IF THE THREE PAULI MATRCIES WORK ON SINGLE QUBIT WHAT W E ARE GETTINGS THIS COLUD BE THE THE RIGHT T COMPLEXITY

so whatw nedd for singel qubit we need a singel qubuit to nalyize is such atht manennrs one is in x axis to confirms as expon neital wit if by pie/2 e exponeet pwoer poe ipie/2 and e power pie/2 minus for x axis is it will be the new gathered values is e pow pie 0 0 e pow pie matrices because the roataiopn si not stopping in it so its stasfy the cubic probelms secodn amtrices y 0 -iepow pie iepow pie 0 third matrcies e 2pie 0 0 -e 2 pie pie coclude for either axises so it gives in such that wayys so that of we rexched to thers means x and ist pow pie and so gates aganst paulis first 2pie.epow ipie 0 0 2pie epowipie second for y 0 -i2pie epowipie+piie/2 i2pie.epowipie+pie/2 0 for z axis 0 2pie e pow 2pie 2pie e pow 2pie 0 then to neutralise them we have to these function iwill work fx = 2pie.epowpie why w pie rotation and ipie is exponent vertical or horizontal limit as we trace horizontal x then horizontal y then horizontal z CONTROLLED GATES :--> we shift the not oeprarion even oevr both the new matricwe sis cretae it works with two matrices or multipel matrices when 2 qunbit and secodn si on not but firts qubit not of the not to work throgh out to wperforms not oeprations \ when firts q bit is 1 on then not oepration on second q ubit then firts q ubit not oepration then second in symemtrci not oeprations then not operation of not opration to eprfomrs to chnegd the controlled gates e powr i pie/4(z1-z2)(x1-x2) e powr ipie/4((x1-x2)(z1-z2) e powr i pie/4(i-(x1-x2)) (i-(z1-z2)) or e powr i pie/4(i-(z1-z2) i-(x1-x2) or e pow i pie/4 (i-(z1-z2))(x1-x2) e pow i pie/4(z1-z2)(i-(x1-x2))

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