## anonymous one year ago Schrodinger's equation question

1. anonymous

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

@IrishBoy123

3. IrishBoy123

that's a very big question post or scan the original:p or add some colour

4. anonymous

Ok! But its not a question but more of a description for the schrodingers eqn|dw:1441577536519:dw|

5. IrishBoy123

@Michele_Laino

6. Michele_Laino

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: $\Large \left\{ \begin{gathered} - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}\psi }}{{d{x^2}}} = E\psi \hfill \\ \\ \psi \left( 0 \right) = \psi \left( a \right) = 0 \hfill \\ \end{gathered} \right.$