(a) With $\langle x\rangle = 0$ we have $\langle x^2\rangle = (\Delta x)^2$ exactly, and likewise $\langle p^2\rangle = (\Delta p)^2$. The uncertainty principle then gives $\Delta p \ge \hbar/2\Delta x$, so

Substituting both into $\langle E\rangle$ produces the stated inequality.

(b) Differentiate: $F' = -2A^2/(\Delta x)^3 + 2B^2\Delta x$, which vanishes when $(\Delta x)^4 = A^2/B^2$, i.e. $\Delta x = \sqrt{A/B}$. At that point $F = A^2\cdot\frac{B}{A} + B^2\cdot\frac{A}{B} = AB + AB = 2AB$.

(Without calculus: $F = \left(\frac{A}{\Delta x} - B\Delta x\right)^2 + 2AB$, and a square is never negative. The minimum is where the square vanishes.)

(c) Match: $A^2 = \hbar^2/8m$ and $B^2 = \frac12m\omega^2$, so $A = \hbar/\sqrt{8m}$ and $B = \omega\sqrt{m/2}$. Then

so $\langle E\rangle \ge \frac12\hbar\omega$.

From Table 6.1, $\psi_0(x) = (a\sqrt\pi)^{-1/2}e^{-x^2/2a^2}$ with $a = \sqrt{\hbar/m\omega}$, so the probability density is

The standard Gaussian of standard deviation $\sigma$ is $\dfrac{1}{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$. Comparing exponents, $\dfrac{1}{2\sigma^2} = \dfrac{1}{a^2}$, so

and the prefactors agree automatically, since both densities are normalized.

Note the factor of $\sqrt2$. The eigenfunction $\psi_0$ has “width” $a$ in the sense that it falls to $1/\sqrt e$ there; the density $|\psi_0|^2$ is narrower by $\sqrt2$, because squaring a Gaussian halves its variance. That is also why $\Delta x = a/\sqrt2$ and not $a$ — the same $\sqrt2$ that made problem 1’s bound come out exactly.

The turning point. A classical oscillator of amplitude $A$ has $E = \frac12m\omega^2A^2$, by Eq. (6.5). Setting that equal to $\frac12\hbar\omega$:

The classical turning point is exactly one oscillator length — which is not a coincidence but the reason $a$ deserves its name.

The probability. Integrating $|\psi_0|^2$ over both forbidden tails:

About 16% — roughly one measurement in six finds the particle somewhere a classical particle of the same energy could not reach. This is not a rare leak like the tunnelling of §5.2a, where $T$ was $10^{-5}$; it is routine, and it happens because the ground state’s own width is the same size as the classical turning point.

For $x > 0$ the equation is unchanged, so any solution must be an oscillator eigenfunction there. The infinite wall forces $\psi(0) = 0$, and only the odd eigenfunctions vanish at the origin — $\psi_1, \psi_3, \psi_5, \dots$

So the spectrum of the half-oscillator is the odd half of the full oscillator’s:

The ground state is $\frac32\hbar\omega$ — three times the full oscillator’s zero-point energy — and the first excited state is $\frac72\hbar\omega$. The sketches are the right-hand halves of $\psi_1$ and $\psi_3$ in Fig. 6.2, with zero drawn to the left of the origin, and renormalized by $\sqrt2$ since only half of each remains.

Note the spacing: adjacent levels are now $2\hbar\omega$ apart, not $\hbar\omega$. Removing half the well doubled the level spacing.

(a) Identical to §6.5’s separation with one dimension removed: each $\hat H_x$ acts only on $x$, so the product passes through and the energies add, giving $(n_x+\frac12) + (n_y+\frac12) = n_x + n_y + 1$ in units of $\hbar\omega$.

(b) Since only $N = n_x + n_y$ enters, level $N$ has a degeneracy degeneracy Several independent eigenfunctions sharing one eigenvalue. It comes from symmetry — a cubical box has it, a box with unequal sides does not — which is why breaking a symmetry splits levels. defined in ch. 4 — open in glossary of $N+1$ — the pairs $(0,N), (1,N-1), \dots, (N,0)$. A linear ladder of degeneracies 1, 2, 3, 4, against the three-dimensional oscillator’s triangular 1, 3, 6, 10.

(c) From Table 6.1, $\psi_0 \propto e^{-x^2/2a^2}$ and $\psi_1 \propto x\,e^{-x^2/2a^2}$, so

using $x^2 + y^2 = r^2$. Then

and $\psi_b = \psi_{1,0} - i\psi_{0,1} \propto r\,e^{-i\phi}e^{-r^2/2a^2}$. The $\phi$ dependence is now a bare $e^{\pm i\phi}$, so

as required.

(a) Each term carries its own phase:

(b) $|\Psi|^2 = \frac12[\psi_0^2 + \psi_1^2 + 2\psi_0\psi_1\cos(\omega t)]$ after using $E_1 - E_0 = \hbar\omega$. Integrating, the first two terms give 1 each and the cross term gives zero by orthogonality, so $\int|\Psi|^2 = \frac12(1+1+0) = 1$ for all $t$.

