# New quantum experiments and its implications

1. Apr 24, 2012

### gespex

Hello everybody,

I just read the following article:
http://arstechnica.com/science/news...-results-of-measurements-taken-beforehand.ars

http://www.nature.com/nphys/journal/vaop/ncurrent/full/nphys2294.html
(This last one I didn't read as I don't have a nature subscription, but including it for those who have)

Anyway, I wonder what the implications of this are. Would this not mean that backward in time communication would be possible, as follows:
1. From two sources, make many pairs of entangled photons
2. Measure two out of each pair; if the correlation is random, assume a 0 has been sent, if high, a 1.
3. Decide to entangle or not entangle the other two photons depending on whether you want to send a 0 or a 1.

Yet it is often claimed that backward in time communication is impossible, even with quantum mechanics, so there's obviously a logic flaw in my logic. So where am I wrong?

Gespex

Last edited by a moderator: May 5, 2017
2. Apr 24, 2012

### billschnieder

Excerpt:
http://arxiv.org/ftp/arxiv/papers/1203/1203.4834.pdf

3. Apr 24, 2012

### StevieTNZ

I don't quite get this part. Perhaps you might want to elaborate?

4. Apr 24, 2012

### zonde

In step 2. correlations between photons from different pairs are always random no matter what Victor does.

What you do in this experiment is postselection. Victor tels at what pairs you should look to see particular correlations. Mind you entanglement correlations in entanglement swapping are of four different types and they all "happen" at the same time i.e. theoretically you can sort uncorrelated photons of Alice and Bob into four subsets where in each subset you can see particular type of correlation (cos^2(a-b); cos^2(a+b); sin^2(a-b); sin^2(a+b)).

5. Apr 25, 2012

### Lord Crc

So from what I understand, if Victor gives Alice and Bob his list of which pair of photons he entangled, Alice and Bob will find that if they compare their measurements they'll see quantum correlation for the photon pairs that is on that list and classical correlation for the pairs that is not on that list. Correct?

6. Apr 25, 2012

### Demystifier

7. Apr 25, 2012

### DrChinese

This is a cool experiment, and a variation on one performed several years ago by the team of Thomas Jennewein, Gregor Weihs, Jian-Wei Pan, and Anton Zeilinger:

http://arxiv.org/abs/quant-ph/0201134

The issue that arises - which is not clear in your references and is often a bit obscured - is that it is fundamentally impossible to entangle particles at will. In reality, only some pairs end up entangled even when you place them in the mode where entanglement is possible. (This is NOT an experiment restriction.) I am not certain what the threshold is or what the percentages are.

So what happens is that it looks like everything is random until you filter to get the subset in which there is entanglement. That filtering process requires classical communication to obtain.

Last edited by a moderator: May 5, 2017
8. Apr 25, 2012

### Lord Crc

So let me see if I got this right. The results Alice and Bob measure do not change regardless of what Victor does. Ie if they all agree up front to do N pairs where Victor does one thing, and then another N pairs where Victor does the other thing, Alice and Bob won't be able to tell the two runs from another, it's all classical correlations.

However if Victor notes which of his pairs ended up entangled and sends this list to Alice and Bob and they look only at this subset of pairs, they will find that these pairs obey quantum correlations.

9. Apr 25, 2012

### DrChinese

That's it exactly.

Now imagine that Alice and Bob are sitting together. They can compare their results right there. A product state will have 75% correlation and an entangled state has 100% correlation (ideal case). So clearly, if you could sense the difference between these, then you could communicate FTL between Alice+Bob and Victor. (Seeing something above 75% could be a 1, while 75% would transmit a 0. Etc.)

I have no idea what the actual rate of entanglement is, but in an ordinary PDC setup it is perhaps 1 in 1,000,000. Perhaps there is someone out there that can chime in to help us understand the proper order of magnitude here.

10. Apr 25, 2012

### gespex

But even if only one in a million pairs show correlation, then certainly we could, theoretically, just send more photons until there is a observable difference (for Alice and Bob) between whether their photons were uncorrelated or correlated? Which would in theory still allow backward in time communication, right?

11. Apr 25, 2012

### DrChinese

The other 999,999 (and this is just a made up number) would be product (separable) state stats, making it difficult to differentiate one result from the other. As I say, I didn't see what the actual frequency of entanglement could be - or more importantly what is the theoretical max.

Most likely, there are multiple entangled states and they produce opposite statistics (or something similar).

12. Apr 25, 2012

### gespex

Fair enough. Thanks!

13. Apr 25, 2012

### zonde

I would say that stating it like "Victor has entangled photons" is confusing (and moreover I would say that it seems like this confusion is intended :grumpy:).

Victor measures two photons together and (theoretically) there are 4 different possible outcomes. Each outcome indicates different entanglement rule for photons of Alice and Bob (usually denoted as $|\Psi^+\rangle\; |\Psi^-\rangle\; |\Phi^+\rangle\; |\Phi^-\rangle$ ).
When you add up probabilities of all 4 rules together you get no correlation at all. If you add up rule 1 and 2 you get product state.

So Victor does not entangle photons but makes measurement how they are supposedly entangled.

14. Apr 25, 2012

### DrChinese

Looking more closely for this setup, I can see why you say this. Victor does choose whether to entangle 1&4 or not (assuming you look at it that way) versus leaving 1&2 and 3&4 entangled.

