Review: Properties of stationary eigenstates
\[
\begin{aligned}
\psi_n \textrm{ are orthonormal}\quad & \int \psi_m(x)^* \, \psi_n(x) \, dx = \delta_{mn}\\
\psi_n \textrm{ form a complete basis}\quad & f(x) = \sum_{n = 1}^\infty c_n \psi_n(x) \qquad \forall f(x)\\
\textrm{Coefficients $c_n$ are given by}\quad & c_n = \int \psi_n(x)^* \, f(x) \, dx\\
\end{aligned}
\]
Coefficients \(|c_n|^2\) give the probability to measure energy as \(E_n\):
\[
\langle \hat{H} \rangle = \int \Psi^* \hat{H} \Psi dx = \sum_{n = 1}^\infty |c_n|^2 E_n
\]
Expectation values for operators \(\hat{x}\), \(\hat{p}\), etc.
\[
\langle \hat{x} \rangle(t) = \int \Psi^* \hat{x} \Psi \, dx = \int x |\Psi|^2 \, dx
\]
Introduction
- Ball-spring problem
- Analog RCL electric circuit
- Many systems are (approximately) harmonic oscillators
- Optical cavity
- 2nd order Taylor approximation \(V(x)\)
- Phonons, vibrations in molecules/matter
- Quantization of light: Photons
Classical harmonic oscillator: parabolic well
- Mass in parabolic well \(V(x) = \alpha mgx^2\)
- Restoring force: \(F = -\frac{dV(x)}{dx} = -2\alpha mgx\)
- Motion via Newton’s equation \(F = m a\):
\[
m a = m \frac{d^2x}{dt^2} = - 2\alpha mgx
\]
Linear equation with constant coefficients
\[
\frac{d^2x}{dt^2} = - 2\alpha g x = -\omega^2 x, \textrm{ with } \omega = \sqrt{2\alpha g}.
\]
Resulting solutions are: \(\quad x \propto \sin(\omega t)\)
Turning points at \(\pm x_\textrm{max}\): \(\quad\alpha m g x_\textrm{max}^2 = \frac{1}{2} m v_0^2\)
Classical harmonic oscillator: ball-spring
- mass attached to a spring
- Restoring force: \(F = -\frac{dV(x)}{dx} = -k x\)
- Motion via Newton’s equation \(F = m a\):
\[
m a = m \frac{d^2x}{dt^2} = -k' x
\]
Linear equation with constant coefficients
\[
\frac{d^2x}{dt^2} = - \frac{k}{m} x = -\omega^2 x, \textrm{ with } \omega = \sqrt{k'/m}.
\]
Resulting solutions are: \(\quad x \propto \sin(\omega t)\)
Turning points at \(\pm x_\textrm{max}\): \(\quad\frac{1}{2} k' x_\textrm{max}^2 = \frac{1}{2} m v_0^2\)
Quantum harmonic osc.: diatomic molecule
- Vibrations approximate harmonic oscillator
- Restoring force: \(F = -\frac{dV(x)}{dx} = -k' x\)
- Schrodinger equation with potential:
\[
V(x) = -\frac{1}{2} k' x^2
\]
\[
\left\{
\begin{aligned}
&\textrm{Quantization energy-levels}\\
&\textrm{Groundstate nonzero energy}\\
&\textrm{Time-evolution }\,\,\rho(x,t) = |\Psi(x,t)|^2\\
\end{aligned}
\right.
\]
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
\[
\frac{1}{2} \frac{\partial^2}{\partial \xi^2}\psi(\xi) - \frac{1}{2} \xi^2 \psi(\xi) = -\frac{E}{\hbar \omega} \psi
\]
Standard method to solve differential equation
STEP 1: Try to find asymptotic solutions
\[
\lim_{\xi\longrightarrow\infty} \quad \Rightarrow\quad \frac{E}{\hbar \omega} \ll \frac{1}{2} \xi^2
\]
\[
\Rightarrow \frac{\partial^2}{\partial \xi^2}\psi(\xi) \approx \xi^2 \psi(\xi) \qquad \Rightarrow \qquad \psi \propto \exp(-\xi^2/2)
\]
STEP 2: Trial solution to hopefully simplify the equation
\[
\psi(x) = \exp(-\xi^2/2) H(\xi), \qquad \textrm{where solutions}\,\,\,H(\xi)\,\,\,\textrm{are yet unknown}
\]
Solving the QM harmonic oscillator
STEP 2: Trial solution to hopefully simplify the equation
Fill in trial solution \(\psi = \exp(-\xi^2/2) H(\xi)\) in the original equation.
