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System with two particles with spin QM

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Two particles with spin ##S_1=1## and ##S_2=1## are in our system. For wavefunction of the first particle we know that ##S_{1z}|\psi _1>=\hbar |\psi _1>## while for the second particle ##S_{2x}|\psi _2>=\hbar |\psi _2>##.
    a) Find wavefunction ##|\psi _1>## in basis with ""good"" (I don't know the right english word but I know am supposed to use CG coefficients) ##S_{1z}## and ##|\psi _2>## in basis with ""good"" ##S_{2x}##.
    b) We measure the squared value of aggregate spin of both particles. What are the possible results and their probabilities?
    c) Calculate the expected value and uncertainty of aggregate spin of both particles.
    d) We measure the ##x## component of aggregate spin of both particles. What are possible results and their values?

    2. Relevant equations
    I will try to use notation ##|s,m>##.

    Also note: $$S_z|\psi >=m\hbar |\psi >$$ $$S_{\pm}|s,m>=\hbar \sqrt{s(s+1)-m(m\pm 1)}|s,m\pm 1>$$ $$S_x=\frac{S_+ +S_-}{2}$$ $$S_y=\frac{S_+ -S_-}{2i}$$

    3. The attempt at a solution
    a)
    Basis vectors for both particles are ##|1,0>,|1,-1>,|1,1>##. Therefore for the first particle $$S_{1z}(\alpha |1,0>+\beta |1,1>+\xi |1,-1>)=\hbar(\alpha |1,0>+\beta |1,1>+\xi |1,-1>)$$ Solving this system leaves me with ##\alpha=\xi =0## while ##\beta =1##. Therefore ##|\psi _1=|1,1>##.
    Things are a bit more complicated for the second particle.The idea is to use ##S_x=\frac{S_+ +S_-}{2}##. I really don't want to write all the equations (because it takes a lot of time for me to put everything in latex), so I hope you don't mind if you make this part a bit shorter. The condition says ##S_{2x}|\psi _2>=\hbar |\psi _2>## where ##\psi _2 =\alpha |1,0>+\beta |1,1>+\xi |1,-1>##. Knowing this, we can calculate $$S_x|1,0>=\hbar \frac{\sqrt 2}{2}|1,1>+\hbar \frac{\sqrt 2}{2}|1,-1>$$ $$S_x|1,-1>=\hbar \frac{\sqrt 2}{2}|1,0>$$ $$S_x|1,1>=\hbar \frac{\sqrt 2}{2}|1,0>$$ This gives me a system saying that ##\beta=\xi ## and ##\alpha =\sqrt 2 \beta## Therefore $$\psi _2 =\alpha (|1,0>+\frac{1}{\sqrt 2}|1,1>+\frac{1}{\sqrt 2} |1,-1>$$ Normalizing the last expression gives me $$\psi _2=\frac{1}{\sqrt 2}|1,0>+\frac{1}{2}|1,1>+\frac{1}{2}|1,-1>$$ Now to transform in the right basis, we simply use Clebsch-Gordan tables:
    For the first particle ##\psi _1=|1,1>## becomes ##|2,2>## and for the second particle
    ##\psi_2=\frac{1}{\sqrt 2}|1,0>+\frac{1}{2}|1,1>+\frac{1}{2}|1,-1>## becomes ##\frac{1}{2}(|2,1>+|1,1>+\frac{1}{\sqrt 6}|2,0>+\frac{1}{\sqrt 2}|1,0>+\frac{1}{\sqrt 3}|0,0>+|2,2>)##

    This should be it for part a).
    b)
    ##S^2|\psi _1>=6\hbar ^2|2,2>## if we know that ##S^2|s,m>=\hbar ^2s(s+1)|s,m>##. Also with probability ##1##.
    Second particle:
    ##S^2|\psi _2>=S^2(\frac{1}{2}(|2,1>+|1,1>+\frac{1}{\sqrt 6}|2,0>+\frac{1}{\sqrt 2}|1,0>+\frac{1}{\sqrt 3}|0,0>+|2,2>))=##
    ##=\frac 1 2 (6\hbar ^2 |2,1>+2\hbar ^2 |1,1>+\frac{6}{\sqrt 6}\hbar ^2 |2,0>+\frac{2}{\sqrt 2}\hbar ^2|1,0>+6\hbar ^2|2,2>)##
    So there are two options: ##6\hbar^2 ## with probability ##\frac {13}{24}## and ##2\hbar ^2## with probability ##\frac 3 8 ##.

    c)
    First particle:
    ##<S^2>=6\hbar ^2##
    ##<S_z>=2\hbar ## while ##<S_x>=<\psi _1|\frac{S_+ +S_-}{2}|\psi _1>=0=<S_y>##

    So ##\delta S=\sqrt{<S^2>-<S>^2}=\hbar \sqrt 2.##
    Second particle:
    This part is easy: ##<S^2>=4\hbar ^2## and ##<S_z>=\hbar ##.
    But to get ##<S_y>## and ##<S_x>## one has to again work with ##S_ {\pm }##. This takes a lot of time and writing. I shall only write the end results:
    ##<S_x>=\hbar## and ##<S_y>=0##.

    Meaning ##\delta S=\hbar \sqrt 2##. Yes, the same as the particle one. Interesting or wrong. I don't know.

    d)
    first particle: Possbile result ##\hbar ## with probability ##1##.
    Second particle: Wush... I would say:
    ##\frac{\sqrt 2}{2}\hbar## with probability ##\frac{7}{16}## and
    ##\hbar## with probability ##\frac{6}{16}## and
    ##\frac{\sqrt 6}{2}\hbar## with probability ##\frac{3}{16}##

    Huh... I know this is a lot of text and a lot of calculating this. But I would highly appreciate it if somebody could check my results. Also I would be happy to hear if some things could be done faster.
     
  2. jcsd
  3. Jan 30, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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