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Energy eigenstates of a particle in a one dimensional box.

  1. Oct 27, 2015 #1

    phoenix95

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    1. The problem statement, all variables and given/known data
    A One dimensional box contains a particle whose ground state energy is ε. It is observed that a small disturbance causes the particle to emit a photon of energy hν=8ε, after which it is stable. Just before emission a possible state of the particle in terms of energy eigenstates {ψ1, ψ2.....} would be
    (a) (√2ψ1-3ψ2+5ψ5)/6

    (b)(ψ2+2ψ3)/√5

    (c)(-4ψ4+5ψ5)/√41

    (d)(ψ12)/√2

    2. Relevant equations
    Energy eigenvalues for a particle in a 1-D potential well

    E=n2h2/8mL2

    3. The attempt at a solution
    I don't know where or how to start. Can someone help me here? Also if someone could suggest me a textbook with similar problems that would be helpful too...

    Thanks
     
    Last edited: Oct 27, 2015
  2. jcsd
  3. Oct 27, 2015 #2

    blue_leaf77

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    One photon transition satisfies ##E_i-E_f = h\nu## where ##E_i##, ##E_f##, and ##\nu## are initial particle's energy, final particle's energy, and the emitted/absorbed photon frequency, respectively. The question already tells you that ##E_f=E_1## and ##h\nu = 8E_1##, calculating ##E_i## should be straightforward. For introductory level books where this kind of problem may usually be found, there are Introduction to QM by Griffith and Quantum Physics by Gasiorowicz.
     
  4. Oct 27, 2015 #3

    phoenix95

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    That leads me to Ei=9E1. One possible energy state would be E3. But all of the answers given are superposition of the energy eigenstates. How can one say that a particular superposition is a possible state?
     
  5. Oct 27, 2015 #4

    Geofleur

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    Can you calculate the highest and lowest possible energies that each state could be found to have? Maybe that would narrow things down.

    Out of the states with the right energy range, for which one is it even possible to measure an energy of 9##\epsilon##?
     
    Last edited: Oct 27, 2015
  6. Oct 27, 2015 #5

    phoenix95

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    I don't know how to do that. Can you please guide me?

    I'm in the middle of the quantum mechanics course. I thought once I had finished potential well theory I could solve this problem, but now I can't. I know little of superposition principle..
     
  7. Oct 27, 2015 #6

    Geofleur

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    First, the coefficients on the eigenfunctions will tell you which energies they correspond to. For example, ##\Psi_2## is the eigenstate with the energy ## E_2 ##. Second, if a particle is in a superposition of eigenstates, then the probability of measuring each energy is the absolute square of the coefficient that is paired with the corresponding eigenstate. For example, if ##\Psi = \frac{1}{\sqrt{2}}\Psi_1 + \frac{1}{\sqrt{2}}\Psi_2 ##, then the probability of measuring ## E_1 ## is ##\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} ##. The probability of measuring the energy corresponding to, say, state ##\Psi_7## in this example is zero, because the "coefficient" for this eigenstate in the expansion is zero.
     
  8. Oct 27, 2015 #7

    phoenix95

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    OK, please let me know if I am wrong...
    Since the probability for the eigenstate ψ5 in (a) is maximum it is most probable and hence the highest possible energy state (a) could have..
    similarly for (b) it is ψ3, for (c) it is ψ5 and for (d) it is ψ2?
    Am I right here?
     
  9. Oct 27, 2015 #8

    Geofleur

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    In the state given in (a), ##\Psi_5 ## corresponds to the highest possible energy in that state and, as long as the coefficient for ##\Psi_5## isn't zero, ##E_5## is a possible measurement outcome.

    Yes, the highest energy eigenstates for (b), (c), and (d) are as you say. Similarly, the lowest energy eigenstates for (b), (c), and (d) are ##\Psi_2##, ##\Psi_4##, and ##\Psi_1##.

    Now, if an eigenstate, say ##\Psi_7##, does not show up in an expansion of some state, what is the probability for measuring the corresponding energy, ##E_7##?
     
  10. Oct 27, 2015 #9

    phoenix95

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    zero.:smile:
     
  11. Oct 27, 2015 #10

    phoenix95

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    (b) because of ψ3?
     
  12. Oct 27, 2015 #11

    Geofleur

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  13. Oct 27, 2015 #12

    phoenix95

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    Wait, could it be that simple? I mean just because state (b) had the eigenstate ψ3 in it (with a finite probability), I can conclude that the initial state is possibly (b)??
     
  14. Oct 27, 2015 #13

    Geofleur

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    It's that simple. No other state could possibly yield the correct energy upon measurement.
     
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