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Eigenvalue of the system and the one of its part

  1. Jan 21, 2014 #1
    Dear all,
    I have a problem about the eigenvalue of the system and the eigenvalue of the part of the system.
    For example,in the theory of the APW method,the space of the primitive cell is divided into muffin-tin (MT) spheres and the interstitial region (IR). In order to gain the eigenvalue and the eigenfunction of the primitive cell,we usually assume the eigenvalue [itex]E^{'}[/itex] of MT ,then determine its eigenfunction ,and then we sovle the eigenvalue [itex]E[/itex] and eigenfunction of the entire primitive cell.
    Here,the eigenvalue [itex]E[/itex] of the system must belong to the the eigenvalue [itex]E^{'}[/itex] of MT (it means [itex]E \subset E^{'}[/itex],mathematically),isn't it ? Can anyone explain the physical meaning of the relationship ?


    Last edited: Jan 21, 2014
  2. jcsd
  3. Jan 21, 2014 #2
    To make it clearer,let us see the secular determinant as follows:
    [tex]DET|H^{ij}+EΔ^{ij}+S^{ij}|=0[/tex] (1)
    where [itex]E[/itex] is the variational energy of the cell (i.e.,the system above I mentioned). In order to determine [itex]E[/itex] , we usually consider (1) as the three terms in the following combinations:
    [tex]H^{ij}_{I}+EΔ^{ij}_{I}[/tex] [tex]H^{ij}_{II}+EΔ^{ij}_{II}[/tex] [tex]S^{ij}[/tex] (2)
    where,I stands for MT,II for IR,and S for the suface term.The first term of the (2) can be written as:
    [tex]H^{ij}_{I}+EΔ^{ij}_{I}=(E^{'}-E)Δ^{ij}_{I}[/tex] (3)
    In order to make this term zero and thus simplify the APW matrix elements, “it was originally proposed (J.C. Slater,1937) that the energy E' be taken to be the same as the characteristic energy which satisfies (1).” (Further details about the APW method may be found in the book by L.T. Loucks,1967)
    ——What does it mean ? In (1), the characteristic energy [itex]E[/itex] is unknown and is need to be solved, but how does [itex]E^{'}= E [/itex] in MT ?
    On the other hand,the suitable trial wave function is determined through the radial Schrödinger quation. There,we must give the trial [itex]E^{'} [/itex] value,then the radial function is solved.But if we let [itex]E=E^{'} [/itex] in MT , why do we solve (1) ? Wouldn't be different between the [itex]E[/itex] in MT and [itex]E[/itex] in IR, so we only need to solve the equation
    [tex]DET|H^{ij}_{II}+EΔ^{ij}_{II}+S^{ij}|=0[/tex] (4)
    which usually appears in some literatures.Furthermore ,it seems that the eigenvalue of the cell belongs to the one of IR and the surface term which I mentioned above.
    Last edited: Jan 21, 2014
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