Why \( \vert \psi(x)\vert ^2\) is not the probability? The Born rule misconception!

|Difficulty level: Medium|

If you think you don’t understand the Born rule in quantum physics you must read this article carefully to clear the Born rule misconception you might have. If you do understand, you should still read this. I bet this article will be fun for everyone if you read the whole story! Anyway, let’s carry on!

The goal of this article is to clear the Born rule misconception, which I have personally observed, challenges many students.

The ‘Born Rule’?

The quite common but misinterpreted form of Born rule is that: if \(\psi(x)\)  represents the analytic wavefunction of a particle defined for a quantum system, \(\vert\psi(x)\vert ^2\) would be the probability of finding the particle at some point \(x\). Yes! That interpretation is not quite correct. Where is the catch? \( \vert\psi(x)\vert ^2\) is not actually the probability but the probability density function. To find the probability, we need to integrate \( \vert\psi(x)\vert^2\) as \(\int_{a}^{b}\vert\psi(x)\vert ^2 dx\) over the region of interest i.e. from point \(a\) to \(b\). What \(\vert\psi(x)\vert ^2\) actually represents is the probability of finding the particle per unit volume about the specified point \(x\).

Calculating the ‘density’

In Cartesian coordinates, for instance, in one dimension the volume element would be \(dx\), in two dimensions it would be \(dxdy\) with the wavefunction as a function of \(x\) and \(y\), of course, and so on. So you must multiply the volume element and integrate throughout the region of interest. This is not very different from the calculation the charge in a region of interest if you are given a charge density. We simply multiply with the volume element, the density and integrate over the region of interest. If you still think this is a non-strict practice and you can anyway do away with the misconception you are carrying, here is an example which can trick you!

The tricky quantum system

There can be cases where the probability density i.e. \(\vert\psi(x)\vert ^2\) can have values greater than one at some specified coordinate. If we interpret that as probability, that would be weird because probability must not be greater than one. This can happen, roughly speaking, for example, if the probability density function is like a normal distribution. One of the examples is the ground state of a quantum harmonic oscillator. Consider a particle of mass \(m\) oscillating with an angular frequency \(\omega\) in influence of a one-dimensional potential \(V(x)=m\omega^2 x^2/2\).

Assuming you understand the standard notation, I won’t explain each term explicitly. Even if you are not much familiar with the analytic solution of the quantum harmonic oscillator, for the moment, take this as an example representing any valid quantum state. That’d work!

The ground state of the quantum harmonic oscillator is given as,

\(\psi _0 (x)=({\frac{1}{\pi {a_0}^2}})^{1/4} e^{-x^2/2a_0^2}\)

and the \(\vert\psi(x)\vert ^2\) can be written as,

\(\vert\psi(x)\vert ^2=({\frac{1}{\pi {a_0}^2}})^{1/2} e^{-x^2/a_0^2}\)

where \(a_0=\sqrt{\hbar/(m\omega)}\) is a constant dependent on the mass of the particle and its angular frequency of oscillation.

If we plot the probability density function with respect to \(a_0\) at the origin \((x=0)\) we see that for some values of \(a_0\) the \(\vert\psi(x)\vert ^2\) can go well above one.

Thus \(\vert\psi(x)\vert ^2\) must not represent probability. For a value of \(a_0=0.2\), for instance, the \(\vert\psi(x)\vert ^2\) would depend on \(x\) in the following way.

Conclusion

It is very clear that the value of probability density, \(\vert\psi(x)\vert ^2\) on the \(y-\) axis goes well beyond unity. The purpose was to show that the misconception leading to incorrect outcomes is indeed possible. Find out more realistic quantum states yourself which can have probability density greater than unity. I bet, it won’t be an easy job! I leave it for you to verify that integrating \(\vert\psi(x)\vert ^2 dx\) for \(x\) over the entire universe would give you nothing else than one, which obviously represents that probability of finding that quantum particle anywhere in the universe is unity.

We hope this example made you think at least and cleared the Born rule misconception for many. I know you do not make mistakes while calculating probabilities, and you do integrate as required. But most people still tend to call the \(\vert\psi(x)\vert ^2\) the probability and not probability density function even in the course lectures and daily conversations. It is important to be aware of the above concept and we must either be precisely correct or realize what the speaker is saying according to the context.

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