Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular momentum (L) in z-axis of hydroxyl radical

  1. Mar 11, 2010 #1
    A hydroxyl radical (O-H*) in the gas phase is found to be in a rotational state with a given wavefunction (Long wavefunction, not needed for my questions).
    Calculate the expected value of Lz (angular momentum in z) for the radical.


    Is this the right equation?
    Lz=mlħ ????

    In which orbital is the radical? How do I figure this out?
    The orbital of the radical (not sure how to find this) would give me the rotational quantum number (l), thus giving me the spacial orientation quantum number (ml), with which I could solve for Lz=mlħ[/QUOTE] But then why does the problem give a wavefunction?

    Information:
    ħ = h/2[tex]\pi[/tex]
    [tex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s[/tex]
    ml = -l, -l+1, -l+2, ... l-2, l-1, l
    H orbital 1s1
    O orbital 1s2 2s2 2p4.
     
  2. jcsd
  3. Mar 11, 2010 #2

    SpectraCat

    User Avatar
    Science Advisor

    Can you please post the wave function?
     
  4. Mar 11, 2010 #3
    Give me a moment and I will have it up.
     
  5. Mar 11, 2010 #4
    [tex]\Psi(\Theta,\varphi)=\sqrt{15/64\pi}(3cos[/tex]2[tex] \Theta-1) + \sqrt{15/64\pi}sin\Thetacos\Thetae[/tex]i[tex]\varphi[/tex][tex] + \sqrt{15/64\pi}sin\Thetacos\Thetae[/tex]-i[tex]\varphi[/tex]
     
  6. Mar 11, 2010 #5

    SpectraCat

    User Avatar
    Science Advisor

    Okay ... so, what does all of that mean? You should be able to identify the angular momentum quantum numbers (l and ml) from looking at that wavefunction. However, I will say that it is a little odd that you are being asked for the Lz expectation value ... the usual convention is that is L specific for orbital angular momentum, for a rotational wavefunction, one usually talks about J and Jz. Furthermore, the hydroxyl radical is one of the more complicated angular momentum cases, with coupling of the spin, orbital, and rotational angular momenta. I am guessing this is a homework assignment (or similar) ... have you covered angular momentum coupling yet? Is the term symbol given for the electronic state of the hydroxl radical?
     
  7. Mar 11, 2010 #6
    No, angular momentum coupling has not been covered in class. This is a Physical Chemistry II class. I've also noticed the complexity of the hydroxyl radical, but this wavefunction is said to express it's "rotational state". I do not understand how I can derive the quantum numbers from the wavefunction.
     
  8. Mar 11, 2010 #7

    SpectraCat

    User Avatar
    Science Advisor

    Ok, I just took a closer look at that wavefunction ... are you sure you reproduced it correctly? I expect that the first "+" is actually supposed to be a multiplication ... but then there should be parentheses around the two sine terms. Is that correct, or is it exactly like you wrote?

    EDIT: No, that's not quite right either, but there is definitely something wrong with that wavefunction ... are the second two terms supposed to have [tex]sin\theta e^{\pm i\varphi}[/tex], instead of what actually appears above?

    In any case, I guess you must have covered spherical harmonics in class, right?
     
    Last edited: Mar 11, 2010
  9. Mar 12, 2010 #8
    Yes, Spherical Harmonics was recently covered. The wavefunction is verbatim. I thought it was also weird that the cos squared didn't have parenthesis around it's term.

    We've also covered Rigid rotors and recently started getting into angular momentums (i.e. orbitials and many-electron systems... molecules)

    There are three terms all added together, the second to last term is
    [tex]\sqrt{15/64\pi} sin\Theta cos\Theta^{+i\varphi}[/tex] <-- positive i*phi exponent

    and the last term is
    [tex]\sqrt{15/64\pi} sin\Theta cos\Theta^{+i\varphi}[/tex] <-- negative i*phi exponent

    Thank you for your help on this. I don't want to learn more in class without understanding this. It all builds on itself.
     
  10. Mar 12, 2010 #9

    SpectraCat

    User Avatar
    Science Advisor

    Please check your tex formatting, and make sure you are reproducing the wavefunction correctly. If what you have posted is *really* what is in your book/homework, then there is a typo there. Based on what you have written, I expect that the correct wavefunction you should be working with is:

    [tex]\Psi(\theta,\varphi)=\sqrt{\frac{15}{64\pi}}\left[\left(3cos^{2}\theta -1\right) + sin\theta cos\theta e^{i\varphi} + sin\theta cos\theta e^{-i\varphi}\right][/tex]

    Anyway .. there are a few questions to ask yourself:

    1) How are the spherical harmonics related to the l and m quantum numbers?

    2) How many distinct spherical harmonics, each with different values of l and m, are represented in the wavefunction you posted, i.e., is that wavefunction a single eigenstate of Lz, or a sum of Lz eigenstates?

