Problem 3: Find the Likely State of a Two-Qubit System



  • s1 = 1 (and s0 = 1) minimizes given cost function (E = -s0/2 - s0*s1)

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  • I will not use Physics or Math or anything complex but logic.

    Given h1 = 0 means that the formula to minimise is: E = (-0.5 * s0) + (-1 * s0 * s1 )

    Now, if I choose negative numbers for s0 or s1 I would revert the minus symbol to a plus which will not lead me to the minimisation of the formula, therefore the answer has to be a positive number.

    The question now is, which number is good between 0 and +1? 

    If I'm greedy thinker I would say that 0 is good since will put E = 0 or E = -0.5 (if s0 would be +1), but can I do better? Of course yes! Why would I totally shut down the right part of the formula (-1 * s0 * s1), by putting s1 = 0, since it could give me a negative number to add to the left hand side of the formula, leading us to a better minima? So s1 must be > 0. 

    Could it be a floating point number? Of course not, since it would lower the negative impact of the right hand side of the E formula. 

    So all that said:

    s1 must be = 1 and for the same reasons s0 must be = 1 also.

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  • the cost function is minimized when s0 and s1 both are equal to 1. See below code for where I have used two methods. Can anyone explain why the CSP method gives a lower energy solution than QUBO method?


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  • Hi Yasas,

    Typically the CSP method is used to find a solution that meets a set of requirements. These requirements are expressed by describing which combinations of variables are valid and which are not. I recommend reading the article Using QUBOs to Represent Constraints to get a better understanding of what's happening behind the scenes.

    Also check out the difference between Ising and QUBO formats - this might help if you're having issues with your code.

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  • The answer to this problem is 1.

    Great answers here - thanks to Jacob, Calogero, and Yasas for participating. Kudos to Jacob, who posted first. 

    Now try Problem 4, posted today!

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