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


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.

– Derivation of the postulates of quantum mechanics from the first principles of scale relativity

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



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


  • 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.