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Expectation value of a SUM using Dirac notation

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a one-dimensional particle subject to the Hamiltonian H with wavefunction [tex]\Psi(r,t) =\sum_{n=1}^{2} a_{n}\Psi _{n}(x)e^{\frac{-iE_{n}t}{\hbar}}[/tex]
    where [tex]H\Psi _{n}(x)=E_{n}\Psi _{n}(x)[/tex] and where [tex]a_{1} = a_{2} = \frac{1}{\sqrt{2}}[/tex]. Calculate the expectation value of the Hamiltonian with respect to [tex]\Psi (x,t)[/tex]? Which energy eigenvalue is the most likely outcome when we measure the energy of particle once?

    2. Relevant equations
    Given in the question.

    3. The attempt at a solution
    [tex]\Psi(r,t) =\sum_{n=1}^{2} a_{n}\Psi _{n}(x)e^{\frac{-iE_{n}t}{\hbar}}[/tex]

    let [tex]\Psi_{1}(r,t) =a_{1}\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}}[/tex] and [tex]\Psi_{2}(r,t) =a_{2}\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}[/tex]

    therefore [tex]\left \langle H \right \rangle = \left \langle \Psi _{1}+ \Psi _{2}|H |\Psi _{1}+ \Psi _{2}\right \rangle[/tex]

    which gives [tex]\left \langle H \right \rangle = (E_{1}+ E_{2}) \left \langle \Psi _{1}+ \Psi _{2}|\Psi _{1}+ \Psi _{2}\right \rangle[/tex]

    but not sure what to do now? is this the best way to do this? the trouble im having is using bra ket notation to work with a sum of wavefunctions.

    any advice on this would really be appreciated!

    thanks in advance
  2. jcsd
  3. Dec 1, 2014 #2
    Are the wavefunctions orthonormal, if so what happens then?
  4. Dec 2, 2014 #3
    [tex](E_{1}+ E_{2}) \left \langle \Psi _{1}+ \Psi _{2}|\Psi _{1}+ \Psi _{2}\right \rangle[/tex]

    Yes, hence the bra ket term above would equal 1. However I need to get the [tex]a_{1} and a_{2}[/tex] out of there. we did a degenerate problem in bra and ket that gave:

    [tex]a_{1}\left \langle \Psi _{1} |H| \Psi _{1} \right \rangle + a_{2}\left \langle \Psi _{2} |H| \Psi _{2} \right \rangle + a_{1}\left \langle \Psi _{1} |V| \Psi _{1} \right \rangle + a_{2}\left \langle \Psi _{2} |V| \Psi _{2} \right \rangle = Ea_{1}\left \langle \Psi _{1}|\Psi _{1} \right \rangle + Ea_{2}\left \langle \Psi _{1}|\Psi _{2} \right \rangle[/tex]

    and was much easier to simplify, for example the exponential terms in my question will not go to 1, because there are 2 different energy levels. how do I deal with these extra terms in bra ket notation?

    many thanks for the reply!
  5. Dec 2, 2014 #4


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    By defining ##\Psi_{1}(r,t)## and then dropping the (r,t) you cause confusion between your ##\Psi_{1}(r,t)## and the eigenfunction of the Hamiltonian ##\Psi_{1}(x)##.

    The "which gives" that follows is not correct:
    $$H \left |\; a_1\Psi _{1}(x)e^{\frac{-iE_{1}t}{\hbar}}+ a_2\Psi _{2}(x)e^{\frac{-iE_{2}t}{\hbar}}\right \rangle = E_1 a_1 \left | \Psi _{1}(x) \right > e^{\frac{-iE_{1}t}{\hbar}} + E_2 a_2 \left | \Psi _{2}(x) \right > e^{\frac{-iE_{2}t}{\hbar}} $$
  6. Dec 2, 2014 #5

    Thanks, I didnt realise you could put everything in a ket like that, it's a little messy, but yes you answered my question. I will work through the problem and see where it goes.

    Many thanks for all the help, really appreciated!
  7. Dec 3, 2014 #6


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    Don't forget to take complex conjugates when you take factors like ##a_1\; e^{\frac{-iE_{1}t}{\hbar}}## outside the bra state
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