PHOT 301: Quantum Photonics

LECTURE 04

Michaël Barbier, Summer (2024-2025)

Overview

For next week

  • Textbook Chapter 2: 2.11, 2.13, 2.14, 2.17, 2.18, 2.25, 2.31, 2.34, 2.41, 2.53

  • Homework documents:

    • phot301_homework_matrices.pdf
    • phot301_homework_system_of_equations.pdf
    • phot301_homework_eigenvalue_equations.pdf

  • Reading (by Thursday 31 July 2025): Chapter 3 of Griffiths

Summary

So far we looked at bound states

  • Infinite well
  • Linear potential well (Electrical field, not seen yet)
  • Harmonic oscillator

Different well potentials lead to different allowed energy levels

Narrower wells \(\longrightarrow\) less energy levels (more spread)

Discrete spectrum of energy-levels

Free particles

Free particle: propagating waves

\[ -\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = E \psi(x), \quad \textrm{ as } V(x) = 0 \]

\[ \Rightarrow \quad \frac{d^2 \psi(x)}{dx^2} = - k^2 \psi(x), \quad \textrm{ with } k = \frac{\sqrt{2 m E}}{\hbar} \]

  • Solutions are unconstrained: all energy values
  • Similar to a very wide well (with infinite walls)

\[ \psi(x) = A e^{ikx} + B e^{-ikx} \]

Free particle solutions

\[ \psi(x) = A e^{ikx} + B e^{-ikx} \]

  • Problem 1: Not normalizable
  • Problem 2: Velocity is half of classical velocity

Problem 1: Not normalizable

\[ \psi(x) = A e^{ikx} + B e^{-ikx} \]

\[ \begin{aligned} |\psi(x)|^2 &= \psi(x)^* \, \psi(x)\\ &= (A^* e^{-ikx} + B^* e^{ikx}) (A e^{ikx} + B e^{-ikx})\\ &= |A|^2 + |B|^2 + (A B^* e^{i2kx} + A^* B e^{-i2kx})\\ &= |A|^2 + |B|^2 + \Re\left(A B^* e^{i2kx}\right)\\ \Rightarrow \int_{-\infty}^{+\infty} |\psi(x)|^2 dx &= (|A|^2 + |B|^2) \, \infty + \int_{-\infty}^{+\infty} \Re\left(A B^* e^{i2kx}\right) dx\\ \end{aligned} \]

The last integral is bounded (oscillating between finite values)

The total integral \(\int |\psi|^2 dx \longrightarrow +\infty\), and therefore doesn’t exist.

Problem 2: Velocity too slow

We can rewrite \(\Psi(x, t) = \psi(x) e^{-iEt/\hbar}\) with \(E = \frac{\hbar^2 k^2}{2m}\):

\[ \begin{aligned} \Psi(x, t) &= A e^{ikx - iEt/\hbar} + B e^{-ikx - iEt/\hbar}\\ &= A e^{ikx - i\hbar k^2 t/2m} + B e^{-ikx - i\hbar k^2 t/2m}\\ &= A e^{ik(x - \frac{\hbar k}{2m}t)} + B e^{-ik(x - \frac{\hbar k}{2m}t)}\\ \end{aligned} \]

  • Both exponentials are a function of \(x \pm v t\), a “wave” moving with velocity \(v\)

Example of a propagating “wave”. It does not change its shape in time.

Problem 2: Velocity too slow

We can rewrite \(\Psi(x, t) = \psi(x) e^{-iEt/\hbar}\) with \(E = \frac{\hbar^2 k^2}{2m}\):

\[ \begin{aligned} \Psi(x, t) &= A e^{ikx - iEt/\hbar} + B e^{-ikx - iEt/\hbar}\\ &= A e^{ikx - i\hbar k^2 t/2m} + B e^{-ikx - i\hbar k^2 t/2m}\\ &= A e^{ik(x - \frac{\hbar k}{2m}t)} + B e^{-ik(x - \frac{\hbar k}{2m}t)}\\ \end{aligned} \]

  • Both exponentials are a function of \(x \pm v t\), a “wave” moving with velocity \(v\)

Real part of a propagating wave \(e^{ik(x - \frac{\hbar k}{2m}t)}\) (Propagating to the right).

