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A basic question on quantum mechanics

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Quantum physics' postulate says that when you measurement a quantity. you have to express the wave function in the basis of that operator's eigenvector. I am confused by the following: If you want to measure a system's energy, [tex] H|\psi>= H|i><i|\psi>[/tex], but how do you find the eigenvector for the energy measurement operator? is every vector an eigenvector of the operator?



    2. Relevant equations



    3. The attempt at a solution
     
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  3. Aug 4, 2009 #2

    LeonhardEuler

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    Not all states are eigenstates of the Hamiltonian. How do you find the eigenstates of the Hamiltonian? Do you know a famous equation in quantum mechanics that looks like an eigenvalue equation for the Hamiltonian?
     
  4. Aug 4, 2009 #3

    Pengwuino

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    I assume you mean the basis? You're looking for the eigenstates of the Hamiltonian. What you've written is a change of basis. Nonetheless, you attempt to find solutions to differential equations based on what your Hamiltonian is. For example, if you have the small harmonic oscillator where [tex]H = \frac{{P^2}}{{2m}} + \frac{{m\omega ^2 X^2 }}{2}[/tex], you're looking to solve the equation [tex]\frac{{\hbar ^2 }}{{2m}}\frac{{d^2 y(x)}}{{dx^2 }} + \frac{{m\omega ^2 x^2 y(x)}}{2} = Ey(x)[/tex]. Those solutions form a basis for the solutions.
     
  5. Aug 4, 2009 #4
    the main reason that I am confused it that, I saw some solutions to some problems where you were given a wave function expressed in position space [tex] \psi (r,t) [/tex]. And they measured the energy using the expectation value formula, [tex] <\psi|H|\psi>/<\psi|\psi> [/tex]. How do they know the basis in position space is eigenvectors of H because they didn't check ,instead they just use the formula. they just differentiate the wavefunction [tex] H= -h^2 \triangle /2m + V(R,t) [/tex]. So my main question is that, how do they know the basis is appropriate for applying the formula directly. sorry if I made it unclear
     
  6. Aug 4, 2009 #5

    LeonhardEuler

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    Ok, I understand your question now. The reason they can use that formula directly is because it applies whether the wave function is an eigenfunction of the Hamiltonian or not. If it is not, then it does not give the energy (which is uncertain), but the expectation of the energy (the average value if the energy was measured many times).
     
  7. Aug 4, 2009 #6
    Ah! awww, thank you. Thanks everyone. That solves the problem ^ ^
     
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