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Why don't all electrons contribute to total orbital angular momentum of an atom?


Quantization of orbital angular momentumFor mesons, or baryons, do sea quarks contribute to the angular momentum of the bound state?Total Angular Momentum of a Hydrogen Atomorbital angular momentum of the silver atomOrbital angular momentum of electronsSpin, orbital angular momentum and total angular momentumTotal orbital and spin angular momentum for a closed shellHow to prepare the silver atoms in Stern–Gerlach experiment with total angular momentum $j_z=+frac{1}{2}hbar,-frac{1}{2}hbar$?Total Orbital Angular Momentum and its relation to $M_L$ & SymmetryOrbital angular momentum quantum numbers - subtracted?






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$begingroup$


There are 47 electrons in a Silver atom, but talking about its orbital angular momentum we only take the outermost valence electron which occupies the 5s orbital. Why don't the remaining inner 46 electrons contribute to the total orbital angular momentum of a silver atom?










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New contributor



ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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  • 2




    $begingroup$
    Are the contributions of all electrons equal in sign?
    $endgroup$
    – DJohnM
    9 hours ago






  • 1




    $begingroup$
    Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
    $endgroup$
    – rob
    8 hours ago


















6












$begingroup$


There are 47 electrons in a Silver atom, but talking about its orbital angular momentum we only take the outermost valence electron which occupies the 5s orbital. Why don't the remaining inner 46 electrons contribute to the total orbital angular momentum of a silver atom?










share|cite|improve this question









New contributor



ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$








  • 2




    $begingroup$
    Are the contributions of all electrons equal in sign?
    $endgroup$
    – DJohnM
    9 hours ago






  • 1




    $begingroup$
    Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
    $endgroup$
    – rob
    8 hours ago














6












6








6


1



$begingroup$


There are 47 electrons in a Silver atom, but talking about its orbital angular momentum we only take the outermost valence electron which occupies the 5s orbital. Why don't the remaining inner 46 electrons contribute to the total orbital angular momentum of a silver atom?










share|cite|improve this question









New contributor



ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




There are 47 electrons in a Silver atom, but talking about its orbital angular momentum we only take the outermost valence electron which occupies the 5s orbital. Why don't the remaining inner 46 electrons contribute to the total orbital angular momentum of a silver atom?







quantum-mechanics angular-momentum quantization






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ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









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asked 9 hours ago









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ROBIN RAJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










  • 2




    $begingroup$
    Are the contributions of all electrons equal in sign?
    $endgroup$
    – DJohnM
    9 hours ago






  • 1




    $begingroup$
    Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
    $endgroup$
    – rob
    8 hours ago














  • 2




    $begingroup$
    Are the contributions of all electrons equal in sign?
    $endgroup$
    – DJohnM
    9 hours ago






  • 1




    $begingroup$
    Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
    $endgroup$
    – rob
    8 hours ago








2




2




$begingroup$
Are the contributions of all electrons equal in sign?
$endgroup$
– DJohnM
9 hours ago




$begingroup$
Are the contributions of all electrons equal in sign?
$endgroup$
– DJohnM
9 hours ago




1




1




$begingroup$
Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
$endgroup$
– rob
8 hours ago




$begingroup$
Silver is perhaps not a great example, since the $s$-wave orbitals (with angular momentum quantum number $ell=0$) don't carry any orbital angular momentum.
$endgroup$
– rob
8 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.



An example using the simplest S-shell.



We have two states available, "up" $|uparrow{} rangle$, and "down" $|downarrow{} rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.



If we are placing a single electron into the S-shell we have 2 states available, either:
$$|uparrow{}rangle text{ & } |downarrow{}rangle$$
Each of these have angular momentum $frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry,
$$|psirangle = frac{1}{sqrt{2}}left(|uparrow{}downarrow{}rangle - |downarrow{}uparrow{}rangleright)$$
This is the singlet configuration which has angular total angular momentum zero.



Importantly there is only one state with total angular momentum $J=0$, two states with $J=frac{1}{2}$, three states with $J=1$ (triplet), and so on.



Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l leq m_l leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.



Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).



The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.



As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
    $endgroup$
    – rob
    8 hours ago










  • $begingroup$
    what about the orbital, Spin and total angular momentum of a Nitrogen atom
    $endgroup$
    – ROBIN RAJ
    38 mins ago














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7












$begingroup$

We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.



An example using the simplest S-shell.



We have two states available, "up" $|uparrow{} rangle$, and "down" $|downarrow{} rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.



If we are placing a single electron into the S-shell we have 2 states available, either:
$$|uparrow{}rangle text{ & } |downarrow{}rangle$$
Each of these have angular momentum $frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry,
$$|psirangle = frac{1}{sqrt{2}}left(|uparrow{}downarrow{}rangle - |downarrow{}uparrow{}rangleright)$$
This is the singlet configuration which has angular total angular momentum zero.



Importantly there is only one state with total angular momentum $J=0$, two states with $J=frac{1}{2}$, three states with $J=1$ (triplet), and so on.



Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l leq m_l leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.



Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).



The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.



As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
    $endgroup$
    – rob
    8 hours ago










  • $begingroup$
    what about the orbital, Spin and total angular momentum of a Nitrogen atom
    $endgroup$
    – ROBIN RAJ
    38 mins ago
















7












$begingroup$

We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.



An example using the simplest S-shell.



We have two states available, "up" $|uparrow{} rangle$, and "down" $|downarrow{} rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.



