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    I'm struggling to understand what this is all about..

    So it arises because of the anti-symmetry requirement for the wave function of fermions, but I can't seem to get my head round it. In the ground state the electrons are obviously anti-aligned in the 1s orbital, is this the singlet state? Where's the triplet state? I think I'm missing something
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    I am by no means an expert in this, but maybe i can help clarify some things

    As helium has 2 electrons, these can interact with each other differently depending on the relative orientations of the spins, and the nature of the specific orbitals each electron is within.
    In a 2 electron system, there are 2 possible ways of pairing the electrons in different orbitals, these are what singlet/triplet states are.

    When you think of this, try to imagine the different electron configurations.
    Firstly remember Hund's rule...The configuration that has the spins parallel is at a lower energy than the configuration with the paired spins (or something like that :P)
    The singlet state is called such because there is only 1 way to achieve it, 1s^2 is the only state of configuration where the individual spin momenta cancel out, so there is no overall spin.
    The triplet state is more complicated, as there are 3 ways to arrange the electrons to achieve it. This is when the individual spin momenta add up to give a non-zero overall spin.
    Going back to ye'olde Hund: The singlet excited state must be of lower energy than the triplet excited state

    One can construct approximate two electron wave-functions for the 1s^12s^1 state by omitting the (negligible) electron-electron interactions. As you said, it requires the anti-symmetry of these electrons. The anti-symmetrised wave functions can either have a symmetric spatial component and an antisymmetric spin component or an antisymmetric spin component along with a symmetric spin component:

    the singlet state is defined as:
    \Psi (\vec{r_1}, \vec{r_2}) = \dfrac{\Psi_{1s}(\vec{r_1})\Psi_  {2s}(\vec{r_2})\Psi_{2s}(\vec{r_  1})}{2} \times (\alpha (\vec{r_1}\Beta(\vec{r_2})-\alpha (\vec{r_2}) \Beta (\vec{r_1}))

    The triplet as:
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    (Original post by The-Spartan)
    I am by no means an expert in this, but maybe i can help clarify some things

    As helium has 2 electrons, these can interact with each other differently depending on the relative orientations of the spins, and the nature of the specific orbitals each electron is within.
    In a 2 electron system, there are 2 possible ways of pairing the electrons in different orbitals, these are what singlet/triplet states are.

    When you think of this, try to imagine the different electron configurations.
    Firstly remember Hund's rule...The configuration that has the spins parallel is at a lower energy than the configuration with the paired spins (or something like that :P)
    The singlet state is called such because there is only 1 way to achieve it, 1s^2 is the only state of configuration where the individual spin momenta cancel out, so there is no overall spin.
    The triplet state is more complicated, as there are 3 ways to arrange the electrons to achieve it. This is when the individual spin momenta add up to give a non-zero overall spin.
    Going back to ye'olde Hund: The singlet excited state must be of lower energy than the triplet excited state

    One can construct approximate two electron wave-functions for the 1s^12s^1 state by omitting the (negligible) electron-electron interactions. As you said, it requires the anti-symmetry of these electrons. The anti-symmetrised wave functions can either have a symmetric spatial component and an antisymmetric spin component or an antisymmetric spin component along with a symmetric spin component:

    the singlet state is defined as:
    \Psi (\vec{r_1}, \vec{r_2}) = \dfrac{\Psi_{1s}(\vec{r_1})\Psi_  {2s}(\vec{r_2})\Psi_{2s}(\vec{r_  1})}{2} \times (\alpha (\vec{r_1}\Beta(\vec{r_2})-\alpha (\vec{r_2}) \Beta (\vec{r_1}))

    The triplet as:
    Thanks I think I understand this a bit better now

    I'm still not sure how to apply it to d) in this question though

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    (Original post by langlitz)
    Thanks I think I understand this a bit better now

    I'm still not sure how to apply it to d) in this question though

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    Im not too sure myself, i am only an A-Level student :P
 
 
 
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