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Measurement for a particle in a box

  1. May 23, 2008 #1
    It is known that a particle in a one dimensional box with walls at (-a/2, +a/2) has energy probabilities:
    P(E1) = 1/3, P(E2) = 1/3, P(E,3) = 1/3, P(En) = 0 for all other n. If the parity of the state is measured and +1 is found, what can you say about the value of the measurement of E sometime later?

    The eigenstates of the particle in a box are root(2/a) sin(n*pi*x/a) for even n and root(2/a) cos(n*pi*x/a) for odd n.

    So I wrote down the initial state as a linear combination of eigenstates of the Hamilitonian corresponding to n = 1, 2, 3 each with coefficient 1/root3. Now, since the measured parity was +1, the wavefunciton must be in an eigenstate of the parity operator with eigenvalue +1. The parity operator and Hamiltonian commute for this system so I can use the Hamiltonian eigenstates. My question is, does the wavefunction now have to be in a linear combination of eigenstates for n = 1 and n = 3 since those were the original even eigenfunctions used in the expression of the wavefunction? Or is it in a linear combination of all even eigenstates of the Hamiltonian?
  2. jcsd
  3. May 23, 2008 #2

    George Jones

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    If |psi> is the state before the parity measurement, then immediately after the parity measurement, the state of system is the (normalized) projection of |psi> into the subspace of all even parity states.

    What is this projection?
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