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Thanks!

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Thanks!

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The overall state of two electrons must be anti-symmetric. The triplet spin state is symmetric, so that must be combined with an anti-symmetric spatial wavefunction. The singlet spin state, on the other hand, is anti-symmetric, so it must be combined with a symmetric spatial wavefunction in order for the overall state of the system to be anti-symmetric.

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In both cases, the state of the system is the product of one symmetric and one anti-symmetric component, giving overall an anti-symmetric state.

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Ok thanks, that makes sense. Would you mind telling me then what the symmetric and antisymmetric spatial wf's look like in this case?The overall state of two electrons must be anti-symmetric. The triplet spin state is symmetric, so that must be combined with an anti-symmetric spatial wavefunction. The singlet spin state, on the other hand, is anti-symmetric, so it must be combined with a symmetric spatial wavefunction in order for the overall state of the system to be anti-symmetric.

In both cases, the state of the system is the product of one symmetric and one anti-symmetric component, giving overall an anti-symmetric state.

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blue_leaf77

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That depends on the Hamiltonian operator of the system.Would you mind telling me then what the symmetric and antisymmetric spatial wf's look like in this case?

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Sure, but isn't the Hamiltonian specified by looking at singlet / triplet states?That depends on the Hamiltonian operator of the system.

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blue_leaf77

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No, the Hamiltonian does not depend on any particular state. You start everything from the Schroedinger equation, in which in turn one needs to know the Hamiltonian operator.

By the way, the separation between the spatial and spin wavefunctions only applies for non-relativistic atoms, i.e. atoms with ##Z<20## because in this situation the spin-orbit coupling can be neglected. For heavy atoms, the Hamiltonian contains terms involving spins, which is due to the spin-orbit effect, and therefore in general ##[H,S_z]\neq 0##. In this case the wavefunctions cannot be separated into spatial and spin parts.

A rather intricate aspect about the spin wavefunctions of many electrons is that one cannot form an anti-symmetrical wavefunction when there are more than two electrons. This means for atoms having more than two electrons, even if the spin orbit effect is not significant, the total wavefunction does not separate into spatial and spin parts. In general, for such systems, the eigenfunctions of spin and of Hamiltonian are combined to form the so-called Slater determinant and the total wavefunction is equal to a Slater determinant (or a linear combination of it - since the eigenfunctions of Hamiltonian is infinitely numerable, there are also infinite possible Slater determinants). It follows from the interchange nature of a determinant, this kind of wavefunction is guaranteed to be antisymmetric.

By the way, the separation between the spatial and spin wavefunctions only applies for non-relativistic atoms, i.e. atoms with ##Z<20## because in this situation the spin-orbit coupling can be neglected. For heavy atoms, the Hamiltonian contains terms involving spins, which is due to the spin-orbit effect, and therefore in general ##[H,S_z]\neq 0##. In this case the wavefunctions cannot be separated into spatial and spin parts.

A rather intricate aspect about the spin wavefunctions of many electrons is that one cannot form an anti-symmetrical wavefunction when there are more than two electrons. This means for atoms having more than two electrons, even if the spin orbit effect is not significant, the total wavefunction does not separate into spatial and spin parts. In general, for such systems, the eigenfunctions of spin and of Hamiltonian are combined to form the so-called Slater determinant and the total wavefunction is equal to a Slater determinant (or a linear combination of it - since the eigenfunctions of Hamiltonian is infinitely numerable, there are also infinite possible Slater determinants). It follows from the interchange nature of a determinant, this kind of wavefunction is guaranteed to be antisymmetric.

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