Physical States as Rays

A pure state is not one phase-chosen Hilbert-space vector. It is a one-dimensional complex subspace, or ray, of a complex Hilbert space $\mathcal H$. For a nonzero vector $\psi$,

$$[\psi] = \{c\psi:c\in\mathbb C^\times\}, \qquad \mathbb C^\times=\mathbb C\setminus\{0\}.$$

The pure-state space is therefore

$$\mathbb P(\mathcal H) = (\mathcal H\setminus\{0\})/\mathbb C^\times.$$

Equivalently, one may normalize a representative and quotient by its remaining $U(1)$ phase, or replace the ray by its rank-one orthogonal projector. These three descriptions are mathematically equivalent but useful for different calculations.

Required background. State Vectors supplies Hilbert-space representatives; the Born Rule supplies squared-overlap and projector probabilities.

Helpful background. Transition Probabilities provides practice calculating the phase-invariant squared overlap used below.

Rays as a projective state space

Define an equivalence relation on nonzero vectors by

$$\psi\sim\phi \quad\Longleftrightarrow\quad \phi=c\psi \text{ for some }c\in\mathbb C^\times.$$

Reflexivity, symmetry, and transitivity follow from the multiplicative group $\mathbb C^\times$. Each equivalence class is a complex line through the origin. The zero vector is excluded because it spans no one-dimensional subspace, cannot be normalized, and carries no probability assignment.

Projective Hilbert space is not a vector space. Adding two equivalence classes is not well defined: independent rescalings of their representatives change the ray of the sum. Superposition is performed with chosen vectors in $\mathcal H$; the final nonzero vector then determines a ray.

The quotient removes two redundancies from a nonzero vector:

1. positive scale, fixed by normalization;
2. global phase, removed by the $U(1)$ quotient.

For finite-dimensional $\mathcal H\cong\mathbb C^d$, the result is the complex projective space $\mathbb{CP}^{d-1}$.

Normalized representatives and U(1)

Every ray contains normalized vectors. If $\psi\ne0$, set

$$\widehat\psi = \frac{\psi}{\|\psi\|}.$$

Any other normalized representative of the same ray has the form

$$\widehat\psi'=e^{i\chi}\widehat\psi, \qquad \chi\in\mathbb R.$$

Thus

$$\mathbb P(\mathcal H) \cong S(\mathcal H)/U(1),$$

where $S(\mathcal H)=\{\psi:\|\psi\|=1\}$. The phase acts freely on the unit sphere: no nontrivial $e^{i\chi}$ fixes a nonzero vector.

This quotient removes a phase multiplying the whole vector. It does not remove relative phases between components. For example,

$$\frac{|0\rangle+|1\rangle}{\sqrt2} \quad\text{and}\quad \frac{|0\rangle+i|1\rangle}{\sqrt2}$$

are different rays, because no single complex scalar turns one vector into the other. Their interference predictions can differ. The operational global-versus-relative phase analysis remains on Rays and Global Phase.

Rank-one projectors as phase-free representatives

Every nonzero $\psi$ defines

$$P_\psi = \frac{|\psi\rangle\langle\psi|} {\langle\psi|\psi\rangle}.$$

This is an orthogonal projector:

$$P_\psi^*=P_\psi, \qquad P_\psi^2=P_\psi, \qquad \operatorname{Ran}P_\psi=\mathbb C\psi.$$

It is invariant under every nonzero rescaling. If $\phi=c\psi$, then

$$P_\phi = \frac{|c|^2|\psi\rangle\langle\psi|} {|c|^2\langle\psi|\psi\rangle} = P_\psi.$$

Conversely, equal rank-one projectors have equal one-dimensional ranges, so $P_\phi=P_\psi$ implies $[\phi]=[\psi]$. There is therefore a bijection

$$\{\text{rays in }\mathcal H\} \longleftrightarrow \{\text{rank-one orthogonal projections on }\mathcal H\}.$$

Projectors are often the cleanest phase-free coordinates for pure states. They also place pure states inside the convex state space of density operators: pure density operators are exactly the rank-one projectors.

