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Very Basic QM

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    Hiya, I have two questions that I would appreciate help with please. I've looked at the course notes but still don't get it

    1) \dfrac{1}{\sqrt{2}}|0 \rangle + \dfrac{1}{\sqrt{2}}|1 \rangle can be distinguished from \dfrac{1}{\sqrt{2}}|0 \rangle - \dfrac{1}{\sqrt{2}}|1 \rangle although they only differ by a phase in the |1 \rangle component. Note that this is a relative phase because they differ in the phase of |1 \rangle relative to that of |0 \rangle. In contrast, |0 \rangle and -|0 \rangle differ by a "global phase". Is it possible to detect the global phase via measurement?
    Suppose we perform the measurement in some arbitrary basis |u \rangle, |u^\perp \rangle. Clearly we can write |0 \rangle = a |u \rangle + b |u^\perp \rangle for some complex numbers a and b.
    a) If the qubit is initially in state |0 \rangle, what is the probability that the outcome is u?
    b) If the qubit is instead initially in state -|0 \rangle, what is the probability that the outcome is u?


    I've looked at the course notes but I've not come across the terms "phase" and don't quite understand what they mean by "relative" and global".
    I thought that since |0 \rangle = a |u \rangle + b |u^\perp \rangle, the probability of measuring u is just |a|^2 but that seems too simple?
    Also is -|0 \rangle = -a |u \rangle - b |u^\perp \rangle; so the probability of measuring u is also |a|^2?


    2) A qubit is either in state |0 \rangle or |\phi \rangle = \cos \theta|0 \rangle + \sin \theta|1 \rangle. Which of the following measurements best distinguishes between these two possibilities?
    a) \cos \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right) |0 \rangle + \sin \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right)|1 \rangle, \sin \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right) |0 \rangle - \cos \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right)|1 \rangle
    b) Standard basis: |0 \rangle, |1 \rangle
    c) |\phi \rangle, |\phi^\perp \rangle
    d)a) \cos \dfrac{\theta}{2}  |0 \rangle + \sin  \dfrac{\theta}{2} |1 \rangle, \sin \dfrac{\theta}{2} |0 \rangle - \cos \dfrac{\theta}{2} \right)|1 \rangle

    Honestly, I have no idea how to approach this

    Thank you for your time
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    (Original post by LogicGoat)
    Hiya, I have two questions that I would appreciate help with please. I've looked at the course notes but still don't get it

    1) \dfrac{1}{\sqrt{2}}|0 \rangle + \dfrac{1}{\sqrt{2}}|1 \rangle can be distinguished from \dfrac{1}{\sqrt{2}}|0 \rangle - \dfrac{1}{\sqrt{2}}|1 \rangle although they only differ by a phase in the |1 \rangle component. Note that this is a relative phase because they differ in the phase of |1 \rangle relative to that of |0 \rangle. In contrast, |0 \rangle and -|0 \rangle differ by a "global phase". Is it possible to detect the global phase via measurement?
    Suppose we perform the measurement in some arbitrary basis |u \rangle, |u^\perp \rangle. Clearly we can write |0 \rangle = a |u \rangle + b |u^\perp \rangle for some complex numbers a and b.
    a) If the qubit is initially in state |0 \rangle, what is the probability that the outcome is u?
    b) If the qubit is instead initially in state -|0 \rangle, what is the probability that the outcome is u?


    I've looked at the course notes but I've not come across the terms "phase" and don't quite understand what they mean by "relative" and global".
    I thought that since |0 \rangle = a |u \rangle + b |u^\perp \rangle, the probability of measuring u is just |a|^2 but that seems too simple?
    Also is -|0 \rangle = -a |u \rangle - b |u^\perp \rangle; so the probability of measuring u is also |a|^2?


    2) A qubit is either in state |0 \rangle or |\phi \rangle = \cos \theta|0 \rangle + \sin \theta|1 \rangle. Which of the following measurements best distinguishes between these two possibilities?
    a) \cos \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right) |0 \rangle + \sin \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right)|1 \rangle, \sin \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right) |0 \rangle - \cos \left( \dfrac{\theta}{2} - \dfrac{\pi}{4} \right)|1 \rangle
    b) Standard basis: |0 \rangle, |1 \rangle
    c) |\phi \rangle, |\phi^\perp \rangle
    d)a) \cos \dfrac{\theta}{2}  |0 \rangle + \sin  \dfrac{\theta}{2} |1 \rangle, \sin \dfrac{\theta}{2} |0 \rangle - \cos \dfrac{\theta}{2} \right)|1 \rangle

    Honestly, I have no idea how to approach this

    Thank you for your time
    I'd recommend reading up about the Bloch sphere. It's the easiest way to represent single qubit states and it makes it easier to understand the phase factors.

    For part 2, Do you know how to distinguish between two non-orthogonal states? You want to measure along a basis that is orthogonal to one of the possible states. In this case you want to measure in the |0>, |1> basis. If you obtain a result corresponding to |1> you know for certain that the original state was not |0>. The probability of successfully distinguishing between two states is given by 1 - |<\psi|\phi>| where psi and phi are the two states you want to distinguish between. If phi and psi are orthogonal the inner product is 0 and you have perfect distinguishability.
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    (Original post by suneilr)
    I'd recommend reading up about the Bloch sphere. It's the easiest way to represent single qubit states and it makes it easier to understand the phase factors.

    For part 2, Do you know how to distinguish between two non-orthogonal states? You want to measure along a basis that is orthogonal to one of the possible states. In this case you want to measure in the |0>, |1> basis. If you obtain a result corresponding to |1> you know for certain that the original state was not |0>. The probability of successfully distinguishing between two states is given by 1 - |<\psi|\phi>| where psi and phi are the two states you want to distinguish between. If phi and psi are orthogonal the inner product is 0 and you have perfect distinguishability.
    Hiya, thanks for replying

    I'm reading about the Bloch sphere now, thanks With regard to part 2, why is the |0>, |1> basis any better than the |\phi \rangle, |\phi^\perp \rangle basis?

    Thanks again for the help
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    (Original post by LogicGoat)
    Hiya, thanks for replying

    I'm reading about the Bloch sphere now, thanks With regard to part 2, why is the |0>, |1> basis any better than the |\phi \rangle, |\phi^\perp \rangle basis?

    Thanks again for the help
    That would be just as good a basis. I must have missed that option
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    (Original post by LogicGoat)
    Hiya, thanks for replying

    I'm reading about the Bloch sphere now, thanks With regard to part 2, why is the |0>, |1> basis any better than the |\phi \rangle, |\phi^\perp \rangle basis?

    Thanks again for the help
    I'm actually going to have to retract that answer for part 2. The strategy I gave you is if you want unambiguous discrimination.
    If you just want the maximum likelihood estimate you want to measure in the |+>, |-> basis so the answer should be d)

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