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Change of basis

  1. Jan 26, 2015 #1
    We are given the vectors la> = (1,0) and lb> = (0,1) and then a Hamiltonian H which is a 2x2 matrix with 2 on the diagonal entires and zero elsewhere. I am asked to now represent H in the basis of the vectors la'> = 1/sqrt(2)(1,1) and lb'> = 1/sqrt(2)(1,-1), which are also eigenvectors of H because of the degeneracy of the eigenvalues of H.
    I always get confused with these basis-change problems. First of all: What does it even mean that a matrix is represented in a particular basis? Why is H necessarily represented in the basis (1,0) and (0,1)?
    Secondly I know that the basis change equation is:
    H' = SHS^-1 (1)
    But my QM also tells me that the elements of H' can be found by taking the inner products <a'lHla'>, <a'lHlb'> etc. How is this equivalent to (1)?
  2. jcsd
  3. Jan 27, 2015 #2


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    See this FAQ post: https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ [Broken]
    Last edited by a moderator: May 7, 2017
  4. Jan 27, 2015 #3
    Okay but Im still not sure why for an orthonormal basis the basis change equation:
    B = SAS^-1
    Reduces to just calculating B using the equations B_ij = <bil A lbj> , where b1,b2,b3... are the new basis vectors of the new basis expressed in the old basis.
  5. Jan 27, 2015 #4


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    I didn't have time to answer that before. Let U be the linear operator that takes the old basis vectors to the new, i.e. ##e_i'=Ue_i##. I'll start by proving that this U must be unitary (assuming that both bases are orthonormal). For all i,j, we have
    $$\langle e_i,e_j\rangle =\delta_{ij}=\langle e_i',e_j'\rangle =\langle Ue_i,Ue_j\rangle =\langle e_i,U^\dagger U e_j\rangle,$$ and therefore
    $$\langle e_i,(I-U^\dagger U)e_j\rangle =0.$$ This implies that for all j and all vectors x, we have
    $$\langle x,(I-U^\dagger U)e_j\rangle =\left\langle\sum_{i=1}^2 x_i e_i,(I-U^\dagger U)e_j\right\rangle =\sum_{i=1}^2 (x_i)^* \langle e_i,(I-U^\dagger U)e_j\rangle =0.$$ In particular, this holds when ##x=(I-U^\dagger U)e_j##. That implies that for all j, we have ##(I-U^\dagger U)e_j=0##. This implies that ##U^\dagger U=I##.

    Now if we define ##S=U^{-1}## and use that ##U^\dagger=U^{-1}##, we have
    H'_{ij}=\langle e_i',He_j'\rangle = \langle Ue_i,HUe_j\rangle =\langle e_i,U^\dagger HUe_j\rangle =(U^\dagger HU)_{ij} =(SHS^{-1})_{ij}
    \end{align} In other words, the components of ##H## in the primed basis are the same as the components of ##SHS^{-1}## in the unprimed basis.
    Last edited: Jan 27, 2015
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