Introduction to Quantum Physics

Abstract

The gestation of Quantum Physics has been very long and its phenomenological foundations were various. Historically, the original idea came from the analysis of the black body spectrum. This is not surprising since the black body, in fact an oven in thermal equilibrium with the electromagnetic radiation, is a simple and fundamental system once the law of electrodynamics are established. As a matter of fact many properties of the spectrum can be deduced starting from the general laws of electrodynamics and thermodynamics; the crisis came from the violation of the equipartition of energy. That suggested to Planck the idea of quantum, from which everything originated. Of course a long sequence of different discoveries, first of all the photoelectric effect, the line spectra for atomic emission/absorption, the Compton effect and so on, gave a compelling evidence for the new theory.

Problems

2.1

A diatomic molecule can be simply described as two point-like objects of mass $m= 10^{-26} \, \mathrm{kg}$ placed at a fixed distance $d=10^{-9}\, \mathrm{m}$. Describe what are the possible values of the molecule energy according to Bohr’s quantization rule. Compute the energy of the photons which are emitted when the system decays from the $(n + 1)$-th to the n-th energy level.

Answer: $E_{n+1}-E_n={(\hbar ^2/2I)}(n+1)^2 - {(\hbar ^2/2I)}n^2 = (2n + 1){\hbar ^2 / (m d^2)} \simeq 1.1\times 10^{-24}(2n+1)\ \mathrm{J}$. Notice that in Sommerfeld’s perfected theory, mentioned in Sect. 2.2, the energy of a rotator is given by $E_n={\hbar ^2n(n+1)/2I}$, so that the factor $2n+1$ in the solution must be replaced by $2n+2$.

2.2

An artificial satellite of mass $m=1\, \mathrm{kg}$ rotates around the Earth on a circular orbit of radius practically equal to that of the Earth itself, i.e. roughly $6370\ \mathrm{km}$. If the satellite orbits are quantized according to Bohr’s rule, what is the radius variation when going from one quantized level to the next (i.e. from n to $n + 1$)?

Answer: If g indicates the gravitational acceleration at the Earth surface, the radius of the n-th orbit is given by $r_n=n^2\hbar ^2/(m^2R^2g)$. Therefore, if $r_n=R$, $\delta r_n \equiv r_{n+1} - r_n \simeq 2\hbar /(m\sqrt{Rg})\simeq 2.6\times 10^{-38}\ \mathrm{m}\, .$

2.3

An electron is accelerated through a potential difference $\varDelta V=10^{8}\ \mathrm{V}$, what is its wavelength according to de Broglie?

Answer: The energy gained by the electron is much greater than $mc^2$, therefore it is ultra-relativistic and its momentum is $p\simeq E/c.$ Hence $\lambda \simeq hc/e\varDelta V\simeq 12.4\times 10^{-15}\ \mathrm{m}$. The exact formula is instead $\lambda = hc/\sqrt{(e\varDelta V+mc^2)^2-m^2c^4} .$

2.4

An electron is constrained to bounce between two reflecting walls placed at a distance $d=10^{-9}\ \mathrm{m}$ from each other. Assuming that, as in the case of a stationary electromagnetic wave confined between two parallel mirrors, the distance d be equal to n half wavelengths, determine the possible values of the electron energy as a function of n.

Answer: $E_n=\hbar ^2\pi ^2n^2/(2md^2) \simeq n^2\ 6.03\times 10^{-20}\ \mathrm{J}\, .$

2.5

An electron of kinetic energy $1 \ \mathrm{eV}$ is moving upwards under the action of its weight. Can it reach an altitude of $1\ \mathrm{km}$? If yes, what is the variation of its de Broglie wavelength?

Answer: The maximum altitude reachable by the electron in a constant gravitational field would be $h=T/mg\simeq 1.6\times 10^{10 }\ \mathrm{m}$. After one kilometer the kinetic energy changes by $\delta T/T\simeq 5.6\times 10^{-8}$, hence $\delta \lambda / \lambda \simeq 2.8\times 10^{-8}$. Since the starting wavelength is $\lambda ={h/\sqrt{2mT}} \simeq 1.2\times 10^{-9}\ \mathrm{m}$, the variation is $\delta \lambda \simeq 3.4\times 10^{-17}\ \mathrm{m}$.

