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Probability in Nth state

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data
    State vector :
    [itex]| \psi \rangle = \sum_{n = - \infty}^\infty a_n| \psi \rangle [/itex]
    where
    [itex]a_n= i\sqrt\frac{3}{5}{\frac{i}{4}}^{|n|/2}e^{-in\pi/2}[/itex]

    Find the probability of the particle in Nth state.
    What is the total probability of the particle in 'N' negative state.

    2. Relevant equations

    [itex]|\langle\psi|\psi\rangle| ^2 = 1[/itex]
    3. The attempt at a solution

    Would the probability of the particle being in Nth state be [itex]|\langle\psi|\psi\rangle| ^2 =
    => |a_n|^2 = [/itex] ?

    [itex]a_n*= -i\sqrt\frac{3}{5}{}\frac{1}{4i}^{|n|/2}e^{+in\pi/2}[/itex]

    If so here is my working...
    [itex] \sum_{n -\infty}^\infty a_n a_n* \delta_{n m} [/itex]

    [itex]\sum_{n -\infty}^\infty ( i\sqrt\frac{3}{5}{\frac{i}{4}}^{|n|/2}e^{-in\pi/2} )(-i\sqrt\frac{3}{5}{}\frac{1}{4i}^{|n|/2}e^{+in\pi/2} )[/itex]

    [itex] |a_n|^2 = \gamma [/itex] [itex] \sum_{n -\infty}^\infty(1/4)^{|n|}
    [/itex]

    where [itex] \gamma = 3/5 [/itex]

    How do I go around evaluating this summation from -infinity to +ve infinity though the power is |n|.

    Evaluating the summation in the range (0,infinty) yields (3/5)+ 0 ....

    Thanks.
     
  2. jcsd
  3. Oct 3, 2014 #2

    BvU

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    Hello QM, welcome to PF :-)


    Could that be ##
    | \psi \rangle = \sum_{n = - \infty}^\infty a_n| \psi_n \rangle## ?

    On the same footing: no, but close: ##|\langle\psi_n|\psi\rangle| ^2##

    So for N you get ##\sum_{n = - \infty}^\infty a_n| \langle \psi_N|\psi_n \rangle|^2 = \delta(N,n) \ a_n^* a_n\ |\langle \psi_N|\psi_N \rangle|^2 = a_N^* a_N##

    Probability for being in N th state with N negative: same procedure (probably dsame answer ?)

    Or are they asking for someting else: the total probability for the particle being in a state with negative N ? Then you have to sum, but fom minus infinity to 0, not from minus infinity to plus infinity, right ?
     
  4. Oct 3, 2014 #3
    Hi,

    Yes, I have been asked to find the total probability of a particle in a negative energy state, so how is this answer any different than the previous one. Both summations add up to 1 i.e (-infty,0) and (0,infty)..

    I thought negative values of N would be discarded due to the^ |N| power, or is my concept of absolute function wrong....

    Edit correction... Using N= -infty gives infinite as an answer, am I right in assuming that this indicates the unattainability of negative states?
     
    Last edited: Oct 3, 2014
  5. Oct 4, 2014 #4

    BvU

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    ##\sum_{-\infty}^{+\infty}## should give 1 After all it is a probability.
    How do you calculate ##|a_n|^2## and how do you calculate the sum ?
     
  6. Oct 4, 2014 #5
    I have calculated [itex]|a_n|^2[/itex] above, I have split summation into two parts..
    [itex]\sum_{-\infty}^{0} (1/4)^{|n|} = (1/4)^{-\infty} + (1/4)^{0} = (\infty) + 1[/itex]
    Similarly for the range [itex] (0,\infty) [/itex].
     
  7. Oct 4, 2014 #6
    Anyone....
     
  8. Oct 4, 2014 #7

    BvU

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    ##({1\over 4})^{|-\infty|} = 0## but in your summation you skipped over all the intermediate terms !
    Why don't you try the easier one first, e.g. ##{1\over 4}+{1\over 16}+{1\over 64} + ... ##

    By the way, ## |a_n|^2 = \gamma \sum_{n -\infty}^\infty(1/4)^{|n|} ## is not correct ! It is simply ##
    \gamma (1/4)^{|n|} ##

    And your concept for the absolute function should be ##|n| = n## for ##n\ge 0## and ##|n|=-n## for ##n \le 0##.

    And it does (fortunately), but you haven't verified that yet. You should do so now.
     
    Last edited: Oct 4, 2014
  9. Oct 5, 2014 #8
    I am failing to link together how
    ##\gamma (1/4)^{|n|}= 1 ##
    Intermediate values of above equation ~ (0.333) for the first few terms.. to meet the unity condition ##(1/4)^{|n|}## must be equivalent to (5/3) since constant is 3/5.

    What am I overlooking...
     
  10. Oct 5, 2014 #9

    BvU

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    Note that there is a summation in front: the SUM over all n must give 1: ##\Sigma\ \gamma (1/4)^{|n|}= 1##
    ##
    {1\over 4}+{1\over 16}+{1\over 64} + ...
    ## are only the terms ##n = 1,2,... \infty##

    You are overlooking the term with n = 0 and the terms with n < 0.

    The sum of the terms with ##n = -\infty ... -3, -2, -1 ## gives you the desired probability

    ##\approx 0.33 ## for the first few terms ? Do you know how to calculate the sum of an infinite geometric series ?
     
  11. Oct 5, 2014 #10
    I have finally realised the flaw in my reasoning. I was not taking the negative values into account, presuming that the summation would blow up at - infinity.

    I really appreciate your input, thanks! .
     
  12. Oct 6, 2014 #11

    BvU

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    Just checking: could you verify ##
    \sum_{-\infty}^{+\infty}\ \gamma (1/4)^{|n|}= 1
    ## and is your final answer to the first question ##\sqrt{3\over 5}\; \left(1\over 4\right)^{|N|} ## and the second -1/3 exactly ?
     
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