Extra Credit 5: The Classical Wave Equation and Separation of Variables

Last time:

Two-slit experiment

• 2 paths to the same point on a screen
• 2 paths differ by \(nλ\)-constructive interference
• 1 photon interferes with itself
• get 1 dot on screen-collapse of “state of system” to a single dot to determine the state of the system, need many experiments, many dots.

Probability amplitude distribution (encodes 10, 01, or 11 where 1 = open, 0 = closed) collapses to single dot due to the act of detection of a photon.

Quantum Mechanics: information about the experimental setup (i.e. “the system”) is “encoded” in results of a sequence of independent experiments.

Musicians know that the sound produced by an instrument reveals

* detailed physical structure of the instrument

e.g. drum head shaped as

Same as for Quantum Mechanics.

Today:

• philosophy
• wave equation
• separation of variables
• boundary conditions — normal modes
• superposition of normal modes: “the pluck”
• cartoons of motion

What do we know so far?

• weirdness
• wave-particle duality
• interference experiment samples probability amplitude (i.e. + or –) distribution
• we can’t see inside microscopic systems
• we do experiments that indirectly reveal structure and mechanism
• patterns-like interference structure – reveal structure and mechanism
• the spectrum contains patterns

1st 1/2 of 5.61 deals with exactly solved problems

• Particle in a Box
• Harmonic Oscillator
• Rigid Rotor
• Hydrogen Atom

These are templates for our understanding of reality

Perturbation Theory will show us how to use the patterns associated with these simple problems to represent and decode reality.

I have been told many times that 5.61 is very difficult because it is very mathematical.

This lecture might be the most mathematical of the entire 5.61 course.

The goal is insight. For chemists, this is usually pictorial and qualitative.

I intend to show the pictures and insights behind the equations.

What are you expected to do when faced with one of the many differential equations in Quantum Mechanics?

1. Know where the differential equation comes from (not derive it)
2. Know standard methods used (by others) to solve it.

* a most common Ordinary Differential Equation

\[\dfrac{\mathrm{d}^{2} f}{\mathrm{d} x^{^{2}}}=kf\]

always 2 linearly independent general solutions for a 2nd order equation.

*find a way to rewrite your equation as one of the well-known solved equations

*separation of variables

What are we looking for?

*general solutions

• nodes (adjacent node spacing is \(λ/2\))
• envelope (related to probability)
• phase velocity

*specific physical system, specific solution

• Boundary conditions
• usually get some sort of quantization from 2nd boundary condition
• “normal modes”
• qualitative sketch: nodes, envelope, frequency for each normal mode

* initial condition: the pluck

• superposition of normal modes
• localization, motion, dephasing, rephrasing

Invert all of this into a description of the physical system.

The Wave Equation

Where does it come from? From Hooke’s Law (for a spring)

\[F=-kx\]

chop string into small segments

segment –1 pulls segment 0 down by force

\[-k[u(x_{0})-u(x_{-1})]\]

segment +1 pulls segment 0 up by force

\[-k[u(x_{0})-u(x_{-1})]\]

The net force is

\[-k [\Delta u_{10}-\Delta u_{0-1}]\]

\[[\Delta u_{10}-\Delta u_{0-1}]=\dfrac{\mathrm{d^{2}}u}{\mathrm{d} x_{2}}\]

\(F = ma\) (units conversion: contains tension and mass of string)

wave equation is

\[\dfrac{\partial^2 u}{\partial x^2}=\dfrac{1}{v^{2}}\dfrac{\partial^2 u}{\partial t^2} \label{WaveEQ}\]

• \(u\) is displacement
• \(v\) is velocity (as you would discover later)

How do we solve Equation \ref{WaveEQ}, which is a second-order, linear, partial differential equation?