2.6

Ozone ($O_3$) is a triatomic molecule made up of three atoms of mass $m \simeq 2.67\times 10^{-26}\ \mathrm{kg}$ placed at the vertices of an equilateral triangle with sides of length l. The molecule can rotate around an axis P going through its center of mass and orthogonal to the triangle plane, or around another axis L which passes through the center of mass as well, but is orthogonal to the first axis. Making use of Bohr’s quantization rule and setting $l = 10^{-10}\ \mathrm{m}$, compare the possible rotational energies in the two different cases of rotations around P or L.

Answer: The moments of inertia are $I_P=m l^2= 2.67\times 10^{-46}\ \mathrm{kg}\,\mathrm{m}^2$ and $I_L=ml^2/2= 1.34\times 10^{-46}\ \mathrm{kg}\,\mathrm{m}^2$. The rotational energies are then $E_{L,n}=2 E_{P,n}=\hbar ^2 n^2/2 I_L \simeq n^2\ 4.2\times 10^{-23}\ \mathrm{J}.$

2.7

A table salt crystal is irradiated with an X-ray beam of wavelength $\lambda = 2.5\times 10^{-10}\ \mathrm{m}$, the first diffraction peak ($d \sin \theta =\lambda$) is observed at an angle equal to $26.3^{\circ }$. What is the interatomic distance of salt?

Answer: $\, d=\lambda /\sin \theta \simeq 5.6\times 10^{-10}\ \mathrm{m}.$

2.8

In $\beta$ decay a nucleus, with a radius of the order of $R=10^{-14}\ \mathrm{m}$, emits an electron with a kinetic energy of the order of $1\ \mathrm{MeV}= 10^{6}\ \mathrm{eV}$. Compare this value with that which according to the uncertainty principle is typical of an electron initially localized inside the nucleus (thus having a momentum $p\sim {\hbar /R}$).

Answer: The order of magnitude of the momentum of the particle is $p \sim \hbar /R \sim 10^{-20}\ \mathrm{N\ s}$, thus $pc \simeq 3\times 10^{-12}\ \mathrm{J}$, which is much larger that the electron rest energy $m_e c^2 \simeq 8\times 10^{-14}\ \mathrm{J}$. Therefore the kinetic energy of the electron in the nucleus is about $pc = 3.15\times 10^{-12}\ \mathrm{J} \simeq 20\ \mathrm{MeV}$.

2.9

An electron is placed in a constant electric field $\mathcal{E}=1000\ \mathrm{V}/\mathrm{m}$ directed along the x axis and going out of a plane surface orthogonal to the same axis. The surface also acts on the electron as a reflecting plane where the electron potential energy V(x) goes to infinity. The behavior of V(x) is therefore as illustrated in the following figure.

Evaluate the order of magnitude of the minimal electron energy according to Heisenberg’s Uncertainty Principle.

Answer: The total energy is given by $\epsilon = p^2/2m + V(x) = p^2/2m + e \mathcal{E} x$, with the constraint $x > 0$. From a classical point of view, the minimal energy would be realized for a particle at rest ($p = 0$) in the minimum of the potential. The uncertainty principle states instead that $\delta p\ \delta x \sim \hbar$, where $\delta x$ is the size of a region around the potential minimum where the electron is localized. Therefore the minimal total energy compatible with the uncertainty principle can be written as a function of $\delta x$ as $E(\delta x)\equiv \hbar ^2/(2 m \delta x^2) + e \mathcal{E} \delta x$ ($\delta x > 0$) and has a minimum

$$ \epsilon _\mathrm{min} \sim \frac{3}{2} \left( \frac{\hbar ^2 e^2 \mathcal{E}^2}{m} \right) ^{1/3} \sim 0.6\times 10^{-4}\ \mathrm{eV}\, . $$

2.10

An atom of mass $M= 10^{-26}\ \mathrm{kg}$ is attracted towards a fixed point by an elastic force of constant $k=1\ \mathrm{N}/\mathrm{m}$; the atom is moving along a circular orbit in a plane orthogonal to the x axis. Determine the energy levels of the system by making use of Bohr’s quantization rule for the angular momentum computed with respect to the fixed point.

Answer: Let $\omega$ be the angular velocity and r the orbital radius. The centripetal force is equal to the elastic one, hence $\omega = \sqrt{k/M}$. The total energy is given by $E = {(1/2)} M \omega ^2 r^2 + {(1/2)} k r^2 = M \omega ^2 r^2 = L \omega$, where $L = M \omega r^2$ is the angular momentum. Since $L = n \hbar$, we finally infer $E_n = n \hbar \omega \simeq n\ 1.05\times 10^{-21}\ {\mathrm{J}} \simeq n\ 0.66\times 10^{-2}\ \mathrm{eV}$.

