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Bell State of two non-orthogonal modes

  1. Feb 25, 2013 #1
    Is this possible? If I have, say a photon and two non-orthogonal polarizations [itex]\mid0\rangle[/itex] and [itex]\mid1\rangle[/itex], can I create a Bell state [itex]\mid10\rangle+\mid 01\rangle[/itex]

    If not, what is the reason?

    Thank you :)
  2. jcsd
  3. Mar 3, 2013 #2

    I understand that I won't get the anticorrelation statistics but does it makes sense to describe a Bell state of non-orthogonal subsystems? That is, can the object [itex]\mid10\rangle+\mid 01\rangle[/itex] exist regardless of what statistics it gives under measurement?
  4. Mar 3, 2013 #3


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    I must confess that I don't understand your notation. If [itex]\mid0\rangle[/itex] and [itex]\mid1\rangle[/itex] are possible states of a single photon, I'd understand what was meant by something like [itex]\frac{\mid0\rangle+\mid1\rangle}{\sqrt{2}}[/itex], but I'm not sure what to make of [itex]\mid10\rangle[/itex] or a "Bell state".
  5. Mar 3, 2013 #4


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    I don't think it makes sense to write it that way, but you can construct such a state. However, it won't have the properties of a Bell state because [itex]\mid10\rangle+\mid 01\rangle[/itex] is not maximally entangled if the two "basis" states are not orthogonal.

    To see how such a state would look like in the usual basis, you can simply decompose one of the "basis" states and calculate the resulting state. For example |1>=a|0>+b|1'> where |1'> is orthogonal to |0>.
  6. Mar 5, 2013 #5
    Thank you for your reply. I don't want to look at it in the usual basis; I want to stay in this non orthogonal basis. Just to understand this from a conceptual point of view.

    I see that calling it a Bell state is probably wrong. But the idea is that we have two non orthogonal polarizations [itex]0[/itex] and [itex]1[/itex].

    Now, the state [itex]\mid10\rangle+\mid 01\rangle[/itex] is a state such that if Alice measures and finds the first photon is polarized along [itex]0[/itex] and Bob sends his photon through a polarizer in direction [itex]1[/itex] and puts a detector after that, then Bob's detector will always register a click.

    Is this correct?
  7. Mar 5, 2013 #6


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    No. If you make a measurement, you have to decompose your state into the (orthogonal) eigenstates of the measurement operator. To analyze your situation, you have to write the state in Alice's basis (0, 1') first. If you don't do this, you get incorrect results. For example, Alice's probability to measure 0 is bigger than 1/2, because the second term contains an additional part a|0>.

    Having written the state in Alice's basis, you can do the state reduction and you will see that the final state is not equal to |01>.

    You probably know this, but just to be clear: your "basis" is not really a basis because its elements are not linearly independent.
    Last edited: Mar 5, 2013
  8. Mar 6, 2013 #7
    Okay I wrote out [itex]\frac{1}{\sqrt{2}}(\mid0,+\rangle\ +\mid +,0\rangle)[/itex] and it turns out to be [itex]\frac{1}{2}(2\mid0,0\rangle\ +\mid 1,0\rangle\ +\mid 0,1\rangle)[/itex]. I see what you mean about the statistics being not so trivial.

    A final query though: I thought that my vectors are indeed linearly independent, though not orthogonal? I can get any state vector I want with combinations of [itex]\mid0\rangle[/itex] and [itex]\mid+\rangle[/itex]. I thought that them being non orthogonal is the problem. Or is there something basic that I am missing?

    Thank you for your replies!
  9. Mar 6, 2013 #8


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    Sorry, I have missed something basic. ;-) You are of course right about this.
  10. Mar 6, 2013 #9
    Thank you kith! I appreciate your help very much :)
  11. Mar 8, 2013 #10


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    You're welcome. :smile:
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