Conservation Laws

A conservation law identifies a quantity whose statistics do not change under the dynamics. For a closed quantum system, the decisive object is not the Hamiltonian alone and not the observable alone, but the combination

$$\mathcal K_A(t) = \frac{\partial A(t)}{\partial t} + \frac{i}{\hbar}[H(t),A(t)].$$

Whenever the relevant derivatives and operator products exist,

$$\frac{d}{dt}\langle A(t)\rangle = \langle\mathcal K_A(t)\rangle.$$

Thus a time-independent observable $A$ is conserved for every state if it commutes with the Hamiltonian at every time. The explicit derivative matters: a deliberately time-dependent observable can be conserved even when $[H,A]\neq 0$, and a time-dependent Hamiltonian need not conserve its own expectation value.

This page owns the dynamical derivation and the practical conservation tests. The relation to unitary symmetry generators and Noether’s theorem is developed in Commutators and Conservation Laws and Generators.

What Does Conserved Mean?

Several claims that sound similar are mathematically different.

Claim	Criterion	What follows
The mean is instantaneously stationary in one state	$\langle\mathcal K_A\rangle=0$ in that state at that instant	Only that first derivative vanishes
The mean is conserved for every initial state	$\mathcal K_A=0$ as an operator or quadratic form	Every evolving state has a constant mean
The full outcome distribution is conserved	$U^\dagger A(t)U=A(t_0)$, with the time arguments understood	Every spectral probability, and hence every well-defined moment, is constant
The system has a sharp conserved value	The state lies in one invariant eigenspace of $A$	Repeated ideal measurements return that eigenvalue with certainty

The second and third statements are equivalent under the usual finite-dimensional or suitably regular spectral assumptions. Neither implies the fourth. A superposition can retain fixed probabilities for several values of a conserved observable.

The closed-system assumptions used below are:

• $H(t)$ is self-adjoint and generates a unitary propagator $U(t,t_0)$;
• $A(t)$ is self-adjoint at each time and differentiable in the required sense;
• the state lies in domains on which the displayed products and expectation values are defined.

Finite-dimensional systems satisfy these domain requirements automatically. Unbounded operators require extra care, discussed below.

Derivation for Pure States

Let the Schrödinger-picture state obey

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

For a possibly time-dependent observable $A(t)$,

$$\langle A(t)\rangle = \langle\psi(t)\rvert A(t)\lvert\psi(t)\rangle.$$

Differentiating all three factors gives

$$\begin{aligned} \frac{d}{dt}\langle A\rangle &= \frac{d\langle\psi\rvert}{dt} A\lvert\psi\rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle \\ &\quad+ \langle\psi\rvert A \frac{d\lvert\psi\rangle}{dt}. \end{aligned}$$

Self-adjointness of $H(t)$ implies

$$\begin{aligned} \frac{d\lvert\psi\rangle}{dt} &= -\frac{i}{\hbar}H\lvert\psi\rangle,\\ \frac{d\langle\psi\rvert}{dt} &= \frac{i}{\hbar}\langle\psi\rvert H. \end{aligned}$$

Substitution yields

$$\begin{aligned} \frac{d}{dt}\langle A\rangle &= \frac{i}{\hbar}\langle HA\rangle - \frac{i}{\hbar}\langle AH\rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle\\ &= \frac{i}{\hbar}\langle[H,A]\rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle. \end{aligned}$$

The Hamiltonian may depend explicitly on time. Both $H(t)$ and $A(t)$ in the commutator are evaluated at the same instant.

The symbol $\partial A/\partial t$ means explicit dependence of the operator assigned to the observable. It does not include the changing state. The total derivative of the expectation value includes both effects.

Density-Operator Form

The same result applies to pure and mixed states. Under closed unitary evolution, the density operator satisfies the von Neumann equation

$$\frac{d\rho}{dt} = -\frac{i}{\hbar}[H,\rho].$$

Using $\langle A\rangle=\operatorname{Tr}(\rho A)$,

$$\begin{aligned} \frac{d}{dt}\langle A\rangle &= \operatorname{Tr}(\dot\rho A)\\ &\quad+ \operatorname{Tr}(\rho\,\partial_t A)\\ &= -\frac{i}{\hbar} \operatorname{Tr}\!\left([H,\rho]A\right)\\ &\quad+ \operatorname{Tr}(\rho\,\partial_t A). \end{aligned}$$

Cyclicity of the trace gives

$$\operatorname{Tr}\!\left([H,\rho]A\right) = -\operatorname{Tr}\!\left(\rho[H,A]\right),$$

and therefore

$$\frac{d}{dt}\operatorname{Tr}(\rho A) = \operatorname{Tr}\!\left[ \rho\left( \frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A] \right) \right].$$

This trace manipulation is immediate for finite matrices. In infinite dimensions, one must ensure that the relevant products are trace class or justify the identity by an appropriate limiting argument. See Trace Rule and Expectation Values for the density-operator framework.

State-Specific Cancellation Is Not a Law

Define

$$\mathcal K_A = \frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A].$$

If $\langle\mathcal K_A\rangle=0$ in one state, the expectation value is stationary at that instant. This need not imply that $\mathcal K_A=0$, that the mean remains constant later, or that any measurement probabilities are conserved.

For example, consider

$$H= \frac{\hbar\Omega}{2}\sigma_z, \qquad A=\sigma_x.$$

Since

$$\frac{i}{\hbar}[H,\sigma_x] = -\Omega\sigma_y,$$

the initial state $\lvert{+x}\rangle$ has

$$\left. \frac{d}{dt}\langle\sigma_x\rangle \right|_{t=0} =0$$

because $\langle\sigma_y\rangle=0$. Nevertheless,

$$\langle\sigma_x\rangle_t = \cos(\Omega t),$$

so $\sigma_x$ is not conserved. The first derivative happens to vanish at the top of the cosine.

By contrast, if $\langle\mathcal K_A\rangle=0$ for every density operator, then $\mathcal K_A=0$ in finite dimensions. More generally, states separate bounded observables, so vanishing expectations for all states imply the operator identity. For unbounded operators the analogous statement is formulated on an appropriate common domain.

Operator Conservation Criterion

An observable is a state-independent constant of motion when

$$\mathcal K_A(t)=0,$$

or explicitly,

$$\frac{\partial A(t)}{\partial t} + \frac{i}{\hbar}[H(t),A(t)] =0.$$

For an observable with no explicit time dependence, this reduces to

$$[H(t),A]=0$$

at every time in the interval of interest. This is sufficient even when Hamiltonians at different times fail to commute:

$$[H(t_1),H(t_2)]\neq 0$$

does not obstruct conservation of a fixed $A$ that commutes with each $H(t)$ separately.

The distinction between these two commutators is important. Pairwise commutation of $H(t)$ controls whether time ordering simplifies; commutation of $A$ with $H(t)$ controls whether $A$ is conserved.

Propagator Form

Let $U(t,t_0)$ satisfy

$$\begin{aligned} i\hbar\frac{\partial U(t,t_0)}{\partial t} &= H(t)U(t,t_0),\\ U(t_0,t_0)&=I. \end{aligned}$$

Write $U=U(t,t_0)$ temporarily and differentiate the pulled-back observable:

$$\frac{d}{dt} \left[ U^\dagger A(t)U \right] = U^\dagger\mathcal K_A(t)U.$$

Therefore $\mathcal K_A=0$ is equivalent to

$$U(t,t_0)^\dagger A(t)U(t,t_0) = A(t_0),$$

subject to the regularity and domain assumptions already stated. Equivalently,

$$A(t) = U(t,t_0)A(t_0)U(t,t_0)^\dagger.$$

This expression explains how explicit time dependence can compensate for dynamical noncommutation. It also connects directly to the Heisenberg Equations of Motion: a Schrödinger-picture dynamical invariant is constant after being pulled back to the reference time.

For a fixed observable $A$, the criterion becomes

$$U(t,t_0)^\dagger A U(t,t_0)=A.$$

That statement is often more robust than a formal commutator equation for unbounded operators.

Conservation of the Full Distribution

Let $P_t(\Delta)$ be the spectral projector of $A(t)$ for a measurable set of outcomes $\Delta$. If

$$A(t)=U(t,t_0)A(t_0)U(t,t_0)^\dagger,$$

then the spectral calculus gives

$$P_t(\Delta) = U(t,t_0)P_{t_0}(\Delta)U(t,t_0)^\dagger.$$

The evolved state is

$$\rho(t) = U(t,t_0)\rho(t_0)U(t,t_0)^\dagger.$$

Hence the Born probability is constant:

$$\begin{aligned} p_t(\Delta) &= \operatorname{Tr}\!\left[ \rho(t)P_t(\Delta) \right]\\ &= \operatorname{Tr}\!\left[ \rho(t_0)P_{t_0}(\Delta) \right]\\ &= p_{t_0}(\Delta). \end{aligned}$$

Thus operator conservation preserves the entire measurement distribution, not only its mean. Whenever the moments exist,

$$\langle A^n\rangle_t = \langle A^n\rangle_{t_0}$$

for every positive integer $n$. In particular, the variance is constant.

Conservation does not mean that the state is an eigenstate of $A$. For

$$\lvert\psi(t_0)\rangle = \sum_a c_a\lvert a\rangle,$$

the phases between components may evolve while the probabilities $|c_a|^2$ remain fixed. A conserved observable can therefore be uncertain.

Time-Independent Hamiltonians and Degeneracy

For a time-independent Hamiltonian,

$$U(t,t_0) = e^{-iH(t-t_0)/\hbar}.$$

A time-independent observable that strongly commutes with $H$ also commutes with $U(t,t_0)$ and is conserved. In a finite-dimensional or discrete pure-point setting, write

$$H= \sum_\alpha E_\alpha P_\alpha,$$

where $P_\alpha$ projects onto the full eigenspace with energy $E_\alpha$. Then

$$[H,A]=0$$

is equivalent to the block structure

$$A= \sum_\alpha P_\alpha A P_\alpha.$$

The observable cannot connect different energy eigenspaces, but it may act nontrivially inside a degenerate eigenspace. Consequently:

• for a nondegenerate finite spectrum, every commuting observable is diagonal in the energy basis;
• with degeneracy, conserved observables need not be functions of $H$;
• the projectors $P_\alpha$ define invariant dynamical sectors, but degeneracy alone does not create a superselection rule.

The last point matters. A coherent superposition of different conserved sectors is allowed unless an additional physical restriction makes their relative phase unobservable.

Energy Conservation

Take $A=H$. If the Hamiltonian is time independent, then

$$\frac{d}{dt}\langle H\rangle = \frac{i}{\hbar}\langle[H,H]\rangle =0.$$

In fact, the full energy distribution is conserved. If $P_H(\Delta)$ is an energy spectral projector, then

$$\operatorname{Tr}\!\left[ \rho(t)P_H(\Delta) \right] = \operatorname{Tr}\!\left[ \rho(t_0)P_H(\Delta) \right].$$

Energy conservation therefore says more than constancy of the mean energy.

For an explicitly time-dependent Hamiltonian, the equal-time commutator still vanishes, but the explicit derivative remains:

$$\frac{d}{dt}\langle H(t)\rangle = \left\langle \frac{\partial H(t)}{\partial t} \right\rangle.$$

For a controlled Hamiltonian

$$H(t)=H_0+\lambda(t)V,$$

the mean power supplied through the control is

$$\mathcal P(t) = \frac{d}{dt}\langle H(t)\rangle = \dot\lambda(t)\langle V\rangle_t.$$

For a differentiable protocol from $t_i$ to $t_f$,

$$\langle H(t_f)\rangle - \langle H(t_i)\rangle = \int_{t_i}^{t_f} \dot\lambda(t)\langle V\rangle_t\,dt.$$

This identity tracks mean energy transferred by the drive. It is not, by itself, a complete operational definition of fluctuating work. The interpretation also depends on where the boundary between the modeled system and its controller is drawn. See Time-Dependent Hamiltonians for the driven propagator.

Example: Particle in a Potential

Consider one-dimensional motion with

$$H= \frac{p^2}{2m}+V(x).$$

The canonical commutator gives

$$[H,x] = -\frac{i\hbar}{m}p,$$

and therefore

$$\frac{d}{dt}\langle x\rangle = \frac{\langle p\rangle}{m}.$$

Likewise,

$$[H,p] = [V(x),p] = i\hbar V'(x),$$

so

$$\frac{d}{dt}\langle p\rangle = -\langle V'(x)\rangle.$$

For a free particle, $V'(x)=0$, so the full momentum distribution is conserved. Position is not conserved even if $\langle p\rangle=0$ in one specially chosen state: its variance can still spread.

In several dimensions, a component $p_j$ is conserved whenever the potential is independent of the corresponding coordinate:

$$\frac{\partial V}{\partial x_j}=0 \quad\Longrightarrow\quad [H,p_j]=0.$$

The symmetry interpretation is the subject of Translation-Invariant Hamiltonians.

Example: A Driven Spin with a Conserved Component

Let

$$H(t)= \frac{\hbar\Omega(t)}{2}\sigma_z.$$

Although the Hamiltonian is time dependent,

$$[H(t),\sigma_z]=0$$

for every $t$. The full $\sigma_z$ distribution is therefore conserved. Yet the mean energy generally is not:

$$\frac{d}{dt}\langle H(t)\rangle = \frac{\hbar\dot\Omega(t)}{2} \langle\sigma_z\rangle.$$

This example cleanly separates conservation of a fixed observable from conservation of energy.

More generally, for

$$H(t)= \frac{\hbar}{2} \boldsymbol\Omega(t) \mathbin{\boldsymbol\cdot} \boldsymbol\sigma,$$

a fixed component $\boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma$ is conserved exactly when

$$\boldsymbol n\times\boldsymbol\Omega(t)=0$$

throughout the evolution interval. The effective field may change in magnitude, but not away from the fixed axis $\boldsymbol n$.

Example: An Explicitly Time-Dependent Invariant

For a free particle,

$$H=\frac{p^2}{2m},$$

define the time-labeled observable

$$I(t)= x- \frac{t-t_0}{m}p.$$

Its explicit derivative is

$$\frac{\partial I}{\partial t} = -\frac{p}{m},$$

while

$$\frac{i}{\hbar}[H,I] = \frac{p}{m}.$$

The two terms cancel:

$$\mathcal K_I=0.$$

Equivalently,

$$U(t,t_0)^\dagger I(t)U(t,t_0)=x.$$

The outcome distribution of $I(t)$ in the evolved state equals the initial position distribution. This does not say that ordinary position is conserved. It says that a specifically time-dependent combination of position and momentum reconstructs the reference-time position observable.

Such operators are often called dynamical invariants. They become especially useful for driven oscillators and shortcut protocols, but their defining equation is the same conservation criterion derived here.

Symmetry Preview

Suppose a continuous unitary transformation is generated by a self-adjoint operator $G$:

$$S(\epsilon)= e^{-i\epsilon G/\hbar}.$$

If the Hamiltonian is invariant,

$$S(\epsilon)H S(\epsilon)^\dagger=H,$$

then differentiating at $\epsilon=0$ gives

$$[H,G]=0.$$

When $G$ has no explicit time dependence, it is conserved. Standard pairings include:

• spatial translations and momentum;
• rotations and angular momentum;
• time translations and energy.

This is the operator-level preview of the quantum Noether pattern. The full statement requires care about active versus passive transformations, projective representations, boundary conditions, and local continuity equations. Those topics remain at their canonical homes in Generators and Commutators and Conservation Laws.

