# I Fubini-Study metric of pure states

1. Nov 18, 2017

### Alex Cros

Hello PF!!

https://en.wikipedia.org/wiki/Fubini–Study_metric (qm section like always )
And can't figure out how to derive:
$$\gamma (\psi , \phi) = arccos \sqrt{\frac{<\psi|\phi><\phi|\psi>}{<\psi|\psi><\phi|\phi>}}$$
I started with
$$\gamma (\psi , \phi) =|| |\psi> - |\phi>||= \sqrt{(<\psi|-<\phi|)(|\psi>-|\phi>)}=...$$
as it is the distance between the two vectors (no?) but don't seem to get anywhere, help!!
NB: I assumed there the vectors are normalized + apologies for the dirac notation format.

2. Nov 18, 2017

### strangerep

That's not what you want. Try thinking in terms of finding the angle between 2 unnormalized vectors. E.g., in a simpler real vector space, what is the dot product of 2 vectors $u \cdot v$ ? Then generalize to a complex vector space.

If they're normalized then the denominator of your 1st expression is 1.

3. Nov 19, 2017

### Alex Cros

But why then is it called the distance between those two points?
I think I know what you are saying: take the inner product and solve for the angle, then you get the arccos, but in the article I sent it was referred as the length. Moreover the ultimate aim of doing this is to derive the ds element by $|\psi> →|\psi + d \psi>$ and finding the length between them two.
I'm confused by the: "the length between two points"

4. Nov 19, 2017

### strangerep

Because angles between vectors in a vector space satisfy the properties of being a metric.
(You should indeed try and prove this assertion. I'll help you if you get stuck.)

It's just an abuse of terminology. The crucial thing is that it defines a metric. When 2 state vectors are maximally "close" to each other it means the angle between them is 0. When 2 state vectors are maximally different from each other, that corresponds to them being orthogonal. I.e., the angle between them is 90deg.

The $ds$ variant is just a special case of the main formula for when the state vectors are only infinitesimally different from each other, hence the angle between them is very small. I find it a bit misleading to call it $ds$ when it's really a small angle, but this usage is quite common.

5. Nov 20, 2017

### Alex Cros

Okay I see now what you mean now, yeah I agree it's quite confusing the notation. Either way I would be very grateful if you could give me a hand with this derivation by quicking off with the first steps please!!
Thanks beforehand!

6. Nov 20, 2017

### Alex Cros

Okay,I think I just proved that (even though its not Fubini-Study metric) $|| |\psi>-|\phi> ||$ is a metric aswell, as it satisfies $d(x,y)≥0 , d(x,y)=0 ↔ x=y, d(x,y)=d(y,x) & d(x,z)...$etc. is this the case? or have I made a mistake?

7. Nov 20, 2017

### strangerep

The final bit just involves the triangle inequality, applied to norms on a vector space.

8. Nov 20, 2017

### strangerep

Heh, well,... actually, I'll let you "quick off" the first steps, and then I'll help if you get stuck.

Hint: start with the case of a real vector space ([wherein you'll need to know about the (spherical) law of cosines]. When you've done the real case, you can generalize the law of cosines to a complex vector space.

9. Nov 21, 2017

### Alex Cros

Okay, I first let you know my original attempt to comment on my procedure:
Let $$dist(\psi,\phi) = || \psi - \phi || = \sqrt{ < \phi, \phi> + < \psi, \psi> - 2 || < \psi, \phi> ||}$$ which is a metric.
now $$dl = dist(\psi+d\psi,\phi) = \sqrt{ < \phi, \phi> + < \psi + d\psi, \psi + d\psi> - 2 || < \psi +d\psi, \phi> ||}$$
expanding $$dl = \sqrt{ \phi * \phi + \psi * \psi + \psi * d\psi + d\psi* \psi +d\psi*d\psi* - 2||\psi * \phi +d\psi * \phi|| }$$
where $*$ denotes complex conjugate on the left

10. Nov 21, 2017

### strangerep

Are you sure about that last term under the square root? Are you sure it's not $2 Re\langle\phi,\psi\rangle$ ?

In any case, I'm reasonably sure that's not the right point to start from if you want an FS metric.

Do you want to derive the finite form of the FS metric (the one in the Wiki page with an arccos, i.e., $\gamma(\psi,\phi)=\arccos \sqrt{\cdots}$), or do you just want to start with that finite form and derive the corresponding infinitesimal version $ds^2 = \cdots$ ?