    3) What is the definition of a quantum mechanical expectation value for an operator, and how can you determine it for a given wavefunction, in particular when that wavefunction is represented a linear combination of eigenstates of the operator in question?
     
    Last edited: Mar 12, 2010
  11. Mar 12, 2010 #10
    You are right about the "e's'" being in the expression. I cannot find how to do this in Tex format.

    1) Oh, I do not know this I will have to look it up. That may put me on the right track.

    2) Is this just simple rearrangement to find 2 spherical harmonic wavefunction (one for each atom) that would satisfy the total wavefunction, or is this thinking off? I'll need to figure out 1, first. That may take some time as this is a new topic for me.

    3) After reading up on expectation values I'm not sure I know what they are either. PErhaps this question is a bit above my level for now.
     
  12. Mar 12, 2010 #11

    SpectraCat

    User Avatar
    Science Advisor

    As far as I can tell, since you have not covered angular momentum coupling, this question cannot have anything to do with atoms or orbital angular momentum .. it is only about the rotational angular momentum of the entire molecule. The wavefunction you supplied gives you everything you need to know about the system, and can yield the answer after some straightforward math (at worst), or just a little logic (if you see the trick).

    Well, you definitely need to know that, because it is what the question is asking for. The definition of the expectation value is one of the postulates of quantum mechanics, and is typically covered fairly early on in PChem II, so I guess you have covered it at some point. The expectation value just gives the average value of an observable (in this case Lz) that would be obtained for a large number of measurements on a given quantum system (in this case, your wavefunction).
     
  13. Mar 12, 2010 #12
    Ya I didn't realize that was what an expectation value was. Confusing terminology... From the table of spherical harmonics I find a few functions that look very similar to the one provided. I've been trying to produce an 2 or 3 expressions that would sum up to this one but I cannot find it.

    http://en.wikipedia.org/wiki/Table_of_spherical_harmonics

    I've also found:
    http://www.phy.duke.edu/courses/100/lectures/Ang_mom/Eq27 [Broken]
    Above describes a wavefunction in terms of theta and phi and the use of tabulated Legendre Polynomials.

    Still trying to piece the information together...

    P.S. I'm missing 20 pages in this chapter (someone pasted another chapter in its place, kind of shady for a $100 book). So thanks for clarifying this, I'm contacting the store I purchased the book from.
     
    Last edited by a moderator: May 4, 2017
  14. Mar 12, 2010 #13
    Alright I think I have it:

    [tex]\sqrt{3/4}[/tex] Y2,0 [tex]\sqrt{1/8}[/tex] Y2,1 [tex]\sqrt{1/8}[/tex] Y2,-1

    However, Y0,0 = [tex]\sqrt{1/4\pi}[/tex], is that represented in the first term?

    So if Lz=mlħ then the average of ml= -1, 0, 1 and the average Lz = 0

    So the vector would be parallel to the Z axis?
     
  15. Mar 12, 2010 #14

    SpectraCat

    User Avatar
    Science Advisor

    Yes, you are making progress. Your first step is to re-write the wavefunction as a sum of the terms you have above.

    No, that is part of the normalization of any spherical harmonic, so you have already accounted for it in the expansion above.

    Well .. you are getting there, but you need to calculate the expectation value properly for the wavefunction. Do you know how to do that? Have you studied Dirac notation, or did you represent the expectation value as an integral in class?

    No, if the projection of the L vector on the z-axis is zero, wouldn't that mean that it is perpendicular to the z-axis?
     
  16. Mar 12, 2010 #15
    No we have no covered Dirac notation. I cannot recall ever doing this in class.

    But the answer is not 0?

    From what I've read the expectation value is the integral of the wavefunction squared? This seems odd to me.
     
  17. Mar 12, 2010 #16

    SpectraCat

    User Avatar
    Science Advisor

    I did not say the answer was not zero ... only that you had not yet presented a complete answer :wink:

    The expectation value for an operator A with wave function psi is given by:

    [tex]\langle \hat{A} \rangle = \frac{\int\psi^{*}\hat{A}\psi d\tau}{\int \psi^{*}\psi d\tau}[/tex]

    where [tex]d\tau[/tex] is the appropriate volume element, the limits of integration cover all space, and the denominator is included to cover the general case of an unnormalized wavefunction (if the wavefunction is normalize, the denominator will be 1 by definition).

    However, you do not need to evaluate any integrals to solve this problem, because the spherical harmonics are eigenfunctions of the Lz operator. Do you understand why this is true?
     
  18. Mar 12, 2010 #17
    Doesn't it have to do with Lz only depending on the quantum number l? I can't visually grasp it. If you have an elegant way of explaining it I would like to hear it.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook