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Homework Help: Finding the probability of energy measurements

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi... My problem says:

    In a given experiment the system, at the time [itex]t = 0[/itex], in the normalized state is given by:

    [tex]\phi(t =0) = \frac{1}{\sqrt{5}}(i\psi_{210} + 2\psi_{211})[/tex]

    What possible outcomes is possible if you do an energy measurement on the system in this state, and with what probability does this occur ?

    2. Relevant equations
    I know that the states [itex]\psi_{21-1}, \psi_{210}[/itex] and [itex]\psi_{211}[/itex] is given by the matrix [itex]H[/itex]:

    [tex]\[ \left( \begin{array}{ccc}
    E_{2} + 2\gamma\hbar^{2} & 0 & 0 \\
    0 & E_{2} + 2\gamma\hbar^{2} & 0 \\
    0 & 0 & E_{2}\end{array} \right)\][/tex]

    Don't know if I need to be telling more ?

    3. The attempt at a solution
    Well, I know the outcome has to be the two last states (In the order I wrote my states), which means: [itex]\psi_{210}[/itex] and [itex]\psi_{211}[/itex].
    This means that I have to calculate the probability for those two, which means: [itex]P(E_{2} + 2\gamma\hbar^{2})[/itex] and [itex]P(E_{2})[/itex].

    My problem is, that I'm not quite sure how to calculate that :/

    My book says:

    [tex]P(j) = \frac{N(j)}{N},[/tex]
    but I have no idea how to make use of that in this case.

    I know the answer should be: [itex]P(E_{2} + 2\gamma\hbar^{2}) = \frac{1}{5}[/itex] and [itex]P(E_{2}) = \frac{4}{5},[/itex]
    but again, not sure how to do it.

    So I was hoping someone could give me some pointers towards this, probably, easy question :)

  2. jcsd
  3. Aug 10, 2010 #2
    I will use the bracket notation...

    You are given a normalized wavefunction [itex]|\Psi\rangle[/itex]. So the probability of measuring a specific eigenstate [itex]|\psi_j\rangle[/itex] is just:

    [tex]P(j) = \left|\langle\psi_j|\Psi\rangle\right|^2[/tex]

    Each eigenstate corresponds to a specific energy. So you know the probability of measuring a specific energy.
  4. Aug 10, 2010 #3
    Ahhh, ofc...
    I think I have it now :)

    Thank you very much.
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