# Modern Physics: Chapter 8 – Normalizing a Wavefunction

I discussed in the previous chapter you the probability of a particle being detected in the continous range $(x_{0}, x_{1})$ can be calculated from its wavefunction using,

• $\int_{x_{0}}^{x_{1}} \psi^{*} \psi dx$

Now, assuming the particle actually exists, if we essentially look for it at every point in the entire universe we’re bound the detect it somewhere right? Hence the probability of finding the particle in the range $(-\infty, \infty)$ should be $1$. In other words,

• $\int_{-\infty}^{\infty} \psi^{*} \psi dx$ must equal $1$.

So, what if we get a wavefunction, say. $g(x)$ where $\int_{-\infty}^{\infty} g^{*} g dx$ doesn’t equal $1$ and equals some other number $k$ instead? Then we can make the function’s probability equal $1$ bu multiplying it with a constant $A$, such that the corrected wavefunction is:

• $\psi(x) = A \times g(x)$

This process is called normalizing the wavefunction. But how do we figure out what the value of $A$ is? Notice that now that $\psi(x) = A \times g(x)$,

• $\int_{-\infty}^{\infty} \psi^{*} \psi dx$
• $\Rightarrow \int_{-\infty}^{\infty} |A|^2 g^{*}(x) g(x) dx$
• $\Rightarrow |A|^2 \int_{-\infty}^{\infty} g^{*}(x) g(x) dx$

Now we already know that $\int_{-\infty}^{\infty} g^{*} g dx = k$, so the above expression simplifies to:

• $\Rightarrow |A|^2 \int_{-\infty}^{\infty} g^{*}(x) g(x) dx \Rightarrow |A|^2 \times k$

Now to ensure $\int_{-\infty}^{\infty} \psi^{*} \psi dx = 1$, we need a value for $A$ such that

• $|A|^2 \times k = 1$  so
• $|A|^2 = \frac{1}{k}$
• $\Rightarrow |A| = \frac{1}{\sqrt{k}}$

# Modern Physics: Chapter 7 – Introduction to Wavefunctions

XKCD/849

So I already mentioned that everything has a matter-wave associated with it. Developing the idea a bit further, we get the concept of a wavefunction which, by convention, is denoted by $\psi$.

Each physical system is described by a state function which determines all can be known about the system. … The wavefunction has to be finite and single valued throughout position space, and furthermore, it must also be a continuous and continuously differentiable function.

Furthermore, a wavefunction is generally a complex function i.e it consists of both a real part and an imaginary part.

• $\psi (x, t)$ – A wavefunction

To get the probably of a particle being at a position $x$, you have to get a real value out of the complex wavefunction. Hence the probably of finding a particle described by a wavefunction $\psi$ is the product of the function $\psi$ at $x$ and its complex conjugate $\psi^{*}$ at $x$.

• $\psi (x_{0}) \psi^{*}(x_{0})$ – probability of finding the particle at $x_{0}$.

To get the probability from a range $x \in (x_{0}, x_{1})$ where $x$ is a continous variable, you can take the integral of $\psi^{*}\psi$ from $x_{0}$ to $x_{1}$:

• $P(x) = \int_{x_{0}}^{x_{1}} \psi^{*}(x) \psi(x) dx$

# Modern Physics: Chapter 6 – The Heisenberg Uncertainty Principle

XKCD/824

One of the best explanations I’ve found of the Heisenberg uncertainty principle is in Volume III of the Feynman Lectures on Physics. Read the following sections (which I’ve linked) to understand the general concept behind it:

Essentially, there is an inverse relation between the width of a wave-packet, $\Delta x$ and the range of wavenumbers of the waves you’ll need to superpose to generate that wave-packet, $\Delta k$. In general, the more localized a wavepacket is, the more waves you need to add to create it. Here’s an animation to show what I mean:

We can describe this relation as:

• $\Delta x \Delta k \sim 1$

Now, since

• $p = \hbar k \Rightarrow \Delta p = \hbar \Delta k$

Multiplying the first relation with $\hbar$, we get

• $\Delta x \hbar \Delta k \sim \hbar$
• $\Rightarrow \Delta x \Delta p \sim \hbar$

or more precisely,

• $\Delta x \Delta p \ge \hbar$

Which is the Heisenberg Uncertainty Principle.

