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Normalization Factor

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A quantum system has a measurable property represented by the observable S with possible eigenvalues nh, where n = -2, -1, 0, 1, 2. The corresponding eigenstates have normalized wavefunctions [tex]\psi_{n}[/tex]. The system is prepared in the normalized superposition state given by

    [tex]\psi = \frac{1}{N}(\psi_{-2}+2\psi_{-1}+4\psi_{1}-6\psi_{2}[/tex]

    Where N is a normalization factor

    (i) Calculate N

    (ii) Write down the probability for each of the following measurements os S: -h, 0, 2h


    3. The attempt at a solution

    Given that the wavefunction is normalised, the sum of the squared moduli of the coefficients equals 1, so

    [tex]\left(\frac{1}{N}\right)^2 = \left(\frac{1}{N}\right)^2\left(+\frac{2}{N}\right)^2+\left(\frac{4}{N}\right)^2+\left(\frac{-6}{N}\right)^2[/tex]

    Which equals

    [tex]\frac{1}{N}^2=\frac{1}{N}+\frac{4}{N}+\frac{16}{N}+\frac{36}{N}[/tex]

    I'm getting a bit lost from here though
     
  2. jcsd
  3. Aug 8, 2007 #2
    The condition you are looking for is [tex] \left< \psi | \psi \right> = 1 [/tex] (or any other value you'd normalize to), I don't understand what you're saying about the "squared moduli of the coefficients" (possible due to my lack in english), but your equations defenitely does not reflect the condition I stated (an addend independent of N is missing).
     
  4. Aug 8, 2007 #3

    George Jones

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    The modulus of a complex number is its "length". What Ben is saying is that if a wavefunction iis written as a linear combination of orthonormal wavefunctions, then the squared "length" of the wavefunction is the sum of the squares of the "lengths" of the coefficients for the linear combination.

    Ben:

    1) you should understand why this is true;

    2) from where did the left side of your second-last equation come;

    3) are you sure about the denominators on the right side of your last equation?
     
  5. Aug 8, 2007 #4
    Hmm... I'm probably missing some vital piece of knowledge here.... My books aren't very explicit in describing this situation... In fact Im finding the whole quantum physics stuff a bit hard to follow... But anyhow

    For the points you raise...

    (i) I understand your point about the squared length of the wavefunction being equal to the sum of the squares

    (ii) I see what you mean about the left hand side of the 2nd equation... I should have it down as

    [tex]\frac{1}{N}\left(\frac{1}{N}+\frac{4}{N}+\frac{16}{N}+\frac{36}{N}\right) = 1[/tex]


    (ii) I'm not sure I understand about the denominators on the right hand side. If we have

    [tex]\frac{1}{N}\left(1+4+16+36) = 1[/tex]


    Isn't this the same as saying

    [tex]\frac{1+4+16+36}{N} = 1[/tex]?

    If I solve for N here I get 57
     
    Last edited: Aug 8, 2007
  6. Aug 8, 2007 #5

    George Jones

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    In your last post, how did you go from the first equation to the second equation?
     
  7. Aug 8, 2007 #6
    I just multiplied out the brackets as you would normally... Something tells me I'm wrong here....
     
  8. Aug 8, 2007 #7

    George Jones

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    What does

    [tex]\frac{2}{3} \times \frac{5}{7}[/tex]

    equal?
     
  9. Aug 8, 2007 #8
    [tex]\frac{2 \times 5}{3 \times 7}=\frac{10}{21}[/tex]


    Am I right in saying that

    [tex]\frac{1}{N}\left(\frac{1}{N}+\frac{4}{N}+\frac{16} {N}+\frac{36}{N}\right) = 1[/tex]

    but wrong in how I've multiplied it out?
     
  10. Aug 8, 2007 #9

    George Jones

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    What does

    [tex]\frac{1}{N} \times \frac{36}{N}[/tex]

    equal?
     
  11. Aug 9, 2007 #10
    Ah.. I see what you mean..

    [tex]\frac{1}{N} \times \frac{36}{N} = \frac{1}{N^2}[/tex]

    Therefore I should have

    [tex]\frac{1}{N^2}+\frac{4}{N^2}+\frac{16}{N^2}+\frac{36}{N^2} = 1[/tex]
     
  12. Aug 9, 2007 #11

    CompuChip

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    No

    Yes.
    So
    [tex]\frac{57}{N^2} = 1[/tex].
    What is [itex]N[/itex] then?
     
  13. Aug 9, 2007 #12
    Sorry... Of course

    [tex]\frac{1}{N} \times \frac{36}{N} = \frac{36}{N^2}[/tex]



    So [tex]\frac{57}{N^2}[/tex]

    has N = 7.5498
     
  14. Aug 9, 2007 #13

    CompuChip

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    You mean [itex]N \approx 7.5498[/itex], or [itex]N = \sqrt{57}[/itex].
    Well done.
     
  15. Aug 9, 2007 #14
    So the probability for each of the measurements S: -h, 0, 2h
    will be simply

    P(-h) = -1/(sqrt(57))
    P(0) = 0
    P(2h) = 4/(sqrt(57))

    Is this right?
     
  16. Aug 9, 2007 #15

    Dick

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    Ouch. Those are amplitudes, not probabilities. Probability can't be negative. What's the relation between amplitude and probability? Wait a minute, they aren't even that. P(-h) should be associated with psi(-1) which has a coefficient of 2, not -1. Similarly for P(2h).
     
    Last edited: Aug 9, 2007
  17. Aug 9, 2007 #16
    Oh dear... Back to the books again I think.... My paper asks for probabilities for each of the measurements and gives an example similar to the answers I just gave.... I can honestly say that quantum stuff really isn't my forte!!

    What should I be looking out for when calculating probabilities?
     
  18. Aug 9, 2007 #17

    Dick

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    P(-h) corresponds to psi(-1), which has a coefficient of 2/sqrt(57), right? As it says in the problem statement, probability is the squared modulus of the coefficient. I.e. 4/57. Check that the sum of ALL the probabilities adds to one. It will help you understand why you normalized the function.
     
    Last edited: Aug 9, 2007
  19. Aug 9, 2007 #18
    Ah... So for P(2h) this corresponds to psi(+2) which has coefficient of -6/sqrt(57) yes? which gives us by modulus square of coefficients 36/57?
     
  20. Aug 9, 2007 #19

    Dick

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    Now you're cooking. What's the sum of probabilities of all four possible states?
     
  21. Aug 9, 2007 #20
    Ok, that would be

    [tex]\frac{1}{57}+\frac{4}{57}+\frac{16}{57}+\frac{36}{57}[/tex]

    which = 1
     
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