Common Mistakes in the Formalism

A quantum calculation can have correct algebra and still answer the wrong question: a component may belong to a different basis, or a probability may refer to a different measurement. The examples below show how to recognize these mistakes and repair them. Throughout the qubit examples, $\lvert0\rangle,\lvert1\rangle$ form an orthonormal basis and $\lvert\pm\rangle=(\lvert0\rangle\pm\lvert1\rangle)/\sqrt2$.

For quick notation help, use the Symbol Map. For checking a finished calculation, use Common Checks and Sanity Tests.

States, components, and phases

A column vector needs a basis. The column $(1,0)^{\mathsf T}$ represents $\lvert0\rangle$ in the ordered basis $(\lvert0\rangle,\lvert1\rangle)$, but $\lvert+\rangle$ in the ordered basis $(\lvert+\rangle,\lvert-\rangle)$. To change coordinates without changing the state, calculate the new components $c'_j=\langle e'_j\vert\psi\rangle$. See Change of Basis.

Global phase and relative phase behave differently. Multiplying the whole ket by $e^{i\alpha}$ leaves every outcome probability unchanged. Changing the phase of only one component can change predictions: $\lvert+\rangle$ and $\lvert-\rangle$ have the same probabilities in the $0,1$ basis but opposite certain outcomes when measured in the $+,-$ basis. See Superposition and Relative Phase.

A superposition is not the mixture with the same populations. Compare

$$\rho_+=\lvert+\rangle\langle+\rvert =\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad \rho_{\rm mix}=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$

Both give $p(0)=p(1)=1/2$, but a $+,-$ measurement gives $p(+)=1$ for $\rho_+$ and $p(+)=1/2$ for $\rho_{\rm mix}$. Coherences are off-diagonal entries in a specified basis; purity itself is basis independent. See Classical Mixtures vs Quantum Superpositions.

Probabilities, averages, and state updates

An amplitude must be squared in the appropriate basis. For a normalized pure state and an orthonormal measurement basis, $p(j)=\lvert\langle j\vert\psi\rangle\rvert^2$. The coefficients in some other basis are not the probabilities for this measurement. For a general state and an outcome effect, use $p(j)=\operatorname{Tr}(\rho E_j)$; see the Born Rule.

An expectation value need not be an allowed outcome. A measurement of $\sigma_z$ has outcomes $+1$ and $-1$. Equal probabilities give $\langle\sigma_z\rangle=0$, although no run returns zero. The mean describes an ensemble of repeated preparations, not an additional spectral value.

Probabilities do not specify the entire measurement. Projectors $P_a$ determine $p(a)=\operatorname{Tr}(\rho P_a)$. If the apparatus performs the ideal Lüders measurement, its conditional state is

$$\rho_a=\frac{P_a\rho P_a}{p(a)},\qquad p(a)>0.$$

Use the full eigenspace projector for a degenerate outcome. Resolving finer states within that subspace and then forgetting which one occurred can remove coherence that the Lüders update preserves. General measurements require an additional state-update model; a POVM alone supplies only probabilities. See Degenerate Measurements and Lüders Rule.

Conditioning is different from ignoring the result. After an ideal projective measurement whose outcome is ignored, the state is $\sum_aP_a\rho P_a$. It is not any one normalized conditional state. Neither expression is generally the unitary evolution of the initial state.

Operators and uncertainty

A zero mean commutator does not imply commuting operators. Since $[\sigma_x,\sigma_y]=2i\sigma_z$, the commutator is nonzero even in a state with $\langle\sigma_z\rangle=0$. In $\lvert+\rangle$, for example, its expectation vanishes. A zero Robertson lower bound therefore need not imply compatible observables. See General Uncertainty Relations.

Matrix functions are not entrywise functions. The exponential $e^{-iHt/\hbar}$ is defined by the operator power series or spectral calculus. Exponentiating each matrix entry independently does not generally give unitary evolution. Diagonalizing $H$ lets you exponentiate its eigenvalues and then transform back; see Functions of Operators.

A differential expression is not a complete operator. To discuss self-adjointness of $-i\hbar\,d/dx$, specify the Hilbert space, the domain of functions, and their boundary conditions. Integration by parts tests a boundary term, but symmetry alone need not imply self-adjointness. See Hermitian vs Self-Adjoint Operators.

Energy–time uncertainty does not permit temporary energy nonconservation. Different energy–time relations concern evolution times, linewidths, or measurement procedures. They do not authorize violations of the conservation law for a time-independent closed-system Hamiltonian. See Energy–Time Uncertainty.

Wavefunctions and subsystems

A probability density must be integrated with its measure. In one-dimensional Cartesian coordinates, $P(a<x<b)=\int_a^b\lvert\psi(x)\rvert^2dx$. A density has units of inverse length and may exceed one. In spherical coordinates the position measure is $r^2\sin\theta\,dr\,d\theta\,d\phi$; omitting it changes normalization. Plane waves on the full line use generalized normalization and are not square-integrable state vectors. See Normalization.

A pure joint state can have mixed subsystem states. For $(\lvert00\rangle+\lvert11\rangle)/\sqrt2$, each reduced state is $I/2$. Neither subsystem can be described by its own pure-state ket. Conversely, mixed local states do not by themselves prove entanglement: a classical mixture of $\lvert00\rangle$ and $\lvert11\rangle$ has the same reductions. See Reduced States and Entangled States.

Tensor order matters. In $\mathcal H_A\otimes\mathcal H_B$, $A\otimes I_B$ acts on the first factor. Writing $I_B\otimes A$ uses a different ordering; keep the order consistent in kets, matrices, and partial traces.

Exercises

1. A normalized qubit has $\langle\sigma_z\rangle=0$. Does this establish that the state is $\lvert+\rangle$?

Solution

No. It establishes only equal probabilities for the two $\sigma_z$ outcomes. Both $\lvert+\rangle$ and $\lvert-\rangle$, as well as the mixed state $I/2$, satisfy it. Measuring $\sigma_x$ distinguishes these three preparations statistically: their means are $+1$, $-1$, and $0$.

2. A particle on $0<x<L$ has $\psi(x)=1/\sqrt L$ inside the interval and zero outside. Find the probability of detecting it in the first quarter of the interval. Why can its density be greater than $1\,\mathrm{m}^{-1}$?

Solution

The probability is $\int_0^{L/4}(1/L)\,dx=1/4$. If $L<1\,\mathrm m$, the density $1/L$ exceeds $1\,\mathrm{m}^{-1}$. Only the dimensionless integral over a region must lie between zero and one. This is a normalized square-integrable state; its discontinuities do not generally place it in the kinetic-energy operator’s domain.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018 — state vectors, probability densities, operators, and uncertainty.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — basis changes, compatible observables, and dynamics.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, Chapters 2 and 6 — states, measurements, and entanglement.