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First question was a little bit ambiguous.

Photons are passed through a linear polarizer that is oriented $\theta$ degrees again the photon passes through another linear polarizer that also have a $\theta$ degree orientation.

Now my question is that why the second polarizer will effect the spin axis angle of the photon.

Actually I am not from a physics background and new to Quantum Cryptography.Please. help.

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Do you mean "polarization" instead of "spin" by any chance? –  ptomato Mar 4 '12 at 20:11
    
The ambiguity between polarisation and spin probably comes from the fact that the two most discussed realization of the qubit are a spin-1/2 and a polarised single photon. And since the polarisation is the manifestation of the spin of the photon, I'm not even sure the confusion is a real error. –  Frédéric Grosshans Mar 5 '12 at 9:57
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If I understood the question correctly, the two polarisers have the same orientation $\theta$. In that case, the second polariser has no effect at all :

  • either the photon is blocked by the first polariser and the second has no effect ;
  • or the photon passes through the first polariser. In that case, it is polarized along $\theta$ and passes through the second polariser.
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Now consider this case for quantum cryptography. If eve can (somehow) use the same angled polarizer as alice used (for every photon alice sending ) then he can get the encryption key without informing others.Then it must be a drawback of quantum cryptography and i think there must be a way to get rid of this drawback otherwise it would be useless to use quantun cryptography. Is there any way? If it is there then how it is achieved? –  fuzzy Mar 5 '12 at 13:25
    
@CMLA : The whole point of quantum cryptography is precisely that Eve cannot have the same angled polariser as Alice, except if she violates the law of quantum physics or if she is extraordinarily lucky. –  Frédéric Grosshans Mar 5 '12 at 13:36
    
But there is a probability that eve uses the same angled polarizer as alice used. And any cryptography approach where there is no guarantee that transaction will be safe is not acceptable. –  fuzzy Mar 5 '12 at 13:48
    
If the cryptographic protocol uses $n$ qubits, this probability decreases exponentially as $2^{-n}$, i.e. at the same speed as random guessing the encrypted bits. –  Frédéric Grosshans Mar 5 '12 at 13:52
    
By the way, modern security proofs of QKD rely on the fact that the situation from Eve's point of view is indistinguishable from random guessing. (By modern, I mean posterior to arxiv.org/abs/quant-ph/0512258 ) –  Frédéric Grosshans Mar 5 '12 at 14:31
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