Superposition and Relative Phase

A superposition is a linear combination of state-vector representatives. If $|u\rangle$ and $|v\rangle$ belong to the same Hilbert space, then

$$|\chi\rangle = a|u\rangle+b|v\rangle$$

is another vector in that Hilbert space for any complex $a$ and $b$. If it is nonzero, it determines a pure-state ray after normalization.

That mathematical statement is simple. Its physical interpretation requires care:

• the terms in a displayed superposition depend on the chosen basis or decomposition;
• the complex coefficients become probability amplitudes only after a measurement context is specified;
• one common phase is global, while phase differences can change interference;
• a coherent superposition is not a statistical mixture;
• writing several terms does not by itself imply entanglement, simultaneous classical properties, or several measurement outcomes at once.

The experimentally meaningful content appears when amplitudes associated with coherent alternatives are combined before taking the squared modulus.

Required background. State Vectors supplies normalized kets and basis amplitudes; Rays and Global Phase separates representative freedom from physical phase relations.

Helpful background. Density Operators provides the trace language used below to distinguish coherent superpositions from statistical mixtures.

The Linear Superposition Principle

A quantum state space is built from a complex Hilbert space $\mathcal H$. Closure under linear combinations means

$$|u\rangle,|v\rangle\in\mathcal H \quad\Longrightarrow\quad a|u\rangle+b|v\rangle\in\mathcal H.$$

More generally, for a finite collection of vectors,

$$|\chi\rangle = \sum_{j=1}^{n}c_j|u_j\rangle$$

is a vector in $\mathcal H$. Infinite sums require convergence in the Hilbert space norm. In an orthonormal basis $\{|e_j\rangle\}$, a state has the expansion

$$|\psi\rangle = \sum_j c_j|e_j\rangle, \qquad \sum_j|c_j|^2=1,$$

where the second statement assumes a countable orthonormal basis and a normalized state.

The zero vector is the one exceptional linear combination: it does not represent a physical state because it cannot be normalized. Destructive cancellation can therefore make a proposed numerator vanish. For example,

$$|u\rangle-|u\rangle=0$$

is not a new state.

The linear structure is kinematic: it defines which state vectors are available. Closed-system dynamics is also linear. If $U$ is a unitary evolution operator, then

$$U\left(a|u\rangle+b|v\rangle\right) = aU|u\rangle+bU|v\rangle.$$

This is why amplitudes can be propagated alternative by alternative and then recombined. It does not mean that every formal decomposition corresponds to a laboratory procedure that independently prepares or manipulates its terms.

Normalizing a Superposition

If $|u\rangle$ and $|v\rangle$ are orthonormal, then

$$\left\lVert a|u\rangle+b|v\rangle \right\rVert^2 = |a|^2+|b|^2.$$

The state is normalized when

$$|a|^2+|b|^2=1.$$

For a nonzero unnormalized combination, the normalized representative is

$$|\psi\rangle = \frac{a|u\rangle+b|v\rangle} {\sqrt{|a|^2+|b|^2}}.$$

This familiar denominator is valid only for orthogonal alternatives. For normalized but nonorthogonal vectors, define

$$s=\langle u|v\rangle.$$

Then

$$\begin{aligned} \left\lVert a|u\rangle+b|v\rangle \right\rVert^2 &= |a|^2+|b|^2 \\ &\quad+ a^*b\,s + b^*a\,s^* \\ &= |a|^2+|b|^2 + 2\operatorname{Re}(a^*bs). \end{aligned}$$

The overlap term depends on relative phase. Consequently, $|a|^2$ and $|b|^2$ are not probabilities for mutually exclusive outcomes unless $|u\rangle$ and $|v\rangle$ are orthogonal outcomes of the relevant measurement.