\[
\frac{1}{2} \frac{\partial^2}{\partial \xi^2}\psi(\xi) - \frac{1}{2} \xi^2 \psi(\xi) = -\frac{E}{\hbar \omega} \psi
\]
(and we multiply equation by \(2\))
\[
\frac{\partial^2}{\partial \xi^2}\left[e^{-\xi^2/2} H(\xi)\right] - \xi^2 e^{-\xi^2/2} H(\xi) = -\frac{2E}{\hbar \omega} e^{-\xi^2/2} H(\xi)
\]
Then calculate 2nd derivative (\(f'(x) = \partial f(x)/\partial x\)):
\[
\begin{aligned}
\left[e^{-\xi^2/2} H(\xi)\right]'' &= \left[- \xi e^{-\xi^2/2} H(\xi) + e^{-\xi^2/2} H(\xi)\right]'\\
&= -e^{-\xi^2/2} H(\xi) + \xi^2 e^{-\xi^2/2} H(\xi) - 2 \xi e^{-\xi^2/2} H'(\xi) + e^{-\xi^2/2} H''(\xi)\\
\end{aligned}
\]
Solving the QM harmonic oscillator
STEP 2: Trial solution to hopefully simplify the equation
\[
e^{-\xi^2/2} H''(\xi) - 2 \xi e^{-\xi^2/2} H'(\xi) = \left(1 - \frac{2E}{\hbar\omega}\right)\, e^{-\xi^2/2} H(\xi)
\]
Divide by \(e^{-\xi^2/2}\)
\[
H''(\xi) - 2 \xi H'(\xi) = \left(1 - \frac{2E}{\hbar\omega}\right)\, H(\xi)
\]
Define dimensionless \(K \equiv \frac{2E}{\hbar\omega}\)
\[
H''(\xi) - 2 \xi H'(\xi) + \left(K - 1\right)\, H(\xi) = 0
\]
New differential equation: Simpler?
Solving the QM harmonic oscillator
STEP 3: Solve by power series expansion
\[
H''(\xi) - 2 \xi H'(\xi) + \left(K - 1 \right)\, H(\xi) = 0
\]
Assume \(H(\xi) = a_0 + a_1 \xi + a_2 \xi^2 + a_3 \xi^3 + \dots = \sum_n a_{j=0}^\infty a_j \xi^j\)
\[
\left\{
\begin{aligned}
H(\xi) &= a_0 + a_1 \xi + a_2 \xi^2 + a_3 \xi^3 + \dots = \sum_{j=0}^\infty a_j \xi^j\\
H'(\xi) &= a_1 + 2 a_2 \xi + 3 a_3 \xi^2 + \dots = \sum_{j=0}^\infty (j+1) a_{j+1} \xi^{j}\\
H''(\xi) &= 2 a_2 + 6 a_3 \xi + \dots = \sum_{j=0}^\infty (j+2)(j+1) a_{j+2} \xi^{j}\\
\end{aligned}
\right.
\]
Then fill in the power series in the equation
Solving the QM harmonic oscillator
STEP 3: Solve by power series expansion
\[
H''(\xi) - 2 \xi H'(\xi) + \left(K - 1 \right)\, H(\xi) = 0
\]
\[
\left\{
\begin{aligned}
H(\xi) &= a_0 + a_1 \xi + a_2 \xi^2 + a_3 \xi^3 + \dots = \sum_{j=0}^\infty a_j \xi^j\\
H'(\xi) &= a_1 + 2 a_2 \xi + 3 a_3 \xi^2 + \dots = \sum_{j=0}^\infty (j+1) a_{j+1} \xi^{j}\\
H''(\xi) &= 2 a_2 + 6 a_3 \xi + \dots = \sum_{j=0}^\infty (j+2)(j+1) a_{j+2} \xi^{j}\\
\end{aligned}
\right.
\]
Then fill in the power series in the equation:
\[
\sum_{j=0}^\infty\left[ (j+2)(j+1) a_{j+2} - 2 j a_{j} + \left(K - 1 \right)\, a_j \right] \xi^{j} = 0
\]
Solving the QM harmonic oscillator
STEP 3: Solve by power series expansion
\[
\sum_{j=0}^\infty\left[ (j+2)(j+1) a_{j+2} - 2 j a_{j} + \left(K - 1 \right)\, a_j \right] \xi^{j} = 0
\]
For every power of \(\xi\) equation needs to be zero
\[
(j+2)(j+1) a_{j+2} - 2 j a_{j} + \left(K - 1 \right)\, a_j = 0
\]
Solve for coefficients:
\[
a_{j+2} = \frac{2 j + 1 - K}{(j+2)(j+1)}\, a_j = 0
\]
If we know \(a_0\) (even series) and \(a_1\) (odd series) we know all \(a_n\)
Solving the QM harmonic oscillator
STEP 3: Solve by power series expansion
One problem: we require
\[
\lim_{\xi\longrightarrow\infty} e^{\xi^2/2} H(\xi) = 0
\]
This is not the case for our infinite series.
