# Quantum State

1. Jun 16, 2008

### Dragonfall

Suppose you were given $$\sum_{i=0}^{2^n}\left| i\right>\left| f(i)\right>$$, is there something you can do so that when you measure in the computational basis, you have a large chance of getting a particular desired $\left| i\right>\left| f(i)\right>$?

2. Jun 16, 2008

### Dragonfall

Is LaTeX not working?

3. Jun 16, 2008

### pmb_phy

That is rather odd. Let me give it a whirl. It works when I do a Preview Post I got the correct result. Let me try posting it

$$\sum_{i=0}^{2^n}\left| i\right>\left| f(i)\right>$$

$\left| i\right>\left| f(i)\right>$

Seems to work for me. Pretty strange indeed!

What do you mean by computational basis? What is |f(i)>? Is it an eigenket? I've never seen that notation before. Usually its written (if you are you referring to what I think you are) as |i>.

Pete

Last edited: Jun 16, 2008
4. Jun 16, 2008

### Dragonfall

I forgot to mention. $f$ is a function from $\{0,1\}^n$ to itself. Suppose I am given the superposition $$\sum_{i=0}^{2^n}\left| i\right>\left| f(i)\right>$$. I want to measure a particular $\left| i\right>\left| f(i)\right>$ with 1 or very high probability. Is it possible to do that by applying some operator to $$\sum_{i=0}^{2^n}\left| i\right>\left| f(i)\right>$$?

5. Jun 16, 2008

### pmb_phy

I'm sorry but I don't understand that notation. Why are you raising a set to the power n??
What does that notation mean? Also you didn't answer my question, i.e. what is |f(i)>? Note that I'm not asking what f(i) is. Does the ket |f(i)> represent an arbitrary quantum state? What do the values f(i) correspond to? Are they eigenvalues of some operator for which the eigenkets are |f(i)>? You do realize that |i> |f(i)> is the tensor product of two state vectors right? The ket |i> I assume is a basis ket for some Hilbert space.
I recommend placing a carrige return before and after the tex label so that the equation doesn't appear inline. Its confusing when it does in that it looks really ugly.
For this to be the case the coeffient must be close to 1. However you don't have any coefficients in that expression. Please check your notation and make sure its what you meant to post. Thanks.

Pete

6. Jun 16, 2008

### Dragonfall

{0,1}^n is the set of all n-length bit-strings. A general state of n qubits is described by $$\frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n}\alpha_i\left| i\right>$$, in the computational basis. Since f is just a permutation of the strings, $$\left| f(i)\right>$$ is just one ket of the said basis.

Last edited: Jun 16, 2008
7. Jun 16, 2008

### pmb_phy

Sorry but I can't help you. All this qubits and computational basis stuff is something I know nothing about. Good luck!

Pete

8. Jun 16, 2008

### Phrak

wait, pmb, come back...

Apparently this deals with spin 1/2 particles. The computational basis means,

|0> = |spin up>
|1> = |spin down>

A ket, like this |1101> = |13> (in hexidecimal), is shorthand for |up> |up> |down> |up>.

Dragonfly. I don't get it either. Is f(i) a digit, n bits long?

9. Jun 16, 2008

### Dragonfall

Yes. Say f(3)=4, then somewhere in the superposition is the term $$\left| 3\right>\left| 4\right>$$. Basically you'd get a superposition of all the input and values of f from 0 to 2^n.

Now I want to somehow extract a PARTICULAR piece of information from that superposition, say f(3). Is this possible? I mean if I just measure the qubits without doing anything to them first then I'd just get a random term, and hence a random value of f.

Now unfortunately I don't actually know any physical interpretation of all this stuff, so I can't refer to it as a "spin-1/2" particle.

Last edited: Jun 16, 2008
10. Jun 17, 2008

### Phrak

Yeah, you could do this I think, but I don't know why. It would throw away the parallelism that you'd be gaining in doing it in the first place.

Let me clean up some of your notation first, if I may. Normally you'd have two registers, x and y. The x register (call it 8 bits long) would be prepared as a superposition of all binary values from 0 to 2^8-1=255. All the binary numbers from 0 to 255 would be equally represented in x. Register y is initialized to 8 bits of binary zero, |00000000>. Usually the word length (how many bits) is implied and just written as y=|0>.

$$\left| x \right> = \frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}\left| i \right>$$

$$\left| y \right> = \left| 0 \right>$$

Note that I fixed the index limit to 2n-1 so that there are 256 numbers respresented, in total.

The idea is to take the register pair $$\left| x \right>\left| 0 \right>$$ to $$\left| x \right>\left| f(x) \right>$$ by applying some quantum gate that will mix |x> and |0> together to get y=f(x) and still leave you |x> so that the operation is reversible. (How, or if this manages to conserve information without mixing with the environment, is the question I'd like answered).

