Quantum states and operators repair

Quantum field theory enlarges the Hilbert spaces, introduces operator-valued distributions, and may admit inequivalent representations, but it still relies on a basic operator grammar: states determine probabilities, observables have spectral data, commutators encode compatibility and dynamics, and picture changes cannot alter predictions. This focused review rebuilds that grammar in finite dimension, where every step can be checked explicitly.

Required background. You should be able to multiply complex matrices, take an adjoint, normalize a vector, and diagonalize a small Hermitian matrix. Review linear and tensor methods if basis changes or adjoints are the obstacle. Finite matrices are used here to repair the reasoning; the final section states what must be added for QFT.

States turn operators into predictions

A pure state is a ray in a Hilbert space. Choose a normalized representative $|\psi\rangle$ with $\langle\psi|\psi\rangle=1$. A mixed state is described by a density operator

$$\rho=\rho^\dagger, \qquad \rho\ge0, \qquad \operatorname{tr}\rho=1.$$

For a self-adjoint observable $A$ with spectral decomposition

$$A=\sum_a aP_a, \qquad P_aP_b=\delta_{ab}P_a, \qquad \sum_aP_a=I,$$

the probability of outcome $a$ is

$$p(a)=\langle\psi|P_a|\psi\rangle$$

for a pure state and $p(a)=\operatorname{tr}(\rho P_a)$ for a mixed state. Degenerate eigenspaces require projectors of rank greater than one; choosing a particular basis inside the eigenspace must not change the probability.

An expectation value is a number,

$$\langle A\rangle_\rho=\operatorname{tr}(\rho A),$$

not another state or operator. A transition matrix element $\langle\phi|A|\psi\rangle$ is different again: it names an initial state, final state, and operator. Keeping these types visible prevents the common mistake of calling a field component, a matrix element, and a measured value the same object. The finite-dimensional spectral theorem and its quantum use are developed in Hall 2013, chs. 2–3.

Pure superposition is not statistical mixing

For a two-level system, let $|0\rangle$ and $|1\rangle$ be eigenstates of $\sigma_z$. Compare

$$|+x\rangle=\frac{|0\rangle+|1\rangle}{\sqrt2}, \qquad \rho_{\mathrm{mix}} =\frac12|0\rangle\langle0|+\frac12|1\rangle\langle1|.$$

Both give equal probabilities for a $\sigma_z$ measurement, but

$$\langle+x|\sigma_x|+x\rangle=1, \qquad \operatorname{tr}(\rho_{\mathrm{mix}}\sigma_x)=0.$$

The off-diagonal terms in $|+x\rangle\langle+x|$ retain phase coherence; the mixture lacks them in this basis. The basis-independent distinction is $\operatorname{tr}\rho^2=1$ for a pure state and less than one for a genuinely mixed finite-dimensional state. A density matrix is not merely an admission that the ket is unknown: it is the complete state used to predict all measurements within the model.

Dynamics must agree in every picture

For a time-independent Hamiltonian $H$, the Schrödinger picture evolves the state,

$$|\psi_S(t)\rangle=e^{-iHt}|\psi_S(0)\rangle, \qquad A_S=A,$$

while the Heisenberg picture fixes the state and evolves the operator,

$$A_H(t)=e^{iHt}Ae^{-iHt}, \qquad \frac{\mathrm dA_H}{\mathrm dt}=i[H,A_H]$$

when $A$ has no explicit time dependence. The observable prediction is the same:

$$\langle\psi_S(t)|A|\psi_S(t)\rangle =\langle\psi_S(0)|A_H(t)|\psi_S(0)\rangle.$$

Take

$$H=\frac\omega2\sigma_z, \qquad |\psi(0)\rangle=|+x\rangle, \qquad A=\sigma_x.$$

Direct evolution gives

$$\langle A\rangle_t=\cos(\omega t).$$

The Heisenberg operator is

$$A_H(t)=\cos(\omega t)\sigma_x-sin(\omega t)\sigma_y,$$

which gives the same answer in $|+x\rangle$. Evolving both the state and this Heisenberg operator would count the same unitary transformation twice.

For a time-dependent Hamiltonian, replace the simple exponential by the time-evolution operator satisfying $i\partial_tU(t,t_0)=H(t)U(t,t_0)$. Time ordering is then part of the definition; exponentials at distinct times cannot be combined as if the Hamiltonians commute.

Symmetries are actions, not labels

Let a continuous unitary transformation be

$$U(\alpha)=e^{-i\alpha Q}$$

with self-adjoint generator $Q$. To first order,

$$U(\alpha)^\dagger A U(\alpha) =A+i\alpha[Q,A]+\mathcal O(\alpha^2).$$

The transformation is a symmetry of a time-independent Hamiltonian when $U^\dagger HU=H$. Infinitesimally this gives $[Q,H]=0$, and the Heisenberg equation then makes $Q$ conserved. These statements depend on what the transformation acts on and on the domain of the operators. A phase convention or basis change can alter components without being a physical symmetry; an antiunitary transformation requires a separate treatment.

Commutators also separate operator statements from state-dependent ones. The Robertson bound

$$(\Delta A)_\psi(\Delta B)_\psi \ge\frac12\left|\langle\psi|[A,B]|\psi\rangle\right|$$

is evaluated in a specified state. A nonzero operator commutator does not force the right-hand side to be nonzero in every state, nor does saturation hold automatically.

