PHOT 301: Quantum Photonics

LECTURE 03

Michaël Barbier, Fall semester (2024-2025)

Infinite deep well

Infinite well: summary

\[ \left\{ \begin{aligned} \psi_n(x) &= \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\\ &\\ E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2\\ &\\ n &= 1, 2, 3, 4, \dots\\ \end{aligned} \right. \]


Plot shows the wave function (\(\psi\), grey), probability (\(|\psi|^2\), color) for first 3 eigenstates

Properties of stationary eigenstates

\[ \begin{aligned} \psi_n \textrm{ are orthonormal}\quad & \int \psi_m(x)^* \, \psi_n(x) \, dx = \delta_{mn}\\ \psi_n \textrm{ form a complete basis}\quad & f(x) = \sum_{n = 1}^\infty c_n \psi_n(x) \qquad \forall f(x)\\ \textrm{Coefficients $c_n$ are given by}\quad & c_n = \int \psi_n(x)^* \, f(x) \, dx\\ \end{aligned} \]

Proof of last property:

\[ \begin{aligned} \int \psi_m(x)^* \, f(x) \, dx & = \int \psi_n(x)^* \, \sum_{n = 1}^\infty c_n \psi_n(x) \, dx\\ & = \sum_{n = 1}^\infty c_n \int \psi_m(x)^* \, \psi_n(x) \, dx = \sum_{n = 1}^\infty c_n \delta_{mn} = c_m \end{aligned} \]

Stationary solution of the TISE

For the infinite well

\[ \psi(x) = \sqrt{\frac{2}{L}} \sum_{n = 1}^\infty c_n \sin\left(\frac{n\pi}{L} x\right) \]

Example state: \[ \left\{ \begin{aligned} & c_1 = 4/5,\\ & c_2 = \sqrt{1 - c_1^2} = 3/5,\\ & n > 2 \longrightarrow c_n = 0\\ \end{aligned} \right. \]

  • How does the wave function (\(\psi\), color) and the probability (\(|\psi|^2\), gray) look?
  • What if we let time evolve?

Infinite well: solution of the TDSE

  • Adding time evolution

\[ \begin{aligned} \Psi(x, t) & = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar}\\ & \textrm{with }\sum_{n = 1}^\infty |c_n|^2 = 1\\ \end{aligned} \]

Coefficients \(|c_n|^2\) give the probability to measure energy as \(E_n\):

\[ \langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \sum_{n = 1}^\infty |c_n|^2 E_n \]

But \(\langle \hat{x} \rangle = \int x \Psi^* \Psi \, dx\) is not constant!

Expand a function in eigenstates

  • Suppose we have a certain function

\[ f(x) = (L/2)^4 - (x-L/2)^4, \quad \textrm{with } x \in [0, L] \]

  • Since \(f(0) = f(L) = 0\) we can expand \(f(x)\) in eigenstates of the infinite well

\[ \begin{aligned} f(x) &= \sqrt{\frac{2}{L}} \sum_{n = 1}^\infty c_n \sin\left(\frac{n\pi}{L} x\right)\\ &\textrm{with} \quad c_n = \int_0^L \psi_n(x)^* \, f(x) \, dx\\ \end{aligned} \]

  • See the exercise sessions for the actual calculation

Harmonic oscillator

Introduction

  • Ball-spring problem
  • Typical analog RCL electric circuit
  • Many systems are approximately harmonic oscillators
    • Classical optics
    • 2nd order Taylor approximation of Potential wells
    • Phonons, vibrations in molecules/matter
    • Quantization of light: Photons

Classical harmonic oscillator

  • mass attached to a spring
  • The spring force counters any deviation: \(F = -k x\)
  • Motion described by Newton’s equation \(F = m a\):

\[ m a = m \frac{d^2x}{dt^2} = -k x \]

This is a linear equation with constant coefficients

\[ \frac{d^2x}{dt^2} = - \frac{k}{m} x = -\omega^2 x \]

with \(\omega = \sqrt{k/m}\).