(c) That same expression has a single time-dependent term, $\psi_0\psi_1\cos\omega t$ — so the density sloshes at exactly $\omega$. The frequency is $(E_1-E_0)/\hbar$, and for this potential adjacent levels are always $\hbar\omega$ apart.

(d) Multiplying by $x$ and integrating, the $\psi_0^2$ and $\psi_1^2$ terms vanish (both are even, $x$ is odd), leaving

so $A = \int\psi_0x\psi_1\,\mathrm dx$, which problem 11 evaluates as $a/\sqrt2 = 0.707a$.

(a) Each level has probability $\frac13$:

(b) $\langle E^2\rangle = \frac13\left(\frac14+\frac94+\frac{25}4\right)(\hbar\omega)^2 = \frac{35}{12}(\hbar\omega)^2$, so

(c) Expanding $|\Psi|^2$ gives one term per pair of levels, oscillating at the difference of their energies over $\hbar$. Pairs $(0,1)$ and $(1,2)$ are adjacent and beat at $\omega$; the pair $(0,2)$ is two rungs apart and beats at $2\omega$. So

$B$ has two terms and $C$ only one — because two of the three pairs are adjacent. And note that $\langle x\rangle$ would keep only the $B$ terms: problem 11 shows $\int\psi_0x\psi_2 = 0$, so the $2\omega$ component is visible in the density but not in the average position, exactly as §6.3 claimed.

Invert Eq. (6.20), exactly as §6.4 did for CO:

With $\mu = 7.46 \times 1.6605\times10^{-27} = 1.239\times10^{-26}\ \mathrm{kg}$ and $2\pi c/\lambda = 3.534\times10^{14}\ \mathrm{s^{-1}}$:

matching the hint’s $1548$.

Worth comparing. §6.4 found $k = 1907\ \mathrm{N\,m^{-1}}$ for CO by the same method. NO’s bond is about 19% softer — and NO has one fewer bonding electron pair than CO, which is exactly the direction chemistry predicts. Two infrared wavelengths, two bond stiffnesses, and a chemical comparison that never required touching either molecule.

The mirror image of §6.6’s raising argument, with the two forms swapped. Start from Eq. (6.31) applied to the $n$-th state:

Apply $[q + \mathrm d/\mathrm dq]$ to both sides:

Compare with Eq. (6.30), which reads $[\dots][\dots]\psi = (2\epsilon + 1)\psi$. Matching constants, $2\epsilon + 1 = 2\epsilon_n - 1$, so $\epsilon = \epsilon_n - 1$.

So $[q + \mathrm d/\mathrm dq]$ steps down one rung — except at $n = 0$, where Eq. (6.40) gives zero and the ladder stops. Try it in the widget above.

Take the norm of the raised state:

Integration by parts moves one operator across, turning $[q - \mathrm d/\mathrm dq]$ into $[q + \mathrm d/\mathrm dq]$ — the sign flips because $\mathrm d/\mathrm dq$ is anti-Hermitian while $q$ is Hermitian. So

and the operator in the middle is exactly Eq. (6.30)‘s, which returns $(2\epsilon_n + 1)\psi_n$. With $\epsilon_n = n + \frac12$:

The lowering case is identical using Eq. (6.31), giving $|b_n|^2 = 2\epsilon_n - 1 = 2n$ — and note it vanishes at $n = 0$, which is Eq. (6.40) once more.

(a) The hint is the key identity: solve the two operator definitions for $q$ and for $\mathrm d/\mathrm dq$ separately —

So $q\psi_n = \frac12(b_n\psi_{n-1} + a_n\psi_{n+1})$: multiplying by $q$ always produces a mixture of the two neighbouring states and nothing else. Taking the inner product with $\psi_m$ and using orthonormality, Eq. (6.41), leaves exactly the stated Kronecker deltas.

This is the selection rule. $\int\psi_m^*q\psi_n = 0$ unless $|m - n| = 1$ — not because of any special property of these functions, but because $q$ is a sum of one raising and one lowering operator.

(b) Apply $q$ twice. $q^2\psi_n$ reaches $\psi_{n-2}$, $\psi_n$ and $\psi_{n+2}$; only the middle survives the inner product with $\psi_n$, and its coefficient is $\frac14(a_{n-1}b_n + b_{n+1}a_n) = \frac14(2n + 2(n+1)) = n + \frac12$. The kinetic term follows the same way, or immediately from Eq. (6.29): since $[-\mathrm d^2/\mathrm dq^2 + q^2]$ has expectation $2\epsilon_n = 2n+1$ and the $q^2$ half is $n + \frac12$, the other half must be $n + \frac12$ too.

That equality is §6.3’s even split — half potential, half kinetic — now derived rather than quoted.

(c) Restore units with $x = qa$ and $p = -i\hbar\,\mathrm d/\mathrm dx$. $\langle x\rangle \propto \int\psi_nq\psi_n = 0$ by part (a) with $m = n$; $\langle x^2\rangle = (n+\frac12)a^2$; and $\langle p^2\rangle = (\hbar^2/a^2)(n+\frac12)$. Hence

which is the result §6.3 stated without proof.