \xi^2)\psi = 0\) and dividing by \(e^{-\xi^2/2}\):
\[h'' - 2\xi h' + (\varepsilon - 1)h = 0\]
This is the Hermite differential equation with parameter \(\lambda = \varepsilon - 1\).
3.4 Power Series Solution
Try \(h(\xi) = \sum_{j=0}^{\infty} a_j \xi^j\). Substituting:
\[ \sum_{j=0}^{\infty}\left[(j+2)(j+1)a_{j+2} - 2ja_j + (\varepsilon - 1)a_j\right]\xi^j = 0 \]
Each coefficient must vanish, giving the recursion relation:
\[ \boxed{a_{j+2} = \frac{2j + 1 - \varepsilon}{(j+1)(j+2)}\, a_j} \]
3.5 The Quantization Condition
For large \(j\), if the series doesn’t terminate, \(a_{j+2}/a_j \to 2/j\), which makes \(h(\xi) \sim e^{\xi^2}\) — causing \(\psi \sim e^{\xi^2/2}\), which is not normalizable.
The series must terminate. This happens when \(2j + 1 - \varepsilon = 0\) for some non-negative integer \(j = n\):
\[ \varepsilon = 2n + 1 \quad \Longrightarrow \quad \frac{2E}{\hbar\omega} = 2n + 1 \]
\[ \boxed{E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n = 0, 1, 2, 3, \ldots} \]
The Energy Spectrum of the Quantum Harmonic Oscillator
\[E_n = \hbar\omega\!\left(n + \tfrac{1}{2}\right), \quad n = 0, 1, 2, 3, \ldots\]
Energy is quantized in steps of \(\hbar\omega\). The spectrum is equally spaced!
4 The Energy Spectrum: Key Features
4.1 Zero-Point Energy
The lowest energy state is \(n = 0\):
\[ E_0 = \frac{1}{2}\hbar\omega \]
This is non-zero! Unlike the classical oscillator (which can have zero energy), the quantum oscillator always has at least this energy. This is the zero-point energy (ZPE).
Why zero-point energy? The uncertainty principle demands that we cannot simultaneously know position and momentum exactly. If the oscillator were at rest at \(x = 0\) with \(p = 0\), that would violate \(\Delta x\, \Delta p \geq \hbar/2\). The zero-point energy is the quantum price of the uncertainty principle.
Physical manifestations of zero-point energy:
- Liquid helium never freezes at atmospheric pressure (even at \(T = 0\)) — the ZPE keeps it liquid
- Casimir effect — two uncharged metal plates attract due to vacuum fluctuations
- Lamb shift — small but measurable correction to hydrogen energy levels
4.2 Equally Spaced Levels
The levels \(E_0, E_1, E_2, \ldots\) are equally spaced with spacing \(\hbar\omega\). This is why:
- Diatomic molecules absorb/emit light at harmonics of \(\omega\)
- A photon of energy \(\hbar\omega\) can excite the oscillator from \(n\) to \(n+1\)
- Laser light at frequency \(\omega\) can be resonantly absorbed
5 The Hermite Polynomials
When the series terminates at \(j = n\), \(h(\xi)\) becomes the Hermite polynomial \(H_n(\xi)\):
\[ H_0(\xi) = 1 \] \[ H_1(\xi) = 2\xi \] \[ H_2(\xi) = 4\xi^2 - 2 \] \[ H_3(\xi) = 8\xi^3 - 12\xi \] \[ H_4(\xi) = 16\xi^4 - 48\xi^2 + 12 \]
Rodrigues formula: \[ H_n(\xi) = (-1)^n e^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2} \]
Generating function: \[ e^{2\xi t - t^2} = \sum_{n=0}^{\infty} \frac{H_n(\xi)}{n!}\, t^n \]
Recurrence relation: \[ H_{n+1}(\xi) = 2\xi H_n(\xi) - 2n H_{n-1}(\xi) \]
Parity: \(H_n(-\xi) = (-1)^n H_n(\xi)\) — so even \(n\) gives even functions, odd \(n\) gives odd functions.