Resulting solutions are:

\[ x \propto \sin(\omega t) \]

Solving the QM harmonic oscillator

The time-independent Schrodinger equation (TISE):

\[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + V(x) \psi(x) = E \psi \]

Potential energy: \(V(x) = \frac{1}{2}m\omega^2 x^2\)

\[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + \frac{1}{2}m\omega^2 x^2 \psi(x) = E \psi \]

Rewrite in dimensionless units: \(\xi = \sqrt{\frac{m\omega}{\hbar}}\)

\[ \frac{1}{2} \frac{\partial^2}{\partial \xi^2}\psi(\xi) - \frac{1}{2} \xi^2 \psi(\xi) = -\frac{E}{\hbar \omega} \psi \]

\(\longrightarrow\) 2nd order linear differential equation

Solving the QM harmonic oscillator

\[ \frac{1}{2} \frac{\partial^2}{\partial \xi^2}\psi(\xi) - \frac{1}{2} \xi^2 \psi(\xi) = -\frac{E}{\hbar \omega} \psi \]

\(\longrightarrow\) Trial solution \(\psi \propto \exp(-\xi^2/2)\)

Substitute \(A_n \exp(-\xi^2/2) H_n(\xi)\) with \(H_n(\xi)\) yet unknown

\[ \frac{d^2 H_n(\xi)}{d\xi^2} - 2\xi\frac{dH_n(\xi)}{d\xi} + \left( \frac{2E}{\hbar\omega} -1 \right) H_n(\xi) = 0 \]


Solutions exist for \(\frac{2E}{\hbar\omega} - 1 = 2n, \qquad n = 0, 1, 2, 3 \dots\)

\[ \longrightarrow \quad \begin{aligned} \psi_n &= A_n \exp(-\xi^2/2) H_n(\xi), \\ E_n &= (n + 1/2) \hbar \omega \,\,\, \textrm{with} \,\, n = 0, 1, 2, \dots\\ \end{aligned} \]

Harmonic oscillator solutions

\[ \begin{aligned} \psi_n &= A_n \exp(-\xi^2/2) H_n(\xi),\\ E_n &= \left(n + \frac{1}{2}\right) \hbar \omega \,\,\, \textrm{with} \,\, n = 0, 1, 2, \dots\\ A_n &= \sqrt{\frac{1}{\sqrt{\pi}2^n n!}} \qquad \xi = \sqrt{\frac{m \omega}{\hbar}} \end{aligned} \]

Hermite polynomials \(H_n(\xi)\)

\[ \begin{aligned} H_0 &= 1\\ H_1 &= 2\xi\\ H_2 &= 4\xi^2 - 2\\ H_3 &= 8\xi^3 - 12\xi\\ \vdots \,\, & \\ H_n(\xi) &= 2\xi H_{n-1}(\xi) - 2(n-1) H_{n-2}(\xi)\\ \end{aligned} \]

Harmonic oscillator solutions

Harmonic oscillator solutions

Alternative (Algebraic) derivation

The time-independent Schrodinger equation (TISE):

\[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + V(x) \psi(x) = E \psi \]

with potential energy: \(V(x) = \frac{1}{2}m\omega^2 x^2\)

\[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + \frac{1}{2}m\omega^2 x^2 \psi(x) = E \psi \]

Operator form:

\[ \frac{1}{2m} \left(\hat{p}^2 + \frac{1}{2}m\omega^2 x^2\right) \psi(x) = E \psi, \qquad \hat{p} = -i\hbar \frac{\partial}{\partial x} \]

This is a sum of squares \(\longrightarrow\) factorize \(u^2 + v^2 = (i u + v)(-iu + v)\)

Ladder operators

Ladder operators \(\hat{a}_-\hat{a}_+ = (i u + v)(-i u + v) = u^2 + v^2\)

\[ \hat{a}_\pm = \frac{1}{\sqrt{2 \hbar m \omega}}\left(\mp i \hat{p} + m\omega x \right), \qquad [\hat{x},\hat{p}] = x\hat{p} - \hat{p}x = i \hbar \]

The product is:

\[ \begin{aligned} \hat{a}_-\hat{a}_+ &= \frac{1}{2\hbar m \omega} (i \hat{p} + m\omega x)(-i \hat{p} + m\omega x)\\ &= \frac{1}{2\hbar m \omega} \left(\hat{p}^2 + (m\omega x)^2 - i m\omega (x\hat{p} - \hat{p}x)\right)\\ &= \frac{1}{2\hbar m \omega} \left(\hat{p}^2 + (m\omega x)^2\right) - \frac{i}{2 \hbar} (x\hat{p} - \hat{p}x)\\ &= \frac{1}{2\hbar m \omega} \left(\hat{p}^2 + (m\omega x)^2\right) + \frac{1}{2}\\ &= \frac{1}{\hbar \omega} \hat{H} + \frac{1}{2}\\ \end{aligned} \]

Ladder operators

Ladder operators \(\hat{a}_-\hat{a}_+ = (i u + v)(-i u + v) = u^2 + v^2\)

\[ \hat{a}_\pm = \frac{1}{\sqrt{2 \hbar m \omega}}\left(\mp i \hat{p} + m\omega x \right), \qquad [\hat{x},\hat{p}] = x\hat{p} - \hat{p}x = i \hbar \]

