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

I Why are total orbital QN l,m zero for closed subshell?

  1. Jul 14, 2016 #1

    goodphy

    User Avatar
    Gold Member

    Hello.

    Here, I'm asking why total orbital quantum number l and total magnetic quantum number m are zero for closed subshell in atom.

    Let me review the addition of angular momentum first: Each electron has its own orbital quantum number li and magnetic quantum number mi. Then for two electrons, total magnetic quantum number is obviously m = m1 + m2. Total orbital quantum number l has possibilities of l = l1 + l2, l1 + l2 - 1, ... , |l1 - l2|. This rule can extends for more electrons.

    In return to closed subshell problem, m should be zero as summation of all individual mi in closed subshell is zero. However, l has several possibility according to rule above. For example, for p (li = 1) subshell, l = 2, 1, 0. Obviously total l can have non-zero value. How is total l for closed subshell zero as literature says?

    And in case of two electrons in p subshell which have m1 = 1 and m2 = 0 (unfilled subshell), m = 1 and l = 2,1,0. Does it mean that there are several possibilities of term symbol to these electronic configuration?
     
  2. jcsd
  3. Jul 15, 2016 #2

    Paul Colby

    User Avatar
    Gold Member

    Been a long time since I've looked at this so I only have hand waving at best. The shells are states that minimize potential for the electron cloud. This means getting as much charge as close to the nucleus as possible while respecting the Pauli exclusion principle. Since the coulomb interaction is so much stronger than the spin-spin one L dominates and spin goes along for the ride.
     
  4. Jul 15, 2016 #3

    goodphy

    User Avatar
    Gold Member

    Hello and thank you to reply this question.

    So...there is no mathematical proof that total orbital quanum number l is zero for closed subshell in atom but it is from physical thinking?
     
  5. Jul 15, 2016 #4

    Paul Colby

    User Avatar
    Gold Member

    The atomic shell model is quite well worked out. As I recall, one solves the single particle wave functions and expands in terms of them. This can get quite computationally involved. I'm not an expert and have never done such calculations. However, in general the higher the angular momentum the further out in radius the bulk of the wave function resides. So, the things close in, lower L values, will fill in first. There was something funny about the order of the f-shell as I recall.
     
  6. Jul 15, 2016 #5

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    If a subshell is fully occupied then there is only one possible value for ##m##, which is zero. The value of ##l## which meets the requirement of having only one ##m## is ##l=0##.
     
  7. Jul 17, 2016 #6
    What is the definition of closed subshell? Would it be two electrons in an orbital such as ##p_z##, which has ##(\ell,m) = (1,0) ##? I would expect something like.
    $$\mid \psi_{2p_z\,\text{full}} \rangle\propto \mid n_{1}=n_2=2,l_1=l_2=1, m_1=m_2=0, \uparrow, \downarrow\rangle- \mid n_{1}=n_2=2,l_1=l_2=1, m_1=m_2=0, \downarrow, \uparrow\rangle $$
    Orbital angular momentum will operate on this state and just give a factor depending on the common ##\ell## value.
    $$L^2\mid \psi_{2p_z\,\text{full}}\rangle = 1(1+1) \mid \psi_{2p_z\,\text{full}}\rangle \neq\mathbf{0} $$
    as for ##p_x## etc.
    You can get into mixing the ##m=\pm1## components but that doesn't seem to cleanly produce a sum state ##L, M=0##. If you look at a table of clebsch gordan coefficients ##L, M=0## corresponds to ##\ell_1=\ell_2 = 1## with ##\mid m_1, m_2 \rangle## thus
    $$\mid L=0, M=0\rangle =\frac{1}{\sqrt{3}}\Big( \mid 1,-1\rangle + \mid -1,1\rangle - \mid 0,0\rangle \Big) $$
    So I would contend that two electrons in this ##L,M=0## state is not the same as what I called ##\mid \psi_{2p_z\,\text{full}}\rangle ## because ##\mid L=0, M=0\rangle ## appears to be a mixture of different sublevels. That is ##\mid 1, 1, m_1, m_2\rangle## with ##m_1\neq m_2## in some cases; you must add electrons with different ##m## values to get the ##L=0## result. Am I missing something here?
     
    Last edited: Jul 17, 2016
  8. Jul 18, 2016 #7

    goodphy

    User Avatar
    Gold Member

    Am...I thought closed subshell is common word in quantum mechanics. It is completely filled subshell specified by n and l. For a given closed subshell nl, all states specified with m and ms under nl are completely filled.

    SO could you please simplify your comments for this case?
     
  9. Jul 18, 2016 #8

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    A subshell is usually denoted by ##nl##. It's said to be a closed subshell when there are ##2(2l+1)## electrons in it.
    The way you wrote the ket state shows that you are using ##L_{1z}## and ##L_{2z}## as the good quantum numbers while these operators do not commute with the total Hamiltonian. You should have used the total orbital angular momentum, ##L## (where ##L=|\mathbf L| = |\mathbf L_1+\mathbf L_2|##), and its z component instead, as the specifier of states. For this reason, people invent the so-called term symbol ##n^{2S+1}L##.
    What is ##L## here, which electron's angular momentum is it?
     
  10. Jul 18, 2016 #9

    goodphy

    User Avatar
    Gold Member

    Hi.

    I think your argument is weak to explain total orbital quantum number l = 0 for closed shell. The essence of your argument is that total magnetic quantum number m has only one value for given states of individual electrons.

    Let's get the case that there are two electrons, (l1 = 1, m1 = -1) and (l2 = 1, m2 = 1) so it is partially filled subshell. In this case, m = -1 + 1 =0 and it is also only one possibility of what m can have. In this case, can I say that total l = 0 because m = 0 and it is only one possibility? Once individual mi are given, there is only one possible m = m1 + m2 + ... + mN. For example, in l = 1 subshell having m1 = 1, m2 = 0, m = m1 + m2 = 1 and it is only possibility of m. But there is no l which can have "only one" m = 1. So can I say that l can not be specified in this case?

    I didn't realize this issue is so much complicated than I expected. I thought It was simple but it looks now that I need to study Slater determinant, rotational transformation, etc. How does textbook just say total l and m for closed subshell is zero and believe it?
     
  11. Jul 18, 2016 #10

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    No, you cannot specify the z-components of orbital angular momenta of each electron because ##L_{iz}## is not a good quantum number as I have pointed out in my previous post. In your example, the outer subshell is of the form ##np^2##. All you know about this subshell is that there are two electrons occupying 6 possible "slots", but you don't know exactly which slots those two occupy. Instead of specifying the state by ##L_{iz}##, you should use the total orbital angular momentum and its z-component.
    My textbook "Physics of Atoms and Molecules" by Bransden and Joachain comes to the conclusion that ##L=0## for closed subshell from the fact that the only possible value of ##m_L = \sum_i m_{Li}## is zero. Even though you don't know the exact values of ##m_{Li}## of each electron, but you do know that they must sum up to zero.
     
  12. Jul 18, 2016 #11

    goodphy

    User Avatar
    Gold Member

    I see.

    I was actually trying to get total orbital number l in uncoupled representation while l can be specified only in coupled representation. I did action of contradiction. Thanks to correct my mind! I appreciate it!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted