= Helstrom measurement for the zero and plus states
{c}
Suppose one <qubit> is prepared in either of the known <pure states> $|0\rangle$ and $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$, with equal prior <probabilities>. Measuring in the <computational basis> seems natural, but a tilted <projective measurement> gives better <quantum state discrimination>. The two <density operators> and their difference are
$$
\rho_0=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad
\rho_+=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix},\qquad
\Delta=\rho_0-\rho_+=\frac12\begin{pmatrix}1&-1\\-1&-1\end{pmatrix}
=\frac12(\sigma_z-\sigma_x).
$$
The <Pauli matrices> give $\Delta^2=I/2$, and the <eigenvalues> of $\Delta$ are $\lambda_\pm=\pm1/\sqrt2$. Their absolute sum gives <trace distance> $D=1/\sqrt2$, agreeing with <trace distance between qubit states>.
The positive <spectral projector> points along the unit <Bloch vector> $\mathbf n=(-1,0,1)/\sqrt2$:
$$
P_+=\frac{I+\mathbf n\cdot\boldsymbol\sigma}{2}
=\frac12\left(I+\frac{\sigma_z-\sigma_x}{\sqrt2}\right),\qquad
P_-=I-P_+.
$$
The outcome labels $+$ and $-$ here refer to the sign of the <eigenvalue> of $\Delta$; they do not denote a <Hadamard basis> measurement. Guess $|0\rangle$ after $P_+$ and $|+\rangle$ after $P_-$. This sign choice is the equal-prior <Helstrom measurement for two pure states>, a case of the <Holevo–Helstrom theorem>. https://cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf[Watrous, The Theory of Quantum Information, Theorem 3.4, equation (3.17)]{ref}
An explicit <orthonormal basis> for this <projective measurement> is
$$
|h_+\rangle=\cos(\pi/8)|0\rangle-\sin(\pi/8)|1\rangle,\qquad
|h_-\rangle=\sin(\pi/8)|0\rangle+\cos(\pi/8)|1\rangle,
$$
with $P_\pm=|h_\pm\rangle\langle h_\pm|$. Thus the real coefficient rotation angle is $\beta=-\pi/8$ relative to the <computational basis>, while the <Bloch vector> of $P_+$ is tilted by $\pi/4$ from the positive $z$ axis toward negative $x$. State-vector and <Bloch sphere> angles differ by a factor of two.
Write $a=(1+1/\sqrt2)/2$ and $b=(1-1/\sqrt2)/2$, so $a+b=1$. The <Born rule> gives the conditional outcome <probabilities>
$$
\begin{array}{c|cc}
\text{input}&P_+&P_-\\\hline
\rho_0&a&b\\
\rho_+&b&a
\end{array}
\qquad
a\simeq0.853553,\quad b\simeq0.146447.
$$
The event $P_+$ has <probability> gap $a-b=1/\sqrt2=D$, attaining the <variational characterization of trace distance>. Both preparations are identified correctly with conditional <probability> $a$, so
$$
P_{\mathrm{success}}^{\mathrm{opt}}=\frac{a+a}{2}
=\frac12\left(1+\frac1{\sqrt2}\right)\simeq0.853553.
$$
For comparison, a $Z$ <projective measurement> in the <computational basis> has conditional outcome <probabilities> $(1,0)$ on $\rho_0$ and $(1/2,1/2)$ on $\rho_+$. Guess zero on outcome zero and plus on outcome one. The average success is
$$
P_{\mathrm{success}}^{Z}=\frac12\left(1+\frac12\right)=\frac34,
$$
and the measured <total variation distance> is only $1/2$. The optimal tilted <projective measurement> gains about $0.103553$ in absolute success <probability> by balancing the two conditional error rates instead of identifying $|0\rangle$ perfectly at the expense of $|+\rangle$.
\Image[/helstrom-zero-plus-measurement.png]
{title=Conditional outcome probabilities and equal-prior success for Z and optimal Helstrom measurements of zero versus plus}
These are statistics for alternative measurements on independently prepared copies. Applying $Z$ first and then the tilted <projective measurement> is a different experiment, because the first <quantum measurement> changes the <density operator>.
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