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anonymous

  • one year ago

Schrodinger's equation question

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  1. anonymous
    • one year ago
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    The symbol H , called the Hamiltonian operator, represents a set of mathematical operations that, when carried out with a particular , yields one of the allowed energy states of the atom.* Thus, each solution of the equation gives an energy state associated with a given atomic orbital. What the does that mean?

  2. anonymous
    • one year ago
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    @IrishBoy123

  3. IrishBoy123
    • one year ago
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    that's a very big question post or scan the original:p or add some colour

  4. anonymous
    • one year ago
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    Ok! But its not a question but more of a description for the schrodingers eqn|dw:1441577536519:dw|

  5. IrishBoy123
    • one year ago
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    @Michele_Laino

  6. Michele_Laino
    • one year ago
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    the equation of Scroedinger, is an equation to eigenvalues: \[\Large \hat H\psi = E\psi \] where we have to diagonalize the hermitian operator \( \large {\hat H}\). We have to compute, both eigenfunctions \( \large \psi \) and eigenvalues \( \large E \), in order to do that we also need of the initial conditions, since the Schroedinger equation above, is at end, a differential equation. What we get by solving the Schroedinger equation, is many or infinite eigenvalues, namely the values for \( \large E \) and many eigenfunctions, namely the functions \( \large \psi\)s, and to each values of \( \large E \) can correspond one, or more than one functions \( \large \psi \). The eigenfunctions \( \large \psi \) are also called the \( \large states \) of our atomic system, for example an atom, an electron, etc... How to get the shape of the Hamiltonian operator \( \large {\hat H}\)? It is simple, we have to write the total energy (Hamiltonian function) of our mechanical system, then we have to make this substitutions: \[\Large x \to x,\quad p \to - i\hbar \frac{d}{{dx}}\] namely we have to go from classical functions, to the corresponding operators. Here, I have in mind a one-dimensional mechanical system, so I need of one cartesian coordinate only. For example, if we take a particle which is confined inside a segment whose length is \( \large a \) and the ends of such segment are located at \( \large x=0, \; x=a \), then, in absence of any interaction, and of any external force, our particle is the free particle, and its total energy, or hamiltonian function, is: \[\Large H = \frac{1}{2}m{v^2} = \frac{{{p^2}}}{{2m}}\] Now we have to go from classical mechanics to quantum mechanics, by the substitution above, then we get the \( \large Hamiltonian \; operator\) like below: \[\Large \hat H = \frac{1}{{2m}}\left( { - i\hbar } \right)\frac{d}{{dx}}\left\{ {\left( { - i\hbar } \right)\frac{d}{{dx}}} \right\} = - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}}\] so the Schroedinger equation, is: \[\Large - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}\psi }}{{d{x^2}}} = E\psi \] which is a second order differential equation. In other words, we have to solve that doifferential equation, using the subsequent initial conditions: \[\Large \left\{ \begin{gathered} - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}\psi }}{{d{x^2}}} = E\psi \hfill \\ \\ \psi \left( 0 \right) = \psi \left( a \right) = 0 \hfill \\ \end{gathered} \right.\]

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