2.11

Compute the number of photons emitted in one second by a lamp of power $10\ \mathrm{W}$, if the photon wavelength is $0.5\times 10^{{-6}}\ \mathrm{m}.$

Answer: The energy of a single photon is $E = h \nu$ and $\nu = c/\lambda = 6\times 10^{14}\ \mathrm{Hz}$, hence $E \simeq 4\times 10^{-19}\ \mathrm{J}$. Therefore the number photons emitted in one second is $2.5\times 10^{19}$.

2.12

A particle of mass $m=10^{-28}\ \mathrm{kg}$ is moving along the x axis under the influence of a potential energy given by $V(x)= v \sqrt{|x|}$, where $v= 10^{-15}\ \mathrm{J\ m^{-1/2}}$. Determine what is the order of magnitude of the minimal particle energy according to the uncertainty principle.

Answer: The total energy of the particle is given by

$$ E= \frac{p^2}{2 m} + v \sqrt{|x|}\ . $$

If the particle is localized in a region of size $\delta x$ around the minimum of the potential ($x = 0$), according to the uncertainty principle it has a momentum at least of the order of $\delta p = \hbar / \delta x$. It is therefore necessary to minimize the quantity

$$ E = \frac{\hbar ^2}{2 m \delta x^2} + v \sqrt{\delta x}\ $$

with respect to $\delta x$, finally finding that

$$ E_{min} \simeq \left( \frac{\hbar ^2 v^4}{m} \right) ^{1/5} \left( 2^{1/5} + 2^{-9/5} \right) \simeq 0.092\ \mathrm{eV}. $$

It is important to notice that our result, apart from a numerical factor, could be also predicted on the basis of simple dimensional remarks. Indeed, it can be easily checked that $(\hbar ^2 v^4/m)^{1/5}$ is the only possible quantity having the dimensions of an energy and constructed in terms of m, v and $\hbar$, which are the only physical constants involved in the problem. In the analogous classical problem $\hbar$ is missing, and v and m are not enough to build a quantity with the dimensions of energy, hence the classical problem lacks the typical energy scale appearing at the quantum level.

2.13

An electron beam with kinetic energy equal to $10\ \mathrm{eV}$ is split into two parallel beams placed at different altitudes in the terrestrial gravitational field. If the altitude gap is $d = 10\ \mathrm{cm}$ and if the beams recombine after a path of length L, say for which values of L the phases of the recombining beams are opposite (destructive interference ). Assume that the upper beam maintains its kinetic energy, that the total energy is conserved for both beams and that the total phase difference accumulated during the splitting and the recombination of the beams is negligible.

Answer: De Broglie’s wave describing the initial electron beam is proportional to $\exp ( \mathrm{i}\,p x/\hbar - \mathrm{i}\,E t/\hbar )$, where $p = \sqrt{2 m E_k}$ is the momentum corresponding to a kinetic energy $E_k$ and $E = E_k + m g h$ is the total energy. The beam is split into two beams, the first travels at the same altitude and is described by the same wave function, the second travels $10\ \mathrm{cm}$ lower and is described by a wave function $\propto \exp ( \mathrm{i}\,p' x/\hbar - \mathrm{i}\,E t/\hbar )$ where $p' = \sqrt{2 m E_k'} = \sqrt{2 m (E_k + m g d)}$ (obviously the total energy E stays unchanged). The values of L for which the two beams recombine with opposite phases are solutions of $(p' - p) L / \hbar = (2 n + 1) \pi$ where n is an integer. The smallest value of L is $L = \pi \hbar / (p' - p)$. Notice that $m g d \simeq 10^{-30}\ J \ll 10\ \mathrm{eV} \simeq 1.6\times 10^{-18}\ \mathrm{J}$ hence $p' - p \simeq \sqrt{2 m E_k} ( m g d / 2 E_k)$ and $L \simeq 2 \pi \hbar E_k / (m g d \sqrt{2 m E_k}) \simeq 696\ \mathrm{m}$. This experiment, which clearly demonstrates the wavelike behavior of material particles, has been really performed using neutrons in place of electrons: that has various advantages, among which that of leading to smaller values of L because of the much heavier mass, as it is clear from the solution. That makes the setting of the experimental apparatus simpler.

2.14

An electron is moving in the $x-y$ plane under the influence of a magnetic induction field parallel to the z-axis. What are the possible energy levels according to Bohr’s quantization rule?