For a general finite set,

$$\left\lVert \sum_j c_j|u_j\rangle \right\rVert^2 = \sum_{j,k}c_j^*c_k \langle u_j|u_k\rangle.$$

The Gram matrix $G_{jk}=\langle u_j|u_k\rangle$ controls normalization. If the vectors are linearly dependent, different coefficient lists can even represent the same vector. This is one reason that coefficients acquire a direct probability interpretation most cleanly in an orthonormal measurement basis.

Superposition Depends on the Chosen Basis

Consider the qubit states

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

Relative to the computational basis $\{|0\rangle,|1\rangle\}$, $|+\rangle$ has two nonzero components. Relative to the Hadamard basis $\{|+\rangle,|-\rangle\}$, the same abstract state is simply

$$|+\rangle = 1|+\rangle+0|-\rangle.$$

Nothing physical happened between these two descriptions. Only the coordinate system changed.

In fact, every normalized pure state can be chosen as the first element of an orthonormal basis. In finite dimensions, extend $|\psi\rangle$ to

$$\{|\psi\rangle,|e_2\rangle,\ldots,|e_d\rangle\}.$$

In that basis, the coordinate column is

$$\begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}.$$

Therefore, “this state is a superposition” is incomplete unless the basis, observable, decomposition, or physical alternatives are identified. Useful statements include:

• the state is a superposition of energy eigenstates;
• two spatial paths contribute coherently to this detector;
• the spin state has nonzero amplitudes for both $S_z$ outcomes;
• a wave packet is expanded in momentum eigenfunctions.

The abstract state does not depend on a basis. Its component list does. The full coordinate transformation rules belong to Change of Basis.

A Basis Change Is Not a Physical Superposition Operation

There are two distinct procedures that are often written with the same unitary matrix.

In a passive change of basis, one keeps $|\psi\rangle$ fixed and changes the basis used to report its components. If $S$ is the overlap matrix, the coordinate column changes while all probabilities remain the same.

In an active unitary operation, one keeps the reference basis fixed and changes the state:

$$|\psi\rangle \longmapsto U|\psi\rangle.$$

For example, saying that

$$H|0\rangle=|+\rangle$$

describes an active Hadamard operation when the computational basis is held fixed. Merely rewriting $|0\rangle$ as

$$|0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt2}$$

is a passive redescription. Confusing these statements turns a coordinate choice into imaginary dynamics.

Coefficients as Measurement Amplitudes

Let $\{|e_j\rangle\}$ be an orthonormal basis associated with a projective measurement and let

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

The coefficient

$$c_j=\langle e_j|\psi\rangle$$

is the amplitude for outcome $j$, and the Born rule gives

$$p(j)=|c_j|^2.$$

This measurement sees the coefficient magnitudes but not their phases. If

$$c_j \longmapsto e^{i\phi_j}c_j,$$

then every $|c_j|^2$ is unchanged. Relative phases become visible only through a different measurement or a physical transformation that recombines the components.

This is not a paradox. One probability distribution rarely determines a quantum state. Measurements in other bases probe other linear combinations of the amplitudes. The systematic calculation procedure is developed in Probability in Different Bases.

Global and Relative Phase

For two orthonormal alternatives, write

$$|\psi\rangle = r_0e^{i\alpha}|0\rangle + r_1e^{i\beta}|1\rangle,$$

where $r_0,r_1\ge0$ and

$$r_0^2+r_1^2=1.$$

Factoring out the first phase gives

$$|\psi\rangle = e^{i\alpha} \left( r_0|0\rangle + r_1e^{i\phi}|1\rangle \right),$$

with

$$\phi=\beta-\alpha.$$

The factor $e^{i\alpha}$ is global and does not change the ray. The remaining $\phi$ is a relative phase between the displayed components. For a qubit it is convenient to set

$$r_0=\cos\frac\theta2, \qquad r_1=\sin\frac\theta2,$$

so that

$$|\psi(\theta,\phi)\rangle = \cos\frac\theta2|0\rangle + e^{i\phi} \sin\frac\theta2|1\rangle.$$

After normalization and global phase are removed, $\theta$ and $\phi$ are the two real parameters of a pure qubit state.

The numerical value assigned to a coefficient phase depends on phase conventions for the basis kets. Observable predictions involve invariant combinations of state, transformation, and measurement amplitudes. The careful distinction between global phase, relative phase, and basis rephasing is developed in Rays and Global Phase.

How Another Basis Reveals Relative Phase

For each real analyzer phase $\gamma$, define

$$|+_\gamma\rangle = \frac{|0\rangle+e^{i\gamma}|1\rangle}{\sqrt2}, \qquad |-_\gamma\rangle = \frac{|0\rangle-e^{i\gamma}|1\rangle}{\sqrt2}.$$

These kets form an orthonormal basis. For

$$|\psi\rangle = r_0|0\rangle+r_1e^{i\phi}|1\rangle,$$

the $+_\gamma$ amplitude is

$$\langle+_\gamma|\psi\rangle = \frac{ r_0+r_1e^{i(\phi-\gamma)} }{\sqrt2}.$$

Its probability is

$$P(+_\gamma) = \frac12 \left[ 1+2r_0r_1 \cos(\phi-\gamma) \right].$$

Similarly,

$$P(-_\gamma) = \frac12 \left[ 1-2r_0r_1 \cos(\phi-\gamma) \right].$$

The cross term is the interference contribution. Its largest possible magnitude is $2r_0r_1$, which reaches one only for equal-magnitude components. If one coefficient vanishes, there are no two amplitudes to interfere and the analyzer phase becomes irrelevant.

For the equal superposition

$$|\psi_\phi\rangle = \frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2},$$

the standard $|\pm\rangle$ measurement corresponds to $\gamma=0$ and gives

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

Thus $\phi=0$ gives certainty for $+$, while $\phi=\pi$ gives certainty for $-$. The computational-basis probabilities remain $1/2$ in both cases.

Worked Qubit Interferometer

A minimal phase-readout sequence uses two Hadamard operations. Start from $|0\rangle$ and apply

$$H|0\rangle = \frac{|0\rangle+|1\rangle}{\sqrt2}.$$

Apply a phase shift to the second component:

$$P_\phi = |0\rangle\langle0| + e^{i\phi}|1\rangle\langle1|.$$

The state becomes

$$P_\phi H|0\rangle = \frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2}.$$

A computational-basis measurement at this point gives equal probabilities and does not reveal $\phi$. Applying the second Hadamard recombines the amplitudes:

$$\begin{aligned} HP_\phi H|0\rangle &= \frac{1+e^{i\phi}}{2}|0\rangle \\ &\quad+ \frac{1-e^{i\phi}}{2}|1\rangle \\ &= e^{i\phi/2} \left( \cos\frac\phi2|0\rangle -i\sin\frac\phi2|1\rangle \right). \end{aligned}$$

The leading factor is global. The output probabilities are

$$P(0)=\cos^2\frac\phi2, \qquad P(1)=\sin^2\frac\phi2.$$

The sequence has converted a relative phase into a measurable population difference. Mach–Zehnder interferometers, Ramsey sequences, and many quantum algorithms use this same amplitude-splitting, phase-accumulation, and recombination pattern.

Multiple Coherent Alternatives

Suppose the amplitude for outcome $a$ is assembled from alternatives $j$:

$$\mathcal A_a = \sum_j \mathcal A_{aj}.$$

The probability is

$$\begin{aligned} p(a) &= \left| \sum_j\mathcal A_{aj} \right|^2 \\ &= \sum_j|\mathcal A_{aj}|^2 \\ &\quad+ \sum_{j\ne k} \mathcal A_{aj}^*\mathcal A_{ak}. \end{aligned}$$

Equivalently, grouping conjugate pairs,

$$p(a) = \sum_j|\mathcal A_{aj}|^2 + 2\sum_{j<k} \operatorname{Re} \left( \mathcal A_{aj}^*\mathcal A_{ak} \right).$$

Each cross term compares a pair of alternatives. Multiplying every $\mathcal A_{aj}$ by one common phase changes none of these products. Changing their relative phases generally changes the probability.

