Rays and Global Phase

A normalized ket is a convenient representative of a pure quantum state, but it is not the physical state by itself. For every real number $\chi$,

$$|\psi'\rangle=e^{i\chi}|\psi\rangle$$

represents the same pure state as $|\psi\rangle$. The phase $e^{i\chi}$ is global because it multiplies the entire ket. It cancels from every measurement probability, every expectation value, and every state-update prediction.

The corresponding physical object is a ray: the equivalence class of all nonzero vectors that differ by an overall nonzero complex factor. After a representative has been normalized, the remaining freedom is precisely a global phase.

The exact equivalence among rays, normalized $U(1)$ classes, and rank-one projectors—together with the transition-probability structure used by Wigner’s theorem—is owned by Physical States as Rays. This page retains the operational global-versus-relative phase treatment.

This statement does not mean that phase is generally irrelevant. A phase between two components or two alternatives is relative, and it can change interference. The reliable distinction is:

Transformation	What changes?	Physical effect
$\lvert\psi\rangle\mapsto e^{i\chi}\lvert\psi\rangle$	one representative of the whole ray	none
$c_j\mapsto e^{i\delta}c_j$ for only part of a superposition	relative relation among alternatives	generally observable
$\lvert e_j\rangle\mapsto e^{i\beta_j}\lvert e_j\rangle$ with coefficients transformed consistently	basis convention	none

Required background. State Vectors supplies normalized Hilbert-space representatives and basis amplitudes; Inner-Product Conventions supplies the bra-slot conjugation used in every phase-cancellation and adjoint identity.

Global Phase in a Basis

Let an orthonormal basis be $\{|e_j\rangle\}$ and write

$$|\psi\rangle = \sum_j c_j|e_j\rangle.$$

A global phase multiplies every coefficient by the same number:

$$|\psi\rangle \longmapsto e^{i\chi}|\psi\rangle = \sum_j e^{i\chi}c_j|e_j\rangle.$$

All coefficient ratios remain unchanged whenever they are defined:

$$\frac{e^{i\chi}c_j}{e^{i\chi}c_k} = \frac{c_j}{c_k}.$$

By contrast, changing only one coefficient generally changes such ratios. For example,

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

do not differ by one overall scalar. They are different rays and can be distinguished by a suitable measurement.

The word “global” always refers to the whole state of the system being described. A phase applied to only one branch, path, energy component, or control sector is not global for that complete state.

Operational Proof for Arbitrary Measurements

The phase cancellation is not restricted to measurements in one basis. Consider any measurement outcome represented by an effect $E_a$, where

$$0\le E_a\le I.$$

For a normalized pure state, its probability is

$$p(a|\psi) = \langle\psi|E_a|\psi\rangle.$$

If $|\psi'\rangle=e^{i\chi}|\psi\rangle$, then the bra transforms with the complex-conjugate phase:

$$\langle\psi'| = e^{-i\chi}\langle\psi|.$$

Therefore

$$\begin{aligned} p(a|\psi') &= \langle\psi'|E_a|\psi'\rangle \\ &= e^{-i\chi}e^{i\chi} \langle\psi|E_a|\psi\rangle \\ &= p(a|\psi). \end{aligned}$$

Because this holds for every effect $E_a$, no POVM can distinguish the two representatives. Projective measurements are included as the special case in which the effects are orthogonal projectors.

The same conclusion holds for expectation values:

$$\langle A\rangle_{\psi'} = \langle\psi'|A|\psi'\rangle = \langle\psi|A|\psi\rangle.$$

It also holds for transition probabilities. Although a transition amplitude changes covariantly,

$$\langle\phi|\psi'\rangle = e^{i\chi}\langle\phi|\psi\rangle,$$

its squared modulus does not:

$$|\langle\phi|\psi'\rangle|^2 = |\langle\phi|\psi\rangle|^2.$$

The canonical rules for amplitudes and probabilities are developed in Probability Amplitudes.

