GHZ entanglement

1. Feb 21, 2012

StevieTNZ

Hi there,

I was looking through Zeilinger's "Dance of the Photons" where I re-read the section of three-photon entanglement. If one photon is measured in H polarisation, the other two take on H - likewise with V polarisation.

What I find interesting is if we measure one photon and its H polarisation, another (delayed) at 45 degrees, and the last at vertical - when the first photon takes on H polarisation, don't the other two? So when the last photon meets the V polarised filter, shouldn't it fail? But according to GHZ results, it should be vertical.

So the other two photons are NOT in H polarisation when the first one meets the polariser? Or the first photon is NOT in H polarisation - evolves as a superposition in accord with Scrodinger equation (no collapse occurs).

Unless I've mis-understood something?

2. Feb 21, 2012

StevieTNZ

I see where I have mis-interpretated it. The author doesn't make it clear under the diagram he is measuring in another polaristion direction before the photon becomes V or H polarised. v' h' and all that stuff just makes it more confusing.

3. Mar 6, 2012

StevieTNZ

If we have the GHZ state: |V>|H>|H>|V> + |H>|V>|V>|H>, if photons 1 and 4 take on polarisation 135, do photons 2 and 3 take on the opposite - 45 polarisation?

4. Mar 6, 2012

lugita15

To find the answer to this, all you need to do is write |H> and |V| in terms of |45> and |135>.

5. Mar 6, 2012

StevieTNZ

So replace |H> with |45> and |V> with |135> ?

Because expanding |H> to include |45> and |135>, I'm not sure if they have to match (i.e. with the other |H>)

6. Mar 6, 2012

lugita15

No, write |H> and |V> as linear combinations of |45> and |135>. Each of them will have an expression that's something like (1/√2)(|45>+|135>). I don't remember the exact expressions off the top of my head.

7. Mar 6, 2012

StevieTNZ

So |45>|135>|135>|45> + |135>|45>|45>|135> ?

8. Mar 6, 2012

lugita15

No, that's not how it's done.

9. Mar 6, 2012

StevieTNZ

How is it done? Do photons 1 and 4 share the same polarisation, likewise with 2 and 3? (e.g. in |H>|V>|V>|H>).

Do we expand it as |45>|45>|45>|45>+|45>|135>|135>|45>+|135>|45>|135>|135> etc.?

10. Mar 6, 2012

lugita15

Yes, that's how it's done. We write |H> and |V> in terms of |45> and |135>, and so we expand the whole quantum state as a sum of products like |45>|45>|45>|45>.

11. Mar 6, 2012

StevieTNZ

"Combinations with odd numbers of |45> do not occur".

But you can write those from |V>|H>|H>|V>+|H>|V>|V>|H>?

12. Mar 6, 2012

lugita15

I told you, just write |H> and |V> in terms of |45> and |135>. Then just expand out the products using the distributive property, like |45>(|45>+|135>)=|45>|45>+|45>|135>