• look for a similar, exactly solved problem
• employ bag of tricks

The most important trick is separation of variables where we ask if

\[u(x,t)=X(x)T(t) \label{ansatz}\]

is a viable solution for Equation \ref{WaveEQ}? If not, then we will get \(u(x,t)=0\). Substitute Equation \ref{ansatz} into Equation \ref{WaveEQ} to get

\[\dfrac{\partial^2 }{\partial x^2}[X(x)T(t)]=\dfrac{1}{v^{2}}\dfrac{\partial^2 }{\partial t^2}[X(x)T(t)] \label{eq20}\]

Since \(T(t)\) is only a function of \(t\) and X(x) is only a function of \(x\), then Equation \ref{eq20} can be simplified

\[T(t) \dfrac{\partial^2 X(x) }{\partial x^2}=X(x) \dfrac{1}{v^{2}}\dfrac{\partial^2 T(t)}{\partial t^2} \label{eq21}\]

move the \(T(t)\) to the right side and \(X(x)\) to the left side of Equation \ref{eq21} to get

\[\dfrac{1}{X(x)}\dfrac{\partial^2 X}{\partial x^2}=\dfrac{1}{v^{2}}\dfrac{1}{T(t)}\dfrac{\partial^2 T}{\partial t^2}\]

\(x\) and \(t\) are independent variables. This equation can only be valid if both sides are equal to a constant (called the separation constant).

\[\dfrac{1}{X}\dfrac{\mathrm{d^2}X }{\mathrm{d} x^{2}}=K \label{space}\]

\[\dfrac{1}{v^{2}}\dfrac{1}{T}\dfrac{\mathrm{d^2}T }{\mathrm{d} t^{2}}=K \label{time}\]

Note we have total not partial derivatives: linear, 2nd-order, and ordinary differential equation.

The Spatial Component \(X(x)\)

The general solutions for the spatial part (Equation \ref{space}) have the form

\[k>0\qquad e^{+kx} , e^{-kx}\qquad let\,K=k^2\]

or

\[k<0\qquad sin kx, cos kx\qquad let\,K=-k^2\]

(always have two linearly independent solutions for 2nd-order equation)

\[k>0\qquad general\,solution\qquad X(x)=Ae^{kx}+Be^{-kx}\]

\[k<0\qquad general\,solution\qquad X(x)=C\sin\,kx-D\cos\,kx\]

also for \(T(t)\) equation

\[\dfrac{\mathrm{d^2} T}{\mathrm{d} t^2}=v^{2}KT\]

\[K>0\qquad (T(t)=E e^{vkt}+Fe^{-vkt}\]

\[K<0\qquad (T(t)=G \sin\, {vkt}+Hcos\,vkt\]

Boundary Conditions

Now look at Boundary Conditions

\[u(0,t)=0\]

\[u(L,t)=0\]

For \(K>0\), try to satisfy boundary conditions

\[X(0)=Ae^{0}+Be^{-0}\]

\[A+B=0\]

\[A=-B\]

\[X(L)=0=Ae^{kL}+Be^{-kL}=A(e^{kL}-e^{-kL})\]

\[e^{kL}-e^{-kL}\]

\[A=0\]

\[u(x,t)=0\]

looks bad. What about \(K< 0\) solutions?

\[X(0)=C\sin\,0+D\cos\,0=0\]

\[D=0\]

\[X(L)=C\sin\,kL+0=0\]

\[kL=n\pi\qquad n=0,1,2,...\]

\["quantization" \qquad k_n=\dfrac{n\pi}{L}\]

pictures are drawn without looking at equation or using a computer to plot them.

• # nodes is n – 1
• nodes are equally spaced at \(x = L/n\), \(λn = 2(L/n)\).
• all lobes are the same, except for alternating sign

Wonderful qualitative picture: cartoon

The Temporal Component \(T(t)\)

Now look at \(T(t)\) equation for \(K < 0\).

\[T(t)=E \sin\,vk_nt+F \cos\,vk_nt\]

with \(\omega_n\equiv vkn\)

\[T(t)=E_nsin\,\omega_nt+F_ncos\,\omega_nt\]

Normal modes

\[u_n(x,t)=(A_nsin\,\dfrac{n\pi}{L}x)(E_nsin\,n\omega{t}+F_ncos\,n\omega{t})\]

The time dependent factor of the \(n^{th}\) normal mode can be rewritten in "frequency, phase " form as

\[E{}'_{n}cos\,[n\omega{t}+\phi_{n}]\]

The next step is to consider the \(t=0\)pluck of the system. This pluck is expressed as a linear combination of the normal modes.

\[u_{pluck}(x,t)=\sum_{n=1}^{\infty}(A_{n}E{}'_{n})sin(\dfrac{n\pi}{L}x)cos(n\omega{t}+\phi_n)\]

There is a further simplification based on the trigonometric formula

\[sin\,a\,cos\,b=\dfrac{1}{2}[sin(a+b)+sin(a-b)]\]

which enables us to write \(u_{pluck}\) as

\[u_{pluck}(x,t)=\sum_{n=1}^{\infty}[\dfrac{{}A_{n}E{}'_{n}}{2}]\left \{ sin(\dfrac{n\pi}{L}{x}+n\omega{t}+\phi_{n})+sin(\dfrac{n\pi}{L}{x}-n\omega{t}-\phi_{n}) \right \}\]

Something wonderful happens now.

