1
$\begingroup$

I was just wondering, whether the common $\langle x \rangle$ symbol for the expectation value of a variable originates in the bra-ket notation of quantum physics? I would think that a fundamental quantum physics equation giving the expectation value of an operator $O$ written in bra-ket notation:

$\langle A|O|A \rangle=\langle O \rangle$

would support this assumption. Or is it just a coincidence?

$\endgroup$

1 Answer 1

2
$\begingroup$

Largely coincidence I believe.   Structures similar to $\langle x, y\rangle$ had been used for inner product, and $\langle X\rangle$ for expected value of random variable $X$, for quite some time before Dirac adopted the bra$\mid$ket notation for vectors and gave meaning to $\langle \mathsf Q\rangle_\psi = \langle \psi\mid \mathsf Q\mid \psi \rangle$ as the expectation value of linear operator $\mathsf Q$ acting on wave function $\psi$.

For instance, when $\mathsf Q$ is the position operator, then the expected position of a particle with wave function $\psi$ on the $x$-axis is:

$$\langle \psi\mid \mathsf Q\mid \psi \rangle = \int_\Bbb R \psi^\dagger(x)\;x\;\psi(x)\operatorname d x$$

Which coincides neatly with the expression for the expectation of a continuous random variable $X$ that has a probability density function $f$.

$$\langle X \rangle = \int_\Bbb R x\;f(x)\operatorname d x$$

$\endgroup$
1
  • $\begingroup$ In any case, funny coincidence it is! :) $\endgroup$ Commented Dec 6, 2015 at 11:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .