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Superposition - basic question

  1. Sep 20, 2009 #1
    Hello, I am brand new to this stuff and am trying to get my head around it all. I've spent considerable time trying to understand this from Landau's book on the subject (chapter 1 of course).

    I bet I'd get more answers by being more brief but I always find that asking the problem carefully sometimes helps me understand the problem better.

    A wavefunction, which completely describes the states of a quantum object, can be decomposed in terms of its eigenfunctions,

    [tex]\psi = \sum_{n} a_{n}\psi_{n}[/tex]

    Eigenvalues (maybe a physical quantity) correspond to the eigenfunctions by

    [tex]\widehat{f}\psi_{n} = f_{n}\psi_{n}[/tex]

    Where [tex]\widehat{f}[/tex] is the operator that corresponds to the quantity in question.

    From this, we see that the value of [tex] a_{n}[/tex] for a given eigenfunction in the decomposition is (somehow) related to the "probability" that the physical quantity [tex]f[/tex] has the value [tex]f_{n}[/tex].

    Given, is

    [tex]\int |\psi_{n}(q)|^{2}dq = 1[/tex]

    and

    [tex]\int |\psi(q)|^{2}dq = 1[/tex]

    How does it follow that [tex]|a_{n}|^{2}[/tex] is the probability of the physical quantity [tex]f[/tex] having the value [tex]f_{n}[/tex]? The reasoning presented in the book is not clear to me - it is a sort of deductive reasoning that seems like guesswork.

    Of course if this is a probability then,

    [tex]\sum |a_{n}|^{2} = 1[/tex]

    I don't understand how this follows from the other things.

    Here is why I am having a problem with this:

    I can see it all works if the following is true:

    [tex]\psi = a_{0}\psi_{0} + a_{1}\psi_{1} + ... + a_{n}\psi_{n}[/tex]

    [tex]|\psi| = \sqrt{|a_{0}\psi_{0}|^{2} + |a_{1}\psi_{1}|^{2} + ... + |a_{n}\psi_{n}|^{2}}[/tex]

    [tex]\int |\psi|^{2}dq = \int |a_{0}\psi_{0}|^{2}dq + \int |a_{1}\psi_{1}|^{2}dq + ... + \int |a_{n}\psi_{n}|^{2}dq[/tex]

    [tex] = |a_{0}|^{2}\int |\psi_{0}|^{2}dq + |a_{1}|^{2}\int |\psi_{1}|^{2}dq + ... + |a_{n}|^{2}\int |\psi_{n}|^{2}dq[/tex]

    Truth of this rests on the truth of two identities for complex numbers.

    [tex]|(a+bi)(c+di)|^{2} = |a+bi|^{2}|c+di|^{2} IDENTITY ONE[/tex]
    According to my calculations this is true.

    [tex]|(a+c) + (b+d)i|^{2} = |a+bi|^{2} + |c+di|^{2} IDENTITY TWO[/tex]
    According to my calculations this is false, unless there is a constraint [tex]2ac = -2bd[/tex].

    What is going on here? Is there some sort of constraint?
     
  2. jcsd
  3. Sep 20, 2009 #2

    jtbell

    User Avatar

    Staff: Mentor

    I think your missing piece of information is that the [itex]\psi_n[/itex] are orthogonal, that is,

    [tex]\int \psi^*_m(q) \psi_n(q) dq = 0[/tex]

    for [itex]m \ne n[/itex].

    This is what lets you go from

    [tex]
    \psi = a_{0}\psi_{0} + a_{1}\psi_{1} + ... + a_{n}\psi_{n}
    [/tex]

    to

    [tex]
    \int |\psi|^{2}dq = \int |a_{0}\psi_{0}|^{2}dq + \int |a_{1}\psi_{1}|^{2}dq + ... + \int |a_{n}\psi_{n}|^{2}dq
    [/tex]
     
  4. Sep 20, 2009 #3
    Thanks I think that makes a lot of sense.

    Basically "cross terms" looking similar to

    [tex]\psi^*_m(q) \psi_n(q)[/tex]

    arise in the expression for [tex]|\psi|^{2}[/tex], and are eliminated when they are integrated, leaving only the terms looking like

    [tex] \psi^*_n(q) \psi_n(q)[/tex]

    Thanks a bunch.
     
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