Module 4: Wavefunctions and Hermite Polynomials
Seeing the Quantum Oscillator
← Module 3 | ↑ Course Index | Module 5 →
1 Getting the Ground State Wavefunction from Ladder Operators
In Module 3, we found that the ground state satisfies:
\[ \hat{a}|0\rangle = 0 \]
In the position representation, \(|0\rangle \to \psi_0(x)\), and the annihilation operator becomes:
\[ \hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right) \]
The condition \(\hat{a}\psi_0 = 0\) gives:
\[ \left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0(x) = 0 \]
This is a simple first-order ODE:
\[ \frac{d\psi_0}{dx} = -\frac{m\omega}{\hbar} x\, \psi_0 \]
Separating variables and integrating:
\[ \ln\psi_0 = -\frac{m\omega}{2\hbar}x^2 + C \quad \Rightarrow \quad \psi_0(x) = A\, e^{-m\omega x^2/2\hbar} \]
Normalizing using \(\int_{-\infty}^{\infty}e^{-\alpha x^2}dx = \sqrt{\pi/\alpha}\) with \(\alpha = m\omega/\hbar\):
\[ \boxed{\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp\!\left(-\frac{m\omega}{2\hbar}x^2\right)} \]
The ground state is a Gaussian. This is the unique minimum-uncertainty state.
2 Generating All Wavefunctions
Using \(|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle\) and translating to position space:
\[ \psi_n(x) = \frac{1}{\sqrt{2^n\, n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\,x\right) e^{-m\omega x^2/2\hbar} \]
Using the oscillator length \(x_0 = \sqrt{\hbar/m\omega}\) and \(\xi = x/x_0\):
\[ \boxed{\psi_n(\xi) = \frac{1}{\sqrt{2^n\, n!\,\sqrt{\pi}\,x_0}}\, H_n(\xi)\, e^{-\xi^2/2}} \]
2.1 The First Five Wavefunctions Explicitly
\[ \psi_0 = \frac{1}{(\pi x_0^2)^{1/4}}\, e^{-\xi^2/2} \]
\[ \psi_1 = \frac{1}{(\pi x_0^2)^{1/4}}\sqrt{2}\,\xi\, e^{-\xi^2/2} \]
\[ \psi_2 = \frac{1}{(\pi x_0^2)^{1/4}}\frac{1}{\sqrt{2}}\,(2\xi^2 - 1)\, e^{-\xi^2/2} \]
\[ \psi_3 = \frac{1}{(\pi x_0^2)^{1/4}}\frac{1}{\sqrt{3}}\,(2\xi^3 - 3\xi)\, e^{-\xi^2/2} \]
\[ \psi_4 = \frac{1}{(\pi x_0^2)^{1/4}}\frac{1}{2\sqrt{6}}\,(4\xi^4 - 12\xi^2 + 3)\, e^{-\xi^2/2} \]
3 Physical Properties of the Wavefunctions
3.1 Number of Nodes
\(\psi_n(x)\) has exactly \(n\) nodes (zeros). This is the node theorem for bound states:
- \(\psi_0\): 0 nodes (the Gaussian never crosses zero)
- \(\psi_1\): 1 node (crosses zero once at \(x = 0\))
- \(\psi_2\): 2 nodes
- \(\psi_n\): exactly \(n\) nodes
Higher-energy states have more rapid oscillations — they “fit more waves” into the potential well.
3.2 Parity
\[ \psi_n(-x) = (-1)^n \psi_n(x) \]
- Even \(n\): even parity (symmetric about \(x = 0\))
- Odd \(n\): odd parity (antisymmetric about \(x = 0\))
This is a consequence of the symmetric potential \(V(-x) = V(x)\).
3.3 Orthonormality
\[ \int_{-\infty}^{\infty} \psi_m^*(x)\psi_n(x)\,dx = \delta_{mn} \]
This follows from the Hermite polynomial orthogonality: \[ \int_{-\infty}^{\infty} H_m(\xi)\, H_n(\xi)\, e^{-\xi^2}\, d\xi = 2^n\, n!\, \sqrt{\pi}\, \delta_{mn} \]
3.4 Completeness
The wavefunctions form a complete set: \[ \sum_{n=0}^{\infty} \psi_n^*(x')\psi_n(x) = \delta(x - x') \]
This means any square-integrable function can be expanded in the \(\psi_n\)’s.
4 Classical Turning Points and Quantum Tunneling
The classical turning points for energy \(E_n = (n + \frac{1}{2})\hbar\omega\) are where \(V(x) = E_n\):
\[ \frac{1}{2}m\omega^2 x_{\pm}^2 = \left(n + \frac{1}{2}\right)\hbar\omega \quad \Rightarrow \quad x_{\pm} = \pm\sqrt{2n+1}\, x_0 \]
Classically, the particle cannot exist beyond \(x_{\pm}\). But quantum mechanically, the wavefunction extends into the classically forbidden region!