Textbook reference: Griffiths Section 2.3; Shankar Section 7.3.
6 The Normalized Wavefunctions
Normalizing \(\psi_n = A_n H_n(\xi)e^{-\xi^2/2}\) using \(\int_{-\infty}^{\infty}H_n^2 e^{-\xi^2}d\xi = 2^n n!\sqrt{\pi}\):
\[ \boxed{\psi_n(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\,x\right) e^{-m\omega x^2/2\hbar}} \]
Let’s write out the first few explicitly (using \(x_0 = \sqrt{\hbar/m\omega}\) and \(\xi = x/x_0\)):
\[ \psi_0(x) = \left(\frac{1}{\pi x_0^2}\right)^{1/4} e^{-x^2/2x_0^2} \]
\[ \psi_1(x) = \left(\frac{1}{\pi x_0^2}\right)^{1/4} \sqrt{2}\,\frac{x}{x_0}\, e^{-x^2/2x_0^2} \]
\[ \psi_2(x) = \left(\frac{1}{\pi x_0^2}\right)^{1/4} \frac{1}{\sqrt{2}}\left(2\frac{x^2}{x_0^2} - 1\right) e^{-x^2/2x_0^2} \]
- \(\psi_n(-x) = (-1)^n \psi_n(x)\)
- Even \(n\): even wavefunctions
- Odd \(n\): odd wavefunctions
- This follows from the parity of \(H_n\) and the symmetry of \(V(x) = V(-x)\).
6.1 Orthonormality
The energy eigenstates form a complete orthonormal set:
\[ \int_{-\infty}^{\infty} \psi_m^*(x)\psi_n(x)\,dx = \delta_{mn} \]
This means: - \(m = n\): \(\int|\psi_n|^2\,dx = 1\) (normalized) - \(m \neq n\): \(\int\psi_m^*\psi_n\,dx = 0\) (orthogonal)
Any quantum state can be written as \(\psi(x) = \sum_n c_n \psi_n(x)\) with \(\sum_n |c_n|^2 = 1\).
7 Comparison: Classical vs. Quantum
| Property | Classical Oscillator | Quantum Oscillator |
|---|---|---|
| Allowed energies | Any \(E \geq 0\) | \(E_n = (n+\frac{1}{2})\hbar\omega\) only |
| Ground state energy | \(E = 0\) (at rest) | \(E_0 = \frac{1}{2}\hbar\omega \neq 0\) |
| Ground state position | \(x = 0\) (definite) | \(\langle x\rangle = 0\), \(\langle x^2\rangle = x_0^2/2\) |
| Level spacing | Continuous | Equal spacing \(\hbar\omega\) |
| Tunneling | Cannot go past turning points | Exponentially small probability beyond |
At large \(n\), \(E_n \approx n\hbar\omega \gg \hbar\omega\), and the quantum energy spacing \(\hbar\omega\) becomes negligible compared to \(E_n\). The energy spectrum looks continuous — this is the classical limit.
Also, for large \(n\), \(|\psi_n(x)|^2\) oscillates rapidly and its envelope matches the classical probability distribution \(P_{\text{cl}}(x) \propto 1/\sqrt{A^2 - x^2}\) (where \(A\) is the classical turning point). This is the Bohr correspondence principle in action.
Before moving on, make sure you can:
- State the canonical quantization rules and write down \(\hat{H}\) for the QHO.
- Write down the TISE for the QHO and explain why solutions must terminate.
- State the energy eigenvalues \(E_n = (n + \frac{1}{2})\hbar\omega\) and explain why \(n=0\) still has energy.
- Write \(\psi_0(x)\) and \(\psi_1(x)\) from memory.
- What is the parity of \(\psi_n\)?
A deep question: The Hermite polynomial \(H_n\) has exactly \(n\) nodes (zeros). Why? (Hint: think about the node theorem for bound-state wavefunctions and the fact that higher eigenstates are orthogonal to lower ones.)