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Is j a nuclear quantum # or subshell notation or both?

  1. Nov 22, 2008 #1
    I can’t seem to find a definitive answer on what makes up the quantum numbers of a nucleus.

    I know that:
    n is an indexing number of l states or the number of sign changes in the wavefunction
    l represents the angular moment of a nucleon
    s represents the spin of a nucleon

    But I can not determine if j is a quantum number. I know vector j is made up two other vectors, e.g., j = l + s.

    I also know that the nuclear shell/subshell notation involves j. For example:

    2d(3/2) where n = 2, l = d and j = (3/2)

    Any thoughts as to which one j is?

    Also is parity a quantum number too? I know it too is often used in the nuclear shell notation.
  2. jcsd
  3. Nov 23, 2008 #2


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    Quantum observables to operators which are constants of motion are often used as "quantum numbers" to label states. And constant of motion are operators which commutes with the Hamiltonian.

    So if you have a spherical symmetric potential, J = L + S is a good quantum number.
  4. Nov 23, 2008 #3
    Gosh, I had to read your comment several times to make sure I extracted the correct meaning. To be sure, I wanted to throw back to you my interpretation of your comment.

    1st Quantum observables would be things like angular momentum, intrinsic spin or spin-orbit coupling.

    2nd Operators are equivalent to functions. (eigenfunction?)

    3rd Constants-of-motion implies steady-state motion with non varying forces or fields. (eigenstate?)

    4th Meeting the first three conditions then the observables would be considered as a "quantum number"

    5th Quantum numbers do label states.

    I did some additional reading and have found that parity is not really a quantum number but only considered a characteristic of the nucleus.

    Is my interpretation of your comment consistent?
    Do you agree with the comment on parity?

    Quite a good response you made Malawi_Glenn and I greatly appreciate your time to help.
  5. Nov 24, 2008 #4


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    Well no the operators are NOT equivalent to functions. Operaters ACTS on functions. And if the operators don't change the function, then the function is an eigenfunction.

    Otherwise you are correct, one label the states with "good" quantum numbers, those which are (quantum) constants of motion, i.e commute with the hamiltonian of the system.

    Now as an excercise, suppose you have an hamiltonian with spherical symmetric potential, show that it commutes with J = L + S. Use the fact that L and S are in different hilbert spaces.

    Suppose your hamiltonian is NOT spherical symmetric, such as the case for elongated nuclei (the ones which rotates). Here J = L + S is not a good quantum number. See for instance the "Nilsson model".

    Parity is a quantum number according to my opinion, since the parity operator commutes with the hamiltonian. That is what me and many others are using as quantum numbers.

    I think this is a quite good introducion to quantum numbers:
  6. Nov 24, 2008 #5
    Quite right,.. operators do act on functions. (don’t know where my head was when I wrote otherwise, hummm)

    I was working my way to the Nilsson model so thanks for the input.

    Ok, I understand and of course accept your logic on parity.

    Does it sound reasonable that 7 quantum numbers are required to describe the quantum state of the nucleus?

    n – A quantum number used for indexing angular momentum states of the nucleon
    l – A quantum number used to quantify the angular momentum state of the nucleon and used to calculate L
    L – A quantum number used to quantify the angular momentum state of the nucleus.
    s – A quantum number used to quantify the nucleon spin state used to calculate j
    j – A quantum number used to quantify the total angular momentum of the nucleon, eg, j = l + s
    J – A quantum number used to quantify the total angular momentum of the nucleus, eg, J = L + S
    parity – A quantum number used to quantify the parity state of the nucleus
  7. Nov 24, 2008 #6


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    The n,l,s,j,parity are for single nucleons. J, parity : for nucleus as a whole.

    What textbook do you use?
  8. Nov 24, 2008 #7
    I’m retired, not in class and don’t have a text to refer to.
    I don’t have a physics background and all information I have picked up has come solely from the Internet.
    There are several very good educational sites and of course Wikipedia.
    Also, Google has scan and made available several physics books which I have consulted.
    I get some books via Inter Library Loan which “The Nuclear Shell Model” by Kris Heyde is in queue.
    However, when I get really stuck I depend on physics.fourms.com
    Everybody at the forum has always been very helpful and I am most grateful for that.

    By the way, in reading your note I didn’t understand how parity could be associated with a single nucleon. Did you mean the family of protons or neutrons? Because parity seems only to be significant when considering in the family of nucleons. I am confused about this point; would it be possible for you to briefly explain that point? Now considering the whole nucleus I know that when all shells are filled they will have a positive parity and it is the parity of the last nucleon that determines the parity of that family of nucleons. I think I read that parity is a (function?) of the wave equation where the angular momentum above the energy line is positive yielding positive parity or below the energy line is negative yielding a negative parity. Likewise, this parity is associated with the family of protons or neutron under consideration. Thus the parity of the whole nucleus is determined by combining the parity of the proton state and neutron state where two odd-odd parities are difficult to determine.
  9. Nov 25, 2008 #8


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    Maybe you should start studying general quantum mechanics before studying nuclear structure physics?
    The parity of a nucleon state in this case is [tex](-1)^l[/tex], just as in the hydrogenic atom which one studies in the first course in QM. Parity of two-nucleon state is: [tex](-1)^{l_1}*(-1)^{l_2} = (-1)^{l_1+l_2}[/tex]. and so on.