It matters that the alternatives are coherent and experimentally indistinguishable at the point of recombination. If another degree of freedom stores perfectly distinguishable which-alternative information, the cross terms disappear from the local detection probability. The physical double-slit realization is treated in the Double-Slit Experiment.

Superposition Versus Statistical Mixture

Compare the coherent pure state

$$|\psi_\phi\rangle = \frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2}$$

with the equal incoherent mixture

$$\rho_{\mathrm{mix}} = \frac12|0\rangle\langle0| + \frac12|1\rangle\langle1|.$$

The pure-state density operator is

$$\rho_{\phi} = |\psi_\phi\rangle\langle\psi_\phi| = \frac12 \begin{pmatrix} 1 & e^{-i\phi}\\ e^{i\phi} & 1 \end{pmatrix}.$$

The mixture is

$$\rho_{\mathrm{mix}} = \frac12 \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}.$$

Both states give

$$P(0)=P(1)=\frac12.$$

They differ in measurements that recombine the basis amplitudes. For the $|\pm\rangle$ basis,

$$\begin{array}{c|cc} & P(+) & P(-) \\ \hline \rho_\phi & \dfrac{1+\cos\phi}{2} & \dfrac{1-\cos\phi}{2} \\ \rho_{\mathrm{mix}} & \dfrac12 & \dfrac12 \end{array}$$

The off-diagonal entries of $\rho_\phi$ encode coherence relative to this basis. They are absent from the mixture. Their numerical values change under a basis change, but the distinction between the two density operators is basis independent: $\rho_\phi$ has rank one and purity one, whereas $\rho_{\mathrm{mix}}$ has rank two and purity $1/2$.

$$\operatorname{Tr}(\rho_\phi^2)=1, \qquad \operatorname{Tr}(\rho_{\mathrm{mix}}^2)=\frac12.$$

This section supplies the essential comparison required to interpret superposition. Ensemble preparation, coherence matrices, basis changes, and phase averaging are developed fully in Classical Mixtures vs Quantum Superpositions.

Phase Randomization and Loss of Interference

Suppose each experimental run prepares $|\psi_\phi\rangle$, but the phase is drawn from a probability density $f(\phi)$. The ensemble state is

$$\bar\rho = \int_0^{2\pi} f(\phi) |\psi_\phi\rangle\langle\psi_\phi| \,d\phi.$$

Its off-diagonal element is proportional to

$$\left\langle e^{i\phi}\right\rangle_f = \int_0^{2\pi} f(\phi)e^{i\phi} \,d\phi.$$

For a uniform phase distribution,

$$f(\phi)=\frac1{2\pi},$$

and

$$\left\langle e^{i\phi}\right\rangle_f=0.$$

The ensemble density operator becomes $I/2$, and the interference fringe vanishes. This is not the same description as one run possessing an unknown but fixed phase unless the operational preparation really is an ensemble over phases. Density operators encode the statistics available from the specified preparation procedure.

Environmental entanglement can suppress local coherence through a related but distinct mechanism. That dynamical process belongs to Decoherence Preview.

Superposition in Wave Mechanics

Wavefunctions inherit the linear structure of state vectors. If $\psi_1(x)$ and $\psi_2(x)$ represent states in the same configuration space, then

$$\psi(x) = c_1\psi_1(x)+c_2\psi_2(x)$$

represents their linear combination, subject to normalization and boundary conditions. The position probability density is

$$\begin{aligned} |\psi(x)|^2 &= |c_1|^2|\psi_1(x)|^2 + |c_2|^2|\psi_2(x)|^2 \\ &\quad+ 2\operatorname{Re} \left[ c_1^*c_2 \psi_1^*(x)\psi_2(x) \right]. \end{aligned}$$

The last term is the local interference term. Its value depends on the relative phase of both the coefficients and the wavefunctions.

