The first is Eq. (7.17), derived in §7.4.

The second. $\hat H = \hat T + \hat V$, and $\hat x$ commutes with $\hat V$ (both are multiplications), so only the kinetic term contributes:

using the identity $[\hat A,\hat B^2] = \hat B[\hat A,\hat B] + [\hat A,\hat B]\hat B$, which is problem 3’s hint in reverse. Note the hint’s form $\hbar^2\partial/m\partial x$ is the same thing, since $\partial/\partial x = i\hat p/\hbar$.

The third. Now $\hat p$ commutes with $\hat T$ (problem 2), so only the potential contributes, and the product rule gives

The significance is two statements, one from each of the chapter’s halves.

By §7.4, no two of position, momentum and energy are compatible — no state has any two of them sharp at once.

By §7.5, neither position nor momentum is a constant of motion. And those last two commutators are not merely non-zero, they are informative: substituting them into Eq. (7.22) gives $\mathrm d\langle x\rangle/\mathrm dt = \langle p\rangle/m$ and $\mathrm d\langle p\rangle/\mathrm dt = -\langle\mathrm dV/\mathrm dx\rangle$ — Newton’s laws, as §7.5 worked out.

(a) $\hat p = -i\hbar\,\partial/\partial x$ and $\hat T = \hat p^2/2m = -(\hbar^2/2m)\,\partial^2/\partial x^2$.

(b) $\hat T$ is built from $\hat p$ alone, and everything commutes with itself, so $[\hat p,\hat p^2] = 0$ identically. Momentum and kinetic energy are always compatible. It is the potential that makes momentum and total energy incompatible, which is exactly what problem 1’s third commutator says: $[\hat p,\hat H] = -i\hbar\,\mathrm dV/\mathrm dx$, zero if and only if $V$ is constant.

(c) $\hat T(A\cos kx) = (\hbar^2k^2/2m)A\cos kx$ — an eigenfunction, with eigenvalue $\hbar^2k^2/2m$. But $\hat p(A\cos kx) = i\hbar kA\sin kx$, which is not a multiple of $\cos kx$. So definite kinetic energy, indefinite momentum.

(d) Yes: $\psi = Ae^{ikx}$, with $\hat p\psi = \hbar k\psi$ and $\hat T\psi = (\hbar^2k^2/2m)\psi$. This is §7.2’s momentum eigenfunction, Eq. (7.11).

(a) Expand $[\hat L_x,\hat L_y]$ into four commutators. Two of them — $[\hat y\hat p_z,\hat x\hat p_z]$ and $[\hat z\hat p_y,\hat z\hat p_x]$ — contain only operators that commute with one another, so they vanish. The two surviving terms are the ones stated.

(b) Take the first: $\hat y$ and $\hat p_x$ pass through everything, so only $\hat p_z$ and $\hat z$ interact —

and likewise $[\hat z\hat p_y,\hat x\hat p_z] = +i\hbar\,\hat x\hat p_y$. Adding,

Its significance: no two components of angular momentum are compatible. A state cannot have $L_x$ and $L_y$ both sharp — so the classical picture of an angular-momentum vector pointing in a definite direction has no quantum counterpart.

(c) Using the hint on each term: $[\hat L_x^2,\hat L_z] = \hat L_x[\hat L_x,\hat L_z] + [\hat L_x,\hat L_z]\hat L_x = -i\hbar(\hat L_x\hat L_y + \hat L_y\hat L_x)$, and $[\hat L_y^2,\hat L_z] = +i\hbar(\hat L_y\hat L_x + \hat L_x\hat L_y)$. These cancel exactly, and $[\hat L_z^2,\hat L_z] = 0$ trivially. Hence $[\hat L^2,\hat L_z] = 0$.

Its significance is the labelling scheme of every atom in the book. The magnitude of the angular momentum is compatible with one of its components. So a state may be labelled by $L^2$ and $L_z$ together — never by two components — and that is precisely the pair chapter 9 uses, alongside the energy, to name hydrogen’s states $(n,\ell,m)$.

Integrate by parts:

The boundary term vanishes because normalizable wave functions go to zero at infinity. And $i\hbar\,\partial\Psi_1^*/\partial x = (-i\hbar\,\partial\Psi_1/\partial x)^* = (\hat p\Psi_1)^*$, so

which is Eq. (7.2).

Both conditions matter. The factor of $-i$ is essential: without it, $\partial/\partial x$ alone is anti-Hermitian, since integrating by parts flips the sign and nothing restores it. And the boundary term must vanish, which is why this proof works for normalizable states and needs care for the plane waves of §7.2.

(a) $\langle A\rangle^* = \int\Psi(\hat A\Psi)^*\,\mathrm dx = \int(\hat A\Psi)^*\Psi\,\mathrm dx = \int\Psi^*\hat A\Psi\,\mathrm dx = \langle A\rangle$ by Eq. (7.2). A number equal to its own conjugate is real.

