Can the Bernstein concentration inequality bound for matrices be improved when we have a sequence of diagonal matrices?

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I am using the Bernstein inequality for the sum of independent random Hermitian matrices presented in the following paper by Joel A.Tropp (page 96, theorem 6.6.1) which states the following:

Consider a finite sequence $\{X_k\}$ of independent, random, Hermitian matrices with dimension $d$.

Assume that $E(X_k) =0$ and $ \lambda_{max}(X_k) \leq L$ for each index $k$. Introduce the random matrix $Y =\sum_{k} X_k$.

Let $v (Y)$ be the matrix variance statistic of the sum: $v(Y)= \|EY^2\| = \|\sum_{k} EX^{2}_{k}\|$

Then for all $t \geq 0$

$$P\{\lambda_{max}(Y) \geq t\} \leq d·\exp \left(\frac{−t^2/2}{v(Y)+Lt/3}\right).$$

For the problem I am working on, the sequence of random matrices $\{X_k\}$ are also diagonal matrices, I am therefore wondering if that information can be used to improve the bound of the inequality.

Also what if each of the matrices where not only diagonal but also only had one non-zero diagonal element?

In other word what if the matrix $X_k$ has a value on the $kk$ diagonal position and zeros everywhere else for all $k \in \{1,..,d\}$.

Thank you