Problem 2: Velocity too slow

We can rewrite \(\Psi(x, t) = \psi(x) e^{-iEt/\hbar}\) with \(E = \frac{\hbar^2 k^2}{2m}\):

\[ \begin{aligned} \Psi(x, t) &= A e^{ikx - iEt/\hbar} + B e^{-ikx - iEt/\hbar}\\ &= A e^{ik(x - \frac{\hbar k}{2m}t)} + B e^{-ik(x - \frac{\hbar k}{2m}t)}\\ \end{aligned} \]


  • This is a function of \(x \pm v t\), a “wave” moving with velocity \(v\)
  • The velocity is \(v_\textrm{quantum} = \frac{\hbar k}{2m} = \sqrt{\frac{E}{2m}}\)
  • Classically the velocity \(v_\textrm{classical} = \sqrt{\frac{2 E}{m}}\) because \(E = \frac{1}{2}m v^2\)
  • The classical velocity is twice the one according to quantum mechanics!

Wave packets

  • Superposition of propagating waves is normalizable
  • Solves the velocity problem as well!
  • Is consistent with uncertainty principle
  • Fourier’s trick to find coefficients

Image from Wikipedia

Localized waves using more k-values

  • Superposition of waves
  • Momentum around main \(\sum_{n=-N}^N k_0 \pm n\delta k\)
  • Plots from top to bottom \(N = 1, 2, 4, 8\)

\[ \psi(x) = \sum_{n=-N}^N e^{i (k_0 + n\delta k) x} \]

Group velocity of the wave packet

Phase velocity \(v\)

\[ v = \frac{\omega}{k} \]

Group velocity:

\[ v_g = \frac{d\omega}{dk} \]

\[ \Psi(x, t) = \sum_{n=-N}^N e^{i (k_0 + n\delta k) x} \]

Phase and group velocity

\[ \begin{aligned} \Psi(x, t) &= \frac{1}{\sqrt{2\pi}}\int \phi(k) e^{i (k x - \omega t)}\\ \end{aligned} \]

Assume \(\phi(k)\) around \(k_0\)

\[ \begin{aligned} \Psi(x, t) &\approx \frac{1}{\sqrt{2\pi}}\int \phi(k_0 + s) e^{i ((k_0 +s) x - (\omega_0 + s\omega_0') t)}\\ &\approx \frac{1}{\sqrt{2\pi}} e^{i (k_0 x - \omega_0 t)} \int \phi(k_0 + s) e^{i s(x - \omega_0' t)}\\ \end{aligned} \]

General wave packets: Fourier transforms

  • Wave packets of any shape
  • Includes all possible k-values (energies)

\[ \Psi(x, t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx - \frac{\hbar k^2}{2m} t)}\, dk \]

  • Time-independent \(\Psi(x, 0) = \psi(x)\):

\[ \Psi(x, 0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{ikx}\, dk \]

\[ \begin{aligned} \textrm{Fourier transform}\qquad\qquad \textrm{Inverse Fourier transform}\\ \phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi(x) e^{-ikx}\, dx, \qquad \psi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{ikx}\, dk \end{aligned} \]

Wave packet time evolution

  • We know \(\Psi(x, 0)\) and

\[ \begin{aligned} \Psi(x, t) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx - \omega(k) t)}\, dk,\\ &\\ \omega(k) &= \frac{\hbar k^2}{2m} \end{aligned} \]

  • Phase velocity of \(k < \langle k \rangle\) slower
  • Phase velocity of \(k > \langle k \rangle\) faster

–> Wave packets broadens in time

Bound states and scattering

Bound states and free particles

  • Bound states have a bounded domain
  • Particle is stuck in a potential valley
  • Free particles can go everywhere
  • Scattering of particles with \(E > V_\infty\)
  • Scattering also called Tunneling when \(E < V_\textrm{max}\)

Intermezzo: The Dirac Delta function

Dirac delta distribution:

\[ \left\{ \begin{aligned} \delta(x \neq 0) &= 0\\ \delta(x = 0) &= +\infty \end{aligned} \right. \]

\[ \int_{-\infty}^{+\infty} \delta(x) = 1 \]

Limit of series of functions:

  • peaked such as sinc(x) or Gaussian
  • limit to infinitely thin and high
  • Area kept normalized