# Modern Physics: Chapter 5 – Fourier Transform I

XKCD – Fourier

To create a truly localized wavepacket we need to superpose not just two but an infinite amount of sinusoidal waves whose wavelengths and amplitudes vary in a continuous spectrum. To do that, we need to learn about what’s called the Fourier integral:

• $f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} a(k) e^{ikx} dk$

where

• $a(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-ikx} dx$

Why an exponential function instead of sine and cosine? This is because we just used Euler’s formula:

• $e^{ix} = cos(x) + isin(x)$

and it’s derivations:

• $cos(x) = \frac{e^{ix} + e^{-ix}}{2}$
• $sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

to replace the sines and cosines with the exponential function.

Essentially, this gives us a framework for expressing practically any function as a superposition of harmonic waves (just plug in the function $f(x)$ in the expression for $a(k)$ and then plug in $a(k)$ into the expression for the Fourier integral). This is known as the Fourier transform.

# Modern Physics: Chapter 4 – Wave Groups

In the last chapter, I discussed matter waves. Now, when one thinks about waves in general, one visualizes plane waves. Essentially something like this:

Now, the mathematical function for a moving plane wave is of the form

• $f(x, t) = Acos(kx - wt)$

where $k = \frac{2 \pi}{\lambda}$ and $w = 2 \pi f$

You can use the following matlab script to generate a moving plane wave:


%Date: March 27, 2015
%Generating Moving Plane Waves

syms f x t;

A = 3;
k = 2;
w = 2;

f(x, t) = A*cos(k*x - w*t);

ix = 0:0.1:10;

for t = 0:0.1:60
tic;
iy = double(f(ix, t));
plot(ix, iy);
pause(0.1 - toc);
end



Since the intensity matter waves describe the probability of finding a particle at a specific point in space, and since particles are limited to specific regions of space (the probability of finding them can’t just be equal at every point in the universe, the earth for example is far far far far far far more likely to be found in its orbit in the solar system than in the andromeda galaxy), plane waves of infinite extend and the same amplitude throughout are an incorrect representation for a moving particle.

The matter wave associated with a localized moving particle is, in fact, a wave packet. This is what a wave-packet looks like:

Wave packets are created by the super-position of waves with similar but not exactly equal wave-numbers and angular frequencies. In other words, the mathematical function for a simple wave-packet created by the super-position of two waves is something like:

$f(x, t) = Acos(k_{1}x - w_{1}t) + Acos(k_{2}x + w_{2}t)$

You can use the following matlab script to generate a moving wave-packet:

%Author: Muhammad A. Tirmazi
%Date: March 27, 2015
%Generating Moving Wave Groups

syms f g h x t;

A = 3;
k1 = 4;
w1 = 4;
k2 = 4.3;
w2 = 4.3;

f(x, t) = A*cos(k1*x - w1*t);
g(x, t) = A*cos(k2*x - w2*t);
h(x, t) = f(x, t) + g(x, t);

ix = 0:0.1:20;

for t = 0:0.1:60
tic;
iy = double(h(ix, t));
plot(ix, iy);
pause(0.1 - toc);
end


The mathematical function describing a simple wave-packet can be simplified by using the identity

• $cos a = cos b = 2 \times cos \frac{1}{2}(a - b) \times cos \frac{1}{2}(a + b)$

This results in a function of the form:

• $f(x, t) = 2Acos(\frac{\Delta x}{2}k - \frac{\Delta w}{2}t) \times cos(\frac{k_{1} + k_{2}}{2}x - \frac{w_{1} + w_{2}}{2}t)$

where $\Delta k = k_{2} - k_{1}$ and $\Delta w = w_{2} - w_{1}$.