    Just for completness, why are you trying to study nuclear structure physics on your own with no physics background?

    I can recommend introductory book: Nuclear Physics: Principles and Applications by J. S. Lilley
    which is cheap and good.
  10. Nov 25, 2008 #9
    Interesting, I have worked my way through the quantum mechanics of the atomic atom and I didn’t recall parity associated with the electron. Although, thinking back I might have read something about electron paring but the topic didn’t seem prevalent in the scheme of atomic quantum numbers which was my focus.

    Thanks for the parity formula; I have seen a few minor references to something similar but nothing as complete as your formula. Thanks for the direction; I now know what to look for in my search. I am going to venture a guess and speculate that the reason parity is considered a quantum number is because


    represents the parity of one nucleon which is ½ the total. And I guess this is the “parity operator that commutes with the Hamiltonian”?.

    My background is engineering; however, I have always wondered about quantum mechanics and elementary particles. I am writing a non-related manuscript and in one of the sections it addresses magnetism. I have learned that magnetism comes in many forms generally from the various atomic magnetic moments (spin, orbital, coupling, etc). However, one form involves the jj coupling of the nucleus and it is this form that has driven me to investigate atomic and nuclear quantum mechanics. And it was not as thought I wasn’t curious about these topics anyway. Besides, my freshman physics instructor late in the semester decided not to present Black Body Radiation and properties of Photoelectric Effect and this has always bother me and by digging into the history of atomic quantum mechanics I quenched my thirst for those topics.

    I will add Nuclear Physics: Principles and Applications by J. S. Lilley to the inter library queue and if they are unable to get it then it is cheap to I will purchase it. Thanks for the reference.

    Again you have been a great help and I certainly don’t want to wear out my welcome because I am in unfamiliar territory and it is easy to get bogged down, with few reliable resources on the Internet that addresses the topics in sufficient detail to resolve amateur questions.

    Thanks again Malawi_Glenn
  11. Nov 26, 2008 #10


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    No it is the parity for wavefunction in a spherical potentinal, i.e the parity of the Spherical Harmonics.

    Other references I can recommend are "Modern Quantum mechanics" by Sakurai, you'll learn alot about angular momentum and symmetries in that one. Start with Sakurai and Lilley :-)

    By they way, here is very good lecture notes about quantum mechanics basics if you want to start right now:
  12. Nov 27, 2008 #11
    The issue of parity is a hazy issue that is starting to clear up for me. I now understand one has parity of a particle, (electron, nucleon and others), one has parity of the nucleus and one has parity conservation & violation from force interactions. The following references were a big help; but, I should note these references were not without questions and I hope you can help me with that.

    Donald Perkins in his book “Particle Astrophysics”
    provides a general definition of particle parity on page 66 in the section 3.3 “The Parity Operation” where he indicates the inversion of spatial coordinates
    (x, y, z) > (-x, -y, -z) is a discrete transformation of the wave amplitude brought about by the parity operator P:
    [tex] P\psi \left(r\right) = \psi\left(-r\right) [/tex]
    And repeating the operation reverts to the original system
    [tex] P^2 = 1 [/tex]
    Thus the enginvalue of P must be +/- 1

    Perkins indicates in a spherically symmetric potential that it has the property V(-r) = V(r) so the bound states by such potential can be parity eignestates. He goes on to indicate for the hydrogen atom the wavefunction in terms of the radial coordinate r and the polar and azimuthal angular coordinates [tex] \theta [/tex] And [tex] \phi [/tex] of the electron with respect to the proton is

    [tex] \chi \left(r, \theta, \phi\right) = \eta\left(r\right)Y_l^m\left(\theta, \phi\right) [/tex]

    Under inversion
    [tex] r \rightarrow -r [/tex],
    [tex] \theta \rightarrow \left(\pi-\theta\right) [/tex]
    [tex] \phi \rightarrow \left(\pi+\phi\right) [/tex],

    produces the following results:

    [tex] Y_l^m \left(\pi- \theta,\pi+ \phi\right) = \left(-1\right)^l \left(\theta, \phi\right) [/tex]


    [tex] P\chi \left(r, \theta, \phi\right) = \left(-1\right)^l \chi \left(r, \theta, \phi\right) [/tex]


    [tex] P = \left(-1\right)^l [/tex]

    PERKINES DID NOT DEFINE [tex] \chi [/tex] or [tex] \eta [/tex] CAN YOU HELP WITH THIS?
    ALSO, I HAVE NOT FOUND HOW [tex] \left(-1\right)^l [/tex] WAS DERIVED. CAN YOU ALSO HELP HERE?

    Perkins continues with an interesting discussion on parity conservations with strong (gluons) and electromagnetic (photon) interactions and parity violation with weak (heavy gauge boson) interactions.