The probability of finding the particle outside the classical turning points is:
\[ P_{\text{tunnel}} = \int_{|x|>x_+}|\psi_n(x)|^2\,dx \]
For the ground state \(n = 0\), \(x_+ = x_0\):
\[ P_{\text{tunnel},0} = 2\int_{x_0}^{\infty}|\psi_0(x)|^2\,dx = \text{erfc}(1/\sqrt{2}) \approx 15.7\% \]
About 16% of the time, the ground-state particle is found in the classically forbidden region!
This is quantum tunneling — the particle “borrows” energy from the uncertainty principle to penetrate the classically forbidden region. This phenomenon is responsible for: - Nuclear fusion in stars (particles tunnel through the Coulomb barrier) - Scanning tunneling microscopy (STM) - Alpha decay of radioactive nuclei
5 Expectation Values in Energy Eigenstates
Using ladder operators (from Module 3):
| Quantity | Value in \(|n\rangle\) |
|---|---|
| \(\langle\hat{x}\rangle\) | \(0\) |
| \(\langle\hat{p}\rangle\) | \(0\) |
| \(\langle\hat{x}^2\rangle\) | \((n + \frac{1}{2})\,x_0^2 = (n + \frac{1}{2})\frac{\hbar}{m\omega}\) |
| \(\langle\hat{p}^2\rangle\) | \((n + \frac{1}{2})\,\frac{\hbar^2}{x_0^2} = (n + \frac{1}{2})m\hbar\omega\) |
| \(\langle V\rangle\) | \(\frac{1}{2}E_n\) |
| \(\langle T\rangle\) | \(\frac{1}{2}E_n\) |
| \(\Delta x\) | \(x_0\sqrt{n + \frac{1}{2}}\) |
| \(\Delta p\) | \(\frac{\hbar}{x_0}\sqrt{n + \frac{1}{2}}\) |
| \(\Delta x\,\Delta p\) | \((n + \frac{1}{2})\hbar\) |
\(\langle T\rangle = \langle V\rangle = \frac{1}{2}E_n\). This is the quantum version of the classical virial theorem. On average, kinetic and potential energies are equal in the QHO.
6 Momentum-Space Wavefunctions
The Fourier transform of \(\psi_n(x)\) gives the momentum-space wavefunction \(\phi_n(p)\):
\[ \phi_n(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi_n(x)\,dx \]
A beautiful result: the momentum-space wavefunctions have the same form as the position-space ones!
\[ \phi_n(p) = \frac{(-i)^n}{(m\omega\pi\hbar)^{1/4}\sqrt{2^n n!}}\, H_n\!\left(\frac{p}{\sqrt{m\omega\hbar}}\right) e^{-p^2/2m\omega\hbar} \]
The momentum-space wavefunctions are also Hermite-Gaussian, with the same Gaussian envelope. This is a special property of the harmonic oscillator — its Hamiltonian is symmetric between \(x\) and \(p\).
7 Hermite Polynomial Reference Table
| \(n\) | \(H_n(\xi)\) | Parity |
|---|---|---|
| 0 | \(1\) | Even |
| 1 | \(2\xi\) | Odd |
| 2 | \(4\xi^2 - 2\) | Even |
| 3 | \(8\xi^3 - 12\xi\) | Odd |
| 4 | \(16\xi^4 - 48\xi^2 + 12\) | Even |
| 5 | \(32\xi^5 - 160\xi^3 + 120\xi\) | Odd |
Recurrence relation: \(H_{n+1} = 2\xi H_n - 2n H_{n-1}\)
Derivative: \(H_n' = 2n H_{n-1}\)
Rodrigues formula: \(H_n(\xi) = (-1)^n e^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2}\)
8 The Classical-Quantum Correspondence
For large \(n\), quantum mechanics must reproduce classical results. Let’s check:
Classical probability: For a classical oscillator with amplitude \(A = x_+\), the particle spends most time at the turning points (where it’s slow). The classical probability density is:
\[ P_{\text{cl}}(x) = \frac{1}{\pi\sqrt{A^2 - x^2}}, \quad |x| < A \]
Quantum probability: \(|\psi_n(x)|^2\) oscillates rapidly for large \(n\).
The coarse-grained (average) quantum probability distribution for large \(n\) approaches \(P_{\text{cl}}(x)\). You can verify this numerically by plotting \(|\psi_{20}(x)|^2\) — the rapid oscillations average out to the classical distribution.
This is the Bohr correspondence principle: quantum mechanics must agree with classical mechanics in the limit of large quantum numbers.
Before moving on, make sure you can:
- Derive \(\psi_0(x)\) by solving \(\hat{a}\psi_0 = 0\) as a first-order ODE.
- Write \(\psi_1(x)\) by applying \(\hat{a}^\dagger\) to \(\psi_0\).
- State the number of nodes, parity, and normalization of \(\psi_n\).
- Compute \(\langle\hat{x}^2\rangle\) and \(\langle\hat{p}^2\rangle\) in \(|n\rangle\) using ladder operators.
- Verify the virial theorem \(\langle T\rangle = \langle V\rangle = E_n/2\).
- Estimate the tunneling probability for the ground state.
Conceptual question: Why is \(\langle\hat{x}\rangle = 0\) for all energy eigenstates? What does this tell you about the symmetry of the potential?