If $\psi_1$ and $\psi_2$ are orthogonal, then

$$\int \psi_1^*(x)\psi_2(x) \,dx=0.$$

This ensures that the integrated cross term vanishes in the normalization integral. It does not require $\psi_1^*(x)\psi_2(x)=0$ point by point. Orthogonal modes can therefore produce position-dependent interference.

Conversely, two wave packets with disjoint support have no local position interference while separated, even if their joint pure state remains a coherent superposition. Bringing the packets back into overlap can reveal the stored relative phase. Absence of fringes in one measurement arrangement is not by itself proof of a statistical mixture.

The relation between the abstract ket and its position representation is developed in Wavefunctions as Representations.

Worked Wave Example on a Ring

Consider a particle on a ring of circumference $L$. For $k=2\pi n/L$ with nonzero integer $n$, the counterpropagating momentum eigenfunctions

$$\phi_{+}(x) = \frac{e^{ikx}}{\sqrt L}, \qquad \phi_{-}(x) = \frac{e^{-ikx}}{\sqrt L}$$

are orthonormal. The equal coherent superposition

$$\psi_\delta(x) = \frac{ e^{ikx}+e^{i\delta}e^{-ikx} }{\sqrt{2L}}$$

has equal probabilities $1/2$ for momenta $+\hbar k$ and $-\hbar k$, independent of $\delta$. Its position density is

$$|\psi_\delta(x)|^2 = \frac1L \left[ 1+\cos(2kx-\delta) \right].$$

The relative phase translates the standing-wave pattern. A uniform incoherent mixture of the two momenta instead gives

$$p_{\mathrm{mix}}(x)=\frac1L,$$

with no fringes. Momentum probabilities alone cannot distinguish the two preparations; position probabilities can.

Relative Phase Under Time Evolution

Let $|E_n\rangle$ be energy eigenvectors of a time-independent Hamiltonian. For an initial superposition

$$|\psi(0)\rangle = \sum_n c_n|E_n\rangle,$$

unitary evolution gives

$$|\psi(t)\rangle = \sum_n c_ne^{-iE_nt/\hbar}|E_n\rangle.$$

The magnitudes $|c_n|$ remain fixed. Define the phase of component $n$ relative to component $m$ by

$$\phi_{nm}(t)=\arg c_n(t)-\arg c_m(t).$$

Then

$$\phi_{nm}(t) = \phi_{nm}(0) - \frac{E_n-E_m}{\hbar}t.$$

If all occupied components share one energy, the entire state acquires only a global phase and its ray is stationary. If distinct energies are present, measurements with matrix elements between those energy sectors can exhibit oscillations, beats, or wave-packet motion.

The canonical discussion of this distinction is Stationary States.

Superposition Is Not Entanglement

Every entangled pure state is a superposition in many product bases, but not every displayed superposition is entangled. Consider two qubits. The product state

$$|+\rangle_A|+\rangle_B = \frac12 \left( |00\rangle+|01\rangle+|10\rangle+|11\rangle \right)$$

contains four terms in the computational product basis and is nevertheless separable.

By contrast, the Bell state

$$|\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2}$$

cannot be factored into one ket for $A$ and one ket for $B$. Its entanglement is a basis-independent property under local basis changes, whereas the number of nonzero terms in a displayed expansion is not.

Superposition concerns linear combinations in one Hilbert space. Entanglement concerns factorization relative to a tensor-product decomposition of a composite Hilbert space. The latter is developed in Entangled States.

When Relative Phase Is Operationally Inaccessible

The statement “relative phase can be observed” assumes access to measurements or operations that coherently mix the relevant alternatives. Suppose

$$\mathcal H = \mathcal H_1\oplus\mathcal H_2$$

and every allowed effect is block diagonal:

$$E = E_1\oplus E_2.$$

For

$$|\psi_\phi\rangle = a|u\rangle + be^{i\phi}|v\rangle,$$

with $|u\rangle\in\mathcal H_1$ and $|v\rangle\in\mathcal H_2$, the cross matrix elements vanish:

$$\langle u|E|v\rangle=0.$$

All allowed probabilities are then independent of $\phi$. This is the operational structure behind a superselection rule. It is stronger than saying one particular apparatus happens not to measure the phase, and it is not implied merely by conservation of a quantity.

