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Measuring Quantum State

  1. Jun 18, 2008 #1
    The other thread was going too off-topic, so I'll repost here. Why is this not possible, without considering FTL signals?

     
  2. jcsd
  3. Jun 18, 2008 #2
    Bob violates the no-clone theorem? Said operator isn't unitary.
     
  4. Jun 19, 2008 #3
    Where does Bob violate the no-cloning theorem?
     
  5. Jun 19, 2008 #4
  6. Jun 19, 2008 #5
    No, [itex]U_f[/itex] is defined only on the computational-basis states |0>,|1>, ...etc. And I forgot to mention it's actually [tex] U_f\mid x\rangle\mid y\rangle =\mid x\rangle\mid y\oplus f(x)\rangle[/tex] where [itex]a\oplus b[/tex] is a bitwise modulo 2 sum.
     
  7. Jun 19, 2008 #6
    If [tex]U_f[/tex] keeps [tex]\left|x\right\rangle[/tex] untouched then the second argument had better not depend on what [tex]\left|x\right\rangle[/tex] was... otherwise the operator can't be unitary.
     
  8. Jun 19, 2008 #7

    gel

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    I'm not 100% sure what you're asking for, but I think the answer to your question is no. Let me rephrase it to see if I've got it right.
    Bob has a 1-1 function f:A->A for a finite set A, and for every x in A, there is a state vector |x> describing a possible state of some system -- say, S. The |x> for x in A form an orthonormal basis for states of S. He then constructs the following state (on two copies of system S).

    [tex]
    \Psi_f=\frac{1}{\sqrt{|A|}}\sum_{x\in A}|x>|f(x)>
    [/tex]

    In your example, you have A={0,1}n, but I don't know if there is any particular reason for this choice other than just wanting to think about a binary string of length n.

    Then, supposing Alice has no knowledge of f other than it being a 1-1 function on A, you want to know if she can compute f(x) for her prefered value of x, simply by making measurements on the system with state [itex]\Psi_f[/itex]? At least, with an arbitrarily small probability of getting the wrong result?
     
    Last edited: Jun 19, 2008
  9. Jun 19, 2008 #8

    gel

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    Well, suppose that [itex]|\Phi_{x,y}>[/itex] is the state if Alice measures f(x) to be y. Then these states would have to be orthogonal for different y and, if the probability of getting the wrong result is less than [itex]\epsilon[/itex] you would need
    [tex]
    | <\Phi_{x,y}|\Psi_f>|^2>1-\epsilon
    [/tex]
    whenever y=f(x) and
    [tex]
    | <\Phi_{x,y}|\Psi_f>|^2<\epsilon
    [/tex]
    whenever y != f(x). I'm fairly certain that you can't make [itex]\epsilon[/itex] very small for all functions f.

    Maybe you can do better than the classical situation though?
     
    Last edited: Jun 19, 2008
  10. Jun 19, 2008 #9

    gel

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    ok, the answer is no.
    For a fixed x in A, you can easily find functions f and g with f(x) != g(x), but where [itex]<\Psi_f|\Psi_g>\not=0[/itex] and therefore can't be distinguished with zero probability.

    If f(x) != g(x) for some f,g then f(y) and g(y) can agree for at most |A|-2 values of y, giving
    [tex]
    |<\Psi_f|\Psi_g>|^2 \le \left(1-2/|A|\right)^2.
    [/tex]
    This is a bit better than the classical case which does not have the square.
     
    Last edited: Jun 19, 2008
  11. Jun 19, 2008 #10

    Hurkyl

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    Sure it can. For example, the operator (given by its action on the chosen basis)

    [tex]| x \rangle | y \rangle \mapsto | x \rangle | y \oplus f(x) \rangle[/tex]

    is linear and has an obvious inverse. (Itself!)
     
  12. Jun 19, 2008 #11

    Hurkyl

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    It seems to me that the answer yes. Alice plugs this qubit into her own computer, and enacts the transformation: (for n unequal to x)

    [tex]| n \rangle | a \rangle |0 \rangle \mapsto | n \rangle | a \rangle |0 \rangle[/tex]
    [tex]| x \rangle | a \rangle |0 \rangle \mapsto | x \rangle | a \rangle |1 \rangle[/tex]

    Then, (and this is where things get iffy for me) can't you use something like Grover's algoritm to increase the amplitude on the state whose last bit is a '1'? If you can do that, then when you measure it, you are highly likely to get the state [itex]| x \rangle | a \rangle |1 \rangle[/itex], from which you can read the value of a/i] to get f(x).

    This method, alas, hinges entirely on my limited recollection of what Grover's algorithm can do for you.
     
  13. Jun 19, 2008 #12
    I will use [itex]| f_{\left|x\right\rangle} \rangle[/itex] to denote the vector onto which [tex]|x\rangle[/tex] is mapped. If I understand you correctly, then the following is assumed:

    [tex]U |x\rangle \otimes |0\rangle = |x\rangle \otimes | f_{|x\rangle}\rangle[/tex].

    Then

    [tex]U \left\{ |u\rangle + |v\rangle \right \} \otimes |0\rangle = \left\{ |u\rangle + |v\rangle \right\} \otimes | f_{|u\rangle + |v\rangle} \rangle = |u\rangle \otimes | f_{|u\rangle + |v\rangle} \rangle + |v\rangle \otimes | f_{|u\rangle + |v\rangle} \rangle [/tex]

    and also

    [tex]U \left\{ |u\rangle + |v\rangle \right \} \otimes |0\rangle = |u\rangle \otimes | f_{ |u\rangle} \rangle + |v\rangle \otimes | f_{ |v\rangle} \rangle[/tex]

    Now, since [itex]\left|u\right\rangle[/itex] and [itex]\left|v\right\rangle[/itex] may be chosen arbitrarily, for the two expressions to be equal, we must have

    [tex]| f_{|u\rangle + |v\rangle} \rangle = | f_{ |u\rangle} \rangle = | f_{ |v\rangle} \rangle[/tex].

    Either I misunderstand your notation or [tex]U[/tex] is not linear.
     
  14. Jun 19, 2008 #13
    Well [tex]
    f_{|u\rangle + |v\rangle}
    [/tex] is not defined unless [tex]
    |u\rangle + |v\rangle
    [/tex] is a basis vector... I'm slightly confused by the contradicting answers. I'll check back a while to hopefully clear this up.
     
  15. Jun 20, 2008 #14

    Hurkyl

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    There is no such thing as [itex]f_{|u\rangle + |v\rangle}[/itex]. What Dragonfall was specifying was the action of U on the chosen basis. (and even then, only on the subset of the basis where the last qubit is zero) The action of U on linear combinations of basis vectors is defined by the linearity relation. It's not defined by assuming f is extended to the entire vector space.
     
  16. Jun 20, 2008 #15
    If Alice measures the first n qubits of [tex]\frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}\mid i\rangle\mid f(i)\rangle[/tex], and gets some random result, say |4>. Is the post-measurement state [tex]\left| 4\right>\left| f(4)\right>[/tex]?
     
  17. Jun 20, 2008 #16


    But you wouldn't even need to do that. Can't you just use Grover's algorithm to search for the exact term you want?
     
  18. Jun 20, 2008 #17

    Hurkyl

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    Yah, I realized after I left for work that if my understanding of Grover's algorithm is right, then I didn't need to bother with the indicator qubit!
     
  19. Jun 20, 2008 #18
    But what's the catch? This seems too easy...
     
  20. Jun 20, 2008 #19

    Hurkyl

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    Either it really is that easy, or I'm misremembering Grover's algorithm. It's up to you to figure out which. :tongue:
     
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