Filters out single point: \(f(a) = \int_{-\infty}^{+\infty} f(x) \, \delta(x-a) \, dx\)

Delta function potential well

The potential energy function:

\[ V(x) = -\alpha\delta(x \neq 0) \]

\[ \left\{ \begin{aligned} V(x \neq 0) = -\alpha\delta(x \neq 0) &= 0\\ V(x \neq 0) = -\alpha\delta(x \neq 0) &= -\alpha\infty \end{aligned} \right. \]

\[ \begin{aligned} -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} -\alpha \delta(x)\psi &= E \psi\\ \end{aligned} \]

Bound states have \(E < 0\). Outside the well:

\[ \begin{aligned} -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} &= E \psi\\ \end{aligned} \]

where \(E < 0\) and thus exponential solutions

Delta function potential well

Bound states have \(E < 0\). Outside the well:

\[ \begin{aligned} -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} &= E \psi\\ \end{aligned} \]

Exponential solutions:

\[ \psi(x) \propto e^{\pm \kappa x}, \qquad \textrm{with}\quad \kappa = \frac{\sqrt{-2mE}}{\hbar} \]

Wave function must be continuous:

\[ \psi(x < 0) = \psi(x > 0) \]

Delta function potential well

Discontinuity: integrate the Schrodinger equation

\[ \begin{aligned} -\frac{\hbar^2}{2m} \int_{-\epsilon}^{+\epsilon} \frac{d^2\psi(x)}{dx^2} \,dx -\alpha \int_{-\epsilon}^{+\epsilon} \delta(x)\psi \, dx &= E \int_{-\epsilon}^{+\epsilon}\psi \,dx\\ &\\ -\frac{\hbar^2}{2m} \frac{d\psi(x)}{dx}\Big|_{-\epsilon}^{+\epsilon} - \alpha\psi(0) &= 0\\ \end{aligned} \]

Then take the limit for \(\epsilon \longrightarrow 0\)

\[ \Delta \left(\frac{d\psi(x)}{dx}\right) = -\frac{2m}{\hbar^2}\alpha\psi(0) \]

We know the derivatives in \(x = \pm 0\):

\[ -B\kappa - (B\kappa) \quad\Rightarrow\quad \Delta \left(\frac{d\psi(x)}{dx}\right) = -2\kappa B \]

Delta function potential well

\[ \left\{ \begin{aligned} \Delta \left(\frac{d\psi(x)}{dx}\right) &= -\frac{2m}{\hbar^2}\alpha\psi(0)\\ \Delta \left(\frac{d\psi(x)}{dx}\right) &= -2\kappa B \end{aligned} \right. \]

Make use of \(\psi(0) = B\)

\[ \begin{aligned} -\frac{2m}{\hbar^2}\alpha B = -2\kappa B\\ \Rightarrow\quad \kappa = \frac{m\alpha}{\hbar^2}\\ \Rightarrow\quad E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2}\\ \end{aligned} \]

The eigenstate with energy \(E\) becomes after normalization:

\[ \psi = B e^{-\kappa |x|} = \sqrt{\kappa} e^{-\kappa |x|} \]

Finite potential barriers/steps/wells

Finite square well

  • Finite square potential well

  • Bound states \(0 < E < V_0\)

  • Scattering states \(E > V_0\)

Schrodinger equation:

\[ - \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = \left(E - V(x)\right) \, \psi(x) \]

Finite well: bound states

For bound states \(0 < E < V_0\)

\[ \textrm{inside: }\quad \psi(-a < x < a) = C \sin(\lambda x) + D \cos(\lambda x) \qquad \lambda = \frac{\sqrt{2 m E}}{\hbar} \]

\[ \textrm{outside:} \quad \begin{aligned} \psi(x < -a) &= A e^{-\kappa x} + B e^{\kappa x} = B e^{\kappa x}\\ \psi(x > a) &= F e^{-\kappa x} + G e^{\kappa x} = F e^{-\kappa x} \end{aligned} \qquad \kappa = \frac{\sqrt{2 m (V_0 - E)}}{\hbar} \]