The Density-Operator Test

The shortest representation-independent test uses the pure-state density operator

$$\rho_\psi=|\psi\rangle\langle\psi|.$$

Global phase disappears exactly:

$$\begin{aligned} \rho_{\psi'} &= e^{i\chi}|\psi\rangle e^{-i\chi}\langle\psi| \\ &= |\psi\rangle\langle\psi| \\ &= \rho_\psi. \end{aligned}$$

Thus $|\psi\rangle$ and $e^{i\chi}|\psi\rangle$ are not merely states that happen to agree for a few observables. They define the same density operator and hence the same prediction for every measurement, channel, and quantum instrument.

For an arbitrary nonzero, not necessarily normalized vector, the ray can be represented by

$$\rho_{[\psi]} = \frac{|\psi\rangle\langle\psi|} {\langle\psi|\psi\rangle}.$$

Under $|\psi\rangle\mapsto\lambda|\psi\rangle$ with $\lambda\ne0$, both numerator and denominator acquire the factor $|\lambda|^2$. Consequently,

$$\rho_{[\lambda\psi]} = \rho_{[\psi]}.$$

This gives a one-to-one correspondence between pure-state rays and rank-one density operators. General mixed states require the broader density-operator language developed in Density Operators.

Rays in Hilbert Space

On the nonzero vectors of a Hilbert space $\mathcal H$, define

$$|\phi\rangle\sim|\psi\rangle \quad\Longleftrightarrow\quad |\phi\rangle=\lambda|\psi\rangle \quad\text{for some }\lambda\in\mathbb C\setminus\{0\}.$$

This is an equivalence relation:

• reflexivity follows by choosing $\lambda=1$;
• symmetry follows because $\lambda^{-1}$ exists;
• transitivity follows because the product of two nonzero scalars is nonzero.

The ray through $|\psi\rangle$ is the equivalence class

$$[\psi] = \left\lbrace \lambda|\psi\rangle: \lambda\in\mathbb C\setminus\{0\} \right\rbrace.$$

Every nonzero member of this class represents the same pure state. If one first imposes normalization, the modulus of $\lambda$ is fixed and only

$$\lambda=e^{i\chi}$$

remains. The normalized representatives of one ray form a $U(1)$ orbit.

The zero vector belongs to no ray because it cannot be normalized and gives no probability distribution. Multiplication by a non-unit scalar preserves the ray but not normalization, so calculations must either renormalize the vector or use $\rho_{[\psi]}$ above.

Projective Hilbert Space

The set of all rays is the projective Hilbert space

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

For a finite-dimensional Hilbert space $\mathcal H\cong\mathbb C^d$, this is complex projective space:

$$\mathbb P(\mathcal H) \cong \mathbb{CP}^{d-1}.$$

A normalized vector in $\mathbb C^d$ has $2d-1$ real parameters after its norm is fixed. Removing one global-phase parameter leaves

$$2d-2$$

real parameters for a pure state. In particular, $\mathbb{CP}^{1}$ is the two-dimensional pure-state space of a qubit and can be represented as the Bloch sphere.

This page uses projective space only to identify the physical pure-state space. Quotient coordinates, geometry, distances, and the relation to the Bloch sphere belong to Projective Hilbert Space.

Global Phase Versus Relative Phase

Consider two nonzero components in a fixed orthonormal basis:

$$|\psi\rangle = |c_0|e^{i\alpha}|0\rangle + |c_1|e^{i\beta}|1\rangle.$$

Factoring out $e^{i\alpha}$ gives

$$|\psi\rangle = e^{i\alpha} \left( |c_0||0\rangle + |c_1|e^{i(\beta-\alpha)}|1\rangle \right).$$

The first factor is global. The phase difference

$$\delta=\beta-\alpha$$

remains inside the superposition and can affect interference. A common phase rotation changes $\alpha$ and $\beta$ by the same amount and leaves $\delta$ fixed. A relative phase shift changes $\delta$.

Dashed arrows show the original coefficients and solid arrows show the transformed coefficients. A common rotation moves both by $\chi$ and preserves their angular separation $\delta$. Rotating only one coefficient changes $\delta$ and can therefore change interference.

The fuller treatment of coherent superpositions, basis dependence, and the distinction from statistical mixtures is in Superposition and Relative Phase.