If we are placing a single electron into the S-shell we have 2 states available, either:
$$|uparrow{}rangle text{ & } |downarrow{}rangle$$
Each of these have angular momentum $frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry,
$$|psirangle = frac{1}{sqrt{2}}left(|uparrow{}downarrow{}rangle - |downarrow{}uparrow{}rangleright)$$
This is the singlet configuration which has angular total angular momentum zero.



Importantly there is only one state with total angular momentum $J=0$, two states with $J=frac{1}{2}$, three states with $J=1$ (triplet), and so on.



Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l leq m_l leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.



Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).



The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.



As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
    $endgroup$
    – rob
    8 hours ago










  • $begingroup$
    what about the orbital, Spin and total angular momentum of a Nitrogen atom
    $endgroup$
    – ROBIN RAJ
    38 mins ago














7












7








7





$begingroup$

We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.



An example using the simplest S-shell.



We have two states available, "up" $|uparrow{} rangle$, and "down" $|downarrow{} rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.



If we are placing a single electron into the S-shell we have 2 states available, either:
$$|uparrow{}rangle text{ & } |downarrow{}rangle$$
Each of these have angular momentum $frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry,
$$|psirangle = frac{1}{sqrt{2}}left(|uparrow{}downarrow{}rangle - |downarrow{}uparrow{}rangleright)$$
This is the singlet configuration which has angular total angular momentum zero.



Importantly there is only one state with total angular momentum $J=0$, two states with $J=frac{1}{2}$, three states with $J=1$ (triplet), and so on.



Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l leq m_l leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.



Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).



The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.



As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.






share|cite|improve this answer











$endgroup$



We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher orbitals would have the angular momentum of only the valence electrons. Now I will demonstrate why a full shell must have zero angular momentum.



An example using the simplest S-shell.



We have two states available, "up" $|uparrow{} rangle$, and "down" $|downarrow{} rangle$. We also have the constraint that these are fermions, meaning any combination has to be entirely antisymmetric when two particles are interchanged.



If we are placing a single electron into the S-shell we have 2 states available, either:
$$|uparrow{}rangle text{ & } |downarrow{}rangle$$
Each of these have angular momentum $frac{1}{2}$. However if we want to add another electron we only have 1 possible state which satisfies antisymmetry,
$$|psirangle = frac{1}{sqrt{2}}left(|uparrow{}downarrow{}rangle - |downarrow{}uparrow{}rangleright)$$
This is the singlet configuration which has angular total angular momentum zero.



Importantly there is only one state with total angular momentum $J=0$, two states with $J=frac{1}{2}$, three states with $J=1$ (triplet), and so on.



Here I will outline the logic of the general proof. Firstly ignoring spin, each single particle orbital $l$ has $2l +1$ states with angular momentum projections ranging from $-l leq m_l leq l$. Secondly not ignoring spin you can place 2 electrons in each orbital with spin up or down. This gives $2(2l + 1)$ single particle states. If we have completely filled this shell that means that we have placed an electron in each single particle orbital.



Now if we count all of the unique ways we can fill all of the orbitals, there is only one way to do this. That means that this is a singlet configuration and not a member of a higher multiplet (such as the triplet with 3 states mentioned above).



The total state is defined in terms of having a total angular momentum and total angular momentum projection. Clearly it has angular momentum projection of $0$ since $sum_{m= -l}^{m = l} m = 0$. However since it is a singlet state it also has total angular momentum $0$, and we can treat it like a "core" with no angular momentum.



As a side note this is used extensively in atomic physics as well as nuclear physics. For nuclear physics we would not be talking about electrons but instead protons and neutrons. Therefore we not only have the choice of spin up or down for each orbital but also between proton and neutron. This gives us 4 particles in each orbital, and is made more rigorous with the idea of "isospin". So far as most nuclear interactions care their interactions are the same so we can treat them as 2 projections of one object, the "nucleon". The total wavefunction must be antisymmetric under interchange in the combined spatial-spin-isospin space. A filled orbital momentum shell would therefor have zero angular momentum as well as zero total isospin.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 8 hours ago

























answered 8 hours ago









TEHTEH

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3567 bronze badges








  • 1




    $begingroup$
    MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
    $endgroup$
    – rob
    8 hours ago










  • $begingroup$
    what about the orbital, Spin and total angular momentum of a Nitrogen atom
    $endgroup$
    – ROBIN RAJ
    38 mins ago














  • 1




    $begingroup$
    MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
    $endgroup$
    – rob
    8 hours ago










  • $begingroup$
    what about the orbital, Spin and total angular momentum of a Nitrogen atom
    $endgroup$
    – ROBIN RAJ
    38 mins ago








1




1




$begingroup$
MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
$endgroup$
– rob
8 hours ago




$begingroup$
MathJax suggestion: hide a $providecommand{ket}[1]{left|{#1}right>}$ somewhere near the top of your answer. Then use $ketuparrow$, $ket{uparrowdownarrow}$, etc., which have nicer-looking kerning than just $|uparrowrangle$.
$endgroup$
– rob
8 hours ago












$begingroup$
what about the orbital, Spin and total angular momentum of a Nitrogen atom
$endgroup$
– ROBIN RAJ
38 mins ago




$begingroup$
what about the orbital, Spin and total angular momentum of a Nitrogen atom
$endgroup$
– ROBIN RAJ
38 mins ago










ROBIN RAJ is a new contributor. Be nice, and check out our Code of Conduct.










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ROBIN RAJ is a new contributor. Be nice, and check out our Code of Conduct.













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