PHOT 301: Quantum Photonics
LECTURE 01
Michaël Barbier, Fall semester (2024-2025)
\[ \vec{F} = m \, \vec{a} = \frac{\partial^2 r_i}{{\partial t}^2} \]
\[ \vec{F} = - \nabla V \]
\[ \left\{\begin{aligned} \nabla \cdot \vec{E} &= \frac{\rho}{\epsilon_0}\\ \nabla \cdot \vec{B} &= 0\\ \nabla \times \vec{E} &= - \frac{\partial \vec{B}}{\partial t}\\ c^2 \, \nabla \times \vec{B} &= \frac{\vec{J}}{\epsilon_0} + \frac{\partial \vec{E}}{\partial t}\\ \end{aligned}\right . \]
\[ \left\{\begin{aligned} \nabla^2 \vec{E} - \frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2} &= 0 \\ \nabla^2 \vec{B} - \frac{1}{c^2}\frac{\partial^2 \vec{B}}{\partial t^2} &= 0 \\ \end{aligned}\right . \]
Vector components \(u := E_i,\, B_i\) obey the wave equation:
\[ \nabla^2 u - \frac{1}{c^2}\frac{\partial u}{\partial t^2} = 0 \]
Light and matter are treated different
Problems:
Quantum mechanics combines both
And solves everything?
\[ i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi \]
where
We will first consider 1D problems:
\[ i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t) \]
\[ P(x \in [a,b]) = \int_a^b |\Psi(x, t)|^2 dx \]
\[ P(x \in [a,b]) = \int_a^b |\Psi(x, t)|^2 dx \]
Copenhagen interpretation:
Typical thought-experiment
Probability density function \(\rho(x)\):
\[ \begin{aligned} \textrm{Expectation value of }x \qquad & \langle x \rangle = \int_{-\infty}^\infty x \,\rho(x)\,dx\\ \textrm{Expectation value of }f(x) \qquad & \langle f(x) \rangle = \int_{-\infty}^\infty f(x) \,\rho(x)\,dx\\ \end{aligned} \]
Probability density function \(\rho(x)\):
\[ \begin{aligned} \textrm{Expectation value of }x \qquad & \langle x \rangle = \int_{-\infty}^\infty x \,\rho(x)\,dx\\ \textrm{Expectation value of }f(x) \qquad & \langle f(x) \rangle = \int_{-\infty}^\infty f(x) \,\rho(x)\,dx\\ \textrm{Variance }\sigma^2 \qquad & \langle (\Delta x)^2 \rangle = \int_{-\infty}^\infty (x - \langle x \rangle)^2 \,\rho(x)\,dx\\ \qquad & \qquad\quad = \langle x^2 \rangle - \langle x \rangle^2\\ \textrm{Standard deviation } \qquad & \sigma = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}\\ \end{aligned} \]
\[ \int_{-\infty}^\infty |\Psi(x, t)|^2 dx = 1 \]
So wave function \(\Psi(x, t)\):
\[ \Longleftrightarrow \int_{-\infty}^\infty |\Psi(x, t)|^2 dx \textrm{ exists and is finite} \]
If \(\Psi(x, t)\) is normalized at \(t = 0\) then it is always normalized.
Follows from the Schrodinger equation (see Griffiths page 15): \[ \frac{d}{dt} \int_{-\infty}^\infty |\Psi(x,t)|^2 dx = 0 \]
What are the particle’s:
Calculate the expectation (average) values:
\[ \textrm{Expectation value of }x \qquad \langle x \rangle = \int_{-\infty}^\infty x \, |\Psi(x,t)|^2 \, dx \]
\[ \textrm{Expectation value of }p = mv \qquad m\frac{d \langle x \rangle}{dt} = -i\hbar \int_{-\infty}^\infty \, \Psi^* \frac{\partial \Psi}{\partial x} \, dx \]
Expectation values are calculated as
\[ \textrm{Position }x \qquad \langle x \rangle = \int_{-\infty}^\infty x \, |\Psi(x,t)|^2 \, dx = \int_{-\infty}^\infty \, \Psi^* \, [x] \, \Psi \, dx \]
\[ \textrm{Momentum }p \quad m\frac{d \langle x \rangle}{dt} = -i\hbar \int_{-\infty}^\infty \, \Psi^* \frac{\partial \Psi}{\partial x} \, dx = \int_{-\infty}^\infty \, \Psi^* [-i\hbar\frac{\partial }{\partial x}]\,\Psi \, dx \]
\[ \textrm{Position operator} \qquad \hat{x} = x \] \[ \textrm{Momentum operator} \qquad \hat{p} = -i\hbar\frac{\partial }{\partial x} \]
\[ i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x} \Psi + V \Psi \]
We have the operators:
\[ \textrm{Position operator} \qquad \hat{x} = x \] \[ \textrm{Momentum operator} \qquad \hat{p} = -i\hbar\frac{\partial }{\partial x} \]
Using operators in the Schrodinger equation:
\[ i \hbar \frac{\partial }{\partial t} \Psi = \frac{1}{2m}[-i\hbar\frac{\partial }{\partial x}]^2 \Psi + V \Psi = \frac{\hat{p}^2}{2m} \Psi + V \Psi = (\hat{T} + \hat{V}) \Psi = \hat{\mathcal{H}} \Psi \]
\[ m \frac{d}{dt}\langle x \rangle, \qquad \frac{d}{dt}\langle p \rangle = - \langle\frac{\partial V(x)}{\partial x}\rangle \]
\[ p = \frac{h}{\lambda} = \frac{2\pi\hbar}{\lambda} \quad (= \hbar k) \]
\[ \sigma_x \sigma_p \ge \frac{\hbar}{2} \]
\[ i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi \]
Lecture 01: Introduction to the Schrodinger equation