Phase of Coefficients and Phase Conventions

Coefficient phases are meaningful only together with a chosen basis convention. Rephase the basis vectors independently:

$$|e_j\rangle' = e^{i\beta_j}|e_j\rangle.$$

The same abstract state must then have coefficients

$$c_j' = e^{-i\beta_j}c_j,$$

because

$$\sum_j c_j'|e_j\rangle' = \sum_j c_j|e_j\rangle.$$

The numerical phase of $c_j/c_k$ therefore changes if the basis kets are rephased separately. This does not make interference conventional. The matrix elements of every measurement and evolution operator transform at the same time, leaving all probabilities invariant.

Accordingly, the phrase “relative phase is observable” is shorthand for a more precise statement: phase-sensitive, convention-independent combinations of state amplitudes, dynamics, and measurement amplitudes can affect observed probabilities. In an interferometer, the beam splitters and propagation phases supply the physical reference against which a path-phase difference is read.

Why Interference Detects Relative Phase

Suppose a detector outcome receives amplitudes $\mathcal A_1$ and $\mathcal A_2$ from two coherent alternatives. The total amplitude is

$$\mathcal A = \mathcal A_1+\mathcal A_2,$$

so the probability contains a cross term:

$$\begin{aligned} |\mathcal A|^2 &= |\mathcal A_1|^2 + |\mathcal A_2|^2 \\ &\quad+ 2\operatorname{Re} \left( \mathcal A_1^*\mathcal A_2 \right). \end{aligned}$$

Multiplying both alternatives by the same $e^{i\chi}$ leaves every term unchanged. Multiplying only the second by $e^{i\delta}$ changes the interference term to

$$2\operatorname{Re} \left( e^{i\delta} \mathcal A_1^*\mathcal A_2 \right).$$

The observable is not the absolute phase of either amplitude. It is the phase-sensitive relation between alternatives in a complete experimental arrangement.

Worked Example: Spin-1/2

Using the $S_z$ eigenbasis, write a general normalized spin-$1/2$ state as

$$|\psi(\theta,\phi)\rangle = \cos\frac\theta2|{+z}\rangle + e^{i\phi} \sin\frac\theta2|{-z}\rangle.$$

An arbitrary global phase $e^{i\chi}$ could multiply the right-hand side, but it would not add another physical coordinate. The probabilities for spin up along the Cartesian axes are

$$\begin{aligned} P(+z) &= \cos^2\frac\theta2, \\ P(+x) &= \frac12 \left(1+\sin\theta\cos\phi\right), \\ P(+y) &= \frac12 \left(1+\sin\theta\sin\phi\right). \end{aligned}$$

The relative phase $\phi$ changes the $x$- and $y$-measurement statistics. A global phase $\chi$ appears in none of them.

For example,

$$|{+y}\rangle = \frac{|{+z}\rangle+i|{-z}\rangle}{\sqrt2}$$

and

$$-i|{+y}\rangle = \frac{-i|{+z}\rangle+|{-z}\rangle}{\sqrt2}$$

are two representatives of the same ray. By contrast,

$$|{+x}\rangle = \frac{|{+z}\rangle+|{-z}\rangle}{\sqrt2}$$

is a different ray. Indeed,

$$|\langle +x|+y\rangle|^2 = \frac12,$$

so a measurement along $x$ distinguishes the two preparations probabilistically. The spin system itself is developed in Spin-1/2 Hilbert Space.

Worked Example: Two-Slit Amplitudes

At a screen position $x$, let the two slit amplitudes be

$$\mathcal A_1(x) = a_1(x)e^{i\varphi_1(x)}, \qquad \mathcal A_2(x) = a_2(x)e^{i\varphi_2(x)},$$

where $a_1,a_2\ge0$. The detection probability is proportional to

$$\begin{aligned} P(x) &\propto |\mathcal A_1(x)+\mathcal A_2(x)|^2 \\ &= a_1^2(x)+a_2^2(x) \\ &\quad+ 2a_1(x)a_2(x) \cos\Delta\varphi(x), \end{aligned}$$

with

$$\Delta\varphi(x) = \varphi_2(x)-\varphi_1(x).$$

Adding the same $\chi$ to both phases leaves $\Delta\varphi$ and the whole pattern unchanged. Inserting a phase shifter in only one path sends

$$\Delta\varphi(x) \longmapsto \Delta\varphi(x)+\delta$$

and shifts the fringes. The physical experiment, including which-way information and loss of coherence, is discussed in the Double-Slit Experiment.