(b) Write $\langle A^2\rangle = \int\Psi^*\hat A(\hat A\Psi)\,\mathrm dx$ and apply Eq. (7.2) with $\Psi_2 = \hat A\Psi$: $= \int(\hat A\Psi)^*(\hat A\Psi)\,\mathrm dx$.

A useful corollary: that integral is $\int|\hat A\Psi|^2 \ge 0$, so $\langle A^2\rangle$ is never negative for any Hermitian operator. Problem 6 needs exactly this.

(c) Apply Eq. (7.2) twice — once to move $\hat A$, once to move $\hat B$ back:

Writing $z = \int\Psi^*\hat A\hat B\Psi$, this says the $\hat B\hat A$ integral is $z^*$. So the anticommutator integral is $z + z^*$, which is real, and the commutator integral is $z - z^*$, which is purely imaginary.

That last fact is why the uncertainty principle has an $i$ in it. $\langle[\hat A,\hat B]\rangle$ is always imaginary, so $|\langle[\hat x,\hat p]\rangle| = |i\hbar| = \hbar$ is a real positive number — and §7.4’s bound $\Delta x\,\Delta p \ge \frac12|\langle[\hat x,\hat p]\rangle|$ makes sense.

(a) Expanding $\int|\alpha + \lambda\beta|^2 \ge 0$ gives

Now substitute the $\lambda$ the problem supplies, $\lambda\int|\beta|^2 = -\int\beta^*\alpha$, and multiply through by $\int|\beta|^2$; what survives is exactly the stated inequality. The trick is that the inequality holds for every $\lambda$, so it holds for the worst one — the value that makes the left side as small as possible.

(b) With $\alpha = \hat A\Psi$ and $\beta = \hat B\Psi$, problem 5(b) identifies $\int|\alpha|^2 = \langle A^2\rangle$ and $\int|\beta|^2 = \langle B^2\rangle$, while $\int\alpha^*\beta = \int(\hat A\Psi)^*\hat B\Psi = \int\Psi^*\hat A\hat B\Psi$ by Hermiticity.

(c) Split $\hat A\hat B = \frac12(\hat A\hat B + \hat B\hat A) + \frac12(\hat A\hat B - \hat B\hat A)$. By problem 5(c) the first expectation is real and the second imaginary, so they are the real and imaginary parts of one complex number — and $|z|^2 = (\mathrm{Re}\,z)^2 + (\mathrm{Im}\,z)^2$ gives the sum of squares.

Discarding the anticommutator term — which is allowed, since it only makes the bound weaker — leaves

which is §7.4’s starting point. With $\hat A = \widehat{\Delta x}$, $\hat B = \widehat{\Delta p}$ and Eq. (7.19), it is $\Delta x\,\Delta p \ge \hbar/2$.

(a) $\hat p_x$ commutes with the kinetic term always (problem 2), so $[\hat p_x,\hat H] = [\hat p_x,\hat V] = -i\hbar\,\partial V/\partial x$. This vanishes for all states exactly when

which says $V$ is unchanged by $x \to x + a$ — translational invariance along $x$.

(b) Identically: $[\hat p_z,\hat H] \propto \partial V/\partial z$, so $p_z$ is conserved when $V$ does not depend on $z$ — invariance under translation along the axis. And $[\hat L_z,\hat H] \propto \partial V/\partial\phi$, so $L_z$ is conserved when $V$ does not depend on $\phi$ — invariance under rotation about the axis.

Note how directly the operator reveals its symmetry: $\hat L_z$ is $-i\hbar\,\partial/\partial\phi$, so it commutes with anything independent of $\phi$. An operator that differentiates with respect to a coordinate commutes with any potential that does not depend on it — that one sentence is the whole of Noether’s theorem in this setting.

(b) is the elegant step. Expand the commutator and use Hermiticity to move $\hat H$ onto the other factor:

Both now contain $\hat H\psi_E = E\psi_E$ or its conjugate, and $E$ is real, so the two terms are $E\int\psi_E^*\,\hat{\mathbf r}\cdot\hat{\mathbf p}\,\psi_E$ each — and they cancel.

This is Eq. (7.22) in disguise: $\langle\hat{\mathbf r}\cdot\hat{\mathbf p}\rangle$ cannot change in a stationary state, so its rate of change — the commutator with $\hat H$ — must average to zero. Combining with part (a) gives the virial theorem.

(c) For the oscillator, $V = \frac12m\omega^2r^2$ so $r\,\mathrm dV/\mathrm dr = m\omega^2r^2 = 2V$, and $2\langle T\rangle = 2\langle V\rangle$, i.e.

which §6.3 already found by a different route — both equal $\frac12E_n$.

For the Coulomb potential, $V = -k/r$ so $r\,\mathrm dV/\mathrm dr = +k/r = -V$, giving

not the $2\langle T\rangle = \langle V\rangle$ the book prints. For hydrogen’s ground state $\langle T\rangle = +13.6\ \mathrm{eV}$, $\langle V\rangle = -27.2\ \mathrm{eV}$, and $E = \langle T\rangle + \langle V\rangle = -13.6\ \mathrm{eV}$ — the Rydberg energy, recovered.