# Can we look upon entanglement as measurement?

1. Sep 3, 2013

### Albert V

Hi,

I have a question about how to interpret the state of an entangled photon

H = horisontal, V = vertical polarization

The global state is |Phi> = (|H>|V> + |V>|H>)/sqrt2.

By density operator formalism:

rho = |Phi><Phi|
= (1/2) ( |H>|V><H|<V| + |H>|V><V|<H| + |V>|H><H|<V| + |V>|H><V|<H| ).

in order to find the state on the right side, trace out the left side :

rho_right = Tr_left(rho) = (1/2) (|H><H|+|V><V|),

How should this state be interpreted? 1) as a photon EITHER in H OR V, or 2) a mixed state (superposition) containing components of both the left and the right side.

This would help me to understand this paper:

http://arxiv.org/abs/1301.1673

2. Sep 3, 2013

### vanhees71

First of all, you start with a two-photon state with entangled photons. It's a pure state given by a normalized state ket $|\Phi \rangle$. The corresponding statistical operator of the pure state is the projection operator $\hat{\rho}_{2 \gamma}=|\Phi \rangle \langle \Phi|$. This gives you the statistical properties on the polarization of both photons according to Borns rule.

Now you want to forget about the fact that you deal with a two-photon state and only know about the polarization state state of one photon. Thus you trace over the other photon's polarization states to end up with your reduced density operator $\hat{\rho}_{\text{right}}=1/2 \mathbb{1}$. This shows that an observer only looking at this photon's polarization measures an totally unpolarized beam of photons, i.e., he finds with probability 1/2 either a horizontally or a vertically polarized photon, and that's it.

3. Sep 3, 2013

### Albert V

Thanks for your answer, but when you say a "totally unpolarized beam of photons" does that imply a statistical mixture of states? Under the discussion: "What is the difference between entangled and "normal" photons?" another Sci Advisor said that:

"When you discard information about the other particle, you are left with a statistical mixture. (mathematically, this amounts to taking a partial trace)"

A statistical mixture of states is not equivalent to a linear superposition of states.

4. Sep 3, 2013