Answer: The electron is subject to the force ${e {\varvec{v}}}{} \wedge {\varvec{B}}$ where ${\varvec{v}}$ is its velocity. Classically the particle, being constrained in the $x-y$ plane, would move on circular orbits with constant angular velocity $\omega = e B/ m$, energy $E = 1/2\ m \omega ^2 r^2$ and any radius r. Bohr’s quantization instead limits the possible values of r by $m \omega r^2 = n \hbar$. Finally one finds $E = 1/2\ n \hbar \omega = n \hbar e B /(2 m)$. This is an approximation of the exact solution for the quantum problem of an electron in a magnetic field (Landau’s levels).

2.15

The positron is a particle identical to the electron but carrying an opposite electric charge. It forms a bound state with the electron, which is similar to the hydrogen atom but with the positron in place of the proton : that is called positronium. What are its energy levels according to Bohr’s rule?

Answer: The computation goes exactly along the same lines as for the proton–electron system, but the reduced mass $\mu = m^2/(m + m) = m/2$ has to be used in place of the electron mass. Energy levels are thus

$$ E_n = - \frac{m e^4}{16 \epsilon _0^2 h^2 n^2}\ . $$

2.16

A particle of mass $M=10^{-29}\ \mathrm{kg}$ is moving in two dimensions under the influence of a central potential

$$ V=\sigma r\ , $$

where $\sigma = 10^5 \ \mathrm{N}.$ Considering only circular orbits, what are the possible values of the energy according to Bohr’s quantization rule?

Answer: Combining the equation for the centripetal force necessary to sustain the circular motion, $m \omega ^2 r = \sigma$, with the quantization of angular momentum, $m \omega r^2 = n \hbar$, we obtain for the total energy, $E = 1/2\ m \omega ^2 r^2 + \sigma r$, the following quantized values

$$ E_n = \frac{3}{2} \left( \frac{\hbar ^2 \sigma ^2}{m} \right) ^{1/3} n^{2/3} \simeq 2\ n^{2/3}\ \mathrm{GeV}\ . $$

Notice that the only possible combination of the physical parameters available in the problem with energy dimensions is $(\hbar ^2 \sigma ^2/{m})^{1/3}$. The potential proposed in this problem is similar to that believed to act among quarks, which are the elementary constituents of hadrons (a wide family of particles including protons, neutrons, mesons ...); $\sigma$ is usually known as the string tension. Notice that the lowest energy coincides, identifying $\sigma = e \mathcal{E}$, with that obtained in Problem 2.9 using the uncertainty principle for the one-dimensional problem.

2.17

The momentum probability distribution for a particle with wave function $\psi (x)$ is given by

$$ \left| \int _{-\infty }^\infty dx {1\over \sqrt{h}}\mathrm{e}^{- \mathrm{i}\,{px/\hbar }}\psi (x) \right| ^2 \equiv |\tilde{\psi }(p)|^2\ . $$

Compute the distribution for the following wave function $\psi (x) = \mathrm{e}^{-a|x|/2} \sqrt{a/2}\$ (a is real and positive) and verify the validity of the uncertainty principle in this case.

Answer: $\tilde{\psi }(p)={(\hbar a)^{3/2}/(\sqrt{4\pi }{(p^2 +{\hbar ^2 a^2/4})})}$ hence

$$ \varDelta ^2_x={a\over 2}\int _{-\infty }^\infty dx\ x^2\mathrm{e}^{-a|x|}={2\over a^2}\, , $$

$$ \varDelta ^2_p={(\hbar a)^3\over 4\pi }\int _{-\infty }^\infty dp\ {p^2\over (p^2+{a^2\hbar ^2\over 4})^2}={a^2\hbar ^2\over 4}\, , $$

so that $\varDelta ^2_x\varDelta ^2_p={\hbar ^2/2}>{\hbar ^2/4}\, .$

2.18

Show that a wave packet described by a real wave function has always zero average momentum. Compute the probability current for such packet.

Answer: From the relation

$$ \tilde{\psi }(p) = \int _{-\infty }^\infty dx {1\over \sqrt{h}} \mathrm{e}^{ - \mathrm{i}\,{px/\hbar }}\psi (x) $$

and $\psi ^* (x) = \psi (x)$ we infer

$$ \tilde{\psi }^* (p) = \int _{-\infty }^\infty dx {1\over \sqrt{h}} \mathrm{e}^{ \mathrm{i}\,{px/\hbar }}\psi (x) = \tilde{\psi }(-p) $$

hence $|\tilde{\psi }(p)|^2 = |\tilde{\psi }(-p)|^2$. The probability distribution function is even in momentum space, so that the average momentum is zero. The probability current is zero as well, in agreement with the average zero momentum, i.e. with the fact that there is not net matter transportation associated to this packet. Notice that the result does not change if $\psi (x)$ is multiplied by a constant complex factor $\mathrm{e}^{ \mathrm{i}\,\phi }$.

