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State vectors and Eigenvalues?

  1. Jun 4, 2014 #1
    If I define a state ket in the traditional way, Say:

    $$|\Psi \rangle =\sum _{ i }^{ }{ a_{ i }|\varphi _{ i }\rangle \quad } $$

    Where $$a_i$$ is the probability amplitude.

    How does:

    $$\hat {H } |\Psi \rangle =E|\Psi \rangle $$ if the states of $$\Psi$$ could possibly represent states that have different energy levels, and one would not be able to factor the energy eigenvalue out of the summation. I know I am missing something big here. Could someone point it out?

    Edit: Does $$ \hat {H} $$ collapse psi to one state phi, and so render only energy eigenvalue of the one phi that remains?

    Thanks,
    Chris Maness
     
    Last edited: Jun 4, 2014
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  3. Jun 4, 2014 #2

    atyy

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    When you write ##\hat {H } |\Psi \rangle =E|\Psi \rangle ##, ##|\Psi \rangle## cannot represent a state with more than one energy. Each possible ##|\Psi \rangle## that is a solution to that equation is an energy eigenstate with a definite energy.

    That equation has many solutions, each corresponding to a different energy. An arbitrary state will be a superposition of energy eigenstates with different energies.

    The equation ##\hat {H } |\Psi \rangle =E|\Psi \rangle ## is the time-independent Schroedinger equation or the the energy eigenstate equation. It is only an intermediate step in the solution of the full time-dependent Schroedinger equation.
     
  4. Jun 4, 2014 #3
    Ok, so $$ \Psi $$ can't be a superposition anymore.

    Does this work for a generalized statement?

    $$ \hat { B } \hat { A } |\Psi \rangle =ab|\Psi \rangle $$ Where if, psi is not collapsed, it is at least degenerate, and where A & B commute.

    Thanks,
    Chris Maness
     
  5. Jun 4, 2014 #4

    Matterwave

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    Statements such as these are eigenvalue problems. In these cases, by definition, ##\Psi## must be an eigenvector of the operator.

    The time-dependent Schroedinger equation is NOT an eigenvalue problem:

    $$H\left|\Psi\right>=i\hbar\frac{\partial}{\partial t}\left|\Psi\right>$$

    And so ACTUAL wave functions do NOT have to be eigen-vectors of any operator. We just use the eigenvalue problems to help us solve the full problem.
     
  6. Jun 4, 2014 #5
    It's the phis that have the different energy levels!
     
  7. Jun 4, 2014 #6

    atyy

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    Yes, in the example you give, if ##|\Psi \rangle## is an eigenstate of A and an eigenstate of B, then it is an eigenstate of AB or BA. This does not mean that ##|\Psi \rangle## is not a superposition. It is a superposition of eigenstates of other operators that do not commute with A or B.

    http://en.wikipedia.org/wiki/Complete_set_of_commuting_observables
    http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect2.pdf (Section 2.7)
     
    Last edited: Jun 4, 2014
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