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Identity operator

  1. Dec 6, 2012 #1
    I was wondering about this: The identity operator writes a vector in the basis that is used to express the identity operator:
    1 = Ʃlei><eil
    But if you are to apply it to a vector in a given basis A should the lei> then be expressed in terms of their own basis or in terms of A?
  2. jcsd
  3. Dec 6, 2012 #2
    It doesn't really matter as long as you can calculate the inner products.
    Maybe an example might help you. (Since I'm not a physicist, I don't know what the correct notation is for bra-ket, but I'll try).

    Take the following basis
    [tex]e_1=(1,0,0),~~ e_2=(1,1,0),~~ e_3= (1,1,1)[/tex]
    Take the inner product
    This formulation of the inner product is going to be dependent on the basis of course.

    Anyway, lets calculate

    where a is given by [itex](1,2,3)[/itex]. Then we get

    So [itex]|e_2><e_2|a>= 3|e_2>=(3,3,0)[/itex].

    Now what if a is given in terms of another basis? So what if


    One possibility is to write a in the normal basis, then we get


    and then calculate as usual:


    Another possibility is to write the inner product in terms of the basis [itex]\{e_1,e_2,e_3\}[/itex]. That is, find a formula for


    and then you can calculate [itex]|e_2><e_2|a>[/itex] without converting to another basis.

    Does this answer your question?
  4. Dec 6, 2012 #3


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    Science Advisor

    Not at all. You just have to be consistent about how you do your inner products. Lets say you have vector |b> expressed as linear combination of |ai> that are the basis of A. In other words, |b> = [itex]\displaystyle\small \sum_i[/itex]bi|ai>, where bi=<b|ai> are the components of |b> in basis A. Now you want to apply unit operator consisting of your |ei> vectors. Fine. You get I|b> = [itex]\displaystyle\small \sum_{ij}[/itex]|ei><ei|aj><aj|b>. The inner products <ei|aj> effectively take care of matching one basis to the other. In matrix form, these will form the transformation matrix from one basis to another.

    Edit: micromass beat me to it.
  5. Dec 6, 2012 #4
    There is one caveat, the completeness of basis only applies to orthonormal basis. If your basis are not mutually orthogonal, the sum of their outer product won't come to unity. But in QM, this is usually satisfied.
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