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Changing basis

  1. Apr 18, 2010 #1

    I'm reading Shankar's Principles of QM and I find it not very clear on how exactly should I change basis of operator. I know how to change basis of a vector so can I treat the columns of operator matrix as vectors and change them? Or is it something else?
  2. jcsd
  3. Apr 26, 2010 #2


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    It is something a little different. Let [tex] v_1 [/tex] denote a vector represented in basis 1. Then to represent this same vector in terms of a different basis, basis 2, we need to find a matrix [tex]T_{1:2} [/tex] that maps any vector representation from basis 1 to basis 2. Thus if we let [tex] v_2 [/tex] denote that vector represented in basis 2, then
    [tex] v_2 = T_{1:2} \, v_1 [/tex].
    This means that
    [tex] v_1 = T_{1:2}^{-1} \, v_2 [/tex]
    so the matrix that maps a vector representation from basis 2 to basis one is
    [tex] T_{2:1}=T_{1:2}^{-1} [/tex].

    Now, if we have a matrix representation of an operator in basis 1, say [tex]A_1 [/tex], then it takes a vector represented in basis 1 and maps it to a different vector represented in basis 1. For our example let
    [tex]y_1 = A_1 v_1 [/tex].
    So if we want to represent y in basis 2 we have,
    [tex]y_2 = T_{1:2} y_1 = T_{1:2} A_1 v_1 = T_{1:2} A_1 T_{2:1} v_2 [/tex].
    Hence, if we want to represent the operator in basis 2, the matrix representation must be
    [tex]A_2 = T_{1:2} A_1 T_{2:1} = T_{1:2} A_1 T^{-1}_{1:2} [/tex],
    and we have
    [tex]y_2 = A_2 v_2 [/tex]
    as required. If you think about what is happening, it should be easy to remember.

    Note that most linear algebra books will cover this.

    Last edited: Apr 26, 2010
  4. Apr 28, 2010 #3
    Thank you very much. Maybe I should do some linear algebra book first and then return to Shankar. Can you advise some good books about the subject?
  5. Apr 28, 2010 #4


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    "Linear algebra done right", by Sheldon Axler.

    But you should start with this post about the relationship between linear operators and matrices.
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