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Homework Help: Probability - Quantum States

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data

    The components of the initial state [tex]|\psi_i>[/tex] of a quantum system are given in a complete and orthonormal basis of three states [tex]|\phi_1>, |\phi_2>, |\phi_3>[/tex] by


    Calculate the probability of finding the system in a state [tex]|\psi_f>[/tex] whose components are given in the same basis by




    3. The attempt at a solution

    Actually, I must admit that I don't really know what I have to do to answer this question. However, while experimenting with possible approaches to a solution I got to




    But is it possible to get a negative probability?

    Also, since the three states are orthonormal, shouldn't they automatically be normalized, and the total probability [tex]\sum{P_i}=1[/tex]?

    Any help here will be greatly appreciated.
  2. jcsd
  3. May 1, 2010 #2
    First, what is the equation that describes the probability of finding the initial state in the final state?

    In your work, you make no mention of the final state. Also, it is impossible to get a negative probability. The [itex]||^2[/itex] mean to multiply the value by its complex conjugate, so you shouldn't get a negative value.
  4. May 3, 2010 #3
    Hi there nickjer! Cheers for your reply.

    I'm not sure actually...could it be



    But how would I find the states from the given info? (I'm sorry if these questions are trivial and elementary, but I'm studying these subjects without the aid of lectures and often struggle to figure these things out by myself).

    Ah, you see, I didn't realise this from the available examples (they obviously skip steps assumed as known).
  5. May 3, 2010 #4


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    Your basis is complete and orthnormal. This means you can express any state as:

    [tex]|\psi>=x_1 |\phi_1>+x_2 |\phi_2>+x_3 |\phi_3>[/tex]

    The conditions given to you tell you what the x's are for the different states. Now, you just need to take the inner product.

    On a side note, the wording of a problem is bad imo. One never finds a particle in some random "state" since the wave-function is not observable. One can only ever observe the observables of that state. Anyways, this is just pedantic...but still...
  6. May 3, 2010 #5


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    Yup. (I assume you meant P equals the modulus squared, not P times that quantity.)
    You have a complete set, so you can say

    [tex]\sum_{j=1}^3 |\phi_j\rangle\langle\phi_j| = 1[/tex]

    If you apply this to [itex]|\psi_i\rangle[/itex], you get

    [tex]|\psi_i\rangle = 1|\psi_i\rangle = \sum_{j=1}^3 |\phi_j\rangle\langle\phi_j|\psi_i\rangle[/tex]

    To evaluate [itex]|\langle\psi_f|\psi_i\rangle|^2[/itex], try inserting a 1 between the bra and ket.
    Last edited: May 3, 2010
  7. May 3, 2010 #6
    Exactly as vela said :)
  8. May 3, 2010 #7
    sorry, wrong topic!
  9. May 4, 2010 #8
    Thanks for the input!

    Typo :smile:

    Thanks for this, it's brought to light a whole new angle on my understanding of these "bras", "kets" and bloomin wave functions! :biggrin:

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