2.19

The wave function of a free particle is

$$ \psi (x)={1\over \sqrt{2Ph}}\int _{-P}^Pdq\ \mathrm{e}^{\mathrm{i}\,qx\over \hbar }\ $$

at time $t = 0$. What is the corresponding probability density $\rho (x)$ of locating the particle at a given point x? What is the probability distribution function in momentum space? Give an integral representation of the wave function at a generic time t, assuming that the particle mass is m.

Answer: The probability density is $\rho (x)=|\psi (x)|^2={\hbar /(\pi P x^2)}\sin ^2{(Px/\hbar )}$ while $\psi (x,t)= {(1/\sqrt{2Ph})}\int _{-P}^Pdq \mathrm{e}^{\mathrm{i}\,\left( {qx}-{q^2 t / 2m}\right) /\hbar }$. The distribution in momentum in instead given by $|\tilde{\psi }(p)|^2 = \varTheta (P^2 - p^2)/2 P$, where $\varTheta$ is the step function, $\varTheta (y) = 0$ for $y < 0$ and $\varTheta (y) = 1$ for $y \ge 0$. Notice that for the given distribution we have $\varDelta ^2_x = \infty$. The divergent variance is strictly related to the sharp, step-like distribution in momentum space; indeed $\varDelta ^2_x$ becomes finite as soon as the step is smoothed.

2.20

An electron beam hits the potential step sketched in the figure, coming from the right. In particular, the potential energy of the electrons is 0 for $x<0$ and $-\mathcal{V}=-300\ \mathrm{eV}$ for $x>0$, while their kinetic energy in the original beam (thus for $x > 0$) is $E_k = 400\ \mathrm{eV}$. What is the reflection coefficient?

Answer: The wave function can be written, leaving aside an overall normalization coefficient which is not relevant for computing the reflection coefficient, as

$$ \psi (x) = b \mathrm{e}^{- \mathrm{i}\,k' x} \,\,\,\, \mathrm{for} \,\,\,\, x < 0 \,\, , \quad \quad \psi (x) = \mathrm{e}^{- \mathrm{i}\,k x} + a \mathrm{e}^{\mathrm{i}\,k x} \,\,\,\, \mathrm{for} \,\,\,\, x > 0 $$

where $k = \sqrt{2 m E_k}/\hbar = \sqrt{2 m (E + \mathcal{V})}/\hbar$ and $k' = \sqrt{2 m (E_k - \mathcal{V})}/\hbar = \sqrt{2 m E}/\hbar$; m is the electron mass and $E = E_k - \mathcal{V}$ is the total energy of the electrons. The continuity conditions at the position of the step read

$$ b = 1 + a \,\, , \quad \quad b k' = (1 - a) k \, ,$$

hence

$$ b = \frac{2}{1 + k'/k} \,\, , \quad \quad a = \frac{1 - k'/k}{1 + k'/k} \, , $$

and

$$ R = |a|^2 = \left( \frac{k - k'}{k + k'}\right) ^2 = \frac{2E + \mathcal{V} - 2 \sqrt{E(E + \mathcal{V})}}{2E + \mathcal{V} + 2 \sqrt{E(E + \mathcal{V})}} = \frac{1}{9} \, . $$

2.21

An electron beam hits the same potential step considered in Problem 2.20, this time coming from the left with a kinetic energy $E=100\ \mathrm{eV}$. What is the reflection coefficient in this case?

