Questions and Answers

0 Like 0 Dislike

George Pau

defining the 3D electron density


In the reference cited, Nanowire uses the NEGF formalism and the 3D density is expressed in terms of that formalism. Could someone tell me what the equation will be like in terms of wavefunctions? I do not think it is very different from the reference but I would like to see what the exact form is like if one decide not to use NEGF formalism.

Thanks, George

Report abuse

Chosen Answer

  1. 0 Dislike

    Saumitra Raj Mehrotra


    I could not understand the reference you refer to. However to answer about calculating 3D electron density.Nanowire mode space approach breaks the 3D problem into 2D Schodinger and 1D NEGF transport problem. On solving 1D NEGF we get 1D electron density (Ne1D).Electron density in terms of wavefunction can now be written as,

    Ne3D=Sum over mode m(Ne1D.{psi_m.psi*_m)

    A more correct representation should be (code being updated), Ne3D=Sum over mode m over mode n


    I have written nth mode here which can be understood as nth subband.

    I hope it helps

    thank, Saumitra

    Reply Report abuse

    Please login to answer the question.

1 Responses

  1. 0 Like 1 Dislike

    Daniel Lee Whitenack

    Hello George,

    If you have the wavefunction (call it phi) for one particle in one dimension, the probability density is defined as the square modulus of phi:

    n(x) = |phi(x)|^2

    Where this should satisfy the condition that if you integrate over all space you just get 1. (phi should be normalized) In 3D, say you have the one particle wavefunction phi(x,y,z). Again the density is interpreted as just |phi(x,y,z)|^2.

    Now for many particles you are going to have the wavefuntion as a function of all of the particle coordinates:


    Where r1, r2, r3 specify the location of particle 1, 2, 3, etc. However, the density is always going to be a funtion of just three spatial coordinates. So now the density is defined as

    n® = N int |phi(r,r2,r3,…)|^2 dr2 dr3 …

    Where N is the number of particles. Therefore if your wavefunction is normalized, the density integrated over all space is N.

    Note I have ignored spin. If you want to include spin, you have to include a sum over all spin states. (these are discrete as opposed to the continuous position)

    Hope that helps at least a little, Daniel

    Reply Report abuse

    Please login to answer the question.