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My question doesn't go much beyond the title: Why does $$\left | \psi \left ( x,t \right ) \right |^{2}$$ give us the probability density of something appearing at a certain location? I understand that $$\left | \psi \left ( x,t \right ) \right |^{2} = \psi \left ( x,t \right )^{*}\psi \left ( x,t \right ) $$ where $\psi \left ( x,t \right )^{*}$ is the complex conjugate, but I still don't understand how multiplying these two variants of the wave equation gives us a probability of a location.


marked as duplicate by ACuriousMind, Kyle Kanos, Martin, Ryan Unger, John Rennie May 27 '15 at 4:51

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    $\begingroup$ possible duplicate of Born's Rule, What is the Reason? $\endgroup$ – Gonenc May 26 '15 at 15:54
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    $\begingroup$ possible duplicate of Born rule and unitary evolution $\endgroup$ – ACuriousMind May 26 '15 at 16:00
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    $\begingroup$ It's literally because that's what the wave function means. Ask yourself "why does electric field give us the force per unit charge on an object?". The answer is that we observe that in some situations charged objects experience force proportional to their charge, so we made up "electric field" to account for it. $\endgroup$ – DanielSank May 26 '15 at 16:06
  • $\begingroup$ @DanielSank, the electric field was originally introduced as force per unit charge; then it is trivial that product of force and charge gives force. However, the function $\psi(x)$ has no such introduction. It is a purely mathematical concept defined by the Schroedinger equation and boundary conditions. There is no probability involved in its definition. The question "why the integral of magnitude squared gives probability" is then very non-trivial. $\endgroup$ – Ján Lalinský May 27 '15 at 16:36
  • $\begingroup$ @JánLalinský This is going to lead us to a discussion of the scientific method and more general ideas about what it means to do science. We should not carry out that discussion in the comments. I will leave my chat window open today in case you'd like to discuss. $\endgroup$ – DanielSank May 27 '15 at 17:08

$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\bra}[1]{\langle #1 \rvert}$As a special instance of the Born rule stating that, given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$, it is an axiom in the standard formulations of quantum mechanics that $$ \lvert\psi(x)\rvert^2 = \lvert \bra{x} \psi\rangle \rvert^2$$ is the probability (density) to find the object at $x$.

  • $\begingroup$ The rule you declared as axiom is not a special instance of the rule that states given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$. The two are separate rules, because the former gives probability as integral of a square, while the latter gives probability as square of an integral. Unfortunately, they are both referred to as the Born rule. $\endgroup$ – Ján Lalinský May 26 '15 at 16:38

It comes from the mathematical framework of Quantum Mechanics. A general expectation value is the result of computing a state $\omega$ over some observable $A$. Mathematically speaking $A$ is a self-adjoint operator from the C*-algebra $\mathfrak A$ of observables and $\omega$ is a state over $\mathfrak A$, i.e. a normalised positive linear functional on $\mathfrak A$. By Riesz-Markov theorem there is a regular probability (which for the moment means that it gives measure 1 on the whole spectrum of $A$) measure $\mu_\omega$ carried by the spectrum of $A$ such that $$\omega(A) = \int_{\sigma(A)}\lambda\ \text d\mu_\omega(\lambda).$$ The probabilistic interpretation stems from the fact that, for any subset $U\subset\sigma(A)$, the number $$\int_U\text d\mu_\omega(A)$$ can then be interpreted as the probability of finding the outcome of a measure of $A$ on the state $\omega$ within the range of values of $U$ (recall that a self-adjoint operator has a spectrum contained in $\mathbb R$).

When the representation of the canonical commutation relation is that of Schrödinger (which is the only one up to isomorphism), the states are the in one-to-one correspondence with the projective Hilbert space $PL^2(\mathbb R)$ (I'm assuming just one degree of freedom for simplicity). In particular, since this is an irreducible representation, every admissible pure state corresponds to a (class, or ray, of a) vector in $L^2(\mathbb R)$, and therefore $$\omega(q) = (\psi_\omega,q\psi_\omega) = \int x|\psi_\omega(x)|^2\text dx.$$ Comparing this expression with the one above coming from the Riesz-Markov theorem one can then interpret $|\psi_\omega(x)|^2$ as a probability density over the spectrum of the position operator $q$, i.e. $\mathbb R$ for translation invariant systems.

  • $\begingroup$ In short, from the formula for expected average $\int \psi^*\hat{A}\psi\,dx$ and its application to coordinate $x$ we infer that probability density $x$ is $|\psi(x,t)|^2$. However, then the question is, why is average expected value calculated the way it is? $\endgroup$ – Ján Lalinský May 26 '15 at 17:53
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    $\begingroup$ @JánLalinský One will always hit an ultimate "why" that can only be answered with "because we want to fit the data coming from the experiments". It is not a mathematical theory, it is a mathematical model for the behavior of nature. $\endgroup$ – anna v May 26 '15 at 18:33
  • $\begingroup$ @JánLalinský what do you mean? Are you asking why the average is computed as $\int\lambda\text d\mu$ or why in quantum mechanics it has that form? The answer to the former is: by definition; the answer to the latter is because that is the way the inner product on a Hilbert space is defined. $\endgroup$ – Phoenix87 May 26 '15 at 22:39
  • $\begingroup$ @annav, the formula for expected average is equally mysterious as the formula used as the Born rule. True, not everything can be given reason, but it is not evident that the Born rule is one of those things. $\endgroup$ – Ján Lalinský May 27 '15 at 5:42
  • $\begingroup$ @Phoenix87, the reason you gave - the rule for calculating average values - is not completely satisfactory, because similarly to the Born rule, it is unclear why that formula is any better than others. It makes no sense to explain the validity of these things in physics by saying they are definitions. $\endgroup$ – Ján Lalinský May 27 '15 at 5:51

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