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

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.

Answer 3

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 4

Well, actually this is a very good and conceptual question.

At undergraduate school and in textbooks it is often mentioned Max Born came up with this idea, although yes, it's a bit unexplained why. Typically, it is taken as a $\textit{postulate}$ of standard Quantum Mechanics, not something that it's computable from first principles.

Digging up a bit, Max Born came up with the idea in an old paper discussing scattering (e.g. https://link.springer.com/article/10.1007/BF01397477 . It's in German, so I aided myself with Google Translator). To attach a physical meaning to his model, he came up - in a footnote - with the idea that the square of the wavefunction could be interpreted as a probability density. Taking into account the already voted as best answer, you can see the influence of Electromagnetism in this definition.

Answer 5

If I remember correctly, it was an educated guess based on early quantum theory as it applied to electro magnetism. When the photoelectric effect was understood, it became natural to understand the intensity of an electro-magnetic wave as being proportional to number of light quanta pass through a surface over time. In this way a wave function, the solution to Maxwell's Equations, can be associated with a position of particles conceptually. De Broglie established a relationship between a particle's momentum and associated wavelength. Energy/momentum to frequency/wavelength relationships invited further comparisons. An analogous wave equation was developed to mimic electromagnetic waves. The intensity of the electromagnetic wave is proportional to the square of the amplitude, hence the need to square* the matter wave function. So the wave function was developed before it was known what it meant. But the various other similarities suggested one could associated a position probability.

Answer 6

I think when particle state is assigned by wave function which is complex, then intensity of the particle or wave is more at amplititude2. This is comming by squaring the wave function as in classical machanics we do. Where intensity is high......it is directly proportional to number of more particles present there. Where particle is more it means particle density is confined in that area...and where particle density is more probability density is more. So psy2 is directly proportional to probability density. Now the intution is very clear. Quantum mechanics told that nothing is particle....all thing in wave states. Particles are also combination of most complex function of frequencies and wavelengths. And in classical physics, we know that Intensity of the wave is given by (Amplitude)2 which is directly proportional to PSY2.