Stationary States and the Energy Zero

If $|E\rangle$ is an energy eigenvector of a time-independent Hamiltonian, then

$$|E,t\rangle = e^{-iEt/\hbar}|E\rangle.$$

At each time this differs from $|E\rangle$ only by a global phase. The state vector evolves, but the ray and all time-independent measurement statistics remain fixed. This is why an energy eigenstate is stationary.

A superposition of distinct energies behaves differently:

$$|\psi(t)\rangle = c_1e^{-iE_1t/\hbar}|E_1\rangle + c_2e^{-iE_2t/\hbar}|E_2\rangle.$$

After one common phase is factored out, the relative phase evolves as

$$e^{-i(E_2-E_1)t/\hbar}.$$

That phase can produce time-dependent interference and expectation values. The complete distinction belongs to Stationary States.

The ray principle also explains the freedom to shift the zero of energy. If

$$H'=H+C I,$$

then for a time-independent $H$,

$$\begin{aligned} U'(t) &= e^{-i(H+CI)t/\hbar} \\ &= e^{-iCt/\hbar}U(t). \end{aligned}$$

Every state evolved by $H'$ differs from the state evolved by $H$ by the same global phase. Closed-system predictions are therefore unchanged. If two control branches experience different offsets, however, the branch-dependent phases become relative phases of the larger controlled system.

A 2π Spin Rotation

For a spin-$1/2$ state, a rotation by $2\pi$ is represented by

$$R(2\pi)=-I.$$

Thus

$$R(2\pi)|\psi\rangle=-|\psi\rangle.$$

The minus sign is a global phase for an isolated spin state, so no measurement on that state alone distinguishes it from the original. This does not make the sign physically empty in every arrangement. If only one arm of an interferometer undergoes the rotation, the minus sign is relative between the two arms and can shift the interference pattern.

The spinor-specific derivation and interferometric interpretation are in Spinors and 2π Rotations.

Can a Reference Reveal Global Phase?

Adding a reference system does not make the global phase of the total state observable. If

$$|\Psi\rangle = |r\rangle\otimes|\psi\rangle,$$

then applying $e^{i\chi}I$ to the second factor gives

$$(I\otimes e^{i\chi}I)|\Psi\rangle = e^{i\chi}|\Psi\rangle,$$

which is still only a global phase of the joint state.

A controlled phase is different. Starting from

$$|\Psi_0\rangle = \frac{|0\rangle_R+|1\rangle_R}{\sqrt2} \otimes|\psi\rangle,$$

suppose the phase is applied only in the $|1\rangle_R$ branch. The result is

$$|\Psi_\chi\rangle = \frac{ |0\rangle_R|\psi\rangle + e^{i\chi}|1\rangle_R|\psi\rangle }{\sqrt2}.$$

This phase is relative between reference branches. Measuring the reference in the $\{|+\rangle_R,|-\rangle_R\}$ basis gives

$$P(+) = \frac{1+\cos\chi}{2},$$

so $\chi$ is observable. What was called a phase on one operation has become a relative phase in the complete state description; the global phase of that complete state remains unobservable.

Time-Dependent Rephasing

At the level of instantaneous physical states,

$$|\psi'(t)\rangle = e^{i\chi(t)}|\psi(t)\rangle$$

describes the same path of rays as $|\psi(t)\rangle$. There is nevertheless a bookkeeping point in the Schrödinger equation. If

$$i\hbar\frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangle,$$

then the rephased representative obeys

$$i\hbar\frac{d}{dt}|\psi'(t)\rangle = H'(t)|\psi'(t)\rangle,$$

where

$$H'(t) = H(t)-\hbar\dot\chi(t)I.$$

Therefore an arbitrary time-dependent rephasing is not generally another solution of the Schrödinger equation with the identical Hamiltonian. It is a change of representative accompanied by an additive scalar shift of the Hamiltonian. The ray trajectory and all physical predictions remain the same.