Reference systems can sometimes enlarge the operational description and turn an apparently inaccessible phase into a relational one. The qualifications and symmetry constraints belong to Superselection Sectors Preview.

What Superposition Does and Does Not Mean

Superposition is a precise statement about vectors and amplitudes. It does not by itself select an interpretation of quantum mechanics.

It is safe to say:

• the state has amplitudes for several outcomes of a specified measurement;
• coherent alternatives contribute to one outcome amplitude;
• relative phase changes interference when the alternatives are recombined;
• linear evolution maps a linear combination to the same linear combination of the evolved vectors.

Extra care is needed with statements such as “the system is in several classical states at once.” Basis states need not represent complete classical configurations, and a quantum state is not a list of simultaneously possessed measurement outcomes. The formalism predicts probability distributions and correlations for specified experiments. Interpretive claims go beyond the linear-algebra statement alone.

A Practical Superposition Audit

When a calculation displays a sum of states or amplitudes, check:

1. Common Hilbert space. Are all terms vectors for the same complete system and compatible boundary conditions?
2. Nonzero vector. Does the proposed combination avoid exact cancellation?
3. Normalization. Are the alternatives orthogonal? If not, include their overlaps in the norm.
4. Specified decomposition. Which basis, observable, path, mode, or energy alternatives define the terms?
5. Global versus relative phase. Can one common phase be factored from the whole ket?
6. Measurement context. Which amplitudes are combined before the squared modulus is taken?
7. Mixture check. Is the preparation coherent, or is it an ensemble with missing or randomized phase information?
8. Composite-system check. Is the issue ordinary basis expansion or nonfactorization across subsystems?
9. Operational access. Can available transformations and measurements mix the alternatives and reveal their phase relation?

Common Mistakes

Calling superposition basis independent

The state is basis independent, but the statement that it has several components refers to a basis or decomposition. Name that structure.

Reading coefficient magnitudes as probabilities in a nonorthogonal set

For nonorthogonal alternatives, overlap terms enter normalization and the alternatives are not mutually exclusive projective outcomes.

Saying global phase makes all phases irrelevant

Only one common phase of the complete state is redundant. Relative phases can change measurements that recombine amplitudes.

Confusing a superposition with a mixture

A ket sum combines amplitudes coherently. A statistical mixture combines probabilities and is represented by a density operator. They may agree in one basis and differ in another.

Inferring entanglement from several terms

Term count changes under basis transformations. Entanglement is failure to factor across a specified subsystem tensor product.

Expecting interference in every measurement

A relative phase can be invisible in the basis that defines the components. Interference appears only when the measurement or dynamics recombines them.

Treating missing fringes as proof of collapse

Disjoint support, averaging, uncontrolled phase drift, distinguishable path records, or environmental entanglement can all suppress a particular interference signal.

Connections

• Rays and Global Phase distinguishes representative freedom from measurable phase relations.
• State Vectors develops the abstract vectors and basis amplitudes combined here.
• Probability Amplitudes gives the canonical amplitude-addition and Born-rule workflow.
• Density Operators supplies the general-state formalism used to separate mixtures from coherent pure states.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, Chapters 2 and 3.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. I, Wiley, 1977, Chapters II and III.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, Chapters I and III.
• L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, Chapters 3 and 4.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, Sections 2.1 and 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 and 2.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Chapters 4 and 5.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955, Chapters I and III.

Exercises

1. Nonorthogonal normalization. Let $|u\rangle$ and $|v\rangle$ be normalized states with

   $$\langle u|v\rangle=\frac12.$$

   Find the normalization factor for

   $$|\chi_\phi\rangle = |u\rangle+e^{i\phi}|v\rangle.$$

   For which $\phi$ is the norm largest and smallest?

