Completeness Relation: 2 Questions Answered

[redtree]

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 $| f(n) \rangle \langle f(n) | = 1$?

 

[Orodruin]

No. The trace of the unit operator is not one so you have a problem there already.

 

[redtree]

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 $| f(n) \rangle \langle f(n) | = 1$?

 

[Orodruin]

No. Same reason.

 

[redtree]

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?

 

[Orodruin]

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 $n$.

 

[redtree]

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$\left(\sum_{j=1}^{m} | \phi_j \rangle \langle \phi |\right)=m$

 

[Orodruin]

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.

 

[redtree]

$c^i c_j = \delta_j^i$ then?

 

[Orodruin]

$c^i c_j = \delta_j^i$ then?

Still no. Still an impossible condition.