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For qubit density operators, the trace distance has a particularly simple geometry. Write their Bloch vectors as $\mathbf{r},\mathbf{s}$ in the Bloch ball:where $I$ is the identity operator and $\mathbf{\sigma }=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ denotes the Pauli matrices. This Bloch vector representation includes both pure states on the Bloch sphere and mixed states in the interior of the Bloch ball. ^[ref]

The Pauli matrix multiplication law gives $[(\mathbf{r}-\mathbf{s})\cdot \mathbf{\sigma }{]}^2=|\mathbf{r}-\mathbf{s}{|}^{2}I$. Hence the two eigenvalues of $\Delta$ are $\pm |\mathbf{r}-\mathbf{s}|/2$, and their absolute sum givesThe trace distance is half the straight-line Euclidean distance between the Bloch vectors, not the angular or surface distance along the Bloch sphere. Opposite Bloch vectors have trace distance one; the centre $I/2$ is at trace distance $1/2$ from every pure state.

For the pure states $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, the Bloch vectors are $\mathbf{r}=(0,0,1)$ and $\mathbf{s}=(1,0,0)$. Their straight-line separation is $\sqrt{2}$, soThese Bloch vectors are perpendicular, but the pure states are not orthogonal: $|⟨0|+⟩|=1/\sqrt{2}$. The Helstrom measurement for the zero and plus states measures along the difference direction $\mathbf{n}=(\mathbf{r}-\mathbf{s})/\sqrt{2}=(-1,0,1)/\sqrt{2}$, rather than along either input Bloch vector.

For $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, a projective measurement of the number of ones in one party's $n$ local qubits gives a common type class of size $d_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The conditional pure state is $d_k^{-1/2}{\sum }_{|x|=k}|x,x,x⟩$. It can be converted exactly and deterministically to a uniform common-label state of any dimension $m\le d_k$: one party measures an $m$-element subset $S$ with Kraus operator ${\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)}$, and all parties relabel their common retained labels after receiving $S$. Choosing $m=2^f$ gives $f$ independent GHZ states. Stirling's formula and the weak law of large numbers give $f/n\to h_2(p)$ in probability. Measuring every local bit separately would instead destroy the coherence and leave a product state.

With phase/parity labels $(u,v)$ encoded by $X^v{Z}^u$, a resource $a|{\Phi }^{+}⟩+b|{\Phi }^{-}⟩$ gives intended-label probability $|a{|}^2$ and phase-label-flip probability $|b{|}^2$ under a Bell-basis measurement. This follows because the local Pauli operator maps the two orthogonal components into the two Bell states whose phase label differs. The resource is a coherent pure state, even though its measured label statistics form a classical noisy channel.

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

For a unit-trace two-dimensional reduced state of a bipartite pure state, determinant $D$ gives eigenvalues $(1\pm \sqrt{1-4D})/2$. Their binary entropy quantifies entanglement. Natural logarithms give a maximum $\ln 2$; base-two logarithms give one bit.

A quantum state on the tensor product of two two-dimensional complex state spaces. A normalized pure state has squared coefficient moduli summing to one. It is a product state exactly when $\alpha \delta =\beta \gamma$, by the two-qubit product-state determinant criterion. The weaker equality of moduli only characterizes independence of the computational-basis outcomes; phases can preserve quantum entanglement despite those independent statistics.

For normalized pure states $|u⟩,|v⟩$, a two-member quantum state ensemble gives the maximally mixed state of a qubit exactly when its members are orthogonal and equally probable. To prove necessity, take expectation values of its density matrix in the two members: $1/2=p+(1-p)|⟨u|v⟩{|}^2\ge p$ and $1/2=1-p+p|⟨u|v⟩{|}^2\ge 1-p$. Hence $p=1/2$ and the overlap vanishes. Sufficiency is the resolution of the identity in an orthonormal basis.

Unambiguous quantum state discrimination allows an inconclusive outcome but forbids a wrong conclusive identification. Two distinct pure states admit such a measurement, whereas two full-rank density operators on the same support do not.

The ideal universal NOT map sends every qubit Bloch vector $\mathbf{s}$ to $-\mathbf{s}$, taking a pure state to its orthogonal pure state. It is a trace-preserving positive map but not a quantum channel: applied to one half of a Bell state, it produces an operator with a negative eigenvalue. A physical approximation can reverse all directions with factor $-1/3$ using the equal mixture of conjugations by the three Pauli matrices. The isotropic qubit map complete-positivity bound proves that a larger uniform reversal is impossible among such isotropic maps.

The Von Neumann entropy of a finite-dimensional density operator measures uncertainty in its eigenvalue distribution:where $p_i$ are the eigenvalues and $0{\log }_{2}0=0$. Base-two logarithms give bits. A pure state has Von Neumann entropy zero, while the maximally mixed state $I/d$ has ${\log }_{2}d$ bits. Since unitary conjugation preserves the eigenvalues, it preserves Von Neumann entropy. ^[ref]

For a bipartite pure state, the Von Neumann entropy of either reduced density operator quantifies its entanglement. That interpretation requires purity of the joint state. A Bell state has one bit of marginal Von Neumann entropy, but so does the separable state $(|00⟩⟨00|+|11⟩⟨11|)/2$. The latter mixture has classical correlations and no entanglement. Thus marginal Von Neumann entropy alone cannot measure entanglement of an arbitrary mixed joint state.

A wavefunction is the position representation of a pure state. In the one-dimensional model used here, $\psi (x,t)$ takes values in the complex numbers, and the Born rule givesThe probability density $\rho$ has units of inverse length; the wavefunction has units of inverse square root of length. A value of $\rho$ at a point is not itself a probability: probabilities come from an integral over a region. For the infinite square well ground state, $\rho (L/2)=2/L$, whereas the probability of the exact single point $L/2$ is zero. The interesting question is how much probability a finite interval contains.

Two wavefunctions differing only by a global phase represent the same pure state. Relative quantum phases between components of a quantum superposition affect matter-wave interference, so retaining the full complex number amplitude is essential. This scalar description covers a particle without active spin degrees of freedom; a general electron wavefunction can have two spin components, with the position probability density obtained by adding the squared modulus of each component.