Register to reply 
Bellstate Measurements 
Share this thread: 
#1
Mar2712, 04:37 PM

PF Gold
P: 794

In entanglement swapping:
If we detect photons 2 and 3 in the following states 135>45> or 45>135> (accompanied by the appropriate coincidence counts), do we project photons 1 and 4 into the same bellstate (so photons 1 and 4 are also found in either 135>45> or 45>135>)? How would we write this four photon GHZ state in terms of V> and H> polarisation? Would it be H>V>H>V> (+ OR ?) V>H>V>H>? http://en.wikipedia.org/wiki/Bell_state We have a list of the four bell states. Taking the first, do we expand as: A>B> = 0>1> + 1>0> ? Which would be H>V> + V>H>? 


#2
Mar2912, 08:13 PM

PF Gold
P: 794

Looking over it being V>H>V>H>  H>V>H>V>:
Would the following basis in 45/135 be the only ones allowed: 45>135>45>135>; 135>135>45>45>; 135>45>135>45>; 45>45>135>135> In all honesty I don't know how these are calculated. Two H's expanded in 135> make a +, because they're both  135. Do we expand over V>H>V>H> then minus that result by what we get for H>V>H>V>? Or is it all done in one go? 


#3
Mar2912, 08:55 PM

PF Gold
P: 794

Okay, I think I've figured it out. Appropriately detecting 2 and 3 in 45>135> etc to create that bell state, and throwing 1 and 4 into the same bell state we get this GHZ state:
V>H>H>V>  H>V>V>H> which allows the following combinations in 45/135 basis: 45>135>45>45> 135>135>45>135> 135>45>135>135> 45>45>135>45> EDIT: corrected last ..> to say 45> rather than 135> 


#4
Apr112, 11:36 PM

PF Gold
P: 1,376

Bellstate Measurements
Let's write 45> as C for short and 135> as D.
So then [itex]H\rangle=\frac{1}{\sqrt{2}}(C+D)[/itex] and [itex]V\rangle=\frac{1}{\sqrt{2}}(CD)[/itex] Now [itex]V_1\rangleH_2\rangleH_3\rangleV_4\rangle  H_1\rangleV_2\rangleV_3\rangleH_4\rangle=[/itex] [itex]=\frac{1}{4}(C_1D_1)(C_2+D_2)(C_3+D_3)(C_4D_4)\frac{1}{4}(C_1+D_1)(C_2D_2)(C_3D_3)(C_4+D_4)[/itex] That results in 32 terms where 8 term will cancel with other 8 terms and 8 terms will add with last 8 terms. So you will have 8 terms like that: [tex]D_1C_2C_3C_4+C_1D_2C_3C_4+C_1C_2D_3C_4D_1D_2D_3C_4C_1C_2C_3D_4+D_1D_2C_3D_4+D_1C_2D_3D_4C_1D_2D_3D_4[/tex] As you can see these are terms with odd number of C and D. If you would swap  for + in the initial expression you would get terms with even number of C and D. 


#5
Apr212, 02:15 AM

PF Gold
P: 794

Okay, I see now. If we start off with photons entangled as H>V>  V>H>, and we detect photons 2 and 3 (one from each pair) in this bell state: http://upload.wikimedia.org/wikipedi...76cca077a2.png
I would assume in this bellstate measuring in 45/135 basis, we would detect 45>135> or 135>45> What is the overall GHZ state now? Is it V>H>V>H> + H>V>H>V> ? 


#6
Apr212, 02:11 PM

PF Gold
P: 794

I'm guessing the GHZ state will be V>H>H>V>  H>V>V>H>, upon projecting 2 and 3 into the bell state http://upload.wikimedia.org/wikipedi...76cca077a2.png (which projects 1 and 4 into the same bellstate).



#7
Apr212, 10:57 PM

PF Gold
P: 1,376




Register to reply 
Related Discussions  
Justify bell state  Quantum Physics  0  
Bell tests: Singlet vs Triplet State  Quantum Physics  2  
Bell Measurement on 3qubit GHZ state?  Quantum Physics  0  
I need clarification about Bell state Measurements  Quantum Physics  3 