[proofplan]
In finite dimension, a density operator is a self-adjoint positive semidefinite operator with trace $1$. We verify exactly these three properties. First we show that each rank-one projection $P_{\psi_j}$ is self-adjoint, positive semidefinite, and has trace $1$. Then we use linearity of adjoints, quadratic forms, and trace to pass these properties to the convex combination $\rho=\sum_{j=1}^n p_jP_{\psi_j}$.
[/proofplan]
[step:Verify the rank-one projections are self-adjoint and positive]
Fix $j \in \{1,\dots,n\}$. The statement defines the [linear map](/page/Linear%20Map) $P_{\psi_j}:H \to H$ by $x\mapsto (x,\psi_j)_H\psi_j$. For all $x,y \in H$, this definition gives
\begin{align*}
(P_{\psi_j}x,y)_H = ((x,\psi_j)_H\psi_j,y)_H = (x,\psi_j)_H(\psi_j,y)_H.
\end{align*}
Using conjugate-linearity in the second variable,
\begin{align*}
(x,P_{\psi_j}y)_H = (x,(y,\psi_j)_H\psi_j)_H = \overline{(y,\psi_j)_H}(x,\psi_j)_H = (\psi_j,y)_H(x,\psi_j)_H.
\end{align*}
Thus $(P_{\psi_j}x,y)_H=(x,P_{\psi_j}y)_H$ for all $x,y \in H$, so $P_{\psi_j}$ is self-adjoint. Also, for every $x \in H$,
\begin{align*}
(P_{\psi_j}x,x)_H = (x,\psi_j)_H(\psi_j,x)_H = |(x,\psi_j)_H|^2 \geq 0.
\end{align*}
Hence $P_{\psi_j}$ is positive semidefinite.
[guided]
Fix $j \in \{1,\dots,n\}$. We need to check the operator-theoretic properties directly from the formula for the linear map $P_{\psi_j}:H\to H$ given by $x \mapsto (x,\psi_j)_H\psi_j$.
Self-adjointness means that for all $x,y \in H$,
\begin{align*}
(P_{\psi_j}x,y)_H = (x,P_{\psi_j}y)_H.
\end{align*}
No row separator is needed in this one-line display. The left-hand side is
\begin{align*}
(P_{\psi_j}x,y)_H = ((x,\psi_j)_H\psi_j,y)_H = (x,\psi_j)_H(\psi_j,y)_H,
\end{align*}
because the [inner product](/page/Inner%20Product) is linear in the first variable. The right-hand side is
\begin{align*}
(x,P_{\psi_j}y)_H = (x,(y,\psi_j)_H\psi_j)_H.
\end{align*}
Since the scalar is in the second argument, conjugate-linearity gives
\begin{align*}
(x,(y,\psi_j)_H\psi_j)_H = \overline{(y,\psi_j)_H}(x,\psi_j)_H.
\end{align*}
By conjugate symmetry, $\overline{(y,\psi_j)_H}=(\psi_j,y)_H$, so the two expressions agree. Therefore $P_{\psi_j}$ is self-adjoint.
Positive semidefiniteness means that $(P_{\psi_j}x,x)_H \geq 0$ for every $x \in H$. Substituting $y=x$ in the same computation gives
\begin{align*}
(P_{\psi_j}x,x)_H = (x,\psi_j)_H(\psi_j,x)_H = |(x,\psi_j)_H|^2 \geq 0.
\end{align*}
Thus $P_{\psi_j}$ is positive semidefinite.
[/guided]
[/step]
[step:Compute the trace of each rank-one projection]
Fix $j \in \{1,\dots,n\}$. Define the orthogonal complement of $\psi_j$ by
\begin{align*}
\{\psi_j\}^{\perp}=\{x\in H:(x,\psi_j)_H=0\}.
\end{align*}
Since $\psi_j$ is a unit vector and $H$ is finite-dimensional, choose a basis of $\{\psi_j\}^{\perp}$ and apply the [Gram-Schmidt orthonormalisation](/theorems/542) procedure inside $\{\psi_j\}^{\perp}$. This gives an [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_m)$ of $H$ with $e_1=\psi_j$, where $m=\dim H$. For a finite-dimensional [Hilbert space](/page/Hilbert%20Space), the trace of a linear operator is the trace of its matrix in any orthonormal basis; equivalently, it is the sum of the diagonal matrix coefficients in that basis. Applying this definition to the orthonormal basis $(e_1,\dots,e_m)$ gives
\begin{align*}
\operatorname{tr}(P_{\psi_j}) = \sum_{k=1}^{m} (P_{\psi_j}e_k,e_k)_H.
\end{align*}
For $k=1$, $P_{\psi_j}e_1=P_{\psi_j}\psi_j=(\psi_j,\psi_j)_H\psi_j=\psi_j$, so $(P_{\psi_j}e_1,e_1)_H=1$. For $k \geq 2$, orthogonality gives $(e_k,\psi_j)_H=0$, hence $P_{\psi_j}e_k=0$. Therefore
\begin{align*}
\operatorname{tr}(P_{\psi_j}) = 1.
\end{align*}
[/step]
[step:Pass the density-operator properties to the convex combination]
The operator $\rho:H \to H$ is linear because it is a finite linear combination of linear maps. Since each $P_{\psi_j}$ is self-adjoint and each $p_j$ is real,
\begin{align*}
\rho^* = \left(\sum_{j=1}^{n} p_jP_{\psi_j}\right)^* = \sum_{j=1}^{n} p_jP_{\psi_j}^* = \sum_{j=1}^{n} p_jP_{\psi_j} = \rho.
\end{align*}
Thus $\rho$ is self-adjoint. For every $x \in H$,
\begin{align*}
(\rho x,x)_H = \sum_{j=1}^{n} p_j(P_{\psi_j}x,x)_H.
\end{align*}
Each coefficient satisfies $p_j \geq 0$, and each term satisfies $(P_{\psi_j}x,x)_H \geq 0$, so $(\rho x,x)_H \geq 0$. Hence $\rho$ is positive semidefinite.
Finally, by linearity of trace and the trace computation above,
\begin{align*}
\operatorname{tr}(\rho) = \operatorname{tr}\left(\sum_{j=1}^{n} p_jP_{\psi_j}\right) = \sum_{j=1}^{n} p_j\operatorname{tr}(P_{\psi_j}) = \sum_{j=1}^{n} p_j = 1.
\end{align*}
This display is also a single-line computation and contains no raw row separator. Therefore $\rho$ is self-adjoint, positive semidefinite, and has trace $1$. By the finite-dimensional density-operator criterion stated at the start of the proof, $\rho$ is a density operator on $H$.
[/step]
