Completeness relation

  • #1
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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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  • #2
Orodruin
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No. The trace of the unit operator is not one so you have a problem there already.
 
  • #3
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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
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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?
 
  • #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##.
 
  • #7
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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##
 
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  • #8
Orodruin
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Given ##c_{i}^* c_{j} = \delta_{ij}## such that:
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.
 
  • #9
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## c^i c_j = \delta_j^i## then?
 
  • #10
Orodruin
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## c^i c_j = \delta_j^i## then?
Still no. Still an impossible condition.
 

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