Spectral sums are the bridge to QFT correlators

Let $H|n\rangle=E_n|n\rangle$, let $|0\rangle$ be a nondegenerate ground state, and define $A(t)=e^{iHt}Ae^{-iHt}$. Inserting the identity gives

$$\begin{aligned} \langle0|A(t)A(0)|0\rangle &=\sum_n\langle0|A(t)|n\rangle\langle n|A|0\rangle\\ &=\sum_ne^{-i(E_n-E_0)t} \left|\langle n|A|0\rangle\right|^2 \end{aligned}$$

for self-adjoint $A$. The frequencies identify excitation energies and the nonnegative coefficients are transition strengths. This is the finite prototype of a QFT spectral representation.

The transfer to QFT needs additional hypotheses: translation invariance, spectrum bounded below, a specified vacuum representation, positivity of the relevant state space, distributional smearing, and an integral over continuous multi-particle spectra. Gauge-fixed fields may live in an auxiliary space where naive positivity does not apply, and non-vacuum states change the spectral weights. Finite-dimensional success therefore supports the operator grammar but does not prove a Källén–Lehmann representation. See Weinberg 1995, §§ 10.2–10.7 for the relativistic spectral setting.

Domains become consequential beyond matrices

Every linear map in finite dimension is bounded and defined everywhere. QFT operators are often unbounded, and fields are generally distributions that must be smeared with test functions before they become operators. For an unbounded $A$, the formal equality $A=A^\dagger$ is incomplete until the domains are specified; a symmetric operator need not be self-adjoint, and a product $AB$ needs a common domain on which it is defined.

Likewise, canonical commutation relations specify an algebra but do not by themselves select a unique representation in an infinite system. Vacuum, thermal, curved-spacetime, and inequivalent phase representations can assign different state-dependent correlators to the same abstract relations. These qualifications are not needed to solve the finite exercises, but they must be restored before exporting a conclusion to QFT.

Exercises

1. Distinguish a coherent state from a mixture

For $|\psi_\varphi\rangle=(|0\rangle+e^{i\varphi}|1\rangle)/\sqrt2$, compute the density matrix, $\langle\sigma_x\rangle$, and $\langle\sigma_y\rangle$. Compare with $\rho_{\mathrm{mix}}=I/2$.

Solution

In the $\sigma_z$ basis,

$$\rho_\varphi =\frac12\begin{pmatrix}1&e^{-i\varphi}\\e^{i\varphi}&1\end{pmatrix}.$$

Therefore

$$\langle\sigma_x\rangle=\operatorname{tr}(\rho_\varphi\sigma_x) =\cos\varphi, \qquad \langle\sigma_y\rangle=\operatorname{tr}(\rho_\varphi\sigma_y) =\sin\varphi.$$

The mixture $I/2$ gives zero for both observables. The two states agree on the $\sigma_z$ probabilities but are distinguished by measurements sensitive to the off-diagonal coherence. Also $\operatorname{tr}\rho_\varphi^2=1$, whereas $\operatorname{tr}(I/2)^2=1/2$.

2. Check the two pictures directly

For the Hamiltonian and initial state used above, calculate the evolved Schrödinger state and obtain $\langle\sigma_x\rangle_t$. Then evaluate the Heisenberg result from $A_H(t)$.

Solution

Since $\sigma_z|0\rangle=|0\rangle$ and $\sigma_z|1\rangle=-|1\rangle$,

$$|\psi_S(t)\rangle =\frac{e^{-i\omega t/2}|0\rangle +e^{i\omega t/2}|1\rangle}{\sqrt2}.$$

The off-diagonal matrix elements of $\sigma_x$ give

$$\langle\psi_S(t)|\sigma_x|\psi_S(t)\rangle =\frac12(e^{i\omega t}+e^{-i\omega t}) =\cos(\omega t).$$

In the Heisenberg picture,

$$\langle+x|A_H(t)|+x\rangle =\cos(\omega t)\langle\sigma_x\rangle -\sin(\omega t)\langle\sigma_y\rangle =\cos(\omega t).$$

The agreement checks that evolution has been assigned to exactly one side of the expectation value.

3. Build a spectral correlator

For the same Hamiltonian, take the lower-energy state $|1\rangle$ as the ground state and $A=\sigma_x$. Compute $C(t)=\langle1|A(t)A(0)|1\rangle$ by inserting a complete basis.

Solution

$A|1\rangle=|0\rangle$, so only the excited state $|0\rangle$ contributes. Its energy difference is $E_0-E_1=\omega$, and the transition matrix element has unit magnitude. Hence

$$C(t)=e^{-i\omega t}.$$

Directly, $A(0)|1\rangle=|0\rangle$, time evolution supplies the relative phase $e^{-i(E_0-E_1)t}$, and the second matrix element returns to $|1\rangle$. A continuum QFT spectrum replaces this single term by pole and multi-particle contributions under the extra hypotheses stated above.

Re-check and return

Choose a two- or three-level Hamiltonian that is not diagonal in your working basis, a normalized pure or mixed state, one observable, and one candidate symmetry generator. Compute its spectral projectors and probabilities, the same time-dependent expectation value in both pictures, the commutator of the generator with the Hamiltonian, and one two-point spectral sum.

The repair is complete when probabilities normalize, projectors reconstruct their operator, both pictures agree, and the symmetry and spectral claims follow from displayed calculations. If only a domain, degeneracy, or state assumption remains implicit, add it and repeat with changed matrices. If the pictures disagree, return to the two-level example and evolve only the state or only the operator.

Then retry the quantum-mechanics diagnostic or return to Readiness. The next Core application is Canonical quantization and the free scalar.

References

• Brian C. Hall, Quantum Theory for Mathematicians, Springer, 2013, doi:10.1007/978-1-4614-7116-5.
• J. J. Sakurai and Jim Napolitano, Modern Quantum Mechanics, third edition, Cambridge University Press, 2021, doi:10.1017/9781108587280.
• Steven Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995, doi:10.1017/CBO9781139644167.