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Homework Help: How do we observe an elextron in a potential well ?

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data

    It's problem on the uncertainty principle. It is as follows: Suppose we have confined an electron in an infinite potential well of length L=1A . Say that this electron is in the ground state of the well. Now, say that we wish to observe the electron with accuracy of 0.2A . What sort of photons do we have to use ? ( wavelength ). What is the uncertainty of the energy and the momentum of this electon after the measurement ? Which are the possible energy states that the electron can occupy after the measurement ?

    2. Relevant equations

    (Δx)(Δp) ≥ [itex]\hbar/2 [/itex]

    3. The attempt at a solution

    My thinking is as follows :

    For starters, by solving the Schrodinger equation for the infinite potential well, we get the expression for the energy states as:

    E=[itex]\frac{h^2n^2}{8m_e L^2}[/itex] (1)

    Also, since this is a steady state, the uncertainty of the energy in this state is Zero. ( Δτ=∞ so ΔΕ =0 ). The ground state can be found for n=1. To observe the electron within a certain area of space, i would have to either change the length of the box, or do something similar. By "bombing" it with a photon of about the same wavelength of the range of space certainty i wish to have in the observation i will excite it in a high energy state, but will be able to determine its location. So, with tha thinking, if i wish to measure the electron with an uncertainty of Δx≤0.2A I should use a photon of λ=0.2A. Now, knowing Δx, i can determine, using the uncertainty principle, the uncertainty of momentum as


    The uncertainty of the momentum of the electron after interraction with that photon. About the uncertainty of the energy after the interraction my thinking is as follows:

    the kinetic energy of the electron is T= [itex]\frac{p^2}{2m}[/itex] so, i could with a rough estimation say that ΔT= [itex]\frac{(Δp)^2}{2m}[/itex] and using the uncertainty principle have a rough estimation of the uncertainty of the electron's energy after the interaction

    ΔΤ= [itex]\frac{\hbar^2}{8m(Δx)^2}[/itex] (1)
    which is the ground state of a new box with L'=Δx. So, the act of interraction, is, by a rough estimation, similar to the confinement of the electron in a more limited potential well.

    As for the possible excited states that we can find the electron after the interraction, that I think i can easily determine by equation of (1) and (2). In the sence that the new energy of the electron should be by all accounts lesser or equal to the energy uncertainty. So, by determining E=ΔT and solving for nmax, rounding to the closest integer, i can find the maximum possible excited state of the electron. So the possible excited states are 1 ≤ n ≤ nmax.

    This exercise is about doing rought estimations using only the most basic quantum mechanical foundations. So, what do you think of my logic ?
  2. jcsd
  3. Jun 14, 2012 #2
    It seems your logic is good.
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