2
$\begingroup$

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 $w$.

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.