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I Confused about slater determinant for excited states

  1. Mar 10, 2016 #1
    Hi all, I have a question about Slater Determinant for excited states.

    Let's say we want to construct approximate (2 level) wavefunction of He in some certain state. Since we have two electrons in two level system with spin in consideration, we can construct total of 4 different wavefunctions corresponding to each of the states.

    So by constructing Slater Determinant, we should be able to achieve wavefunctions for each of these 4 different states with anti-symmetry in consideration. However, in order to achieve the "correct" wavefunction of a state (for example, excited singlet state), we sometimes need linear combination of the Slater Determinant. However, in cases like ground state, only single Slater Determinant is required.

    QUESTION:
    How do we know whether a wavefunction constructed simply from Slater Determinant is "correct" or "not correct"? In another words, how do we know if we need linear combination of Slater Determinant or not?
     
    Last edited: Mar 11, 2016
  2. jcsd
  3. Mar 10, 2016 #2
    for the construction of possible wave function- the He wave function should have antisymmetric nature- and the wave function has space and spin descriptions - for Pauli principle to hold perhaps you can have only four combinations of say Psi (space Symmetric/antisymmetric) and Psi(spin symmetric/antisymmetric) combinations.
     
  4. Mar 11, 2016 #3
    I'm sorry, I accidentally wrote 6 states because I forgot to omit ground state. I have edited the OP.
     
  5. Mar 11, 2016 #4
    If some body writes a slater determinant -it means that its a representation of all the states (possible) on the basis of single particle states of the physical system and the elements of the determinant form a linear combination conforming to the properties of physical system-

    as you said in previous post as to 'how you limited the six combinations to four only"
    I think thats the way unphysical states can be done away with leading to correct description
    but it will remain an approximation as electrons may not free to be represented by single particle wave functions -as small perturbations may come/enter due to realistic picture in the Hamiltonian.
     
  6. Mar 11, 2016 #5
    Well I don't really plan to get into Configuration Interaction, yet. I just need very approximate vague picture of the system. When you look through lecture notes of these wavefunction stuff, even if they are doing the very approximate method in deriving wavefunction, they would often give slater determinant representation of the wavefunction, and then tells you that for excited states, you need to combine these slater determinant to get the "correct" form of wavefunction, but never tells you why and how you should do this.

    Some lecture notes are very inconsistent, telling you that Slater Determinant can get you the wavefunction you are looking for, and yet gives a "correct" wavefunction of two level Helium that was derived by linear combination of Slater Determinant without telling you that they did.

    I am really confused when I should use linear combination of Slater Determinants. Sure, if I actually use two electron Hamiltonian, there are one wavefunction that gives energy of singlet state and three wavefunction that gives energy of triplet state and that energy difference between singlet and triplet is two times the exchange energy.. But it is only after I derive such energy that I know that I have used the "correct" wavefunction (because I know as a prior knowledge that result of energy would be this way). Without deriving energy, I cannot determine whether or not the wavefunction I got (and the linear combination of these) are "correct".
     
  7. Mar 11, 2016 #6

    blue_leaf77

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    It is correct if it exactly solves the Schroedinger equation of the atom being considered.
    A state constructed from a single Slater determinant is an exact solution of the Schroedinger equation with the interaction potential removed, but it is not if this potential is retained. Therefore, you need an (infinite) linear combinations of Slater determinants.
     
  8. Mar 11, 2016 #7
    Well of course that is true. But I am only considering approximate solution without Configuration Interactions.
     
  9. Mar 11, 2016 #8

    DrDu

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    You can get a singlet in two ways: Either two electrons with antiparallel spin occupy the same orbital or they occupy different orbitals. In the first case, you can describe the system with a single Slater determinant, in the second case you need two.
    What is probably confusing you is the following: As long as you don't care about electron correlation, a description in terms of a single determinant is correct even for excited states with single electrons in different orbitals. These single determinant states will neither be pure singlet nor triplet states, but this doesn't matter because singlet and triplet are degenerate as long as electron correlation is not taken into consideration.
     
  10. Mar 11, 2016 #9
    Okay, now I am beginning to understand my own confusion.

    So when I see a lecture note that gives two level Helium state energies, and they use the wavefunction in which the appropriate linear combination of Slater Determinant has already been done, are they taking into account of Configuration Interaction without telling the readers that they did? I've seen several lecture notes online that uses the "correct" wavefunctions before they introduce the concept of Configuration Interaction.

    So when taking in account of Electron Correlation (and therefore Configuration Interactions), is the simplest way to find the constant used for linear combination the variational method?
     
  11. Mar 11, 2016 #10

    DrDu

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    This is about symmetry. Symmetry adapted wavefunctions comprise in most cases more than one Slater determinant. Without correlation singlet and triplet are degenerate, but we know from symmetry considerations (here rotational symmetry of the electronic wavefunction), that singlet and triplet will split appart once correlation is taken into account. As the relevant Slater determinants are energetically degenerate, we are free to chose symmetry adapted linear combinations which will remain eigenfunctions once correlation is taken into account and degeneracy lifted.
     
  12. Mar 12, 2016 #11
    Hmm. So it was about Symmetry Adapted Linear Combination? I haven't really studied symmetry and wavefunctions (I have absolutely no quantum chemical/physical background in my study. I am studying this alone because I have no one else to refer to). Is it possible if you could briefly and qualitatively explain this?

    How would symmetry apply with the two level Helium case? Did the two level (two orbitals) of helium must be specified what type of orbital symmetry it had, before I would actually start getting on with wavefunction and combination and stuff? Because if symmetry was how I would know what Slater determinant has to be combined and what not, then I would need symmetry information on the Helium orbitals prior to doing anything.

    However, looking through lecture notes, this is the first time I have ever heard about symmetry needing to be considered for two level Helium case.
     
  13. Mar 12, 2016 #12

    DrDu

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    I didn't wanted to distract you from your main line of study. Group theory is quite a formidable topic to learn on its own.
    In the context of your question I meant symmetry in the sense that singlet and triplet are just names given to the angular momentum quantum number which itself is a consequence of rotational symmetry. That symmetry adapted wavefunctions are usually composed of several Slater determinants is a general feature which pertains to any kind of symmetry.

    If you really want to delve into this a good reference is Michael Tinkham, Group theory and quantum mechanics.
    A good reading involving quantum chemistry and also some group theory relevant in this respect is the book by Roy McWeeny, Methods of molecular quantum mechanics.
     
  14. Mar 14, 2016 #13
    When you say "angular momentum quantum number", do you mean total angular momentum quantum number? If so, then orbital and spin momentum must be taken into account, right?

    I think I'm starting to understand this a little more. but I still have some holes that's keeping me from nodding my head. I'm sorry that I am so stupid.

    How can I apply total angular momentum quantum number to knowing how to combine Slater determinants for specific states? Sorry, I am just clueless.
     
  15. Mar 14, 2016 #14

    DrDu

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    In nonrelativistic quantum mechanics, spin and orbital angular momentum are decoupled. This means that both span on themselves a representation of the rotation group which may then be combined to form a representation of total angular momentum.
    Slater determinants are but poorly suited to reflect this structure, as orbitals and spins are mingled in them.
    There are methods to construct first representations of both the spin and orbital wavefunctions of appropriate symmetry and combine them afterwards into a wavefunction for full angular momentum. This is called the unitary group approach in quantum chemistry and the book by McWeeny has some information on it.
    Warning, this is quite specialized stuff, and easily may lead you astray!

    Maybe this talk is also helpful
    http://www.univie.ac.at/columbus/workshops/argonne2005/pdf/Shavitt_Argonne_UGA.pdf
     
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