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Inner product

  1. Apr 25, 2008 #1

    oww

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    Hello everyone.

    I have a question that i think belongs here or in the linear algebra section; but now i put it here, maybe someone will move the post if it's in the wrong place.

    The question:

    I have a normalized (1D) state in the x-axis in the harmonic oscillator

    [tex]\Psi(x,t=0)=1/\sqrt{45}(6u_0(x)\chi_+ +(2+i)u_1\chi_- -2u_1(x)\chi_+[/tex]

    where [tex]u_n[/tex] is the n'th normalized eigenstate of the 1D HO and [tex]\chi_{+/-}[/tex] are the normalized eigenspinors in z-direction. Now i must find the possibility of getting the result +h/2 when i measure [tex]S_y[/tex]. My idea was to take the inner product with the desired state and square the coefficient:
    [tex]prob(h/2)=(\left\langle \chi^y_+ | \Psi\right\rangle)^2[/tex]
    but how do i do that when i have both spinors(vectors) and functions of x at the same time? If anyone could do the calculation stepwise, I would appreciate it very much!

    Best regards
    Oistein
     
  2. jcsd
  3. Apr 25, 2008 #2

    malawi_glenn

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    Homework Helper

    I would not care about the spatial part, the spin "lives" in another hilbert space than the spatial wave functions [itex] u_n(x) [/itex]

    So just do [tex] = | \langle \chi^y_+ | \Psi \rangle |^2 [/tex]
     
    Last edited: Apr 25, 2008
  4. Apr 25, 2008 #3
    Your Hilbert space comprises the following states
    [tex] \left|u_0, +\right\rangle_y, \left|u_0, -\right\rangle_y, \left|u_1, +\right\rangle_y, \left|u_1, -\right\rangle_y, \dots [/tex]
    I have changed into the y basis for convenience. You'll have to do the same, somehow.

    The probability of getting something is equal to the sum of the probabilities of all the independent ways of getting something. The possible ways of getting +h/2 from [tex]S_y[/tex] by finding the state to be in the following states:
    [tex] \left|u_0, +\right\rangle_y, \left|u_1, +\right\rangle_y \dots [/tex]

    In other words,
    [tex]P(+) = P(u_0,+) + P(u_1,+) + P(u_2,+) ... [/tex]
    or
    [tex]P(+) = |\left\langle \psi | u_0,+ \right\rangle |^2 + |\left\langle \psi | u_1,+ \right\rangle |^2 + |\left\langle \psi | u_2,+ \right\rangle |^2... [/tex]

    Note that the projection operator I'm considering is
    [tex]\mathcal{P} = \sum_n \left|u_n, +\right\rangle \left\langle u_n, +\right|[/tex],
    and that
    [tex]\left\langle \psi \right| \left(\mathcal{P} \left|\psi\right\rangle\right) = P(+)[/tex]

    To convince yourself that what I'm doing is Kosher, what is the probability that measuring L_z on a three-state system with wf [tex]\tfrac{1}{\sqrt{3}}\left|-1\right\rangle + \tfrac{1}{\sqrt{3}}\left|0\right\rangle + \tfrac{1}{\sqrt{3}}\left|1\right\rangle[/tex] will yield a value greater than -h?. Clearly it is 2/3. You can also calculate this by finding the projection operator for the possibilities in consideration.
     
  5. Apr 26, 2008 #4

    reilly

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    Diagonalize sigma y to get the unitary matrix(rotation, 90 degrees about the z axis) to convert toe the y basis.. This transformation does not effect the spatial part of the wave function, and allows you to express the wave function in eigenstates of sigma y. The rest is standard QM -- just keep the new spin up states. I've just said what Ibrits did, in a slightly different form.
    Regards,
    Reilly Atkinson
     
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