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Coefficients of a Superposition of State Vectors?

  1. Jul 29, 2015 #1
    1. The problem statement, all variables and given/known data
    (f) At t = 0, a particle of mass m trapped in an infinite square well of width L is in a superposition of the first excited state and the fifth excited state, ψs(x, 0) = A (3φ1(x) − 2iφ5(x)) , where the φn(x) are correctly-normalized energy eigenstates with energies En. Which of the following values of A give a properly normalized wavefunction? ##\frac { 1 }{ \sqrt { 5 } }## ##\frac { i }{ 5 } ## ##\frac { i }{ \sqrt { 13 } }## ##\frac { 1 }{ 13 } ## None of these

    2. Relevant equations
    A normalized wave function: ##A^{ 2 }\sum _{ i }{ a^{ * }_i a_i } =1##.

    3. The attempt at a solution
    Since, ##\sum _{ i }{ a^{ * }_{ i }a_{ i } } =9-4=5##, ##A=\frac { 1 }{ \sqrt { 5 } } ##.

    Does this look correct?

    Thnx,
    Chris
     
  2. jcsd
  3. Jul 29, 2015 #2

    blue_leaf77

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    That does not look correct:wink:.
    The basic definition of linear combination (or superposition) is the sum of all weighted basis states, that means if there is a negative sign in front of a particular term, that sign must belong to the expansion coefficient of that corresponding term.
     
  4. Jul 29, 2015 #3
    I have a negative sign there. a1*a1=9 because it is real. (-2i)*=2i, so (-2i)*x(2i)=-4 for index 5. I am not sure I follow your concern.

    Chris
     
  5. Jul 29, 2015 #4

    blue_leaf77

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    Shouldn't it be (-2i)*x(-2i) = 4 ?
     
  6. Jul 29, 2015 #5
    That would be (-2i)*x(-2i)=(2i)x(-2i)=4. Oh, ok. Simple mistake. I should have wrote it out explicitly.

    It is now ##A=\frac{1}{\sqrt{13}}## which is not one of the choices. That always makes me feel insecure :D

    Chris
     
  7. Jul 29, 2015 #6

    blue_leaf77

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    The coefficient ##A## is in general complex, thus instead of ##A^2## you should write it ##|A|^2##. Among the choices, which will give you ##|A|^2 = \frac{1}{13}##?
     
  8. Jul 29, 2015 #7
    But it is asking for ##A##, not ##{|A|}^2##. Is it not?

    Thanks,
    Chris
     
  9. Jul 29, 2015 #8
    I emailed the prof who wrote this test. Yes, I see that A is in general complex, so i/sqrt(13) would satisfy ##{|A|}^2=\frac{1}{13}##. Man, I am so rusty. A lot of brain sweat for what seemed to be such an easy question at first.

    Thanks,
    Chris
     
  10. Jul 29, 2015 #9

    blue_leaf77

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    Yes that's right.
     
  11. Aug 5, 2015 #10
    On part g) of this question:
    (g) Given the wavefunction ψs, what is the probability of measuring the energy to be E6 at t = 0? Circle one:

    The correct answer would be zero, correct? That is because the coefficient in front of a state vector at level 6 would be zero. It is not included in the original wave function correct?

    Thanks,
    Chris
     
  12. Aug 5, 2015 #11

    blue_leaf77

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    Yes the probability should be zero.
     
  13. Aug 5, 2015 #12
    This one should be zero too, no? The probability of finding a particle and a particular point would would be infinitesimal. You have to integrate over some range on the axis to have a finite probability.

    (h) Given the wavefunction ψs, what is the probability density of finding the particle in the middle of the box at time t = 0?

    0 3 9 9 Undetermined

    Thanks for helping me. I have no answers to this material, and I want to make sure I have these concepts down solid.

    Thanks,
    Chris
     
  14. Aug 5, 2015 #13

    George Jones

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    Are "probability" and "probability density" the same thing?
     
  15. Aug 5, 2015 #14
    Good point. Density is per unit length in a 1d scenario. Let me go back and work on it some more.

    Chris
     
  16. Aug 5, 2015 #15
    I would be taking the derivative of ##|\psi|^2## and evaluating it at L/2, correct? I get zero because there are sine functions in both terms after finding mod squared, and taking the derivative. I am using his version of the infinite square well, where x1=0 and x2=L. See: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/exams/MIT8_04S13_exam1.pdf

    My ##|\psi|^2## is: ##|\psi |^{ 2 }=\frac { 1 }{ 13 } \left[ 9\frac { 2 }{ L } { sin }^{ 2 }(\frac { 2\pi }{ L } x)+4\frac { 2 }{ L } { sin }^{ 2 }(\frac { 6\pi }{ L } x) \right] ##

    Thanks,
    Chris
     
  17. Aug 5, 2015 #16
    Never, mind. I don't know why I was thinking of taking the derivative. Mod squared is the probability density. The units are correct for that too. However, this mod square evaluated at x=L/2 should still be zero because the sine functions evaluated at multiples of ##\pi## are equal to zero.

    Correct?

    Thanks,
    Chris
     
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