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How do we know E is energy in the time-independent Schrodinger eq

  1. Jan 26, 2014 #1
    Hi everyone,

    One approach to solve the Schrodinger equation is to use separation of variables: the solution is composed of a time dependant and space dependant component. When we go through the math, we get a time dependent LHS equal to a space dependant RHS, which means they must both be equal to a separation constant E.

    We know that E is the total energy of the system. Could someone explain why that is? I know the answer has to do with the Hamiltonian operator, but I have trouble understanding this concept and explaining it to others.

    Thank you.
     
  2. jcsd
  3. Jan 26, 2014 #2

    bhobba

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    You need to see a proper derivation of Schrodinger's equation based on symmetry ie the probabilities of the outcomes of observations are coordinate system independent.

    For that see chapter 3 of Ballentine - Quantum Mechanics - A Modern Development:
    https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

    He actually derives the FORM of the equation using that and you can see why its the Hamiltonian, and even why the Hamiltonian has that form. QM is in fact the basis of Classical Mechanics, not the other way around, and explains the stuff you simply must accept classically such as why the Hamiltonian has the form it does. Mathematically the reason is QM is much richer in the symmetries it allows.

    That symmetry is the underlying essence of much of physics is one of our greatest discoveries and QM is a prime example of this rather striking and totally unexpected fact:
    http://www.pnas.org/content/93/25/14256.full

    Thanks
    Bill
     
    Last edited: Jan 27, 2014
  4. Jan 26, 2014 #3

    WannabeNewton

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    What does the Hamiltonian of a classical system correspond to when the configuration space coordinates have no explicit time dependence and the external potential is velocity independent?

    Moreover, why do we relate the eigenvalues of ##\hat{P} = -i\hbar \nabla## to momentum? What correspondence do we make between ##\hat{P}## and the momentum from classical mechanics? In other words, what is the relationship between observables and self-adjoint operators in the general framework?
     
  5. Jan 27, 2014 #4

    atyy

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    An eigenvector of an operator is a vector that when acted on by the operator results in the vector being scaled by a constant. The constant is an eigenvalue of the operator. An operator has many eigenvectors and each has its corresponding eigenvalue. The time-independent Schroedinger equation is the eigenvalue equation of the Hamiltonian: Hψ = Eψ, because it says that when H acts on ψ, it only multiplies it by E. So E is an eigenvalue of H.

    In quantum mechanics, to each observable we associate an operator. When you measure an observable, the measured value is an eigenvalue of the operator associated with the observable.

    The Hamiltonian operator is the operator associated to the observable we call energy. When you measure the energy, the observed value E is an eigenvalue of the Hamiltonian operator.

    Apart from being the operator associated to the observable we call energy, the other function of the Hamiltonian is to govern the time evolution of the wave function, via the time-dependent Schroedinger equation. An easy way to see that H is the operator for the energy is to observe that H = p2/2m + V is the classical energy of a particle in a potential. In quantum mechanics p and V are operators, but an elementary way to see that it is ok to call ∂/∂x the operator for momentum p, and to call the operator V the potential energy V is to demonstrate that their averages satisfy the familiar classical equations for momentum and potential energy: Ehrenfest's theorem.

    If you really want to check that we get the correct classical limit, you can take the classical limit of Schroedinger's equation to get the classical Hamilton-Jacobi equation, or check that you get the classical trajectory in the classical limit of the path integral.
     
    Last edited: Jan 27, 2014
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