Transition probability is ray data

For nonzero $\psi$ and $\phi$, define

$$\mathcal P([\psi],[\phi]) = \frac{|\langle\psi|\phi\rangle|^2} {\|\psi\|^2\|\phi\|^2}.$$

Independent nonzero rescalings cancel, so this is a function of rays rather than representatives. In projector language,

$$\mathcal P([\psi],[\phi]) = \operatorname{Tr}(P_\psi P_\phi).$$

To verify the identity, use

$$P_\psi P_\phi = \frac{ |\psi\rangle \langle\psi|\phi\rangle \langle\phi| }{ \|\psi\|^2\|\phi\|^2 }$$

and the rank-one trace rule $\operatorname{Tr}(|u\rangle\langle v|)=\langle v|u\rangle$.

The transition probability distinguishes two important cases:

$$\mathcal P([\psi],[\phi])=1 \quad\Longleftrightarrow\quad [\psi]=[\phi],$$

by equality in Cauchy–Schwarz, and

$$\mathcal P([\psi],[\phi])=0 \quad\Longleftrightarrow\quad \langle\psi|\phi\rangle=0.$$

Thus identity and orthogonality are intrinsic projective relations. More generally, the Fubini–Study distance can be written as

$$d_{\mathrm{FS}}([\psi],[\phi]) = \arccos\sqrt{\mathcal P([\psi],[\phi])},$$

up to a conventional overall scale. The detailed geometry belongs to the Projective Hilbert Space treatment; here the invariant prepares the symmetry theorem.

A qubit ray as a concrete anchor

For $\mathcal H=\mathbb C^2$, a normalized representative can be written

$$|\psi\rangle = \cos\frac{\theta}{2}|0\rangle +e^{i\varphi}\sin\frac{\theta}{2}|1\rangle.$$

Its projector is

$$P_\psi = \frac12(I+\mathbf r\cdot\boldsymbol\sigma),$$

where

$$\mathbf r = (\sin\theta\cos\varphi, \sin\theta\sin\varphi, \cos\theta)$$

is a unit Bloch vector. Hence $\mathbb{CP}^1$ is represented by the Bloch sphere. Two qubit rays satisfy

$$\operatorname{Tr}(P_{\mathbf r}P_{\mathbf s}) = \frac{1+\mathbf r\cdot\mathbf s}{2}.$$

Identical Bloch vectors give probability one; antipodal vectors give orthogonal rays, not equivalent rays.

Mixed states are not rays

A mixed state is represented by a positive trace-one operator $\rho$ that need not have rank one. If

$$\rho^2\ne\rho,$$

there is no vector ray whose projector equals $\rho$. The set of density operators is convex, whereas projective Hilbert space is not: a probabilistic mixture

$$pP_\psi+(1-p)P_\phi$$

is generally a mixed density operator rather than another ray.

This matters for reduced states. A subsystem of an entangled pure state can be mixed even though the composite state is a ray in the tensor-product Hilbert space. The phrase “the state is a ray” must therefore identify the system and restrict the claim to pure states.

Superselection qualification

If a superselection rule decomposes the Hilbert space into sectors,

$$\mathcal H = \bigoplus_\alpha\mathcal H_\alpha,$$

the observable algebra may contain no operation that detects relative phase between distinct sectors. Operational pure states are then represented sectorwise by rays $\mathbb P(\mathcal H_\alpha)$, rather than by every ray of the unrestricted direct sum as a physically coherent pure state.

This is not a failure of projective geometry. It says that the physical state space is determined jointly by the Hilbert space and the observable algebra. The unrestricted ray postulate is appropriate only when no superselection restriction has been imposed.

Why ray language prepares symmetry theory

A physical symmetry acts first on pure states, hence on $\mathbb P(\mathcal H)$. It should preserve the operational invariant $\mathcal P([\psi],[\phi])$. Unitary and antiunitary operators both induce such transformations:

$$[\psi]\longmapsto[U\psi].$$

The converse is Wigner’s theorem: under its bijectivity and preservation hypotheses, every transition-probability-preserving ray map has a unitary or antiunitary lift, unique up to an overall phase. The projective branch continues with Transition-Probability Preserving Maps, which isolates the hypothesis before the theorem is proved.