Phase Choices Over Parameter Space

For one isolated ray, choosing a normalized ket is simply choosing a phase representative. For a smoothly parameterized family $|\psi(R)\rangle$, one may choose a different phase at every point:

$$|\psi(R)\rangle \longmapsto e^{i\chi(R)}|\psi(R)\rangle.$$

Quantities such as $i\langle\psi|d\psi\rangle$ depend on this choice, while appropriate closed-loop phases and curvatures can be gauge invariant. This is not a failure of the global-phase principle. It is the geometry of making phase choices consistently across many rays.

The bundle language is developed in U(1) Bundles and Quantum Phase, and its adiabatic physical application is developed in Berry Phase.

What the Ray Statement Does Not Say

• It does not say that every phase in a state expansion is unobservable.
• It does not identify coherent superpositions with statistical mixtures.
• It does not permit dropping normalization in probability calculations.
• It does not say that amplitudes are directly observable complex numbers.
• It does not make basis-phase conventions into physical transformations.
• It does not imply that a phase applied to one branch is global for a larger composite state.
• It does not remove geometric or topological phase effects, which concern comparisons along paths or between alternatives.
• It does not mean a rephased vector satisfies the same dynamical equation unless the Hamiltonian is transformed consistently.

A Practical Phase Audit

When a phase appears in a calculation, ask the following questions in order:

1. What is the complete state space? Include path, spin, control, ancilla, and reference degrees of freedom that participate in the experiment.
2. Does one scalar multiply the entire ket? If yes, it is global for that state.
3. Can the phase be factored from every component? If not, it is relative within the chosen decomposition.
4. Was only the basis rephased? Transform state coefficients and operators consistently before assigning physical meaning.
5. What probability changes? An observable claim should be expressible as a change in $\operatorname{Tr}(\rho E)$ or another invariant prediction.
6. Is time evolution involved? A time-dependent phase convention shifts the Hamiltonian by a scalar multiple of the identity.
7. Is a family of rays involved? Local phase choices may lead to geometric connections even though each individual ray is phase invariant.

Common Mistakes

Saying “phase does not matter”

Only a common phase multiplying the complete state is redundant. Relative phases can affect interference, spin measurements, oscillations, and quantum control.

Comparing entries instead of rays

Two coordinate columns can look different and still differ by one common nonzero scalar. Test proportionality of the complete vectors, not equality of individual entries.

Ignoring normalization after rescaling

$2|\psi\rangle$ lies on the same ray as $|\psi\rangle$, but it is not a normalized representative. The ordinary Born formula assumes normalization unless an explicit norm denominator is included.

Treating basis rephasing as state evolution

Changing $|e_j\rangle$ to $e^{i\beta_j}|e_j\rangle$ while transforming coefficients and operators consistently changes coordinates, not physics.

Calling a controlled phase global

A phase conditioned on one path or control state is relative in the composite state. It can be converted into a population difference by recombination.

Confusing a stationary ray with a constant ket

An energy eigenket generally acquires $e^{-iEt/\hbar}$. Its ray is stationary even though the chosen vector representative is time dependent.

Connections

• State Vectors develops normalized kets and basis coefficients.
• Physical States as Rays gives the precise projective-state and rank-one-projector equivalence used by the rigorous representation spine.
• Superposition and Relative Phase owns the detailed analysis of coherence and interference.
• Probability Amplitudes develops composition and interference of amplitudes.
• Density Operators gives the phase-invariant language for general states.

References

• J. Anandan and Y. Aharonov, “Geometry of quantum evolution,” Physical Review Letters 65, 1697–1700 (1990), https://doi.org/10.1103/PhysRevLett.65.1697.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, Chapters 2–3.
• I. Bengtsson and K. Życzkowski, Geometry of Quantum States, 2nd ed., Cambridge University Press, 2017, Chapters 4–5.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, Chapters I and III.
• A. Messiah, Quantum Mechanics, Vol. I, North-Holland, 1961, Chapters II and IV.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, Sections 2.1–2.2.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995, Chapters 2 and 4.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, Chapters 1–2.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955, Chapters I and III.

Exercises

1. Arbitrary POVM and state update. Let $|\psi'\rangle=e^{i\chi}|\psi\rangle$. Prove that the two representatives give the same probability for every POVM effect $E_a$. If outcome $a$ is implemented by a Kraus operator $M_a$, show that the normalized post-measurement rays also agree.

