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I Solving Schrodinger's equation

  1. Jan 9, 2017 #1
    What are the key things to look when solving Schrodinger equation for the particular system like Hydrogen atom
     
  2. jcsd
  3. Jan 9, 2017 #2

    BvU

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    Textbooks !

    Corny, I know. Could you be more specific ? Where are you in the curriculum, what brings you to this question and what kind of answer do you expect ?
     
  4. Jan 9, 2017 #3
    All I know so far is to solve partial differential equation, and to solve Schrodinger equation for the particle in a box situation, now for the atom is a different story.
     
  5. Jan 9, 2017 #4
    For Hydrogen? A quick search on the internet would easily lead you to some good information. Hydrogen atom is one of the simplest things to solve the Schrodinger's equation, and also one of the most explained on the internet.

    This link may help you:
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html
     
  6. Jan 9, 2017 #5

    stevendaryl

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    In my experience from a small number of examples, solving the (time-independent) Schrodinger equation involves two steps:
    1. Figuring out what the asymptotic form of the wave function is. That is, in the limit as [itex]x \rightarrow \pm \infty[/itex] for one-dimensional problems, or the limit as [itex]r \rightarrow \infty[/itex] for three-dimensional problems.
    2. Working out the solution close-in (when [itex]x[/itex] or [itex]r[/itex] is small) as a modification of the asymptotic solution.
    For example, in the case of a particle in a box with impenetrable walls, the asymptotic value of the wave function is zero. So this constrains the wave function inside the box to go to zero at the walls. In the example of a box with finite constant potential outside the box, the asymptotic form is [itex]e^{-\kappa |x|}[/itex] for an appropriate value of [itex]\kappa[/itex]. In the example of the harmonic oscillator, the asymptotic form is [itex]e^{-\lambda x^2}[/itex], for an appropriate value of [itex]\lambda[/itex]. In the example of the hydrogen atom, the asymptotic form is [itex]\frac{e^{-\kappa r}}{r}[/itex] for some appropriate value of [itex]\kappa[/itex].

    Once you have the asymptotic form, [itex]\psi_{asymp}(x)[/itex], you make the guess that the full [itex]\psi(x)[/itex] has the form:

    [itex]\psi(x) = u(x) \psi_{asymp}(x)[/itex]

    Plug that into the Schrodinger equation to get a modified equation for [itex]u(x)[/itex]. At this point, you try to look for series solutions fo [itex]u(x)[/itex]; that is, solutions of the form [itex]u(x) = \sum_j a_j x^{j+a}[/itex]. The quantization of the energy levels then comes out of the requirement that [itex]u(x)[/itex] must be sufficiently well-behaved as [itex]x \rightarrow \infty[/itex] that you don't spoil the asymptotic form. For a lot of the cases where the Schrodinger equation is exactly solvable, this means that the series for [itex]u(x)[/itex] actually terminates after a finite number of terms, but I don't think that's true with all possible potentials.
     
  7. Jan 9, 2017 #6
    I made a post above, but now I think that the OP is not very fond of solving partial differential equation itself. It might be therefore important that we give him good information on how to solve them in the first place. In the end, quantum physics is almost all about solving a mathematical problem with some conditions.

    Mathworks website (creator of MATLAB) has Differential Equations and Linear Algebra video series that teaches basics of solving differential equations.


    What you should keep in mind when solving such system, is:
    1) Use spherical coordinate for 3-dimensional case since it makes solving Schrodinger Equation much much easier. In an atom, we have electron orbitals that expand in a spherical way with the nucleus at the origin. Using Cartesian coordinate is extremely difficult and impractical.
    2) You would most likely have to separate variables. In the case of hydrogen atom, this is possible. Specifically, you will probably have to separate the wavefunction into radial distance (typically written in "r") and angles (azimuthal angle "θ" and polar angle "φ"). The wavefunction should look something like this:
    [tex]\psi (r,\theta ,\varphi ) = R_{nl}(r)Y_{lm}(\theta ,\varphi )[/tex]
    The first part includes generalized Laguerre Polynomials and the latter part is Spherical Harmonics.
    3) If you are working on time-dependent solution in non-relativistic Schrodinger equation, the result is still simple enough since variable separation can separate time and spatial coordinates. You just have to solve them separately and then place them back together.

    Unfortunately, I don't really know about Dirac equation since I, as a chemist, barely use relativistic limits to the Schrodinger equation. I hope someone will post something about that.
     
  8. Jan 9, 2017 #7

    stevendaryl

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    In general, what you're saying is good advice: If there is spherical symmetry, then you should use spherical coordinates. I found out the hard way that there are a few exceptions to the general principle, though. The one that comes to mind is the 3-D harmonic oscillator: [itex]H = \frac{-\hbar^2}{2m} \nabla^2 + \frac{K}{2} r^2[/itex]. You can solve it using spherical coordinates, but it's actually easier to use cartesian coordinates, and assume that the wave function has the form: [itex]X(x) Y(y) Z(z)[/itex], then [itex]X, Y, Z[/itex] each satisfy the equation for a one-dimensional harmonic oscillator.
     
  9. Jan 9, 2017 #8
    You are absolutely right. I left that out because I wasn't thinking about harmonic oscillators. Thanks.

    EDIT:
    I actually have a question. When solving for transition dipole moments, I think it is generally easier if we use Cartesian coordinates. I wonder if that remains true when Spin-Orbit coupling is quite strong like in the 4f-4f transition in Lanthanides. In such cases, the spin angular momentum and orbital angular momentum is coupled, and the transition is between these differently coupled states. I would assume that spherical coordinate would be better. I haven't done it myself so I don't know. What do you think?

    Also, I would want to know about zero-field splitting of triplet states, which also have to do with Spin-Orbit coupling. In the weak limit, I guess using Cartesian coordinate makes sense since in the wavefunction of weak SO coupling, the real representation of the wavefunction (which is a linear combination of complex wavefunction of spherical harmonics) like x, y, z for p-orbitals, and xy, yz, zx, x2-y2, and z2 for d-orbitals are "good quantum numbers". I wonder if that stands true in the case of more "relativistic" cases where SO coupling is strong and the "good quantum number-ness" is broken and complex spherical harmonics must be used.
     
    Last edited: Jan 9, 2017
  10. Jan 9, 2017 #9

    ZapperZ

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    When things like this appear and no context is given, I always scratch my head and have more questions.

    For example, for most students, by the time they have to solve something like this, they have done undergraduate E&M courses, and thus, have solved E&M problems in different geometries using different coordinate systems. Thus, the skill of first looking at the geometry of the problem and knowing the proper coordinate system to choose should already be almost second nature. After all, in E&M, didn't we deal with many problems, each having different geometries and having to choose the appropriate coordinate system for each of those?

    So this is the source of my puzzlement, because why are you having a difficult time switching coordinate systems and not seeing why you need to use, say, a spherical polar coordinates to solve a problem with spherical symmetry? If you have had E&M, I'm trying to find the source of your problem in making the transition into a similar QM problem. If you haven't taken E&M, then that's a different problem entirely.

    Zz.
     
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