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2 Quant. questions (expanding an abitrary state, transition amplitude?)

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data

    1) The position operator x has a continuous spectrum of engenvalues {x'} with corresponding eigenkets {|x'>}. Expand this state in terms of the position eigenstates.

    2) Suppose the transition amplitudes c_a',b' = <a'|b'> are known where {|b'>} are the eigenstates of a Hermition operator. Write cos(A) in terms of {|b'>}.

    2. Relevant equations

    -

    3. The attempt at a solution

    1) Is this correct? Let the arbitrary state be |x>, then:

    |x> = integral(dx' * |x'><x'| * |x>) ?

    2) I have no idea how one would write cos(A) in terms of {|a'>}, let alone in terms of {|b'>}??

    Any help appreciated!
     
  2. jcsd
  3. Jun 21, 2011 #2
    Normally you would normalize the position eigenstates such that <x'|x> = δ(x-x'), so that you get |x> = ∫ |x'> <x'|x> dx' = ∫ |x'> δ(x-x') dx' = |x>, which is a tautology. I don't quite get the reasoning behind the first question, expand a position eigenstate in terms of position eigenstates? Maybe it was meant momentum eigenstates.

    For the second part, use that cos(A) = ½ [ exp(i A) + exp(-i A) ], which is defined by the series expansion of the exponential. Then you can use that cos(A) = cos(A) Σ_{a'} |a'><a'| = Σ_{a'} ( cos(A) |a'> ) <a'| and so on.
     
  4. Jun 21, 2011 #3

    lanedance

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    For the first i would ask what state though? it doesn't specifically say position eigenstate, is a state referenced somewhere? ro could you assume soem arbitrary state [itex] |\phi> [\itex]
     
  5. Jun 21, 2011 #4
    So we can write:

    cos(A) = sum_a' [(cos(A) |a'>) <a'|] = sum_a' [sum_b' [|b'><b'| (cos(A) |a'>) <a'|] ] = sum_a' [sum_b' [cos(A) <b'|a'> <a'|b'>] ] = sum_c [ c^2 * cos(A)],

    since <b'|a'> = <a'|b'>* = <a'|b'> = c (given in the question), since A,B hermitian? OR should I do it like this?

    cos(A) = sum_a' [(cos(A) |a'>) <a'|] = sum_a' [sum_b' [|b'><b'| (cos(A) |a'>) <a'|] ] = sum_a' [sum_b' [<b'|cos(A)|b'> |a'> <a'|] ] = sum_b' [<b'|cos(A)|b'>]

    Both methods and results seem correct, but if I write it in terms of c, then I don't end up writing cos(A) in terms of {|b'>} as it asks...



    As for the first question, I forgot to write it all out: "The position operator x has a continuous spectrum of engenvalues {x'} with corresponding eigenkets {|x'>}. The momentum operator p has a continuous set of eigenvalues {p'} with corresponding eigenkets {|p'>}. An ensemble of particles is in an arbitrary state |alpha>. Write out this state in terms of the position eigenstates."

    Is the answer to this still just: |alpha> = integral [dx' delta(x-x')* |x'>], like you wrote grey_earl?

    Thanks very much for your replies so far, it has been a great help to my study :)
     
  6. Jun 22, 2011 #5

    lanedance

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    no, that is only if the state is a position eigenstate |x>, replace it with alpha
     
  7. Jun 22, 2011 #6
    Ah I see, thanks for that lancedance :) There's more parts to this question that I might post here in case anyone sees this in the next 12 hours or so before my exam I'm studying for :)...

    (Continuing on from the arbitrary state |alpha>...) Suppose we make a measurement that selects only particles in the range x'-A/2 to x'+A/2, where A is some small real number. What is the normalized state of these particles immediately after this measurement?

    So the new state after the measurement would be (call it beta)... |beta> = integral_lower bounds x'-A/2, upper bounds x+A/2 [dx' |x'><x'|alpha>] = integral[dx' |x'> dirac(alpha-x')]

    But have I found the "normalized state"? what do i have to do if i haven't?



    Thanks so much for all help so far :)
     
  8. Jun 22, 2011 #7
    Before introducing the sum over b', you should apply cos(A) on the state |a'>, that's why I put it in parentesis. Coming from the definition, you get cos(A) |a'> = Σ_{n=0}^∞ (-1)^n/(2 n)! A^(2n) |a'> = Σ_{n=0}^∞ (-1)^n/(2 n)! (a')^(2n) |a'> = cos(a') |a'>. Then write |a'> = Σ_(b') |b'> <b'|a'> and an analog expression for <a'| to get the final result. Note that you cannot change |b'><a'| to <a'|b'>, since the first is an operator, but the second is a number.

    No, since <x'|α> is not equal to a δ distribution. You just have |α> = ∫ |x'><x'| dx' |α> = ∫ <x'|α> |x'> dx' = ∫ α(x') |x'> dx'.

    :) Any time again!
     
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