    In a different reference Sidney Yip in his MIT Open Courseware presents a discussion in his lecture “22.101 Applied Nuclear Physics, Fall 2004”
    where in his section ‘Prediction of Ground-State Spin and Parity’ on page 11 he presents the following rules which have a bearing on parity:
    1- Angular momentum of odd-A nuclei is determined by the angular momentum of the last nucleon in the species (neutron or proton) that is odd.
    2- Even-even nuclei have zero ground-state spin, because the net angular momentum associated with even N and even Z is zero, and even parity.
    3- In odd-odd nuclei the last neutron couples to the last proton with their intrinsic spins in parallel orientation.

    In the last rule Yip explains the determination of odd-odd nuclei spin but does not explain the parity associated with an odd-odd nucleus.


    This last question may be answered by the parity formula posted by malawi_glenn where he indicates
    (-1)^{l_1}*(-1)^{l_2} = (-1)^{l_1+l_2}
    If sub 1 and sub 2 represents the angular momentum of two different particles, specifically, a proton and a neutron then it would be a simple matter of doing the math to determine an odd-odd nuclei parity.

  13. Nov 28, 2008 #12


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    [tex] \chi \left(r, \theta, \phi\right) = \eta\left(r\right)Y_l^m\left(\theta, \phi\right) [/tex]

    Here eta(r) is an ARBITRARY function depending on r only.

    And xhi(r,theta,phi) is an Arbitraryt function of r, theta and phi.

    You can google for "Sperical Harmonics" and make the parity transformation:
    [tex] \theta \rightarrow \left(\pi-\theta\right) [/tex]
    [tex] \phi \rightarrow \left(\pi+\phi\right) [/tex]
    And see how they changes the spherical harmoic with number L. you result will be (-1)^L

    The parity of odd-odd nuclei is simple (-1)^L_1 (the unpaired neutron) TIMES (-1)^L_2 (the unpaired proton)
    Say for instance you have one unpaired neutron in 1d and one unpaired proton in 1p. The neutron has pairity (-1)^2 = 1 and the neutron (-1)^1 = -1, so total parity is -1.
    Antother good book which I think will interesst you is "Particle Physics" by martin.
  14. Nov 30, 2008 #13
    I think I have a grasp of the parity fundamentals and I thank you for your help; but, before I move on to other subjects I would like to get your perspective (or others) as to why you think parity is a quantum number. I have thought about this subject more than once and I don’t have enough information to draw the same determination. I understand parity, like spin, comes in two states; furthermore, parity of a nucleon or other particles is determined by the angular momentum l of that particle. However, parity is conserved and does not seem to affect the energy level of the particle from strong and electromagnetic interaction and I think I can extend that to nucleon interactions as well. Thus, if the energy state of the nucleon does not change by these interactions how is it considered a quantum number? This lack of energy change is what blocks me from understanding why parity could be a quantum number. Now if parity is deemed a quantum number because it only manifests two states, like spin, I could almost understand except by definition I though quantum meant integer or half integer increments of Plank’s or Dirac’s constant. I think you can see where I am coming from so I hope you can help me understand your perspective on nucleonic parity.
  15. Dec 1, 2008 #14


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    If Parity commutes with the hamiltonian, hence it is a quantum number.

    Parity is a property of a state, which can be measured. Quantum numbers is more than what is related to energy.

    Measuring e.g photon angular distributions from decays, one knows which energy levels was involved and hence comparison with theory can be made. Parity is conserved in reactions, if you know final state of nucleus and all parities of the photons from it (imagine a gamma cascade from an unknown excited initial state), it helps you alot to build up a level diagram of the energy levels and make comparison with theories.

    Imainge you had a hamiltonian with no spin-orbit interaction. You would have the same states as in the "usual" shell model, but the energy levels would be degenerate.
  16. Dec 1, 2008 #15

    Vanadium 50

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    Let me repeat some advice you got earlier - learning QM first will make understanding nuclear physics a lot easier. It would be difficult to learn calculus before algebra, right? Same thing - a solid foundation is necessary for good understanding.
  17. Dec 1, 2008 #16
    Thanks for the info malawi_glenn. Before I move on there is one final question about parity. You indicated “parity is a property of state”, does parity have any significance with respect to the Pauli Exclusion Principle? In other words, if a nucleon had a specific state with n, l, s, j and plus parity then could another like nucleon occupy the same specific state n, l, s, j except with negative parity. Or is parity not a state considered by the Pauli Exclusion Principle.
  18. Dec 1, 2008 #17


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    Well a state defined by n,L,s,j have also its parity defined uniquely as (-1)^L

    So to answer your question: No
  19. Dec 1, 2008 #18
    malawi_glenn, you can disregard my previous note. I remembered that parity of a particle is determined by the angular momentum thus a particle with n, l, s, j state can not have two parities, which was my question. Thus, parity does not provide an additional state to the nucleon with respect to the Pauli Exclusion Principle.
    That should wrap it up for me and I extend a gracious thanks to you for sharing of your wealth of knowledge and your patience. Also, thanks for the book tip on "Particle Physics" by Martin. I will be looking forward to reading his text.
  20. Dec 1, 2008 #19


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    Cool I really recomend that book, and those books I recomend earlier.

    Remember that parity plays a great role in spectroscopy.

    Don't hesitate to ask more question here.

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