PHOT 301: Quantum Photonics

LECTURE 02

Michaël Barbier, Fall semester (2024-2025)

Solving differential equations

Differential equations are only useful for a handful of people:

mathematicians (not relevant for all of them)
scientists
engineers
3rd year students of photonics engineering

\(\longrightarrow\) Please remember your course on differential equations
Or consult some resources such as “Notes on Diffy Qs”, Jiri Lebl

Stationary Solutions & Energy levels

Solving the 1D Schrodinger equation

The Schrodinger equation was given by:

\[ i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t) \]

The complex wave function \(\Psi(x, t)\) is not observable
Potential energy: \(V \rightarrow V(x,y,z,t)\)
\(\hbar = \frac{h}{2\pi} = 1.055 \times 10^{-34}\) J s
Probability to find particle in \(x\) at time \(t\) given by \(|\Psi(x, t)|^2\):

\[ P(x \in [a,b]) = \int_a^b |\Psi(x, t)|^2 dx \]

Solving the 1D Schrodinger equation

\[ i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t) \]

How do we solve this equation for given \(V(x,t)\) ?

Assume \(V(x, t)\) independent of time: \(V(x) \leftarrow V(x, t)\)
Solve by separation of the variables \(\Psi(x,t) = \psi(x) \phi(t)\)

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

Solving the 1D Schrodinger equation

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

\[ \Rightarrow i \hbar \psi(x) \frac{\partial \phi(t)}{\partial t} = -\phi(t)\frac{\hbar^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x)\phi(t) \]

Divide the equation by \(\Psi(x, t) = \psi(x) \phi(t)\)

\[ \Rightarrow i \hbar \frac{1}{\phi(t)} \frac{\partial \phi(t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) \]

\(\longrightarrow\) the left hand side depends only on \(x\) and the right hand side only on \(t\).

\[ \Rightarrow i \hbar \frac{1}{\phi(t)} \frac{\partial \phi(t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{1}{\psi(x)} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) = \textrm{constant } E \]

Time-dependence & Stationary equation

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

\(\longrightarrow\) System of 2 ordinary differential equations:

Time-dependency (left)	x-dependency (right)
\(\frac{d \phi(t)}{d t}=-\frac{i}{\hbar}E \phi(t)\)	\(-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{d x^2} + V(x) \psi(x) = E \psi(x)\)

IF we can solve both equations \(\Longrightarrow\) \(\Psi(x,t) = \psi(x) \phi(t)\) is a solution

Time evolution

Solving the equation for \(\phi(t)\)

\[\frac{d \phi(t)}{d t}=-\frac{i}{\hbar}E \phi(t)\]

1st order differential equation with general solution:

\[ \phi(t) = C \exp(- i E t /\hbar) \]

Full solution of the form (C is absorbed):

\[ \Psi(x,t) = \psi(x) \phi(t) = \psi(x) \exp(- i E t /\hbar) \]

Notice that the probability \(|\Psi(x, t)|^2 = |\psi(x)|^2\) is independent of \(t\)

Time-independent equation

Time-independent Schrodinger equation (TISE):

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

Or we can write

\[ \hat{H} \psi = E \psi \quad \textrm{with Hamiltonian }\,\, \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{d x^2} + V(x) \]

The expectation value of \(\hat{H}\) is:

\[ \langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \int \Psi^* E \Psi dx = E \int |\Psi|^2 dx = E \int |\psi|^2 dx = E \]

General Solution of the TDSE

From the theory of differential equations:
- The general solution is a linear superposition of solutions \(\{\psi_n(x)\} = \psi_1(x), \psi_2(x), \psi_3(x), \dots\)
- Independent solutions
- Separate energies \(\{E_n\}\) for corresponding \(\{\psi_n(x)\}\)
- Solutions form an infinite and complete basis

\[ \Psi(x, t) = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar} \]

Notice: General probability \(|\Psi(x, t)|^2\) does depend on time

General Solution of the TDSE

\[ \Psi(x, t) = \sum_{n = 1}^\infty c_n \psi_n(x) \, e^{-iE_n t/\hbar} \]