We can also flip the ladder operators:

\[ \begin{aligned} \hat{H} & = \left(\hat{a}_-\hat{a}_+ - \frac{1}{2}\right) \hbar \omega\\ \hat{H} & = \left(\hat{a}_+\hat{a}_- + \frac{1}{2}\right) \hbar \omega\\ \end{aligned} \]

Stationary Schrodinger equation becomes:

\[ \begin{aligned} \hat{H}\psi = \hbar \omega \left(\hat{a}_+\hat{a}_- + \frac{1}{2}\right) \,\psi = E\,\psi \\ \end{aligned} \]

Ladder operators generate solutions

If \(\psi(x)\) is a solution, the \(\hat{a}_+\psi(x)\) is another solution:

\[ \hat{H} \psi(x) = E \psi \Rightarrow \hat{H} (\hat{a}_+ \psi(x)) = (E + \hbar \omega) (\hat{a}_+ \psi(x)) \]

If \(\psi(x)\) is a solution, then \(\hat{a}_- \psi(x)\) is another solution:

\[ \hat{H} \psi(x) = E \psi \Rightarrow \hat{H} (\hat{a}_- \psi(x)) = (E - \hbar \omega) (\hat{a}_- \psi(x)) \]

Ladder operators generate solutions

Since energy \(E > 0\) operating with \(\hat{a}_-\) leads at some point to:

\[ \hat{a}_-\psi_0 = 0 \]

The leads to the following differential equation \[ \begin{aligned} \frac{1}{\sqrt{2\hbar m \omega}}\left(\hbar \frac{d}{d x} + m \omega x\right) \, \psi_0(x) &= 0\\ \Rightarrow \frac{d \psi_0(x)}{d x} &= - \frac{m \omega}{\hbar} \, x \, \psi_0(x)\\ \Rightarrow \int \frac{d \psi_0(x)}{\psi_0(x)} \, dx &= - \frac{m \omega}{\hbar} \, \int x \, dx\\ \Rightarrow \ln(\psi_0(x)) &= - \frac{m \omega}{2\hbar} \, x^2 + C \\ \Rightarrow \psi_0(x) &= A \, e^{- \frac{m \omega}{2\hbar} \, x^2} \\ \end{aligned} \]

Ladder operators generate solutions

\[ \begin{aligned} \Rightarrow \psi_0(x) &= A \, e^{- \frac{m \omega}{2\hbar} \, x^2} \\ \end{aligned} \]

Normalization requires \(\int |\psi_0(x)|^2 = 1\)

\[ \int_{-\infty}^{\infty} |\psi_0(x)|^2 \, dx = |A|^2 \, \int_{-\infty}^{\infty} e^{- \frac{m \omega}{\hbar} \, x^2} = |A|^2 \, \sqrt{\frac{\pi\hbar}{m\omega}} \]

where we used the identity

\[ \int_{-\infty}^{\infty} e^{- a x^2} dx = \sqrt{\frac{\pi}{a}} \]

This results in the solution:

\[ \psi_0(x) = \left(\frac{m \omega}{\pi \hbar}\right)^{1/4} e^{-\frac{m\omega}{2\hbar} x^2} \]

Solutions with the ladder operators

Other solutions \(\psi_n(x)\) can now be generated:

\[ \psi_n(x) = A_n \, (\hat{a}_+)^n \, \psi_0(x), \quad \textrm{with} \quad E_n = \left(n + \frac{1}{2}\right) \hbar \omega \]

The normalization factor \(A_n\) can be calculated

\[ \psi_n(x) = \frac{1}{\sqrt{n!}} \, (\hat{a}_+)^n \, \psi_0(x), \quad \textrm{with} \quad E_n = \left(n + \frac{1}{2}\right) \hbar \omega \]

And operating with a single ladder operator:

\[ \hat{a}_+ \psi_n = \sqrt{n+1} \psi_{n+1}, \qquad \hat{a}_- \psi_n = \sqrt{n} \psi_{n-1} \]

Summary

  • Infinite well
    • Eigenstates evolve different in time
    • Pure eigenstates are stationary for finite expectation energy \(\langle \hat{H} \rangle\)
    • Mixing of eigenstates leads to non-constant \(\langle \hat{x} \rangle\), i.e. a nonzero velocity
  • Harmonic oscillator
    • Energy levels equally spaced \(\quad E_n = \hbar \omega (n + 1/2)\)
    • Nonzero ground energy \(\,\,E_0 = \frac{1}{2}\hbar\omega\)
    • Solutions proportional with Hermite polynomials \(\,H_n(x)\)
    • Alternative algebraic method
    • Ladder operators (Algebraic method)

Lecture 03: The time-independent Schrodinger equation (ctu’d)