However, we still have a problem with this function. A simple superposition of two slightly different waves generates wave-pulses that repeat at a constant interval. To represent a localized moving particle, we need a localized wave-packet. Essentially, we need something like this:

# Modern Physics: Chapter 3 – Making (Non)Sense of the Double Slit Experiment

So let’s get this straight. An electron gives a wave-life diffraction pattern in the double slit experiment. It clearly shows wave behavior. However, as soon as you put detectors in front of the slits and try to detect the electron, its pattern changes to a bullet-like particle pattern. So is an electron a wave or a particle? Well… it’s both and neither.

Wave-Particle Duality

What we have understood till now is that an electron seems to behave like a way till it gets detected. After detection it behaves like a particle. A workable hypotheses might be to consider an electron (and everything else) as both a wave and a particle. A question arises, though. We already know how to calculate the particle properties of the electron such as momentum ( $p = mv$ ), but if it’s a wave, how do we calculate its wavelength?

Matter Waves

The French Physicist, Louis de Broglie postulated that the wavelength of the ‘matter wave’ associated with a particle (such as an electron, proton or even you, me, the earth, Jupiter and chickens) can be calculated using the following relation:

• $\lambda = \frac{h}{p}$

where $\lambda$ is the wavelength of the associated matter wave, $p$ is the momentum and $h$ is what’s called the Planck constant.

# Modern Physics: Chapter 2 – Double Slit Experiment II

XKCD – Nerd Sniping

In the last chapter I mentioned that the maxima and minima formed with electrons in the double slit experiment were probability maxima/minima. In fact, the intensity of the wave displayed as the result of the experiment is proportional to the probability of finding the electron at a given position.

The wave, in this case, is a function of the position. Let’s call it $\psi (x)$ . Since $I \propto A^2$ , the probability of finding the electron at a given position $x$ is proportional to $|\psi^2(x)|$. $\psi(x)$ is called a wavefunction.

One Electron at a Time

If you send one electron at a time, it gets detected at one point on the screen. However, if you keep detecting the positions of the individual electrons coming per unit time and record them, you seen find the same pattern emerging.

Here’s a nice animation to show you what I mean: LINK

Detectors in Front of Slits

If you had detectors in front of the slits (even if the detectors are very gentle), the interference pattern disappears. Surprisingly, as soon as it gets detected at the slits, the electron starts giving the expected bullet-like particle pattern on the screen instead of the wave pattern:

# Modern Physics: Chapter 1 – Double Slit Experiment I

The results of the double slit experiment challenge the classical concept of the electron being a particle and the concept of nature being divided into particles and waves. Essentially you place a source of whatever you want to conduct the double slit experiment with behind a barrier that only has two narrow slits and then place a detector/screen behind that barrier and record where and when the source gets detected on the screen.

Bullets

If you conduct the experiment with bullets, you should get the following result:

Where $P_{1}(x), P_{2}(x) and P_{12}(x)$ are the probabilities of finding the particle at a given position x when the first slit is open, the second slit is open and both slits are open respectively. In the case of bullets, it is apparent that:

• $P_{12}(x) = P_{1}(x) + P_{2}(x)$

But life isn’t that simple in general. If you conduct the same experiment with waves, you get a somewhat different result.

Waves

If you repeat the pattern with waves, you get the following results:

when $I_{1}(x), I_{2}(x) and I_{12}(x)$ are the intensities of the waves at a given position x when the first slit is open, the second slit is open and both slits are open respectively. Over here,

• $I_{12}(x)$ is clearly not simply equal to $I_{1}(x) + I_{2}(x)$.

This is because the waves interfere with each other either constructively or destructively at different places forming interference maxima and minima at different points.

Electrons

Electrons are clearly particles right? The pattern we get by conducting the experiment with electrons will obviously be the same as the one formed by bullets, right? Wrong! When conducting the experiment with electrons, you get a result more like the one we got with waves than the one we got with bullets.

Interference maxima and minima are formed. Note that unlike the experiment with waves, these are probability maxima/minima, not intensity maxima/minima.

But… electrons were particles, right? Wrong. There’s more to the story than that. To quote Shakespeare,

There are more things in heaven and earth, Horatio,
Than are dreamt of in your philosophy.

~ William Shakespeare – Hamlet