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Trying to plot de Broglie standing wave.

  1. Jun 23, 2008 #1
    Does anyone know how to plot the electron standing wave around a nucleus in Cartesian coordinates? I have tried adding a sine/cosine function to the circle equation but the image does not look correct and I also don’t know how to increase the wave count when n > 1.
  2. jcsd
  3. Jun 23, 2008 #2


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    Staff: Mentor

    Exactly what function (formula) are you trying to plot, and what makes you think the image does not "look correct"?
  4. Jun 23, 2008 #3
    I wish to recreate the Electron Standing Wave as described by de Broglie. This is where the electron exhibits its wave property. There must be exactly n number of wavelengths to fit a specific energy level and circumference otherwise the wave becomes destructive.

    When I added the cosine function to the radius of a circle the harmonic wave did not look correct. For example. Consider a harmonic wave going in a straight line. Assume a vector is pointing in the same direction of the wave movement. What you get is a straight vector pointing from left to right and a perfect harmonic wave moving in the positive x axis.

    When I add the cosine function to the radius and compute the circle then I do get a harmonic wave that follows the circumference but the direction of the wave vector is still in the x direction instead of along the circumference line. Consequently, the wave is distorted from examples I have seen AND I don’t know how to add additional wavelengths to the circumference.
  5. Jun 23, 2008 #4
    The whole standing wave argument for the atom is very heuristic and not really the way to go (solving the Schrodinger equation is). For example, the lowest state of hydrogen is spherically symmetrical, and there is no "wave". However, if you're interested, look at the n=2, m_l = +/- 1 solutions here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html The relevant angular coordinate is [tex]\phi[/tex].
  6. Jun 23, 2008 #5
    Gosh, now that you point that out I don’t ever remembering a de Broglie example illustrating the first orbit. If I understand correctly, this would also apply to the first orbit of all n values because the first orbit is always circular for each different n energy level.

    If this then be true how does Schrodinger’s equation address the first orbit of any energy level. His equation describes harmonic waves? So how does he handle the circular orbits?

    I’ll work on the hyperphysics equations this evening and see if I can get a plot for the other orbits.
  7. Jun 23, 2008 #6
    Like I said, in the Bohr model, which is based on de Broglie's ideas, things are very sketchy and very classical-mechanics-ish. There it doesn't make sense for angular momentum to be zero, so that orbital isn't even considered. Schrodinger's equation only has harmonic solutions (by that, we presumably mean solutions to laplace's equation), when there is no potential. For the hydrogen atom, there is a Coulomb potential, so solutions aren't "free plane waves". However, they can be built up from linear combination of plane waves, by completeness.
  8. Jun 23, 2008 #7
    Yeah, I have always been a little uncomfortable with the idea, in non-classical physics, that the circular orbits don’t have angular momentum.

    Although, back to the circular orbits and standing waves, for any circular orbit and particularly the first circular orbit in Hydrogen I have seen a standing wave made up of 2 wavelengths instead of just one. I don’t know of any restriction that limits the first circular orbit of hydrogen to have just one wavelength thus making it some what awkward to create a standing wave.

    Oh, BTW the hyperphysics example is in polar coordinates. The only plotting tool I have available only works in Cartesian coordinates. Is there some free plotting tool on the Internet that plots in polar coordinates that I could download that can handle this type of equations?
  9. Jun 23, 2008 #8
    Actually, now that I think about it any number of wavelengths will fit into a circular orbit. I think there are only two restriction. First, there has to be more than one wavelength to form a standing wave no matter what the shape of the orbit is. Second, the wavelengths have to be some integer of 2pi. For example n=1, 2, 3, etc for n2pi.
  10. Jun 23, 2008 #9
    I guess there is a third constraint and perhaps the most important constraint,… the energy level. Any electron orbit only allows a specific energy level. Once the electron gains that specific energy level then it can assume the orbit, circular or not. Now the standing wave has to have a wavelength that meets the first, second and third conditions for it not to be destructive and for it to be at the right energy level so the length will form a standing wave.
  11. Jun 23, 2008 #10
    This is great and all, but it isn't actually correct. I mean, an electron is spread out all over space. It's not like a runner on a race track confined to a single lane. I wouldn't spend too much time trying to visualize Bohr orbits or de Broglie waves since these are hold-overs from when quantum mechanics was not fully understood.