Answer: In this case we write:

$$ \psi (x) = \mathrm{e}^{ \mathrm{i}\,k' x} + b \mathrm{e}^{- \mathrm{i}\,k' x} \,\,\,\, \mathrm{for} \,\,\,\, x < 0 \,\, , \quad \quad \psi (x) = a \mathrm{e}^{\mathrm{i}\,k x} \,\,\,\, \mathrm{for} \,\,\,\, x > 0 \, , $$

where again $k = \sqrt{2 m (E + \mathcal{V})}/\hbar$ and $k' = \sqrt{2 m E}/\hbar$ with $E = E_k$ being the total energy. By solving the continuity conditions we find:

$$ b = \frac{k'/k - 1}{1 + k'/k} \,\, ; \,\,\,\,\,\,\, R = |b|^2 = \left( \frac{k' - k}{k + k'}\right) ^2 = \frac{2E + \mathcal{V} - 2 \sqrt{E(E + \mathcal{V})}}{2E + \mathcal{V} + 2 \sqrt{E(E + \mathcal{V})}} = \frac{1}{9} \; . $$

We would like to stress that the reflection coefficient coincides with that obtained in Problem 2.20: electron beams hitting the potential step from the right or from the left are reflected in exactly the same way, if their total energy E is the same, as it is in the present case. In fact this is a general result which is valid for every kind of potential barrier and derives directly from the invariance of the Schrödinger equation under time reversal: the complex conjugate of a solution is also a solution. It may seem a striking result, but it should not be so striking for those familiar with reflection of electromagnetic signals in the presence of unmatching impedances.

Notice also that there is actually a difference between the two cases, consisting in a different sign for the reflected wave. That is irrelevant for computing R but significant for considering interference effects involving the incident and the reflected waves. In the present case interference is destructive, hence the probability density is suppressed close to the step, while in Problem 2.20 the opposite happens. To better appreciate this fact consider the analogy with an oscillating rope made up of two different ropes having different densities (which is a system in some sense similar to ours), and try to imagine the different behaviors observed if you enforce oscillations shaking the rope from the heavier (right-hand in our case) or from the lighter (left-hand in our case) side. As an extreme and easier case, you could think of a single rope with a free end (one of the densities goes to zero) or with a fixed end (one of the densities goes to infinity): the shape of the rope at the considered endpoint is cosine-like in the first case and sine-like in the second case, exactly as for the cases of respectively the previous and the present problem in the limit $\mathcal{V} \rightarrow \infty$.

2.22

An electron beam hits, coming from the right, a potential step similar to that considered in Problem 2.20. However this time $-\mathcal{V}=-10\ \mathrm{eV}$ and the kinetic energy of the incoming electrons is $E_k= 9\ \mathrm{eV}$. If the incident current is equal to $J= 10^{-3}\ \mathrm{A}$, compute how many electrons can be found, at a given time, along the negative x axis, i.e. how many electrons penetrate the step barrier reaching positions which would be classically forbidden.

Answer: The solution of the Schrödinger equation can be written as

$$ \psi (x) = c\ \mathrm{e}^{p'x/\hbar } \,\,\,\, \mathrm{for} \,\,\,\, x < 0 \,\, , \quad \quad \psi (x) = a\ \mathrm{e}^{\mathrm{i}\,p x/\hbar } + b\ \mathrm{e}^{- \mathrm{i}\,p x/\hbar } \,\,\,\, \mathrm{for} \,\,\,\, x > 0 $$

where $p = \sqrt{2 m E}$ and $p' = \sqrt{2 m (\mathcal{V} - E)}$. Imposing continuity in $x = 0$ for both $\psi (x)$ and its derivative, we obtain $c = { 2 a}/{(1 + \mathrm{i}\,p'/p)}$ and $b = a {(1 - \mathrm{i}\,p'/p)}/{(1 + \mathrm{i}\,p'/p)}$. It is evident that $|b|^2 = |a|^2$, hence the reflection coefficient is one. Indeed the probability current $J(x) = - {\mathrm{i}\,\hbar }/{(2 m)} (\psi ^*\partial _x\psi - \psi \partial _x\psi ^*)$ vanishes on the left, where we have an evanescent wave function, hence no transmission. Nevertheless the probability distribution is non-vanishing for $x < 0$ and, on the basis of the collective interpretation, the total number of electrons on the left is given by

$$ N = \int _{-\infty }^0 |\psi (x)|^2 dx= |c|^2 \hbar /(2 p') = \frac{2 |a|^2 \hbar p^2}{p' (p^2 + p'^2)\, }. $$

The coefficient a can be computed by asking that the incident current $J_\mathrm{el} = e J = e |a|^2 {p}/{m} \equiv 10^{-3}\ \mathrm{A}$. The final result is $N \simeq 1.2$.

2.23

An electron is confined inside a cubic box with reflecting walls and an edge of length $L = 2\times 10^{-9}\ \mathrm{m}$. How many stationary states can be found with energy less than $1\ \mathrm{eV}$? Take into account the spin degree of freedom, which in practice doubles the number of available levels.

