# Inequality for quantum probability

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?
• 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. – Dvij Mankad Aug 18 at 22:21
• 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. – Nobody-Knows-I-am-a-Dog Aug 19 at 7:23