    As for plotting wavefunctions, you can always convert the expressions in spherical coordinates to cartesian coordinates. However, there are plenty of tools on the web to visualize wavefunctions. Here is a good one: http://www.falstad.com/qmatom/
  12. Jun 23, 2008 #11
    Hi Ibrits, thanks for your comments. I agree and fully understand that the Schrodinger equation and electron orbital’s are the end-game of atomic quantum physics. I was writing a paper and I wanted to describe some of the events leading up to Schrodinger and illustrate orbital standing waves. I also agree and understand that an electron is only a name for a continuous wave that has collapsed because of an observation. So I believe we are on the same page.

    There were some comments made earlier that I let slide because I wanted to stay on point. But perhaps now is a good a time as any to address them. The Bohr atom was introduced in 1913 by N. Bohr. The Bohr model was enhanced by A.J.W. Sommerfeld in 1916 when he introduced and explained elliptical orbits to better match spectral lines. There was little improvement beyond Sommerfeld’s work. Then in 1923/24 de Broglie introduced his particle wave equation and it is my understanding that this was the dawn of the new quantum physics where the Bohr quantum physics was now considered the ‘old quantum physics. In 1925 Heisenberg introduced his matrix mechanical model and in 1926 Schrodinger introduced his wave equation. I understand that Hesienberg’s matrix better described the electron but was very cumbersome to work with so researchers gravitated to Schrodinger’s equation and everything else is history.
  13. Jun 23, 2008 #12
    Mmm, I would lump de Broglie's under old quantum physics also... but maybe that's just me. Copenhagen was probably the definitive point, where "everything was understood". I don't know if the probablistic interpretation was known to Schrodinger originally.

    In any event, Eisberg and Resnick is a good source for this kind of stuff, as is any "history of QM" book.
  14. Jun 23, 2008 #13


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    Staff: Mentor

    No, it wasn't. I think Schrödinger originally thought of [itex]\psi[/itex] or [itex]|\psi|^2[/itex] as representing a charge or matter density of some kind. People often associate the probability-density interpretation with Max Born, but it was actually Wolfgang Pauli who first stated it in late 1926, several months after Schrödinger came up with his wave equation in the spring of that year.

    Born did introduce the idea of probability into QM, but he was talking about what we now know as expanding a general wave function in terms of energy eigenfunctions:

    [tex]\psi = \sum c_i \psi_i[/tex]

    He interpreted [itex]|c_i|^2[/itex] as the probability of finding the system with the energy associated with [itex]\psi_i[/itex].
  15. Jun 24, 2008 #14
    To draw a picture of n squiggles around a loop

    [tex] r = R + r_{1}sin \left( n \theta) \right) [/tex]

    Pick a value of [tex]r_{1}[/tex] to be about one quater of [tex]R[/tex] , the diameter of the circle.

    Vary theta from 0 to 2 pi, or 0 to 360 if your sine function takes degrees.

    I don't know how you can piece together some functions to plot this in Cartesian coordinates. However, if you can make a drawing parametrically; that is, if your drawing package takes some variables like theta1 and theta2 and draws a line from (x1,y1) to (x2,y2), then

    [tex]x = r cos \left( \theta \right)[/tex]
    [tex]y = r sin \left( \theta \right)[/tex]

    Substitue the variable r from the first equation.
    Last edited: Jun 24, 2008
  16. Jun 24, 2008 #15
    EXCELLENT, A big thanks to Phrak! Your formula worked perfectly. To get the bottom half, just multiply the whole equation by -1 and plot. Actually, I downloaded the 300 pound guerrilla… SAGE (for free) and used its polar plotting routine where I could add both the positive and negative parts of the equation AND add a circle to represent the Bohr orbit. It looks great!
    Much appreciation for solving my problem with a very simple equation.
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