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:$$A=\frac{1}{n} \sum_{r, \, q \, = \, 1}^n (-1)^{r+q}|r \rangle \langle q|$$ where {$|r \rangle$} is an orthonormal basis. I know that $(\langle r |q \rangle =\delta_{r, \, q})$ but I don't know what $(|r \rangle \langle q|)$ truly means! In particular, I know the completeness relation: $$I=\sum_{j=1}^{n} |e_j \rangle \langle e_j|$$ where is again an orthonormal basis; but I also know 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 from 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 has the second sum gone? (rhow can this step be legitimate?)

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

Answer 1

Here's one solution:

$$ \begin{align} \operatorname{tr}\left[\frac{1}{n} \sum_{r, q = 1}^n (-1)^{r+q}|r \rangle \langle q| \right] &= \frac{1}{n}\operatorname{tr}\left[ \sum_{r, q = 1}^n (-1)^{r+q}|r \rangle \langle q| \right] \\ & = \frac{1}{n} \sum_{r, q = 1}^n \operatorname{tr}\left[(-1)^{r+q}|r \rangle \langle q|\right] \\ & = \frac{1}{n} \sum_{r, q = 1}^n (-1)^{r+q}\operatorname{tr}\left[|r \rangle \langle q|\right] \\ & = \frac{1}{n} \sum_{r, q = 1}^n (-1)^{r+q}\langle q|r\rangle \\ &= \frac{1}{n} \sum_{r, q = 1}^n (-1)^{r+q}\delta_{q,r} = \frac{1}{n} \sum_{r = 1}^n (-1)^{r+r} = \frac nn = 1. \end{align} $$

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General proof that $\operatorname{tr}(\langle x| y \rangle) = \langle y | x \rangle$: first, we need a definition of trace in bra-ket notation. For a finite-dimensional Hilbert space $\mathcal H$ and an operator $A : \mathcal H \to \mathcal H$, we define the trace of $A$ to be the quantity $$ {\rm tr}(A) = \sum_{r=1}^n \langle r|A|r \rangle $$ I claim that this quantity is the same regardless of which orthonormal basis is chosen. Indeed, suppose that $|\psi_1\rangle,\dots,|\psi_n\rangle$ is another orthonormal basis. It follows that $$ \begin{align} \sum_{r=1}^n \langle \psi_r|A|\psi_r \rangle &= \sum_{r=1}^n \left\langle \psi_r\left|\left(\sum_{s=1}^n |s\rangle \langle s|\right) A \left(\sum_{t = 1}^n |t\rangle \langle t|\right)\right|\psi_r \right\rangle \\ & = \sum_{r,s,t = 1}^n \langle \psi_r | s \rangle \langle s|A|t \rangle \langle t | \psi_r \rangle \\ & = \sum_{r,s,t = 1}^n \langle t | \psi_r \rangle \langle \psi_r | s \rangle \langle s|A|t \rangle \\ & = \sum_{s,t = 1}^n \left \langle t \left |\sum_{r = 1}^n |\psi_r \rangle \langle \psi_r| \right| s \right \rangle \langle s |A| t\rangle \\ & = \sum_{s,t = 1}^n \langle t|s \rangle \langle s |A| t\rangle = \sum_{s = 1}^n \langle s|A|s \rangle = \operatorname{tr}(A). \end{align} $$ With that, it is easy to see that for any $|x\rangle,|y\rangle$, we have $\operatorname{tr}(|x \rangle \langle y |) = \langle y|x \rangle$. First, we note that in the case that $|x \rangle$ is the zero-vector, this holds trivially. For the case that $|x \rangle$ is non-zero, construct an orthonormal basis $|\psi_1\rangle,\dots,|\psi_n\rangle$ such that $\psi_1$ is parallel to $|x \rangle$. That is, we have $|\psi_1\rangle = \frac{1}{\sqrt{\langle x|x \rangle}}|x \rangle$, and $\langle \psi_r|x \rangle = 0$ for $r \neq 1$. We have $$ \begin{align} \operatorname{tr}(|x \rangle \langle y |) &= \sum_{r = 1}^n \langle \psi_r |x \rangle \langle y|\psi_r \rangle \\ & = \langle \psi_1|x\rangle \langle y| \psi_1 \rangle \\ & = \left[\frac 1{\sqrt{\langle x|x\rangle}}\right]^2 \langle x|x \rangle \langle y|x \rangle = \langle y| x \rangle, \end{align} $$ which is what we wanted.

Answer 2

NOTE) This answer will act as a direct comment to @BenGrossmann.

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I have just two points to discuss about: in your (second-and-third)-last steps are you implying that $\left[ \text{tr($|r \rangle\langle q|$)} = \langle q|r \rangle \delta_{q,r}\right]$? In this case one can omit the sum wrt $(q)$, since the only surviving terms are the ones with $(q=r)$. Still i will remain with: $$\left[ \text{tr($A$)} = \frac{1}{n} \sum_{r=1}^n (-1)^{r+r} \langle r|r \rangle\right] = 1$$ that is different from what the book shows: $$\left[ \text{tr($A$)} = \frac{1}{n} \sum_{r=1}^n (-1)^{r+r} |r \rangle \langle r| \right] = 1$$ Again i think that it is just a typo (and with your answer it seems clearer than ever). What do you think about? The second thing that i want to clarify is if you know some other smart trick to do the job.. maybe something that relies on property of orthonormal basis and bra-ket notation only. Thank you.