UNLESS the series terminates: \(\qquad \exists j: \quad 2 j + 1 - K = 0\)
Solutions exist for \(K - 1 = \frac{2E}{\hbar\omega} - 1 = 2n, \qquad n = 0, 1, 2, 3 \dots\)
\[
\longrightarrow \quad \left\{
\begin{aligned} \psi_n &= A_n \exp(-\xi^2/2) H_n(\xi), \\
&\\
E_n &= (n + 1/2) \hbar \omega \,\,\, \textrm{with} \,\, n = 0, 1, 2, \dots\\
\end{aligned}
\right.
\]
Harmonic oscillator solutions
\[
\left\{
\begin{aligned}
\psi_n &= A_n \exp(-\xi^2/2) H_n(\xi),\\
E_n &= \left(n + \frac{1}{2}\right) \hbar \omega \,\,\, \textrm{with} \,\, n = 0, 1, 2, \dots\\
A_n &= \sqrt{\frac{1}{\sqrt{\pi}2^n n!}} \qquad \xi = \sqrt{\frac{m \omega}{\hbar}}
\end{aligned}
\right.
\]
Hermite polynomials \(\, H_n(\xi)\,\) (our even/odd power series)
\[
\begin{aligned}
H_0 &= 1\\
H_1 &= 2\xi\\
H_2 &= 4\xi^2 - 2\\
H_3 &= 8\xi^3 - 12\xi\\
\vdots \,\, & \\
H_n(\xi) &= 2\xi H_{n-1}(\xi) - 2(n-1) H_{n-2}(\xi)\\
\end{aligned}
\]
Harmonic oscillator solutions
Solutions for the wave function:
\[
\psi_n(x) = \color{green}{\frac{1}{\sqrt{2^n \, n!}} \, \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}} \, \color{red}{H_n\left(\sqrt{\frac{m\omega}{\hbar}} \,x\right)} \, \color{blue}{e^{-\frac{m\omega}{2\hbar}\,x^2}}
\]
- \(\color{blue}{\textrm{Gaussian}}\)
- \(\color{green}{\textrm{Normalization}}\)
- \(\color{red}{\textrm{Hermite polynomials}}\,\,\,\color{red}{H_n(\xi)}\)
\[
\begin{aligned}
\color{red}{H_0} &\color{red}{= 1}\\
\color{red}{H_1} &\color{red}{= 2\xi}\\
\color{red}{H_2} &\color{red}{= 4\xi^2 - 2}\\
\color{red}{H_3} &\color{red}{= 8\xi^3 - 12\xi}\\
\color{red}{\vdots} \,\, & \\
\end{aligned}
\]
Harmonic oscillator solutions
\[
\textrm{Wave function}\quad \psi(\xi) \qquad\qquad\qquad \textrm{Probability density function}\quad |\psi(\xi)|^2
\]
![]()
High energy solutions
Classically \(x \in \left[-\sqrt{\frac{2E}{m\omega^2}}, \,\,\sqrt{\frac{2E}{m\omega^2}}\right]\quad\) and \(\quad|\psi_n|^2 \,\, \rightarrow\,\, \rho_\textrm{class.}(x) = \frac{1}{T}\,\frac{1}{v(x)}\)
![]()
Potential parameters & solutions
- Energy-levels \(E_n \propto \omega\), \(\qquad\) Width parabola scales with \(1/\omega\)
\[
E_n = \left(n + \frac{1}{2}\right) \hbar \omega \,\,\, \textrm{with} \,\, n = 0, 1, 2, \dots\\
\]
![]()
Energy infinite well vs. harmonic oscillator
Infinite well:
- \(E_n \propto n^2\)
- \(E_n \propto 1/L^2\)
\[
E_n = \frac{\hbar^2 \pi^2 n^2}{2mL^2}\\
\]
Quantum H.O.:
- equidistant \(E_n\)
- \(E_n \propto \omega\)
- \(\textrm{width} \propto 1/\omega\)
\[
E_n = (n + 1/2) \hbar\omega
\]
Alternative (Algebraic) derivation
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:
\[
\frac{1}{2m} \left(\hat{p}^2 + m^2\omega^2 x^2\right) \psi(x) = E \psi, \qquad \hat{p} = -i\hbar \frac{\partial}{\partial x}
\]
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
- Single eigenstates are stationary for finite expectation energy \(\langle \hat{H} \rangle\)
- Superposition 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)
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)
