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I Completeness relation

  1. Apr 21, 2017 #1
    Two questions regarding the completeness relation:

    First: I understand that the completeness relation holds for basis vectors such that ## \sum_{j=1}^{m} | n_{j} \rangle \langle n_{j} | =\mathbb{I}##. Does it also hold for unit-normalized sets of state vectors as well, where ## | \phi_{j} \rangle = c_{j} |n_{j}\rangle ##, because ##\sum_{j=1}^{m} |c_{j}|^2=1 ##, such that ## \sum_{j=1}^{m} | \phi_{j} \rangle \langle \phi_{j} | =\mathbb{I}##? I assume it does not hold for non-unit-normalized sets of state vectors.

    Second: For the continuous case, can one define a function ##f(n)=\int_{\mathbb{R}} \phi_{n} dn ##, unit-normalized such that ##\int_{\mathbb{R}} |f(n)|^2 dn =1##, such that ## | f(n) \rangle \langle f(n) | = 1 ##?
     
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  3. Apr 22, 2017 #2

    Orodruin

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    No. The trace of the unit operator is not one so you have a problem there already.
     
  4. Apr 22, 2017 #3
    Thank you for your response. You are right. I made a correction on the first question; the second one still holds:

    First: I understand that the completeness relation holds for basis vectors such that ## \sum_{j=1}^{m} | n_{j} \rangle \langle n_{j} | =\mathbb{I}##. Does it also hold for normalized sets of state vectors as well, where ## | \phi_{j} \rangle = c_{j} |n_{j}\rangle ##, where ##\sum_{j=1}^{m} |c_{j}|^2=\mathbb{I} ##, such that ## \sum_{j=1}^{m} | \phi_{j} \rangle \langle \phi_{j} | =\mathbb{I}##?.

    Second: For a continuous case, can one define a function ##f(n)=\int_{\mathbb{R}} \phi_{n} dn ## , unit-normalized such that ## \int_{\mathbb{R}} |f(n)|^2 dn =1##, such that ##| f(n) \rangle \langle f(n) | = 1 ##?
     
  5. Apr 22, 2017 #4

    Orodruin

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    No. Same reason.
     
  6. Apr 22, 2017 #5
    Not entirely sure to what you are referring. Trace for ## | f(n) \rangle \langle f(n) |## is 1 if n is a scalar. If n a vector, ##| f(n) \rangle \langle f(n) | =\mathbb{I}. ##. Is that the concern?
     
  7. Apr 22, 2017 #6

    Orodruin

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    The trace is always a scalar. The trace of the identity operator in an ##n## dimensional vector space is ##n##. Obviously, your "normalised set of vectors" has trace 1, not ##n##.
     
  8. Apr 22, 2017 #7
    Given ##c_{i}^* c_{j} = \delta_{ij}## such that:
    Where ## | \phi_j \rangle = c_j |n_j \rangle ##:
    ##\sum_{j=1}^{m} | \phi_j \rangle \langle \phi | = \sum_{j=1}^{m} c_j| n_j \rangle \langle n | c_j ^* ##
    ##\sum_{j=1}^{m} c_j| n_j \rangle \langle n | c_j ^* = \sum_{j=1}^{m} | n_j \rangle |c_j|^2 \langle n | ##
    ## \sum_{j=1}^{m} | n_j \rangle |c_j|^2\langle n | = \sum_{j=1}^{m} | n_j \rangle \langle n | = \mathbb{I}##

    Thus:
    Trace##\left(\sum_{j=1}^{m} | \phi_j \rangle \langle \phi |\right)=m##
     
    Last edited: Apr 22, 2017
  9. Apr 23, 2017 #8

    Orodruin

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    This is not compatible with your claim that ##\sum_{i = 1}^m |c_i|^2 = 1##, nor is it compatible with ##\sum_{i=1}^M |c_i|^2 = \mathbb I## (actually, there is no difference between those two statements since the latter applied to a state is the same as multiplying the state by the former). If you are trying to ask whether you can use any complete orthonormal basis to make a completeness relation, the answer is yes by definition.

    Edit: It is also impossible to find ##c_i## such that the relation ##c_i^* c_j = \delta_{ij}## holds. As the components of a matrix, the matrix on the LHS has rank 1 while the one on the RHS has rank m.
     
  10. Apr 23, 2017 #9
    ## c^i c_j = \delta_j^i## then?
     
  11. Apr 23, 2017 #10

    Orodruin

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    Still no. Still an impossible condition.
     
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