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Quantum mechanics angular distribution function

  1. Jan 6, 2014 #1
    1. Radial and angular distribution functions for an orbital

    Find the most probable value of theta and r for a 2pz orbital

    2. Relevant equations

    [itex]\psi _{2p_{z}} = N \textrm{cos}(\theta) r exp (-r/2)[/itex] in units of [itex]a_0[/itex]

    3. The attempt at a solution

    Most probable r is when [itex] \textrm{d/d}r (P(r))=0 \Rightarrow r_\textrm{max} = 4[/itex] where [itex] P(r)=r^2 |R(r)|^2[/itex]

    The most probable value of theta is the maximum of the theta function squared weighted by sin(theta), the volume of a thin conical shell.

    [itex]P(\theta)=\textrm{sin}(\theta) |\Theta (\theta)|^2 \Rightarrow \theta_\textrm{max} = \textrm{arcsin}(\pm1/\sqrt{3})[/itex]

    Cheers.
     
  2. jcsd
  3. Jan 6, 2014 #2

    Simon Bridge

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    Interesting, your most probable value of theta appears to have a probability of zero.
    Wouldn't the expectation value play a role in this somewhere?

    Anyway... what was your question?
     
  4. Jan 6, 2014 #3

    vela

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    Why would it? The OP is asked to find where the probability density attains a maximum, not the average value of r or ##\theta##.
     
  5. Jan 6, 2014 #4

    Simon Bridge

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    @Vela: that would be one reading of the question - however, it does not actually ask for the maximum of the probability distribution in so many words - it asks for the most probable value. i.e what value would one expect to find the particle near? Mind you - it's probably the same place.

    The maximum value is as probable as any other after all.

    If they wanted the maximum, then why not say so?
    OTOH, this is a very common wording for questions about averages.

    However - context is everything.
    If previous assignment questions have used that phrasing to mean "maximum" then fine.
    Which is why I asked the OP whether the expectation value has a role to play in this
    ... prompts an explanation.

    There's another question I'm saving up for after the answer to this one.
     
    Last edited: Jan 6, 2014
  6. Jan 6, 2014 #5

    vela

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    I'll have to disagree with you here. Most probable unambiguously means the place where there probability density attains a maximum. I have never seen it used to refer to the expectation value. Indeed, many books make a point to note that these two values are not the same. See, for example, http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html.
     
  7. Jan 6, 2014 #6

    Simon Bridge

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    In the hyperphysics example, I see "most probable" used in a loose summary at the end, while the body used the terms "maximum probability" in the context of a probability distribution. These are easily synonymous though.

    The question comes as homework - from a class.
    So lets see how working teachers cover the topic:

    I've seen it done both ways - here is a counter example from a lecture:
    http://lcbcpc21.epfl.ch/Course/aimf/lecture6_06.pdf
    ... slide 1:
    . The most
    likely position of the particle is the average over this probability
    distribution (expectation value).​

    And another one, from a slightly more advanced course:
    http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/ProbabilitiesExpectationValues.pdf
    p37 the "most likely" position for a particle in a box is computed using the expectation value.

    Another introductory lesson - this time for the hydrogen atom:
    http://spiff.rit.edu/classes/phys314/lectures/hyd_probs/hyd_probs.html
    So, where is the electron most likely to be? We can answer this question in two ways.
    One way is to compute the expectation value of the radius, ... Another way is to look at a graph of the quantities in question. The peak of the curve showing probability as a function of radius defines the MOST LIKELY position for the electron.​
    ... which is how I normally teach it.
    You need context to work out what counts as an appropriate answer.
    Technically - a particle, on a single measurement, is most likely to be found in regions near a local maximum.
    A small region around the expectation value may even have zero probability.
    That's really just a central value.

    I can also find course-notes which agree with hyperphysics.
    It's curious that few of the course notes actually discuss the distinctions properly.

    ... so what does the person who set the question mean by "most probable"?
    Is it safe to assume, blind, that they are using the same approach as hyperphysics on this?
    Perhaos the question was taken from the same sorts of text-books you are used to?
    Or are we better advised to ask the person who attended the classes?

    Probably OP has chosen that particular approach because past examples from class have done the same.
    Still: I don't want to second guess the teacher at this stage.

    I was kinda hoping to have this discussion with OP.
     
    Last edited: Jan 6, 2014
  8. Jan 8, 2014 #7

    vela

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    I'd say this person is flat out wrong.

    I don't see where it says this on page 37. In fact, it says,
    Note that this does not necessarily correspond to where the probability is a maximum. In fact, for, say n = 2, the particle is most likely to be found in the vicinity of x = L/4 and x = 3L/4.​
    It points out the most likely value is not given by the expectation value.

    Frankly, I think anyone who teaches that "most likely" means "expectation value" does a disservice to his or her students. Sure, you can point out to students that sometimes people erroneously conflate the two, but to suggest that it's correct in some contexts is just wrong. All that does is propagate a common misconception.
     
  9. Jan 8, 2014 #8

    Simon Bridge

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    Wrong stuff still gets taught and wrong ideas promulgate and teachers do do disservice to their students.
    I'm sorry, I didn't realize this was controversial.

    Conflating a central value with a likely one is common enough IRL to be worth checking for.
    It would be nice to just tell students "this is the one thing everyone means when they ask" but that is not true.
    Hopefully the result of this discussion is that I don't have to have it with as many others ;)

    Shall we agree to agree and get back to OPs actual question?
    [edit] oh wait: didn't ask a question ...

    What do you think of the "thin conical shell"?
    I suspect, but do not know for sure, that OP is asking if the calculation is correct.
     
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