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Mean value of energy <E> for a QM state?

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data
    If the system is in a state

    |ψ> = 1/sqrt(6) |v1> + 1/sqrt(3) |v2> - i/sqrt(2) |v3>

    with Hamiltonian satisfying H|vj> = (2-j)a|vj>

    Find the mean value of energy <E> and the root mean square deviation √(<E2> - <E>2 ) that would result from making a number of measurements of the energy of the system in state |ψ>


    2. Relevant equations
    <E> = <ψ|E|ψ>

    for a free particle E = p2 / 2m
    3. The attempt at a solution

    To find the mean value of Energy <E> is it just eigenvalues (a, 0 -a) multiplied by the probability of it being in the corresponding state P = (1/6, 1/3, 1/2)?

    = a/6 - a/2 = -2a/6.
    For E2 then do you just have the same, but the eigenvalues squared multiplied by the probabilities?
     
  2. jcsd
  3. May 31, 2016 #2
    except divide the energy by two? as its a mean, so it would be -a/6 ?
     
  4. May 31, 2016 #3

    BvU

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    Why divide by 2 ? What would be the mean value for the identity operator 1 ?
     
  5. May 31, 2016 #4
    1, so do you just find the mean of the eigenvalues? So 0...
     
  6. Jun 1, 2016 #5

    BvU

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    For the identity operator you get ##\ {1\over 6} + {1\over 3} + {(-i^*)(-i)\over 2} = 1 \ ##. No dividing by 2. The state is properly normalized.

    For <H> you get ##\ {1\over 6} (a) + {1\over 3} (0) + {(-i^*)(-i)\over 2} (-a) = {\displaystyle -a\over 3} \ ## as you did. No dividing by 2 either.

    So for <H2> ...
     
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