Solution

The outcome probabilities are

$$\begin{aligned} p'(a) &= \langle\psi'|E_a|\psi'\rangle \\ &= e^{-i\chi}e^{i\chi} \langle\psi|E_a|\psi\rangle \\ &= p(a). \end{aligned}$$

For a nonzero-probability outcome, the two unnormalized conditional vectors are related by

$$M_a|\psi'\rangle = e^{i\chi}M_a|\psi\rangle.$$

Their norms are equal, so normalization preserves the same global factor. They therefore define the same post-measurement ray. Outcomes of zero probability have no normalized conditional state for either representative.

2. Same ray or different ray? For normalized orthogonal kets $|0\rangle$ and $|1\rangle$, classify each pair.

   $$\begin{array}{ll} \text{(a)} & \dfrac{|0\rangle+i|1\rangle}{\sqrt2}, \quad \dfrac{-i|0\rangle+|1\rangle}{\sqrt2}, \\ \text{(b)} & \dfrac{|0\rangle+|1\rangle}{\sqrt2}, \quad \dfrac{|0\rangle-|1\rangle}{\sqrt2}, \\ \text{(c)} & |0\rangle, \quad 3e^{i\pi/7}|0\rangle. \end{array}$$

Solution

In (a), multiplying the first ket by $-i$ gives the second, so they are the same ray. In (b), matching the $|0\rangle$ coefficient would require the overall scalar to be $1$, but then the $|1\rangle$ coefficients disagree. They are different rays. In (c), the vectors are nonzero scalar multiples and therefore lie on the same ray, although the second vector must be normalized before the ordinary Born rule is applied.

3. Spin phase tomography. Consider

   $$|\psi_\phi\rangle = \frac{|{+z}\rangle+e^{i\phi}|{-z}\rangle}{\sqrt2}.$$

   Compute $P(+x)$ and $P(+y)$. Which values of $\phi$ give the $+x$, $-x$, $+y$, and $-y$ rays? Explain why an additional global phase cannot be inferred from these measurements.

Solution

Using

$$|{+x}\rangle = \frac{|{+z}\rangle+|{-z}\rangle}{\sqrt2}, \qquad |{+y}\rangle = \frac{|{+z}\rangle+i|{-z}\rangle}{\sqrt2},$$

one obtains

$$P(+x) = \frac{1+\cos\phi}{2}, \qquad P(+y) = \frac{1+\sin\phi}{2}.$$

Thus $\phi=0,\pi,\pi/2,3\pi/2$ give the $+x$, $-x$, $+y$, and $-y$ rays, respectively. Multiplying the whole state by $e^{i\chi}$ multiplies every measurement amplitude by the same phase and leaves both probabilities unchanged. No collection of spin measurements can infer $\chi$.

4. Ray projector from an unnormalized vector. Let

   $$|v\rangle = \begin{pmatrix} 1+i\\ 2 \end{pmatrix}.$$

   Compute $\rho_{[v]}=|v\rangle\langle v|/\langle v|v\rangle$. Verify explicitly that replacing $|v\rangle$ by $(2-i)|v\rangle$ leaves the result unchanged.

Solution

The norm squared is

$$\langle v|v\rangle = |1+i|^2+|2|^2 = 6.$$

Therefore

$$\rho_{[v]} = \frac16 \begin{pmatrix} 2 & 2+2i\\ 2-2i & 4 \end{pmatrix}.$$

If $\lambda=2-i$, then

$$|\lambda v\rangle\langle\lambda v| = |\lambda|^2|v\rangle\langle v|, \qquad \langle\lambda v|\lambda v\rangle = |\lambda|^2\langle v|v\rangle.$$

The common factor $|\lambda|^2=5$ cancels in the quotient.

5. Unequal two-path amplitudes. At one detector, take $\mathcal A_1=a$ and $\mathcal A_2=be^{i\delta}$ with $a,b\ge0$. Derive the detection probability and show that its fringe visibility under a variable $\delta$ is

   $$\mathcal V = \frac{2ab}{a^2+b^2}.$$

   What happens under a common phase? What happens if $\delta$ is uniformly random from trial to trial?

