Clarifying Bra-Ket Notation: Orthonormal Bases

Question

I was asked to find the trace of $(A \in M_{n \times n})$, the matrix that can be written in the form:$$A=\frac{1}{n} \sum_{r, \, q \, = \, 1}^n (-1)^{r+q}|r \rangle \langle q| \quad ,$$ where {$|r \rangle$} is an orthonormal basis. I know that $(\langle r |q \rangle =\delta_{r, \, q})$ but I don't know what truly means $(|r \rangle \langle q|)$! In particular, I know the completeness relation: $$I=\sum_{j=1}^{n} |e_j \rangle \langle e_j| \quad ,$$ where is again an orthonormal basis; but, I know also that not every basis has that property.. Instead, it seems the case; in fact the solution gives me: $$\text{tr}(A)=\frac{1}{n} \sum_{r \, = \, 1}^n (-1)^{r+r}|r \rangle \langle r| = 1$$ Aside the abuse of notation, I think that the right solution should be: $$\require{cancel}\text{tr}(A)=\text{tr$\left(\frac{1}{n} \sum_{r \, = \, 1}^n (-1)^{r+r}|r \rangle \langle r|\right)$}=\frac{1}{n} \, \text{tr$\left(\sum_{r \, = \, 1}^n \cancel{(-1)^{2r}}|r \rangle \langle r|\right)$} = \frac{1}{n} \, \text{tr$\left(I\right)$}=\frac{1}{\cancel{n}}\cdot \cancel{n}=1$$ at this point what is really unclear to me is the step: $$\left[A=\frac{1}{n} \sum_{r, \, q \, = \, 1}^n (-1)^{r+q}|r \rangle \langle q|\right] \longmapsto \left[ \text{tr}(A)= \text{tr$\left(\frac{1}{n}\sum_{r \, = \, 1}^n |r \rangle \langle r|\right)$} \right]$$ where the second sum has gone? (how can this step be legitimate?)

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Note: I already have been able to show that such a matrix is idempotent $(A^2 = A)$.

Answer 1

$\underline{\text{Hint:}}$ you can use the formula $$\text{tr}A=\sum_{p=1}^n\langle p|A|p\rangle$$ which essentially picks up all the diagonal elements of the operator's $A$ matrix form. Substituting $A$ (and taking into consideration that you are working with an orthonormal basis, $\langle r|s\rangle=\delta_{rs}$) will yield the result you have derived, which I believe is true...

Answer 2

The trace is a number, and the sum of the diagonal elements so take the diagonal elements $\langle e_j\vert A\vert e_j\rangle$ and sum over them: \begin{align} \hbox{Tr}(A)&=\sum_{j}\langle e_j\vert A\vert e_j\rangle\\ &=\sum_{j}\sum_{r,q}A_{rq}\langle e_j\vert r\rangle\langle q\vert e_j\rangle\\ &= \sum_{j}\sum_{r,q}A_{rq}\delta_{jr}\delta_{qj} \end{align} where $\langle e_j\vert r\rangle=\delta_{jr}$ has been used.

Answer 3

$$\require{cancel}\text{tr}(A) = \sum_{k=1}^n \langle k|A| k \rangle = \sum_{k=1}^n \frac{1}{n} \langle k \left| \sum_{r,q} (-1)^{r+q} |r \rangle \langle q | \right| |k \rangle = \frac{1}{n} \sum_{k,r,q} (-1)^{r+q} \langle k | r \rangle \langle q | k \rangle = \frac{1}{n} \sum_{k=1}^n \cancelto{1}{(-1)^{2k}}=\frac{1}{\cancel{n}} \cdot \cancel{n} = 1$$