Why is $|\Psi|^2$ the probability density?

Question

I am starting with Quantum Mechanics, learning online. I can't seem to find the reason for $|\Psi|^2$ being the probability density of finding an electron. They've just taken it for granted everywhere. I am learning all this math but I am not able to fully grasp the intuitive idea behind all of it. If anyone could explain this with proper reasoning, I would be grateful.

Answer 1

Short answer

The reason why a physical quantity such as probability is given by $|\Psi|^2$ rather than some other function of $\Psi$ is geometry, namely Pythagoras' theorem. If you have a vector which points from the origin to the $(\hat x,\hat y,\hat z)$ coordinates $(x,y,z)$, then the length $\ell$ is given by $\ell^2=x^2+y^2+z^2$.

Why is this the definition of length? If you rotate your coordinates or move the vector to another place, then what we call the length shouldn't change. So $\ell$ is called the length because the form of $\ell$ (the sum of squares) is the only quantity that is constant even if you rotate or move the vector (or move or rotate your coordinates).

Longer answer

Quantum mechanics is linear which means if you have 2 (or more) mutually exclusive states/outcomes which we write symbolically using Dirac notation as $\left|A\right>$ and $\left|B\right>$, then any linear combination is also a valid state i.e. $$\left|\psi\right>=\psi_a\left|A\right>+\psi_b\left|B\right>,$$ where $\psi_a$ and $\psi_b$ are numbers. But because of this linearity we can also represent $\left|\psi\right>$ in a different basis rather than $\left|A\right>$ and $\left|B\right>$, but the "length" of $\psi$ should not change. The only functional form for the "length" $\|\psi\|$, that doesn't change if we change (or "rotate") our basis functions/states is the sum of squares, just like in the case of the geometrical length of a vector, i.e. $$\|\psi\|^2\propto|\psi_a|^2+|\psi_b|^2+\cdots.$$

Now if you say a state should be normalized to 1 (i.e. $\|\psi\|^2=1$), then you now have a group of positive terms that sum to one no matter how you describe your state. Note that $\sqrt{\psi_a^2+\psi_b^2+\cdots}$ or $\psi_a^2+\psi_b^2+\cdots$ is invariant (equals 1) but $\sqrt{\psi_a^2}+\sqrt{\psi_b^2}+\cdots$ (or any other form) does not. In addition terms like $\psi_a\equiv\left<A|\psi\right>$ are a measure of how close the state $\left|\psi\right>$ is to the state $\left|A\right>$ so the probability that $\left|\psi\right>$ is measured in state $\left|A\right>$ should be some function of $\psi_a$. Combine these 2 facts gives $|\psi_a|^2$ as the only possible form for this probability.

Answer 2

Let us consider the famous double-slit experiment with photons. With the usual set-up, we denote the number of photons passing through by $N$ and we will denote the number of photons which hit the film between $y$ and $y + \Delta y$ by $n(y)$. The probability that a photon will be detected between $y$ and $y+ \Delta y$ at a time $t$ is given by: \begin{equation} P_y(y,t) \equiv \displaystyle\lim_{N\to \infty} \left( \frac{n(y)}{N}\right) \end{equation} If we consider this from an classical electromagnetic point of view, then the above quantity is known as the intensity $I$ of the electromagnetic wave which is well known to be proportional to: \begin{equation} I(y,t) \propto \left| \psi(y,t) \right|^2 \Delta y \end{equation} where $\psi$ denote the wave function of the electromagnetic wave. (Note that this equation can be derived from Maxwell's equations.) From the above two equations it is easy to see that the probability density is given by: \begin{equation} P(y,t) \propto \left| \psi(y,t) \right|^2 \end{equation} A more detailed discussion can be found in pages 18 to 24 of these notes.

Answer 3

I think the answer is "because it works". Early in the development of QM, that interpretation was given to the wave function, and over the decades it has proven to be a useful interpretation. It works. Additionally, the fact that it is possible to define an associated current, and that there is a quantum mechanical expression that guarantees that $|\Psi|^2$ is conserved when that current is taken into account, supports the interpretation.