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Simultaneous diagonalization

  1. Dec 22, 2009 #1
    hello,
    i am having some trouble understanding simultaneous diagonalization. i have understood the proof which tells us that two hermitian matrices can be simultaneously diagonalized by the same basis vectors if the two matrices commute. but my book then shows a proof for the case when the matrices are both degenerate. it is basically going over my head, will anyone please help me with a good proof/ sketch?
    my book uses block diagonal elements for the proof which i cannot understand....:confused:
    thank you.
     
  2. jcsd
  3. Dec 22, 2009 #2
    What do you mean by "the matricies are both degenerate"?
     
  4. Dec 23, 2009 #3
    i mean that multiple eigenvectors for the same eigenvalue.
    ie. the characteristic equation has multiple roots like 3,3,1 etc.
    the characteristic equation of both matrices have this property.

    am i clear now,sir?
     
  5. Dec 23, 2009 #4

    Avodyne

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    Let [itex]|i\rangle[/itex] be an eigenstate of [itex]A[/itex] with eigenvalue [itex]a_i[/itex]. Then, sandwiching [itex]AB-BA=0[/itex] between [itex]\langle i|[/itex] and [itex]|j\rangle[/itex], we get

    [tex](a_i-a_j)\langle i|B|j\rangle = 0.[/tex]

    So, if [itex]a_i\ne a_j[/itex], then [itex]B_{ij}\equiv \langle i|B|j\rangle = 0[/itex]. If all the eigenvalues of [itex]A[/itex] are different, then [itex]a_i\ne a_j[/itex] whenever [itex]i\ne j[/itex], and so all the off-diagonal elements of [itex]B_{ij}[/itex] are zero. Thus [itex]B[/itex] is diagonal in the same basis as [itex]A[/itex].

    But if two or more eigenvalues of [itex]A[/itex] are the same (say [itex]a_1=a_2=\ldots=a_N[/itex]) then we do not know anything about [itex]B_{ij}[/itex] for [itex]i,j = 1,\ldots,N[/itex].

    Let [itex]b[/itex] be the [itex]N\times N[/itex] hermitian matrix with matrix elements [itex]B_{ij}[/itex], [itex]i,j = 1,\ldots,N[/itex]. We can diagonalize this matrix with a unitary transformation, [itex]b=UdU^\dagger[/itex], where [itex]d[/itex] is diagonal. (It does not matter whether or not any of the diagonal elements of [itex]d[/itex] are equal.) Now define new basis states [itex]|\tilde i\rangle[/itex]; these are the same as the old basis states for [itex]i>N[/itex], and for [itex]1\le i\le N[/itex],

    [tex]|i'\rangle = \sum_{j=1}^N U_{ij}|j\rangle.[/tex]

    Now the [itex]|i'\rangle[/itex] states are eigenstates of both [itex]A[/itex] and [itex]B[/itex]. For [itex]i=1,\ldots,N[/itex], the eigenvalue of [itex]A[/itex] is [itex]a_1[/itex], and the eigenvalue of [itex]B[/itex] is [itex]d_i[/itex].
     
  6. Dec 28, 2009 #5
    many many thanks avodyne. i am just starting this subject, so i am inexperienced . but i thought that the first step will give us (ai*-aj).
    also please tell me how i can be sure that the new basis,' i ' prime will still remain the basis of A after the unitary passive transformation that diagonalises B?
    please dont be angry with me. i am slow and take much time to understand things.:frown:
     
  7. Dec 28, 2009 #6

    Avodyne

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    Since A is hermitian, the eigenvalues are real.
    We are making the transformation only among the [itex]N[/itex] states that all have the same eigenvalue of [itex]A[/itex]. In this subspace, [itex]A[/itex] can be written as the number [itex]a_1[/itex] (the eigenvalue of [itex]A[/itex]) times the [itex]N\times N[/itex] identity matrix [itex]I[/itex]. A unitary transformation in this subspace therefore leaves [itex]A[/itex] unchanged.
     
  8. Dec 28, 2009 #7
    understood. thanks.:biggrin:
     
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