Let $H$ be a separable Hilbert space for a quantum mechanical system then $$w (x, y) = {{\langle y \mid x\rangle\langle x \mid y \rangle} \over \langle x \mid x \rangle\langle y \mid y \rangle}$$ is the corresponding probability function providing the probability to find state $x$ when measuring state $y$.

Which inequalities and, more generally speaking, abstract properties are known that $w$ must satisfy?

$w(x,x) = 1$ is obvious as well as $w(x,y) = w (y,x)$ and independence of phase factors and scaling factors. I suppose there also should be something connecting $w(x,y)$, $w(y,z)$ and $w(x,z)$ to an inequality, intuitively something like a triangle inequality.

Update: I am particularly interested in restrictions on probability functions $w: S \times S \to [0,1]$, where $S$ may be taken to be the projective space of Hilbert space rays, which are necessary for $w$ to have the above mentioned form. This question, of course, is way too general. This is particularly true, if I even would allow $S$ to be something more general. So I want to start "small".

$w$ has something to do with "angles". "Angles" allows to define a metric in ${\Bbb S}^n$ and satisfy a triangle inequality. So I am looking for something like that for $w$.


The Cauchy-Schwarz-Bunyakovsky inequality says that $$ |\langle x\vert y\rangle|^2\le \langle x\vert x\rangle\,\langle y\vert y\rangle. $$ In other words $\omega(x,y)\le 1$. Is that what you want?

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    $\begingroup$ I think the OP is interested in some relation between $|\langle x| z \rangle|$ and $|\langle x| y \rangle||\langle y| z \rangle|$. I am not sure if such a relationship is well known in a nice form. $\endgroup$ – Dvij Mankad Aug 18 at 22:21
  • $\begingroup$ Not really. Some properties are really...um...easy, Such as $w(x,x)=1$ or $w(x,y)=w(y,x)$. Also $w(x,y) \leq 1$ is obvious for a probability function. On the other hand, there must be more deep properties restricting $w$. Something like being a metric. Like being, in a certain sense, quadratic in x and not linear. Such kind of thins. $\endgroup$ – Nobody-Knows-I-am-a-Dog Aug 19 at 7:23

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