For a symmetry group, choosing one lift for each group element can introduce phase multipliers. The ray action may therefore lift to a projective unitary representation rather than an ordinary unitary representation. Removing the state-vector phase does not remove this group-level cocycle question.

Scope relative to Core Formalism

This page owns the precise equivalence among rays, normalized phase classes, and rank-one projectors, together with the transition-probability structure needed for Wigner’s theorem. Core Formalism retains the operational treatment of global versus relative phase and the extended geometry of projective Hilbert space. Those roles are complementary, not duplicate derivations.

Common pitfalls

Calling a normalized vector the physical state without qualification. A normalized vector is a representative. Multiplying it by a global phase does not change the ray.

Quotienting relative phase. Only a scalar multiplying the entire vector is removed. Relative phases generally change the ray and interference.

Adding rays as if projective space were linear. Choose representatives, form a nonzero superposition in $\mathcal H$, and then take its ray.

Calling every density operator a ray. Only rank-one projectors represent pure rays. Mixed and reduced states require the full convex density-operator space.

Ignoring superselection. The physically available ray space may be a sectorwise union rather than the projective space of an unrestricted direct sum.

Exercises

1. Test ray equality

Determine which pairs represent the same ray:

$$(1,i),\ (2,2i); \qquad (1,i),\ (i,1); \qquad (1,0),\ (-i,0).$$

Solution

The first pair differs by the scalar $2$ and the third by $-i$, so each pair represents one ray. If $(i,1)=c(1,i)$, then $c=i$ from the first component but $ci=-1\ne1$ in the second, so the middle pair represents distinct rays.

2. Prove projector invariance

Show that $P_{c\psi}=P_\psi$ for every $c\ne0$, and prove the converse: equal rank-one projectors imply equal rays.

Solution

The numerator and denominator both acquire the factor $|c|^2$, which cancels. If $P_\phi=P_\psi$, their ranges agree. Those ranges are respectively $\mathbb C\phi$ and $\mathbb C\psi$, so $\phi=c\psi$ for some nonzero $c$.

3. Derive the projector trace formula

For nonzero $\psi$ and $\phi$, prove

$$\operatorname{Tr}(P_\psi P_\phi) = \frac{|\langle\psi|\phi\rangle|^2} {\|\psi\|^2\|\phi\|^2}.$$

Solution

Multiply the rank-one operators:

$$P_\psi P_\phi = \frac{\langle\psi|\phi\rangle |\psi\rangle\langle\phi|} {\|\psi\|^2\|\phi\|^2}.$$

Using $\operatorname{Tr}(|\psi\rangle\langle\phi|)=\langle\phi|\psi\rangle$ gives the squared magnitude.

4. Qubit transition probability

Use $P_{\mathbf r}=(I+\mathbf r\cdot\boldsymbol\sigma)/2$ and $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$ to derive

$$\operatorname{Tr}(P_{\mathbf r}P_{\mathbf s}) = \frac{1+\mathbf r\cdot\mathbf s}{2}.$$

Solution

Expand the product. Terms linear in one Pauli matrix have zero trace, while

$$\operatorname{Tr}[(\mathbf r\cdot\boldsymbol\sigma) (\mathbf s\cdot\boldsymbol\sigma)] = 2\mathbf r\cdot\mathbf s.$$

Dividing the total trace by four gives the result.

5. A mixture is not a ray

Let $\psi$ and $\phi$ be orthogonal normalized vectors and $\rho=pP_\psi+(1-p)P_\phi$ with $0<p<1$. Show that $\rho$ is not a rank-one projector.

Solution

Orthogonality gives $P_\psi P_\phi=0$, so

$$\rho^2 = p^2P_\psi+(1-p)^2P_\phi.$$

This differs from $\rho$ because $p^2\ne p$ and $(1-p)^2\ne1-p$ for $0<p<1$. Hence $\rho$ is mixed and does not represent a ray.

References

• N. P. Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Springer, 2017.
• M. Pankov, Wigner-Type Theorems for Hilbert Grassmannians, Cambridge University Press, 2020.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.