Answer: Energy levels in a cubic box are $E_{n_x,n_y,n_z}={\pi ^{2}\hbar ^{2} ( n_x^{2}+n_y^{2}+n_z^{2} )/(2m L^2)}$, where $m = 0.911\times 10^{-30}\ \mathrm{kg}$ and $n_x, n_y, n_z$ are positive integers. The constraint $E < 1\ \mathrm{eV}$ implies $n_x^2 + n_y^2 + n_z^2 < 10.7$, which is satisfied by 7 different combinations ((1,1,1), (2,1,1), (2,2,1) plus all possible different permutations). Taking spin into account, the number of available levels is 14.

2.24

When a particle beam hits a potential barrier and is partially transmitted, a forward going wave is present on the other side of the barrier which, besides having a reduced amplitude with respect to the incident wave, has also acquired a phase factor which can be inferred by the ratio of the transmitted wave coefficient to that of the incident one. Assuming a thin barrier, describable as

$$ V(x)=v\ \delta (x)\ , $$

and that the particles are electrons of energy $E = 10\ \mathrm{eV}$, compute the value of v for which the phase difference is $-{\pi /4}$.

Suppose now that we have two beams of equal amplitude and phase and that one beam goes through the barrier while the other goes free. The two beams recombine after paths of equal length. What is the ratio of the recombined beam intensity to that one would have without the barrier?

Answer: On one side of the barrier the wave function is $\mathrm{e}^{\mathrm{i}\,\sqrt{2mE}{x/\hbar }}+{a}\ \mathrm{e}^{-\mathrm{i}\,\sqrt{2mE}{x/\hbar }}$, while it is ${b}\ \mathrm{e}^{\mathrm{i}\,\sqrt{2mE}{x/\hbar }}\$ on the other side. Continuity and discontinuity constraints read $1+a=b\$ and $b-1+a=\sqrt{2m/E}{v b / \hbar },$ from which $b={(1+\mathrm{i}\,\sqrt{m/ 2E}{v / \hbar })^{-1}}$ can be easily derived. Requiring that the phase of b be $-\pi /4$ is equivalent to $\sqrt{m/2E}{v/\hbar }=1$, hence $v\simeq 2.0\times 10^{-28}\ \mathrm{J}\ \mathrm{m}$.

With this choice of v the recombined beam is ${}[1+{1/(1+\mathrm{i}\,)}]\mathrm{e}^{\mathrm{i}\,\sqrt{2mE}{x/\hbar }}$. The ratio of the intensity of the recombined beam to that one would have without the barrier is $|[1+{1/(1+\mathrm{i}\,})]/2 |^2= {5/8}$.

2.25

If a potential well in one dimension is so thin to be describable by a Dirac delta function:

$$ V(x)=- VL\delta (x) $$

where V is the depth and L the width of the well, then it is possible to compute the bound state energies by recalling that for a potential energy of that kind the wave function is continuous while its first derivative has the following discontinuity:

$$ \lim _{\epsilon \rightarrow 0}\left( \partial _x\psi (x+\epsilon )-\partial _x\psi (x-\epsilon )\right) =-{2m\over \hbar ^2}VL\psi (0)\ . $$

What are the possible energy levels?

Answer: The bound state wave function is

$$ a\ \mathrm{e}^{- \sqrt{2 m B} x / \hbar } \,\,\,\, \mathrm{for} \,\,\,\, x > 0 \quad \quad \mathrm{and} \quad \quad a\ \mathrm{e}^{ \sqrt{2 m B} x / \hbar } \,\,\,\, \mathrm{for} \,\,\,\, x < 0 $$

where the continuity condition for the wave function has been already imposed. $B = |E|$ is the absolute value of the energy (which is negative for a bound state). The discontinuity condition on the first derivative leads to an equation for B which has only one solution, $B = {m V^2 L^2}/{(2 \hbar ^2)}$, thus indicating the existence of a single bound state.

2.26

A particle of mass m moves in the following one-dimensional potential:

$$ V(x)=v(\alpha \delta (x-L)+\alpha \delta (x+L)-{1\over L} \varTheta (L^2-x^2))\ , $$

where $\varTheta$ is the step function, $\varTheta (y)=0$ for $y<0$ and $\varTheta (y)=1$ for $y>0$. Constants are such that

$$ {2mvL\over \hbar ^2}=\left( {\pi \over 4}\right) ^2\, . $$

Find the conditions on $\alpha >0$ for the existence of bound states.

