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Entropy. Quantum statistic

  1. Jun 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate entropy for density matrix with eigenvalues ##0## and ##1##.



    2. Relevant equations
    ##S=-\lambda_1 \ln \lambda_1-\lambda_2 \ln \lambda_2##
    where ##\lambda_1## and ##\lambda_2## are eigenvalues of density matrix.


    3. The attempt at a solution
    How to calculate this when ##\ln 0## is not defined?
     
  2. jcsd
  3. Jun 2, 2013 #2
    I'm no expert, but...

    What is the value of ##\displaystyle \lim_{\lambda_1\rightarrow 0}\left[\lambda_1\ln{\lambda_1}\right]##?

    I think that is the only way to get a numerical answer here: make it approach what you want.
     
  4. Jun 2, 2013 #3
    For a density matrix ρ with eigenvalues only 0 and 1, we have [itex] ρ = ρ^{2} [/itex]. This is true only for pure states and thus we know the Von Neumann entropy must be zero. To calculate it numerically I would guess the approach Mandelbroth suggested is valid.
     
  5. Jun 2, 2013 #4
    What is interpretation of that. For pure state entropy is zero. Why?
    ##-1\ln 1-\lim_{\lambda \rightarrow 0}\lambda \ln \lambda=-\lim_{\lambda \rightarrow 0}\lambda \ln \lambda ##
    How to calculate this limit?
     
  6. Jun 2, 2013 #5
    To calculate the limit let [itex] t = 1/x [/itex]. Then you have [tex] \lim_{t\to\infty}=\frac{log(1/t)}{t} = \frac{\infty}{\infty}.[/tex] Then use L'Hospital's Rule and you will get the answer. Otherwise, you could just type it into wolfram alpha.
     
  7. Jun 2, 2013 #6
    Tnx a lot! And physically why entropy of pure state is zero?
     
  8. Jun 2, 2013 #7
    It is easy to see mathematically why the entropy of a pure state is zero. However, why its true physically seems a much harder question, one I'm not sure I know how to answer.
     
  9. Jun 2, 2013 #8

    Mute

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    Homework Helper

    Entropy is in some sense a measure of our uncertainty about the state a system. If a system is in a mixed state, is our uncertainty big or small? What about when it is in a pure state?
     
  10. Jun 2, 2013 #9
    Pure state is minimum uncertainty so it makes sense Entropy would be zero.
     
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