Solution

The squared norm is

$$\begin{aligned} \langle\chi_\phi|\chi_\phi\rangle &= 1+1 + e^{i\phi}\langle u|v\rangle + e^{-i\phi}\langle v|u\rangle \\ &= 2+\cos\phi. \end{aligned}$$

Therefore

$$|\psi_\phi\rangle = \frac{ |u\rangle+e^{i\phi}|v\rangle }{\sqrt{2+\cos\phi}}.$$

The norm is largest at $\phi=0$ modulo $2\pi$, where its square is $3$, and smallest at $\phi=\pi$ modulo $2\pi$, where its square is $1$. The phase affects normalization because the alternatives overlap.

2. One state in two bases. Express $|0\rangle$ and $|+\rangle$ in both the computational basis and the Hadamard basis. Explain why counting nonzero coefficients cannot define a basis-independent amount of superposition.

Solution

In the computational basis,

$$|0\rangle = 1|0\rangle+0|1\rangle, \qquad |+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt2}.$$

Using

$$|0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt2},$$

the same states in the Hadamard basis are

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

Each state has one nonzero component in a basis containing that state and two nonzero components in the other basis. The term count is therefore a property of the representation, not of the abstract ray alone.

3. Phase readout. Verify directly that

   $$HP_\phi H|0\rangle = e^{i\phi/2} \left( \cos\frac\phi2|0\rangle -i\sin\frac\phi2|1\rangle \right).$$

   Find the output probabilities for $\phi=0,\pi/2,\pi$, and $3\pi/2$.

Solution

After the first two operations,

$$P_\phi H|0\rangle = \frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2}.$$

The final Hadamard gives

$$HP_\phi H|0\rangle = \frac{1+e^{i\phi}}2|0\rangle + \frac{1-e^{i\phi}}2|1\rangle.$$

Using

$$1+e^{i\phi} = 2e^{i\phi/2}\cos\frac\phi2,$$

and

$$1-e^{i\phi} = -2ie^{i\phi/2}\sin\frac\phi2,$$

gives the stated result. Since the leading factor is global,

$$P(0)=\cos^2\frac\phi2, \qquad P(1)=\sin^2\frac\phi2.$$

Thus the pairs $(P(0),P(1))$ are

$$\begin{array}{c|c} \phi & (P(0),P(1)) \\ \hline 0 & (1,0) \\ \pi/2 & (1/2,1/2) \\ \pi & (0,1) \\ 3\pi/2 & (1/2,1/2) \end{array}$$

The two equal-probability cases have different output relative phases and can be distinguished by an additional phase-sensitive measurement.

4. Three-path interference. Let

   $$|\psi\rangle = \frac{ |0\rangle+e^{i\phi}|1\rangle+e^{i\chi}|2\rangle }{\sqrt3}$$

   and

   $$|f_0\rangle = \frac{|0\rangle+|1\rangle+|2\rangle}{\sqrt3}.$$

   Compute $P(f_0)$ and identify its pairwise phase differences.

Solution

The amplitude is

$$\langle f_0|\psi\rangle = \frac{1+e^{i\phi}+e^{i\chi}}3.$$

Therefore

$$\begin{aligned} P(f_0) &= \frac19 \left|1+e^{i\phi}+e^{i\chi}\right|^2 \\ &= \frac19 \left[ 3 +2\cos\phi +2\cos\chi +2\cos(\phi-\chi) \right]. \end{aligned}$$

The three cross terms compare paths $(0,1)$, $(0,2)$, and $(1,2)$. A common phase added to all three amplitudes leaves every phase difference unchanged.

5. Pure superposition versus mixture. For

   $$|\psi_\phi\rangle = \frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2},$$

   calculate $\langle\sigma_x\rangle$ and $\langle\sigma_y\rangle$. Compare them with the equal mixture of $|0\rangle$ and $|1\rangle$. Which two observables can determine $\phi$ up to the usual $2\pi$ periodicity?

