Physics Help Forum Probability Distribution of time dependant Sx operator

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 Nov 26th 2016, 08:14 PM #1 Junior Member   Join Date: Nov 2016 Posts: 1 Probability Distribution of time dependant Sx operator I have a time dependant wave function, which is described as: Lamda(t) = 1/sqrt(2) * (exp (iuBt / 2) * ket (up) + exp (-iuBt/2) * ket (down) ) Where ket (up) and ket (down) are the eigenstates of Sx (the spin operator in the x-direction). I need to find the probability of measuring either up or down as a function of time - I thought this could be achieved by squaring lambda t, but I unable to do this. Could anyone give guidance? My attempt so far has been to write Ket up & down as 2x1 column vectors, add them together and using the definition of Cos and Sin, I was able to write it as (2cos(uBt/2), 2isin(uBt/2)) - however, this is impossible to square as it is a 2x1 matrix... I feel like this attempt might not be necessarily on the correct tracks, although mathematically it is sound. Thanks Last edited by BertTomas; Nov 26th 2016 at 08:25 PM.
Nov 26th 2016, 09:47 PM   #2

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 Originally Posted by BertTomas I have a time dependant wave function, which is described as: Lamda(t) = 1/sqrt(2) * (exp (iuBt / 2) * ket (up) + exp (-iuBt/2) * ket (down) ) Where ket (up) and ket (down) are the eigenstates of Sx (the spin operator in the x-direction). I need to find the probability of measuring either up or down as a function of time - I thought this could be achieved by squaring lambda t, but I unable to do this. Could anyone give guidance? My attempt so far has been to write Ket up & down as 2x1 column vectors, add them together and using the definition of Cos and Sin, I was able to write it as (2cos(uBt/2), 2isin(uBt/2)) - however, this is impossible to square as it is a 2x1 matrix... I feel like this attempt might not be necessarily on the correct tracks, although mathematically it is sound. Thanks
You are mixing some ket and wavefunction notation, but let's see what we can do. I'm going to define:
$\displaystyle | \lambda (t) > = \frac{1}{\sqrt{2}} e^{i \mu Bt/2} | S_x; + > + \frac{1}{\sqrt{2}} e^{-i \mu Bt/2} | S_x; - >$

The probability of measuring a + spin in the x direction will therefore be $\displaystyle | < S_x; + | \lambda (t) > |^2$.

-Dan
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 Tags dependant, distribution, operator, probability, time