How to prove that the trace of a density matrix is $1$?

Question

Equation 2 gives the following proof:

$$ \text{Tr}[\rho] = \sum_x \langle x\vert \rho\vert x\rangle = \sum_x \langle x\vert \sum_i p_i\vert \psi_i\rangle \langle \psi_i\vert\vert x\rangle = \sum_i p_i\sum_x \vert \langle \psi_i\vert x\rangle \vert^2 = \sum_i p_i = 1. $$

I wonder how they got from $\sum_x \langle x\vert \sum_i p_i\vert \psi_i\rangle \langle \psi_i\vert\vert x\rangle$ to $\sum_i p_i\sum_x \vert \langle \psi_i\vert x\rangle \vert^2$. When I did the math, I got $\sum_x \sum_i p_i \langle x\vert \vert \psi_i\rangle \langle \psi_i\vert \vert x\rangle$. Is it correct?

Answer 1

I wonder how they got from $\sum_x \langle x\vert \sum_i p_i\vert \psi_i\rangle \langle \psi_i\vert\vert x\rangle$ to $\sum_i p_i\sum_x \vert \langle \psi_i\vert x\rangle \vert^2$.

When I did the math, I got $\sum_x \sum_i p_i \langle x\vert \vert \psi_i\rangle \langle \psi_i\vert \vert x\rangle$. Is it correct?

Yes. The meanings of the following three things are all the same: $$ \langle x||\psi_i\rangle \equiv \langle x|\psi_i\rangle \equiv \psi_i(x) $$

Similarly: $$ \langle \psi_i||x\rangle \equiv \langle \psi_i|x\rangle = \langle x|\psi_i\rangle^* = \psi_i(x)^* $$

So you can re-write: $$ \langle x||\psi_i\rangle\langle \psi_i||x\rangle = \langle x|\psi_i\rangle\langle x|\psi_i\rangle^* \equiv |\langle x|\psi_i\rangle|^2\;, $$ by definition of the absolute square: $$ |z|^2 = z z^* $$