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Infinite Square Well and Energy Eigenstate question

  1. Apr 15, 2012 #1
    Hi all, just studying for my final exam and needed a little clarification on this.

    Our prof did an example: Consider a particle of mass m moving in the nth energy eigenstate of a one-dimensional infinite square well of width L. What is the uncertainty in the particle's energy?

    He said the answer was that there was no uncertainty simply because it was in this nth energy eigenstate. I don't really understand why this is the case. I thought maybe it had to do with this postulate he mentioned earlier: In any measurement of the observable associated with the operator Q, the only values that will ever be measured are the eigenvalues qn, which satisfy the eigenvalue equation Q*phi = q*phi. Can someone please help me with this?
  2. jcsd
  3. Apr 16, 2012 #2
    Whenever a system is in an eigenstate of the Hamiltonian, there is a definite value of energy--that's the definition of an eigenstate. However, the system can also be in a superposition of different eigenstates, at which point there will be an uncertainty in the energy--it has a nonzero probability of having the energy of any of the component eigenstates. Does that make sense?
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