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.
2 Answers
$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\bra}[1]{\langle #1 \rvert}$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 $x$.
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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.
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