Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 ##?
     
  2. jcsd
  3. Apr 22, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    Still no. Still an impossible condition.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Completeness relation
  1. Completeness relation (Replies: 4)

  2. Completeness relation (Replies: 1)

  3. Completeness relation (Replies: 7)

Loading...