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.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/73329/2451 and links therein. $\endgroup$ – Qmechanic Apr 10 '14 at 15:26
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    $\begingroup$ It's to a different kind of question, but perhaps this answer of mine can help somewhat. $\endgroup$ – Wouter Apr 10 '14 at 15:31
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    $\begingroup$ See this paper by Scott Aaronson. He considers other ways of calculating the probability density and shows that the 2-norm is the only physically reasonable one. $\endgroup$ – John Rennie Apr 10 '14 at 16:37
  • $\begingroup$ @JohnRennie thanks, but that paper looks way beyond my scope right now. $\endgroup$ – user42733 Apr 10 '14 at 16:51
  • $\begingroup$ @Wouter- That was a very interesting answer, very intuitive. Thanks. $\endgroup$ – user42733 Apr 10 '14 at 16:53

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.

  • $\begingroup$ I was looking for a more intuitive answer, but I think this is the one which makes the most sense out of all. Thanks a lot! $\endgroup$ – user42733 Apr 10 '14 at 16:55
  • $\begingroup$ @ParthVader no problem. Also, be warned: quantum mechanics in general is notorious for being counter-intuitive. $\endgroup$ – Hunter Apr 10 '14 at 17:03
  • $\begingroup$ How is that expression for intensity derived from Maxwell's Equations? Perhaps this argument would work better if it were about electron interference. That would avoid mixing E&M with quantum mechanics, and smooth out the discussion. $\endgroup$ – garyp Apr 10 '14 at 17:09
  • $\begingroup$ @garyp I'm not sure what you mean. I deliberately chose photon interference so that we would mix E&M with QM. From this we can postulate the well-known expression for the probability density, which then carries over to other particles (such as electrons) due to the wave-particle duality. This gives the OP the logic behind the derivation of the probability density. In the end, of course, this is valid because experiments have shown that this is true. $\endgroup$ – Hunter Apr 10 '14 at 17:18
  • $\begingroup$ @Hunter Sure, I completely understand your point of view. My small quibble is that you have to take the classical wave equation as the quantum mechanical wave function for the photon. This is justifiable, but I think it puts the discussion into the realm of quantum field theory, which I would argue is more complexity than needed here. $\endgroup$ – garyp Apr 10 '14 at 17:31

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.

  • $\begingroup$ +1 Though this is probably a better answer, I am not accepting it because I am not sure if I fully understand it (nothing wrong with your explanation, I have just started QM so it is difficult to get the intuition behind all this). Thanks for the help! $\endgroup$ – user42733 Apr 10 '14 at 20:01
  • $\begingroup$ But why should the probabilities be invariant under rotations? $\endgroup$ – Luke Pritchett Oct 5 '16 at 2:20
  • $\begingroup$ A ray in the Hilbert space just means the particle exists. Or stated another way, the probabilities of all possible outcomes should sum to one. A rotation can represent an active transformation (the quantum state changes), or a passive transformation (you simply change the representation or measurement basis). In either case, the probabilities should still sum to one. $\endgroup$ – Punk_Physicist Oct 10 '16 at 21:54

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.

  • $\begingroup$ Are you saying that 'probability density' is just an interpretation of $|\Psi|^2$? It could've been interpreted as something else too? $\endgroup$ – user42733 Apr 10 '14 at 16:59
  • $\begingroup$ Yes, but the alternatives are not as intuitively satisfying. Certain features of the interpretation have troubled many people. Most notably, the notion that the wave function "collapses" to a well-defined value upon measurement. How can that be if $|\Psi|^2$ represents the probability density of finding the particle? The arguments and alternatives are hard to understand, IMO. Certainly, most people pay little attention to them, leaving the problems to specialists and philosophers. One alternative is to give $\Psi$ no meaning at all, attaching meaning only to expectation values. $\endgroup$ – garyp Apr 10 '14 at 17:25
  • $\begingroup$ That is what I should do too while studying QM, I guess haha. $\endgroup$ – user42733 Apr 10 '14 at 17:38

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