After the gate operation is preformed, y = f(x). The register pair become:

$$\left| x \right> \left| y \right> = \left| x \right> \left| f(x) \right>$$

We're both learning this material :: so I had to think a bit to come up with a "method". It would be do-able, just as much as any of this is do-able, given the current technology. Begin with the pair |3> |y> by thowing away the |x> superposition and preparing x = |3> without effecting |y>. Apply a gate operation to obtain |3> |f(3)>. This looks a lot like an inverse gate where |x> has been prepared in a singular state.

Eventually you should need something like this to extract classical information, I think. Maybe that's what you're asking. If so, I've been curious too. If we're lucky someone like Hirkyl will step in and clear up some uncertainty.

No worries.

Last edited: Jun 17, 2008
11. Jun 17, 2008

### pmb_phy

This can't be done in the form you wrote the equation since it isn't normalized. I.e. since there are no coefficients in front of the kets this doesn't represent a physical state. If you were to rewrite this as you did above when you used $\alpha_i$ and applied the ket <4|<3| then the only surviving term would be $\alpha_3$. The square of this value is the probability of the system being measured to be in the |3>|4>.

Pete

12. Jun 17, 2008

### Dragonfall

I don't think this is possible. Suppose Alice and Bob shared the entangled EPR pair $$\frac{\left| 00\right> +\left| 11\right>}{\sqrt{2}}$$. If Alice were able to CHOOSE which state she can measure, then she'd be able to communicate with Bob faster than light.

13. Jun 17, 2008

### pmb_phy

In the physics literature people have presented arguements that faster than light communication is possible with entangled states. Would you like references?

Pete

14. Jun 17, 2008

### Dragonfall

Really? I'd sure love some references. How does that reconcile with relativity?

15. Jun 17, 2008

### Dragonfall

I'm just going over what I'm asking again since I wasn't clear before:

Bob takes n+1 qubits, all initiated at |0> (the first n being input registers and the last is an output register), and sends the first n through an n-fold Hadamard gate. This results in the superposition

$$\mid\Psi\rangle =\frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}\mid i\rangle\mid 0\rangle$$

Bob then chooses some function f bounded above by 2^n-1, then finds a unitary operator which implements the function:

$$U_f\mid x\rangle\mid 0\rangle =\mid x\rangle\mid f(x)\rangle$$

Bob then sends $U_f\mid\Psi\rangle$ to Alice. Alice now has n+1 qubits with loads of information about the function f. Alice would like to learn a PARTICULAR VALUE of f (not a random one). Can she do it?

16. Jun 17, 2008

### pmb_phy

Here are the ones that I picked up on my journey's to a research library

Can EPR-correlations be used for the transmission of superluminal signals? P. Mittelstaedt, Ann. Phys (Leipzig) 7 (1998), 7-8, 710-715
Superluminal signal velocity, G. Nimtz, Ann. Phys (Leipzig) 7 (1988), 7-8, 618-624
Bell's theorem: Does quantum mechanics contradict relativity?, L.E. Ballentine, Am. J. Phys. 55(8), August 1987
Faster than Light?, Raymond Y. Chiao, Paul G. Kwiat and Aephraim M. Steinberg, Scientific American, August 1993
re - How does that reconcile with relativity? - I can send you any and all of the papers in e-mail to you if you'd like? I haven't found time to study this phenomena yet. Actually I'm reviewing quantum mechanics again and plan on studying this in detail when I get to EPR and Bell's theorem. I'll try to look the articles over in the near future since this is a subject of great interest to me.

Best wishes

Pete

17. Jun 17, 2008

### Phrak

So *that's* what this is all about. Too cool. Unfortunately, I'm not in a position to understand it well enough. But to extract y(x) for a given x, Alice needs place x in a definite state--destroy the superposition of x. Bob doesn't know what Alice did to x does he?

There is one other point, though. You do realize that your f(x) is one bit wide as you've stated it. n bits for |x>, leaving one for |y>.

18. Jun 18, 2008

### Dragonfall

Is it possible to find these online?

19. Jun 18, 2008

### Cthugha

Well, the Nimtz paper can be found on Arxiv http://xxx.lanl.gov/abs/physics/9812053".

However none of these papers is really worth your time as they are either papers, which explain small loopholes, which were still left at the time these papers were written, papers, which explixitly say, that the velocity exceeding c is not a signal velocity like the Chiao paper or papers, which are on the brink to crackpottery, because they also discuss a velocity exceeding c, which is no signal speed, but do not explicitly tell that like in the Nimtz paper.

The Nimtz stuff has also been discussed several dozen times in this forum. I am a bit amazed, that it is still quoted as a reference from time to time.

Last edited by a moderator: Apr 23, 2017
20. Jun 18, 2008

### pmb_phy

Not that I know of. I scanned all the papers that I have in my filing cabinet and placed them into PDF files. If you don't want me to e-mail them to you then I could try to upload them onto one of my web sites. I could then post the URL to the papers and you could download them yourself. It'd be easier for me to e-mail them to you though.

Pete