Solution

The coherent probability is proportional to

$$P(\delta) = |a+be^{i\delta}|^2 = a^2+b^2+2ab\cos\delta.$$

Its extrema are

$$P_{\max}=(a+b)^2, \qquad P_{\min}=(a-b)^2.$$

Hence

$$\mathcal V = \frac{P_{\max}-P_{\min}} {P_{\max}+P_{\min}} = \frac{2ab}{a^2+b^2}.$$

Multiplying both amplitudes by the same $e^{i\chi}$ leaves $P$ unchanged. If $\delta$ is uniformly random, $\langle\cos\delta\rangle=0$ and the averaged probability is $a^2+b^2$; phase averaging removes the interference term.

6. Shift of the energy zero. Let $H'=H+CI$ for real constant $C$. Without assuming that $|\psi(0)\rangle$ is an energy eigenstate, prove that evolution under $H'$ produces the same ray at every time as evolution under $H$. Show that expectation values of a time-independent observable $A$ are equal.

Solution

Because $CI$ commutes with $H$,

$$e^{-iH't/\hbar} = e^{-iCt/\hbar}e^{-iHt/\hbar}.$$

Thus

$$|\psi'(t)\rangle = e^{-iCt/\hbar}|\psi(t)\rangle,$$

which is the same ray for every $t$. Consequently,

$$\begin{aligned} \langle A\rangle' &= \langle\psi(t)| e^{iCt/\hbar}A e^{-iCt/\hbar} |\psi(t)\rangle \\ &= \langle\psi(t)|A|\psi(t)\rangle. \end{aligned}$$

The scalar phase commutes with every operator and cancels.

7. Controlled phase is relative. For

   $$|\Psi_\chi\rangle = \frac{|0\rangle_R|\psi\rangle +e^{i\chi}|1\rangle_R|\psi\rangle}{\sqrt2},$$

   compute the probabilities of $|+\rangle_R$ and $|-\rangle_R$. For which values of $\chi$ is the reference certainly $+$ or certainly $-$? Explain why this does not measure the global phase of $|\psi\rangle$.

Solution

Using

$$|0\rangle_R = \frac{|+\rangle_R+|-\rangle_R}{\sqrt2}, \qquad |1\rangle_R = \frac{|+\rangle_R-|-\rangle_R}{\sqrt2},$$

the state becomes

$$|\Psi_\chi\rangle = \frac{1+e^{i\chi}}{2} |+\rangle_R|\psi\rangle + \frac{1-e^{i\chi}}{2} |-\rangle_R|\psi\rangle.$$

Therefore

$$P(+) = \frac{1+\cos\chi}{2}, \qquad P(-) = \frac{1-\cos\chi}{2}.$$

The result is certainly $+$ when $\chi=0$ modulo $2\pi$ and certainly $-$ when $\chi=\pi$ modulo $2\pi$. The experiment reads a relative phase between the two reference branches. Rephasing $|\psi\rangle$ globally multiplies the entire joint ket by that phase and leaves these probabilities unchanged.

8. Time-dependent phase convention. Suppose $|\psi(t)\rangle$ solves the Schrödinger equation with Hamiltonian $H(t)$. Derive the Hamiltonian obeyed by $|\psi'(t)\rangle=e^{i\chi(t)}|\psi(t)\rangle$. Then specialize to $\chi(t)=-Ct/\hbar$ and interpret the result.

Solution

Differentiate the rephased ket:

$$\frac{d}{dt}|\psi'\rangle = i\dot\chi e^{i\chi}|\psi\rangle + e^{i\chi}\frac{d}{dt}|\psi\rangle.$$

Multiplication by $i\hbar$ and use of the original Schrödinger equation gives

$$\begin{aligned} i\hbar\frac{d}{dt}|\psi'\rangle &= -\hbar\dot\chi|\psi'\rangle + H(t)|\psi'\rangle \\ &= \left(H(t)-\hbar\dot\chi I\right) |\psi'\rangle. \end{aligned}$$

Thus

$$H'(t)=H(t)-\hbar\dot\chi(t)I.$$

For $\chi(t)=-Ct/\hbar$, one has $-\hbar\dot\chi=C$, so

$$H'=H+CI.$$

Changing the energy zero and applying the corresponding time-dependent global phase are two descriptions of the same ray evolution.