• A single normal mode is a standing wave. No left-right motion, no “breathing”
• A superposition of 2 or more normal modes with different values of n gives more complicated motion. For two normal modes, where one is even-n and the other is odd-n, the time-evolving wavepacket will exhibit left-right motion. For two normal modes where both are odd or both even, the wavepacket motion will be “breathing” rather than left-right motion.

Here is a crude time-lapse movie of a superposition of the \(n = 1\) and \(n = 2\) (fundamental and first overtone) modes.

The period of the fundamental is \(T=\dfrac{2\pi}{\omega}\). We are going to consider time-steps of \(T/8\).

The time-lapse movie of the sum of two normal modes can be viewed as moving to the left at \(t =–T/4\), close to the left turning point at \(t =–T/8\), at the left turning point but dephased at \(t=0\), moving to the right at \(t =+T/8\). It will reach the right turning point but dephased at \(t =T/2\).

In Quantum Mechanics you will see wavepackets that exhibit motion, breathing, dephasing, and rephrasing. The “center of the wavepacket” will follow a trajectory that obeys Newton’s laws of motion.

If we generalize from waves on a string of waves on a rectangular drumhead,

The separable solution to the wave equation will have the form

\[u(x,y,t)=X(x)Y(y)T(t)\]

There will be two separation constants, and we will find that the normal mode frequencies are

\[\omega_{nm}=v\pi[\dfrac{n^2}{a^2}+\dfrac{m^2}{b^2}]^{\dfrac{1}{2}}\]

This is a more complicated quantization rule than for waves on a string, and it should be evident to an informed listener that these waves are on a rectangular drum head with edge lengths \(a\) and \(b\).

NON-Lecture

The underlying unity of the \(e^{kx}\), \(e^{-kx}\) and \(sin\,kx\), \(cos\,kx\) solutions to

\[\dfrac{\mathrm{d}^{2}y}{\mathrm{d} x^{2}}=k^{2}y\]

Let's take a step back and look at the two simplest 2nd-order ordinary differential equations:

\[\dfrac{\mathrm{d}^{2}y}{\mathrm{d} x^{2}}=+k^{2}y\rightarrow y(x)=Ae^{kx}+Be^{-kx}\]

and

\[\dfrac{\mathrm{d}^{2}y}{\mathrm{d} x^{2}}=-k^{2}y\rightarrow y(x)=Csin\,kx+Dcos\,kx\]

The solutions to these two equations are more similar than they look at first glance.

Euler's formula

\[e^{\pm i\theta}=\cos\,{\theta}\pm{i}\sin\,{\theta}\]

or

\[\dfrac{1}{2}(e^{i\theta}+e^{-i\theta})=\cos\theta\]

\[\dfrac{i}{2}(e^{-i\theta}-e^{i\theta})=\sin\theta\]

So we can express the solution of the second differential equation in (complex) exponential form to bring out its similarity to the solution of the first differential equation:

\[\begin{align} y(x)&=C\sin kx+D\ cos kx \\[5pt] &=\dfrac{i}{2}C(e^{-ikx}-e^{ikx})+\dfrac{1}{2}D(e^{ikx}+e^{-ikx}) \end{align}\]

rearrange

\[y(x)=\dfrac{1}{2}(D-iC)e^{ikx}+\dfrac{1}{2}(D+iC)e^{-ikx}\]

The \(sin\,\theta\), \(cos\,\theta\)and \(e^{i\theta}\), \(e^{-i\theta}\)forms are two sides of the same coin. Insight. Convenience. What do we notice? The general solutions to a 2nd-order differential equation consist of the sum of two linearly independent functions, each multiplied by an unknown constant.