Finite well: bound states

Since \(\psi(x)\) is continuous in \(x = -a\) and \(x = a\):

\[ \begin{aligned} B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\ F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\ \end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a \]

\[ \begin{aligned} B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + D \sin(\lambda a)\\ -F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\ \end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a \]

Finite well: bound states

\[ \begin{aligned} B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\ F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\ \end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a \]

\[ \begin{aligned} B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + \lambda D \sin(\lambda a)\\ -F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\ \end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a \]

Add the first two equations and subtract the others:

\[ \begin{aligned} (B + F) e^{-\kappa a} &= 2 D \cos{\lambda a}\\ (B + F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 D \sin{\lambda a}\\ \end{aligned} \]

And divide them \(\Leftrightarrow B \neq -F\)

\[ \begin{aligned} \frac{\kappa}{\lambda} &= \tan{\lambda a}\\ \end{aligned} \]

Finite well: bound states

\[ \begin{aligned} B e^{-\kappa a} &= -C \sin(\lambda a) + D \cos(\lambda a)\\ F e^{-\kappa a} &= C \sin(\lambda a) + D \cos(\lambda a)\\ \end{aligned} \qquad \psi(x) \textrm{ continuous in } x = -a, a \]

\[ \begin{aligned} B \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) + \lambda D \sin(\lambda a)\\ -F \kappa e^{-\kappa a} &= \lambda C \cos(\lambda a) - \lambda D \sin(\lambda a)\\ \end{aligned} \qquad \frac{d\psi(x)}{dx} \textrm{ continuous in } x = -a, a \]

Now subtract the first two equations and add the others:

\[ \begin{aligned} (B - F) e^{-\kappa a} &= -2 C \sin{\lambda a}\\ (B - F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 C \cos{\lambda a}\\ \end{aligned} \]

And divide them \(\Leftrightarrow B \neq F\)

\[ \begin{aligned} \frac{\kappa}{\lambda} &= -\cot{\lambda a} \qquad \textrm{ if } B \neq F\\ \frac{\kappa}{\lambda} &= \tan{\lambda a} \qquad \textrm{ if } B \neq -F \textrm{ from before} \\ \end{aligned} \]

Finite well: bound states

\[ \begin{aligned} \frac{\kappa}{\lambda} &= -\cot{\lambda a} \qquad \textrm{ if } B \neq F\\ \frac{\kappa}{\lambda} &= \tan{\lambda a} \qquad \textrm{ if } B \neq -F \textrm{ from before} \\ \end{aligned} \]

  • The relations are inconsistent!
  • Only possible if one of them is invalid: \(B = F\) or \(B = -F\)

\[ B = F \quad \Rightarrow \quad \begin{aligned} (B - F) e^{-\kappa a} &= -2 C \sin{\lambda a}\\ (B - F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 C \cos{\lambda a}\\ \end{aligned} \quad \Rightarrow C = 0, \psi(x) = D \cos(\lambda x) \]

\[ B = -F \quad \Rightarrow \quad \begin{aligned} (B + F) e^{-\kappa a} &= 2 D \cos{\lambda a}\\ (B + F) \frac{\kappa}{\lambda} e^{-\kappa a} &= 2 D \sin{\lambda a}\\ \end{aligned} \quad \Rightarrow D = 0, \psi(x) = C \sin(\lambda x) \]

Finite well: bound states (wave function)

Symmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= D \cos(\lambda x)\\ \psi(x > a) &= B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)

Asymmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= C \sin(\lambda x)\\ \psi(x > a) &= -B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)

Finite well: bound states

Symmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= D \cos(\lambda x)\\ \psi(x > a) &= B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)

Asymmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= C \sin(\lambda x)\\ \psi(x > a) &= -B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)

Let’s first obtain the energy from \(\kappa\) and \(\lambda\):

\[ \kappa^2 = \frac{2m}{\hbar^2}(V_0 - E), \qquad \lambda^2 = \frac{2m}{\hbar^2}(E) \]

Define \(z = \lambda \, a\) and \(z_0 = a\, \frac{\sqrt{2mV_0}}{\hbar}\)

\[ \Longrightarrow \quad \kappa^2 + \lambda^2 = \frac{2m}{\hbar^2}(V_0) = z_0^2/a^2 \quad \Longrightarrow \quad \kappa^2 a^2 = z_0^2 - z^2 \]