Approximate Conservation

Exact commutation is an idealization in many effective models. If $\mathcal K_A(t)$ is bounded and

$$\lVert\mathcal K_A(t)\rVert \le \varepsilon(t),$$

then every normalized state satisfies

$$\left| \frac{d}{dt}\langle A\rangle \right| \le \varepsilon(t).$$

Integration gives

$$\left| \langle A\rangle_{t_f} - \langle A\rangle_{t_i} \right| \le \int_{t_i}^{t_f}\varepsilon(t)\,dt.$$

This bound converts a small commutator into a quantitative time window for near-conservation. For unbounded operators, a global operator norm may not exist; state-dependent bounds, energy cutoffs, or estimates on a controlled domain are then more appropriate.

Approximate conservation should always be stated together with its scale and regime. A quantity can drift slowly over laboratory times while failing to be conserved asymptotically.

Boundaries of the Closed-System Formula

Open dynamics

If the reduced state satisfies a master equation

$$\frac{d\rho}{dt}=\mathcal L_t(\rho),$$

then the adjoint generator determines observable evolution:

$$\frac{d}{dt}\langle A\rangle = \left\langle \frac{\partial A}{\partial t} + \mathcal L_t^\dagger(A) \right\rangle.$$

The Hamiltonian commutator is only one part of $\mathcal L_t^\dagger$. Dissipative terms can destroy a closed-system conservation law, or preserve one because of a special structure. The canonical open-system treatment begins with the Lindblad–GKSL Equation.

Unbounded operators and domains

For $x$, $p$, differential Hamiltonians, and other unbounded operators, the expression $[H,A]$ is defined only where both $HA$ and $AH$ make sense. A formal commutator that vanishes on a convenient set of test functions need not imply that the self-adjoint operators strongly commute.

The robust time-independent condition is that their spectral projectors commute, or equivalently that the unitary evolution preserve the domain of $A$ and satisfy

$$U(t)^\dagger A U(t)=A$$

there. Boundary conditions can decide whether this statement is true. This is one reason an operator’s domain is part of its physical definition.

Effective descriptions

A conserved operator in a truncated or effective Hamiltonian may fail to be exactly conserved in the underlying theory. Conversely, an approximation can accidentally break an exact microscopic symmetry. Conservation claims should identify the Hamiltonian, Hilbert space, boundary conditions, and approximation order to which they apply.

Practical Checklist

1. Specify the closed-system Hamiltonian $H(t)$ and its domain.
2. Decide whether the observable $A(t)$ has explicit time dependence.
3. Compute $\partial A/\partial t$ and $[H,A]$ separately.
4. Form $\mathcal K_A=\partial_tA+(i/\hbar)[H,A]$.
5. Decide whether the claim concerns one state or every state.
6. For a full distribution, check the propagator or spectral-projector relation.
7. Identify external drives, environment terms, boundary fluxes, or approximations that can exchange the quantity with degrees of freedom outside the model.

Common Mistakes

• Omitting $\partial A/\partial t$ because the calculation is being done in the Schrödinger picture.
• Finding $\langle[H,A]\rangle=0$ in one state and claiming the operator commutator vanishes.
• Treating a zero first derivative at one instant as conservation for all time.
• Saying that a conserved observable must have a sharp value.
• Proving only that the mean is constant when the intended claim concerns the full outcome distribution.
• Assuming a time-dependent Hamiltonian conserves energy because $[H(t),H(t)]=0$.
• Confusing $[H(t),A]=0$ with the unrelated condition $[H(t_1),H(t_2)]=0$.
• Calling an approximately conserved quantity exact without specifying the error scale and time interval.
• Manipulating commutators of unbounded operators without checking domains and boundary conditions.
• Applying the closed-system formula directly to dissipative reduced dynamics.

Cross-Links

• Hamiltonians
• Time-Evolution Operator
• Time-Dependent Hamiltonians
• Stationary States
• Commutators
• Expectation Values
• Projective Measurement
• Heisenberg Equations of Motion
• Ehrenfest Theorem
• Constants of Motion
• Translation-Invariant Hamiltonians

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955, chs. 2–3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, chs. 4 and 7.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 2.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, chs. 3–4.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chs. 7–9.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, rev. ed., Academic Press, 1980, chs. 7–8.
• H. R. Lewis, Jr. and W. B. Riesenfeld, “An exact quantum theory of the time-dependent harmonic oscillator and of a charged particle in a time-dependent electromagnetic field,” Journal of Mathematical Physics 10, 1458–1473 (1969), doi:10.1063/1.1664991.

Exercises

1. Starting from the von Neumann equation, derive the expectation-value identity for a mixed state and a time-dependent observable.

Solution

Begin with

$$\langle A\rangle = \operatorname{Tr}(\rho A).$$

The product rule gives

$$\frac{d}{dt}\langle A\rangle = \operatorname{Tr}(\dot\rho A) + \operatorname{Tr}(\rho\,\partial_tA).$$

Insert

$$\dot\rho = -\frac{i}{\hbar}[H,\rho].$$

Then

$$\begin{aligned} \operatorname{Tr}(\dot\rho A) &= -\frac{i}{\hbar} \operatorname{Tr}\!\left((H\rho-\rho H)A\right)\\ &= -\frac{i}{\hbar} \operatorname{Tr}\!\left(\rho AH-\rho HA\right)\\ &= \frac{i}{\hbar} \operatorname{Tr}\!\left(\rho[H,A]\right), \end{aligned}$$

where cyclicity of the trace was used in the second line. Therefore

$$\frac{d}{dt}\langle A\rangle = \operatorname{Tr}\!\left[ \rho\left( \frac{\partial A}{\partial t} + \frac{i}{\hbar}[H,A] \right) \right].$$

2. Let $H=(\hbar\Omega/2)\sigma_z$ and $A=\sigma_x$. Show that $d\langle A\rangle/dt=0$ initially in the state $\lvert{+x}\rangle$, but that $A$ is not conserved.

Solution

Using $[\sigma_z,\sigma_x]=2i\sigma_y$,

$$\frac{i}{\hbar}[H,A] = -\Omega\sigma_y.$$

The state $\lvert{+x}\rangle$ has $\langle\sigma_y\rangle=0$, so

$$\left. \frac{d}{dt}\langle\sigma_x\rangle \right|_{t=0} =0.$$

However, the propagator is

$$U(t)= e^{-i\Omega t\sigma_z/2},$$

and direct evolution gives

$$\begin{aligned} \langle\sigma_x\rangle_t &= \cos(\Omega t),\\ \langle\sigma_y\rangle_t &= \sin(\Omega t). \end{aligned}$$

The mean changes for generic $t$. A state-specific zero derivative at one instant is weaker than the operator condition $[H,A]=0$.

3. Consider $H(t)=\hbar\Omega(t)\sigma_z/2$. Determine which of $\sigma_x$, $\sigma_y$, and $\sigma_z$ are conserved, and find $d\langle H\rangle/dt$.

Solution

The commutators are

$$\begin{aligned} [H,\sigma_x] &= i\hbar\Omega(t)\sigma_y,\\ [H,\sigma_y] &= -i\hbar\Omega(t)\sigma_x,\\ [H,\sigma_z] &=0. \end{aligned}$$

Thus only $\sigma_z$ is conserved for a generic nonzero drive. The energy obeys

$$\frac{d}{dt}\langle H\rangle = \left\langle \frac{\partial H}{\partial t} \right\rangle = \frac{\hbar\dot\Omega(t)}{2} \langle\sigma_z\rangle.$$

The conserved value of $\langle\sigma_z\rangle$ makes the energy change directly proportional to the changing field magnitude.

4. For a free particle, prove that $I(t)=x-(t-t_0)p/m$ is a dynamical invariant and interpret the result.

Solution

With $H=p^2/(2m)$,

$$\frac{\partial I}{\partial t} = -\frac{p}{m}.$$

Also,

$$[H,x] = -\frac{i\hbar}{m}p, \qquad [H,p]=0,$$

so

$$\frac{i}{\hbar}[H,I] = \frac{p}{m}.$$

Therefore

$$\frac{\partial I}{\partial t} + \frac{i}{\hbar}[H,I] =0.$$

Equivalently,

$$U(t,t_0)^\dagger I(t)U(t,t_0)=x.$$

The distribution of the time-dependent combination $I(t)$ in the evolved state is the initial position distribution. Ordinary position itself is not conserved.

5. Suppose $A$ is time independent and strongly commutes with a time-independent Hamiltonian $H$. Prove that every spectral probability of $A$ is conserved.

Solution

Strong commutation implies that every spectral projector $P_A(\Delta)$ commutes with

$$U(t,t_0)=e^{-iH(t-t_0)/\hbar}.$$

For $\rho(t)=U\rho(t_0)U^\dagger$,

$$\begin{aligned} p_t(\Delta) &= \operatorname{Tr}\!\left( U\rho(t_0)U^\dagger P_A(\Delta) \right)\\ &= \operatorname{Tr}\!\left( \rho(t_0)U^\dagger P_A(\Delta)U \right)\\ &= \operatorname{Tr}\!\left( \rho(t_0)P_A(\Delta) \right). \end{aligned}$$

Thus $p_t(\Delta)=p_{t_0}(\Delta)$ for every measurable outcome set $\Delta$.

6. Let

$$H= \begin{pmatrix} 0&0&0\\ 0&E&0\\ 0&0&E \end{pmatrix}, \qquad E\neq 0.$$

Find the most general Hermitian matrix $A$ satisfying $[H,A]=0$. Explain why it need not be a function of $H$.

Solution

The one-dimensional zero-energy eigenspace cannot be coupled to the two-dimensional $E$ eigenspace. Inside the degenerate $E$ eigenspace, any Hermitian action is allowed. Hence

$$A= \begin{pmatrix} a&0&0\\ 0&b&z\\ 0&z^*&c \end{pmatrix},$$

where $a,b,c$ are real and $z$ is complex.

A function $f(H)$ has the more restrictive form

$$f(H)= \begin{pmatrix} f(0)&0&0\\ 0&f(E)&0\\ 0&0&f(E) \end{pmatrix}.$$

It is proportional to the identity within each degenerate energy block. Choosing $z\neq0$ or $b\neq c$ gives a conserved observable that is not a function of $H$.

7. A closed system is driven by $H(t)=H_0+\lambda(t)V$. Derive the change in mean energy over a smooth protocol and evaluate it for a state that remains an eigenstate of $V$ with eigenvalue $v$.

Solution

Since $[H,H]=0$,

$$\frac{d}{dt}\langle H(t)\rangle = \left\langle \frac{\partial H}{\partial t} \right\rangle = \dot\lambda(t)\langle V\rangle_t.$$

Therefore

$$\Delta\langle H\rangle = \int_{t_i}^{t_f} \dot\lambda(t)\langle V\rangle_t\,dt.$$

If the state remains an eigenstate of $V$ with eigenvalue $v$, then $\langle V\rangle_t=v$ and

$$\begin{aligned} \Delta\langle H\rangle &= v\int_{t_i}^{t_f}\dot\lambda(t)\,dt\\ &= v\left[\lambda(t_f)-\lambda(t_i)\right]. \end{aligned}$$

8. Let $A$ be bounded and suppose $\lVert\mathcal K_A(t)\rVert\le\varepsilon$ throughout an interval of duration $T$. Bound the change in $\langle A\rangle$ for an arbitrary normalized state.

Solution

For a normalized state,

$$\left| \frac{d}{dt}\langle A\rangle \right| = |\langle\mathcal K_A\rangle| \le \lVert\mathcal K_A\rVert \le \varepsilon.$$

Integrating from $t_i$ to $t_f=t_i+T$ gives

$$\left| \langle A\rangle_{t_f} - \langle A\rangle_{t_i} \right| \le \int_{t_i}^{t_f}\varepsilon\,dt = \varepsilon T.$$

The estimate is uniform over all normalized states. It establishes near-conservation only on times for which $\varepsilon T$ is small relative to the physically relevant scale of $A$.