One can proof that \(|c_n|^2\) is the probability to measure energy as \(E_n\) (Griffith’s Chapter 3):

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

Potential energy function V(x)

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

Potential energy \(V(x)\) is linked to force \(F = - \frac{\partial V}{\partial x}\)

\(\Longrightarrow\) if \(V(x)\) is a constant corresponds to zero force

\(\Longrightarrow\) A linear \(V(x)\) corresponds to a constant force

\(\Longrightarrow\) A parabolic \(V(x)\) corresponds to a linear force (like a spring)

Square potential energy well

Infinite well

Inside the well a particle can exist
Outside the well the potential is infinite

\[ \left\{ \begin{aligned} V(x < 0) & = \infty \\ V(0 < x < L) & = 0 \\ V(x > L) & = \infty \\ \end{aligned} \right. \]

Task: solve the stationary Schrodinger equation for \(V(x)\)

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

Infinite well: Solution in the well

Particles outside would have infinite energy
Wave function \(\psi(x)\) should be zero outside
Assume \(\psi(0) = \psi(L) = 0\) \(\longleftarrow\) \(\psi(x)\) ctu
Inside the well \(V(x) = 0\):

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

General solution:

\[ \psi(x) = A \cos(k x) + B \sin(k x) \]

Infinite well: Solution in the well

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

\[ \psi(x) = A \cos(k x) + B \sin(k x) \]

with \(A\) and \(B\) complex numbers
\(k = \sqrt{2mE / \hbar^2}\) a complex number

Apply BC’s \(\psi(0) = \psi(L) = 0\):

\[ \psi(0) = 0 \quad\Rightarrow\quad A = 0 \\ \]

\[ \psi(x) = B \sin(k\,x) \\ \]

Infinite well: Energies

\[ \psi(x) = B \sin(k\,x) \\ \]

Apply the other BC: \(\quad\psi(L) = 0\):

\[ k_n = \sqrt{2mE_n/\hbar^2} = n \pi / L \\ \]

\[ \Longrightarrow \left\{ \begin{aligned} \psi_n(x) &= A_n \sin\left(\frac{n\pi x}{L}\right)\\ E_n &= \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2 \end{aligned} \right. \]

Infinite well

\[ \psi(x) = B \sin(k\,x) \\ \]

Apply the other BC: \(\quad\psi(L) = 0\):

\[ k_n = \sqrt{2mE_n/\hbar^2} = n \pi / L \\ \]

Infinite well: Normalization

\[ \begin{aligned} \psi_n(x) &= A_n \sin\left(\frac{n\pi x}{L}\right)\\ \end{aligned} \]

Obtain \(A_n\) from normalization \(\int|\psi|^2 = 1\)

\[ 1 = \int_0^L |A_n|^2 \left|\sin\left(\frac{n\pi x}{L}\right)\right|^2 dx = \frac{|A_n|^2 L}{2} \]

\[ \Longrightarrow \quad |A_n|^2 = \frac{2}{L} \Rightarrow |A_n| = \sqrt{\frac{2}{L}} \]

\[ \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \]

Infinite well: summary

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

Eigenenergies and eigenstates

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

Lowest state \(n = 1\) we call ground state
Higher states \(n > 1\) are excited states
Parity of wave functions is either:
- Even (\(n = 1, 3, 5, \dots\))
- Odd (\(n = 2, 4, 6, \dots\))

Properties

\[ \begin{aligned} \textrm{Eigenstates} \quad &\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\\ \end{aligned} \]

The eigenstates are orthonormal:

\[ \int \psi_m(x)^*\, \psi_n(x) dx = \delta_{nm} \]

Eigenstates form a complete basis
Every \(f(x)\) we can expand as a series:

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

A more complex example

Ammonia molecule has two possible geometries

The ammonia molecule \(NH_3\) has two possible geometries

Experiments tell that \(NH_3\) flips between states

Possible by quantum tunneling