Because either state (entanglement swapped or not) has both a + and and - mode, what Alice and Bob see are 50% correlated at all times and that never varies.

15. Apr 25, 2012

### zonde

Okay, Victor obviously is changing his measurement arrangements so we have to be clear about this part too.

Victor can perform measurement in two different configurations - optical modulators on/off. He has 4 detectors and that means that there are 6 different ways how he can record double-click.

So lets write down all options.
Victor makes measurement with optical modulators ON
Possible measurement outcomes (in pairs):
- both detectors b'' or c'' give double-click - Alice and Bob in entangled state $|\Phi^+\rangle$
- one detector in b'' and one in c'' with the same polarization - Alice and Bob in entangled state $|\Phi^-\rangle$
- one detector in b'' and one in c'' with the different polarization - discarded (HV and VH separable states).

Victor makes measurement with optical modulators OFF
Possible measurement outcomes (in pairs):
- both detectors b'' or c'' give double-click - discarded (should not appear at all for ideal setup as constructive interference for two inputs forms at different output ports).
- one detector in b'' and one in c'' with the same polarization - Alice and Bob in HH and VV separable states.
- one detector in b'' and one in c'' with the different polarization - discarded (HV and VH separable states).

So Victor choses if he will measure how Alice's and Bob's photons are entangled ($|\Phi^+\rangle$ or $|\Phi^-\rangle$ with other two options discarded) or what polarization they have (HH or VV with HV and VH discarded) i.e. his choice is about what information he will find out.

16. Apr 26, 2012

### gespex

Hmm I have to admit I'm a bit confused here. Could you explain to me a bit more about Victor's setup? So what happens with the two photons (I understand they are called 1 and 4?) that arrive at Victor?

17. Apr 26, 2012

### zonde

Victor's photons are called 2 and 3.
First photons 2 and 3 go trough Mach-Zehnder interferometer where Victor can change phase between two arms. Then each output of Mach-Zehnder interferometer is measured with polarization beam splitter and two detectors. Look at Figure 2: Experimental setup from the paper. It is after references so you might have missed it.

Link to the paper one more time: http://arxiv.org/abs/1203.4834

18. Apr 26, 2012

### StevieTNZ

Would someone be able to write out the evolution of the photons to get to that state (above), in a way similiar to what the authors have done for the bell-states they detect (pg 14 of the pre-print).

19. Apr 28, 2012

### San K

SPDC is stimulated by random vacuum fluctuations, and hence the photon pairs are created at random times. The conversion efficiency is very low, on the order of 1 pair per every 10^12 incoming photons.

20. Apr 28, 2012

### zonde

I would like to find that out too. For me derivation in paper (pg 14) is too short to follow it so I won't try to offer derivation for HV and VH separable states.

21. Apr 28, 2012

### al onestone

Hey dr Chinese, this guy believes in FTL or superluminal communication. Do you think he's onto something?
Isn't it grand how many snags a person might come across when trying to create a thought experiment which could do such things. Problems with understanding that induced coherence usually does not occur when dealing with spontaneous processes. I recently had a snag with understanding that complete entanglement in position does not allow well defined momentum for the individual components. You remember.
There is alot of literature out there which will indicate the possibility of something miraculous to physics, but the miraculous never really manifests does it?
I think that indicating the Austrian study on entanglement swapping was a little beyond this thread. You should be spending your time preparing for the real "instant communication protocol", but it won't be sent superluminally, you'll just have to wait.

Am I getting this thread straight? Are we really considering signals being sent "backwards" in time. Please, first attempt to master the sending of signals in a manner that is simply "faster" than light, then move onto the heavyweights like "backwards" in time signalling.

My thought experiments never were so bold, and I even proffessed to have instantaneous signalling.

22. Apr 28, 2012

### StevieTNZ

I've emailed a few of the authors regarding this. Hopefully they get back to me soon.

I gather the two bell-state evolutions in the paper are in superposition of being both (so 1/2 |equations for first bell state> + 1/2 |equations for second bell state>) up until one bell-state is detected at the end?

23. Apr 28, 2012

### lugita15

Can you elaborate on this? Why can't you have a quantum system whose creation operators only produce entangled particle pairs? Or is this just a practical limitation?

24. Apr 28, 2012

### gespex

Nope, I don't think you're getting this thread straight. I don't believe in either instant communicating or sending messages into the past. In fact, my possible ignorance on science swings the other way: I am still convinced that there is a hidden variable theorem for all of quantum mechanics. In fact, I like the way you put it:
Because I couldn't agree with you more. That was kind of my point in this thread: the article I read *seems* to imply that it can be done, I definitely don't believe it can be done, so, as I quote from my first post:
[quote ]
[...] so there's obviously a logic flaw in my logic.
[/quote]

I was simply trying to understand my misunderstanding of the article. Because I am open to be proven wrong, but Bell's Theorem simply doesn't cut it for me.

25. Apr 29, 2012

### San K

Even if Victor could entangle particle at will (which I gather is fundamentally impossible) FTL information is still impossible because:

Alice and Bob still need to compare to find out if Victor entangled the pair or not.

Is that correct?

Side note: Per technical/nerd speak the name should have been Charlie or Chimp or Csomething than Victor. Perhaps Ma et. al. choose Victor because its easier to relate by having one slant of V point towards Alice and other towards Bob.