Bell-state Measurements


by StevieTNZ
Tags: bellstate, measurements
StevieTNZ
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#1
Mar27-12, 04:37 PM
PF Gold
P: 779
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 bell-state (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>?
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StevieTNZ
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#2
Mar29-12, 08:13 PM
PF Gold
P: 779
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?
StevieTNZ
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#3
Mar29-12, 08:55 PM
PF Gold
P: 779
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>

zonde
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#4
Apr1-12, 11:36 PM
PF Gold
P: 1,376

Bell-state 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}}(C-D)[/itex]
Now [itex]|V_1\rangle|H_2\rangle|H_3\rangle|V_4\rangle - |H_1\rangle|V_2\rangle|V_3\rangle|H_4\rangle=[/itex]
[itex]=\frac{1}{4}(C_1-D_1)(C_2+D_2)(C_3+D_3)(C_4-D_4)-\frac{1}{4}(C_1+D_1)(C_2-D_2)(C_3-D_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_4-D_1D_2D_3C_4-C_1C_2C_3D_4+D_1D_2C_3D_4+D_1C_2D_3D_4-C_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.
StevieTNZ
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#5
Apr2-12, 02:15 AM
PF Gold
P: 779
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 bell-state 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> ?
StevieTNZ
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#6
Apr2-12, 02:11 PM
PF Gold
P: 779
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 bell-state).
zonde
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#7
Apr2-12, 10:57 PM
PF Gold
P: 1,376
Quote Quote by StevieTNZ View Post
I would assume in this bell-state measuring in 45/135 basis, we would detect |45>|135> or |135>|45>
Why do you think so?


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