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Find normalised linear combinations that are orthogonal

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm a little weary of posting this in this forum. If I post it in the math section it will be answered in about 30 min whereas here it might take about 5 hours, but we'll see.
    Screenshot2012-08-19at124734AM.png
    Screenshot2012-08-19at124700AM.png
    Screenshot2012-08-19at124709AM.png
    2. Relevant equations
    3. The attempt at a solution
    Number one, I'm not exactly sure how they get from
    [tex]\int \phi_1^* (c_1\phi_1 + c_2\phi_2)d\tau = c_1 + c_2 d [/tex]

    I think it's because

    [tex] \phi^*\phi = 1[/tex] but I'm not sure.
    Number two, I don't understand the following step:
    [tex] \int (c_1\phi_1 + c_2 phi_2)*(c_1 \phi_1 + c_2\phi_2)d\tau = c_1^2 + c_2^2 + 2dc_1c_2 [/tex]

    why does [itex]\phi[/itex] disappear?
    I figure that it must have something to do with the fact that [itex]\phi_1[/itex] is orthogonal which means it = 0
    Number three, I can't get step 3. I put the equations as follows:
    [tex]
    c_1 + c_2d = 0
    c_1^2 + c_2^2 + 2 dc_1c_2 = 1
    c_1 = -c_2d[/tex]
    therefore
    [tex]
    (-c_2d)^2 + c_2^2 + 2d(-c_2d)c_2 = 1
    (-c_2d)^2 + c_2^2 - 2(c_2d)^2 = 1
    [/tex]
    And then I can go no further.
    Number four, I don't understand how
    [tex]
    \int (\phi_1 + \phi_2)*(c_1\phi_1 + c_2\phi_2)d\tau [/tex]

    simplifies to
    [tex]
    (c_1 + c_2)(1+d)
    [/tex]
    Number five, what do they mean by 2 and 5 gives
    [tex]
    \frac{(\phi_1 - \phi_2)}{\sqrt{2-2d)}}
    [/tex]
    As you can see I'm real clueless with regards to this stuff. I've got a private tutor lined up but I won't be able to meet with him until sometime next week.
     
    Last edited by a moderator: Aug 19, 2012
  2. jcsd
  3. Aug 19, 2012 #2
    From the first sentence of the problem: "[itex]\phi_1[/itex] and [itex]\phi_2[/itex] are normalized eigenfunctions..."

    What is the definition of normalized? Specifically, how does the fact that [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are normalized relate to the integrals [itex]\int \phi_1^*\phi_1 d\tau[/itex] and [itex]\int \phi_2^*\phi_2 d\tau[/itex]?

    Answer that question and that should tell you everything you need to know to work through the derivation.
     
  4. Aug 19, 2012 #3
    Normalized means = 1. I already knew that. Still clueless.
     
  5. Aug 19, 2012 #4
    What, specifically, is equal to 1?

    Edit: Hint - Your statement in the original post that [itex]\phi_1^* \phi_1 = 1[/itex] is not quite correct. What is the correct equation?
     
    Last edited: Aug 19, 2012
  6. Aug 19, 2012 #5
    [tex]\phi_1 \phi_2[/tex]
     
  7. Aug 19, 2012 #6
    Not true. Nothing in the problem set up indicates [itex]\phi_1 \phi_2 =1[/itex] in general. The correct relation between the two is [itex]\int \phi_1^* \phi_2 d\tau = d[/itex].

    So, I think I'm seeing several issues here that are causing you problems. First, you need to be a lot more careful about the difference between [itex]\phi[/itex] and [itex]\phi^*[/itex]. The two are not the same, and which terms have the * and which terms don't matters.

    Second, you need to go back and review the definition of normalization. I think you have assumed that it means something that it doesn't mean, and that's why you can't evaluate the integrals above.

    Third, slow down, take a bit more time, and don't skip any steps when you are working through the problem. Also, when replying on this thread, be as specific as you can; it'll make it easier to diagnose where your misunderstanding is coming from.

    To work through each of the integrals in the derivation above, you need to know three different quantities:

    [itex]\int \phi_1^* \phi_1 d\tau = ?[/itex]
    [itex]\int \phi_2^* \phi_2 d\tau = ?[/itex]
    [itex]\int \phi_1^* \phi_2 d\tau = ?[/itex]

    Each of the integrals in the derivation can be broken up into pieces containing those three terms.
     
  8. Aug 19, 2012 #7
    I think I'll just wait until I meet with that private tutor. My problems are just so overwhelming. I think maybe I zipped through the calc, linear algebra and DE prerequisites too quickly. Maybe I'm trying QM without sufficient knowledge.
     
  9. Aug 19, 2012 #8
    Actually, the math in this particular derivation is quite simple, once you have all of your definitions straight. That's why I was asking you about what normalization meant. I don't think this particular derivation is beyond you, I just think you need to slow down, take your time with it, and review the basics.

    As another hint: for a normalized (1d) wave function in the position representation, [itex]\phi(x)[/itex], the probability of finding the particle between the points [itex]x_1[/itex] and [itex]x_2[/itex] is given by [itex]P[x_1<x<x_2]=\int_{x_1}^{x_2} \phi^*(x) \phi(x) dx [/itex].

    If I take [itex]x_1[/itex] to [itex]-\infty[/itex] and [itex]x_2[/itex] to [itex]\infty[/itex], what's the probability of finding the particle between [itex]x_1[/itex] and [itex]x_2[/itex]? That is,

    [itex]P[-\infty<x<\infty]=\int_{-\infty}^{\infty} \phi^*(x) \phi(x) dx = ?[/itex]

    Don't try to solve the integral, just use common sense. What's the probability of finding the particle if we search the entirety of the real line?
     
  10. Aug 19, 2012 #9
    as for the odds of finding a particle between negative infinity and plus infinity it has to be 1, nothing can exist beyond the edge of infinity since there is no edge to infinity. I certainly hope the math is not beyond me.
     
  11. Aug 19, 2012 #10
    Exactly. [itex]P[-\infty<x<\infty]=\int_{-\infty}^{\infty} \phi^*(x) \phi(x) dx = 1[/itex]. That is the definition of what it means for a wave function to be normalized.

    So, what does that statement imply for the three integrals needed for this problem?

    [itex]\int \phi_1^* \phi_1 d\tau = ?[/itex]
    [itex]\int \phi_2^* \phi_2 d\tau = ?[/itex]
    [itex]\int \phi_1^* \phi_2 d\tau = ?[/itex]
     
  12. Aug 19, 2012 #11
    I'm guessing they equal 1.
     
  13. Aug 19, 2012 #12
    Not all of them. :) One of the three is given to you in the problem. The other two equal 1.
     
  14. Aug 19, 2012 #13
    [itex]\int \phi_1^* \phi_2d\tau = 0[/itex] because it's orthogonal, right?
     
  15. Aug 19, 2012 #14
    Nope. Actually, that's the point of the problem. [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are not orthogonal, but rather integrate to a constant, [itex]\int \phi_1^* \phi_2 d\tau = d[/itex]. The question is, given that the are not orthogonal, how would you go about constructing orthogonal functions from them.

    That is, you are trying to find some [itex]\bar{\phi}_2 \equiv c_1 \phi_1 + c_2 \phi_2[/itex] such that [itex]\int \phi_1^* \bar{\phi}_2 d\tau = 0[/itex].
     
  16. Aug 19, 2012 #15
    I have to leave now and I will return to this problem after about 20 hours have passed. I appreciate you helping me.
     
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