Solution

Using

$$\sigma_x = |0\rangle\langle1|+|1\rangle\langle0|,$$

and

$$\sigma_y = -i|0\rangle\langle1| +i|1\rangle\langle0|,$$

one finds

$$\langle\sigma_x\rangle_{\psi_\phi} = \cos\phi, \qquad \langle\sigma_y\rangle_{\psi_\phi} = \sin\phi.$$

For $\rho_{\mathrm{mix}}=I/2$, both expectation values vanish. Measuring $\sigma_x$ and $\sigma_y$ estimates both cosine and sine and therefore determines $\phi$ modulo $2\pi$. Neither observable alone distinguishes all phases.

6. Standing-wave phase. For the ring state

   $$\psi_\delta(x) = \frac{ e^{ikx}+e^{i\delta}e^{-ikx} }{\sqrt{2L}},$$

   derive the position density. Show that increasing $\delta$ by $\Delta\delta$ translates the pattern by

   $$\Delta x=\frac{\Delta\delta}{2k}.$$

   Verify normalization over the ring for allowed nonzero $k=2\pi n/L$.

Solution

Expanding the squared modulus gives

$$\begin{aligned} |\psi_\delta(x)|^2 &= \frac1{2L} \left( 2 +e^{i(\delta-2kx)} +e^{-i(\delta-2kx)} \right) \\ &= \frac1L \left[ 1+\cos(2kx-\delta) \right]. \end{aligned}$$

Keeping the cosine argument fixed under $\delta\mapsto\delta+\Delta\delta$ requires

$$x\mapsto x+\frac{\Delta\delta}{2k}.$$

For $k=2\pi n/L$ with nonzero integer $n$,

$$\int_0^L\cos(2kx-\delta)\,dx=0.$$

Hence

$$\int_0^L|\psi_\delta(x)|^2\,dx=1.$$

7. Energy beating. Let

   $$|\psi(0)\rangle = \frac{|E_1\rangle+e^{i\phi_0}|E_2\rangle}{\sqrt2},$$

   where $E_1\ne E_2$. For

   $$A = |E_1\rangle\langle E_2| + |E_2\rangle\langle E_1|,$$

   calculate $\langle A\rangle(t)$.

Solution

Time evolution gives

$$|\psi(t)\rangle = \frac{ e^{-iE_1t/\hbar}|E_1\rangle + e^{i\phi_0}e^{-iE_2t/\hbar}|E_2\rangle }{\sqrt2}.$$

The relative phase of the second component with respect to the first is

$$\phi(t) = \phi_0- \frac{E_2-E_1}{\hbar}t.$$

Since $A$ exchanges the two energy states,

$$\langle A\rangle(t) = \cos\phi(t) = \cos\left( \phi_0- \frac{E_2-E_1}{\hbar}t \right).$$

The energy-basis populations remain $1/2$, while this phase-sensitive observable oscillates at angular frequency $|E_2-E_1|/\hbar$.

8. Many terms do not imply entanglement. Expand

   $$|+\rangle_A|+\rangle_B$$

   in the computational basis and compare it with $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Use coefficient factorization to show that the first state is a product and the second is not.

Solution

The first state is

$$|+\rangle_A|+\rangle_B = \frac12 \left( |00\rangle+|01\rangle+|10\rangle+|11\rangle \right).$$

Its coefficient matrix is

$$C_+ = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix},$$

which has rank one and factors as

$$C_+ = \frac1{\sqrt2} \begin{pmatrix}1\\1\end{pmatrix} \frac1{\sqrt2} \begin{pmatrix}1&1\end{pmatrix}.$$

For the Bell state,

$$C_{\Phi^+} = \frac1{\sqrt2} \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix},$$

which has rank two and cannot be written as one column times one row. The first state has more displayed terms in this basis but no entanglement; the second has fewer terms and is entangled.