Finite well: bound states

Symmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= D \cos(\lambda x)\\ \psi(x > a) &= B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)

\[ \Longrightarrow \tan z = \sqrt{\frac{z_0^2}{z^2} - 1} \]

Asymmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= C \sin(\lambda x)\\ \psi(x > a) &= -B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)

\[ \Longrightarrow -\cot z = \sqrt{\frac{z_0^2}{z^2} - 1} \]

where we made use of:

\[ \kappa^2 a^2 = z_0^2 - z^2 \Longrightarrow \kappa/\lambda = \sqrt{z_0^2/z^2 - 1} \]

Finite well: E for symmetric bound states

Symmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= D \cos(\lambda x)\\ \psi(x > a) &= B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = B \quad \& \quad \kappa = \lambda \tan \lambda a\)

\[ \Longrightarrow \tan z = \sqrt{\frac{z_0^2}{z^2} - 1} \]

Finite well: E for asymmetric bound states

Asymmetric solutions

\[ \left\{ \begin{aligned} \psi(x < -a) &= B e^{\kappa x}\\ \psi(-a < x < a) &= C \sin(\lambda x)\\ \psi(x > a) &= -B e^{-\kappa x}\\ \end{aligned} \right. \]

with \(F = -B \quad \& \quad \kappa = -\lambda \cot \lambda a\)

\[ \Longrightarrow -\cot z = \sqrt{\frac{z_0^2}{z^2} - 1} \]

Asymmetric energies approximately “shifted” to the right

Finite well: graphical energy-levels

  • Symmetric and asymmetric solutions
  • Energy levels at intersections of red curve with blue/green curves.
  • Energies close to infinite well energies (vertical asymptotes):

\[ E_n = \frac{\hbar^2 z^2}{2m a^2} \lessapprox E^\infty_n = \frac{\hbar^2}{2m}\frac{n^2\pi^2}{(2a)^2} \]

  • \(z_0\) is intersection of red curve with x-axis
  • Increasing \(z_0 \, \Longrightarrow \,\) more energy-levels
  • Decreasing \(z_0 \, \Longrightarrow \,\) less energy-levels
  • \(z_0 = \frac{\sqrt{2mV_0}}{\hbar} \, a \propto \sqrt{V_0} a\)

Finite well: numerical solutions

Numerically solving for \(z\):

\[ \begin{aligned} \tan z &= \sqrt{z_0^2/z^2 - 1},\\ -\cot z &= \sqrt{z_0^2/z^2 - 1}\\ \end{aligned} \]

Gives us:

  • Energy levels \(E_n = \frac{\hbar^2\lambda^2}{2m} = \frac{\hbar^2 z^2}{2m a^2}\)
  • Values for \(C\) or \(D\) as function of \(B\):

\[ \begin{aligned} B &= D \cos(\lambda a) e^{\kappa a}\\ B &= -C \sin(\lambda a) e^{\kappa a}\\ \end{aligned} \]

  • Normalization \(\longrightarrow\) final unknown \(B\)

Scattering

Probability density current

  • Probability \(\rho(x,t) = |\Psi(x,t)|^2\)
  • Continuity equation

\[ \frac{\partial \rho(x,t)}{\partial t} = \]

\[ \frac{\partial \rho(x)}{\partial x} \]

  • Change in \(\rho(x,t)\) leads to probability current

\[ \frac{\partial \rho(x)}{\partial x} \]

\[ \frac{\partial \rho(x)}{\partial x} \]

Scattering at a barrier

Scattering at a barrier

Scattering at a single potential step

  • Scattering states \(E > V_0\)
  • Classically no reflection
  • \(R\): Reflection coefficient, probability to reflect
  • \(T\): Transmission probability/coefficient
  • Total probability one:

\[ T + R = 1 \]

Schrodinger equation (TISE):

\[ - \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = \left(E - V(x)\right) \, \psi(x) \]

Scattering at a single potential step

  • Scattering states \(E > V_0\)
  • Classically no reflection
  • Reflection coefficient \(R\): probability to reflect
  • Transmission probability/coefficient \(T\)
  • Total probability one:

\[ T + R = 1 \]

Schrodinger equation (TISE):

\[ - \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = \left(E - V(x)\right) \, \psi(x) \]