# Entangled Photons

1. Jan 28, 2012

### StevieTNZ

When you create a pair of entangled photons, are they simultaneously in a superposition of V and H, and 45 and 135 polarisations?

2. Jan 28, 2012

### Antientrophy

This is were Heisenberg's uncertainty principle comes in. The more precisley you try to measure the state of one photon, the less likely you will understand or reveal the state of the other.

3. Jan 28, 2012

### questionpost

Doesn't them being entangled imply your not measuring them?

4. Jan 28, 2012

### StevieTNZ

I'm not sure I follow. Before we make a measurement of one of the entangled photons, what superposition of polarisations is it (V and H AND/OR 45 and 135)?

Also, when I measure one photon and the polarisation turns out to V, I know the other photon has taken on V polarisation. I don't see where the uncertainty comes in there?

5. Jan 28, 2012

### questionpost

They are uncertain because they aren't being measured.
All mass particles in an entangled state are uncertain, but their states such as with spin are uncertain (or indistinguishable from each other) but always opposite since they are occupying the same quantum state and therefore have to have opposite spin. I don't know about photons though, because non-mass particles don't have to occupy opposite spins to exist in the same quantum state.

6. Jan 28, 2012

### StevieTNZ

Yes, but you usually can describe the superposition they're in when not being measured (hence wave functions). Uncertainty is a property of measurement results.

7. Jan 28, 2012

### questionpost

8. Jan 28, 2012

### StevieTNZ

9. Jan 28, 2012

### questionpost

Polarization is associated with vector state which is uncertain...there should be a chance of either measuring a photon having either, which actually happens without entanglement, since if you just pass a photon beam through a let's say a prism, there's a 50% chance of it polarizing one way and a 50% chance of it going the other way, which is why it separates into two beams of light.

Last edited: Jan 28, 2012
10. Jan 29, 2012

### Ken G

Wouldn't that depend on the history of how the photons were created? Let's say you annihilate a positron and electron to make the photon pair, then I should think the photon pair will inherit the conserved attributes of the initial pair. But one thing is clear-- you cannot talk about an entangled system in language that refers to superpositions of states for the individual particles. An entangled system is a superposition of two-particle states, where the two-particle states are tensor products of single particle states. Is it not so?

11. Jan 29, 2012

### kith

The definition of entanglement is that you can't find a basis, where the state is a product state wrt to the entangled properties. So if a state is entangled in one basis, it also is in every other.

[Note that this assumes that the entangled properties are the same for both particles, which means they can be described by the same basis / Hilbert space. In general, the polarization of a particle could also be entangled with other degrees of freedom of the other particle. Then you'd have to use "set of bases" instead of "basis" in the sentence above. But this doesn't change the bottom line that entanglement is always there.]

Last edited: Jan 29, 2012
12. Jan 31, 2012

### StevieTNZ

How would we model this situation:

Entangled pair of photons, each sent to a polarising beam splitter, then each subsequently sent to another polariser beam splitter orientated at 45 degrees.

=1/2 |VV>+|HH> (superposition of V and H polarisations)
evolves into (remember, according to QM no collapse has occurred)
=1/2 |45> [1/2 |VV>+|HH>] + |135> [1/2 |VV>-|HH>] (superposition of 45 and 135 degrees, as well as including the fact the system is still in superposition of V and H polarisations)

But, if we measure the polarisation at 45 degrees, it doesn't necessarily mean that each photon will be both pass/pass or fail/fail. There could be pass/fail, fail/pass, pass/pass, fail/fail. So I guess I've modelled it wrong, or something. Not sure how it is done.

13. Jan 31, 2012

### Ken G

It seems to me that if the initial state is 1/root(2)*(|VV> + |HH>), then if both beams encounter a V filter, the entanglement is broken before encountering any subsequent filters, because each entangled pair either fails to pass the V filter, or if they do, the state becomes |VV>, which is |V>|V> so is no longer entangled. To maintain the superposition, you could write the state after the first filter as 1/root(2)*(|pass>|pass> + |fail>|fail>), but the |fail>|fail> will also fail every subsequent filter because it never passed the first one.

14. Feb 4, 2012

### StevieTNZ

How do we right the combination of two photons that were entangled, and then reaching polarisers? Say the photons first travel to a polarising beam splitter.

Photons that are V polarised and then becoming 45, or 135 degrees polarised have the plus sign before the state, and H polarised photons then becoming 45 degrees have plus signs before the state, but minus if its H then 135. But you can get a combination of H polarised photons, one becoming 45 the other 135. |H,45>-|H,135> - how can we combine those two so they're in the same | >, and when we add that to the complete picture do we put it in with a + or - before?