Answer: The potential is such that $V(-x) = V(x)$: in this case the lowest energy level, if any, corresponds to an even wave function. We can thus limit the discussion to the region $x > 0$, where we have $\psi (x)=\cos kx$ for $x<L$ and $\psi (x)=a\mathrm{e}^{-\beta x}$ for $x>L$, with the constraint $0<k<\sqrt{2mv/(\hbar ^2 L)}={\pi /(4L)}$, since $\beta =\sqrt{{2mv/(\hbar ^2 L)}-k^2}$ must be real in order to have a bound state, hence $kL<{\pi /4}$. Continuity and discontinuity constraints, respectively on $\psi (x)$ and $\psi '(x)$ in $x = L$, give: $\tan kL={(\beta +{2mv\alpha /\hbar ^2})/ k}$. Setting $y \equiv kL$, we have

$$ \tan y={\sqrt{{\pi ^2\over 16}-y^2} +{\pi ^2\over 16}\alpha \over y}\ . $$

The function on the left hand side grows from 0 to 1 in the interval $0< y <{\pi / 4}$, while the function on the right decreases from $\infty$ to $\alpha {\pi / 4}$ in the same interval. Therefore an intersection (hence a bound state) exists only if $\alpha <{4/\pi }$.

2.27

An electron moves in a one-dimensional potential corresponding to a square well of depth $V= 0.1\ \mathrm{eV}$ and width $L=3\times 10^{-10}\ \mathrm{m}$. Show that in these conditions there is only one bound state and compute its energy in the thin well approximation, discussing also the validity of that limit.

Answer: There is one only bound state if the first odd state is absent. That is true if $y = {\sqrt{2 m V} L}/({2 \hbar }) < {\pi }/{2}$, which is verified in our case since, using $m = 0.911\times 10^{-30}\ \mathrm{kg}$, one obtains $y \simeq 0.243 < \pi /2$.

Setting $B \equiv -E$, where E is the negative energy of the bound state, B is obtained as a solution of

$$ \tan \left( \frac{L \sqrt{2 m (V - B)}}{2 \hbar } \right) = \sqrt{\frac{B}{V - B}} \; . $$

The thin well limit corresponds to $V \rightarrow \infty$ and $L \rightarrow 0$ as the product V L is kept constant. Neglecting B with respect to V we can write

$$ \tan \left( \frac{\sqrt{2 m V L^2 }}{2 \hbar } \right) = \sqrt{\frac{B}{V}} \; . $$

In the thin well limit $V L^2 \rightarrow \mathrm{constant} \cdot L \rightarrow 0$, hence we can replace the tangent by its argument, obtaining finally $B = {m V^2 L^2}/({2 \hbar ^2}) \simeq 0.59\times 10^{-2}\ \mathrm{eV}$, which coincides with the result obtained in Problem 2.25. In this case the argument of the tangent is $y \sim 0.24$ and we have $\tan 0.24 \simeq 0.245$; therefore the exact result differs from that obtained in the thin well approximation by roughly 4 %.

2.28

An electron moves in one dimension and is subject to forces corresponding to a potential energy:

$$ V(x)=\mathcal{V} [-\delta (x) +\delta (x-L)] \, . $$

What are the conditions for the existence of a bound state and what is its energy if $L= 10^{-9}$ m and $\mathcal{V}= 2\times 10^{-29}\ \mathrm{J}\ \mathrm{m}$?

Answer: A solution of the Schrödinger equation corresponding to a binding energy $B \equiv -E$ can be written as

$$ \psi (x) = \mathrm{e}^{\sqrt{2mB}x/\hbar } \,\,\,\,\,\,\, \mathrm{for} \,\,\,\,\,\,\, x < 0\, , $$

$$ \psi (x) = a \mathrm{e}^{\sqrt{2mB}x/\hbar }+ b \mathrm{e}^{-\sqrt{2mB}x/\hbar } \,\,\,\,\,\,\, \mathrm{for} \,\,\,\,\,\,\,\,\,\,\,\, 0<x<L\, , $$

$$ \psi (x) = c \mathrm{e}^{-\sqrt{2mB}x/\hbar } \,\,\,\,\,\,\,\,\, \mathrm{for} \,\,\,\,\,\,\,\,\, L<x\, . $$

Continuity and discontinuity constraints, respectively for the wave function and for its derivative in $x=0$ and $x=L$, give: $a+b=1,$ $a-b-1=-\sqrt{2m/B}{\mathcal{V}/\hbar },$ $a \mathrm{e}^{\sqrt{8mB}L/\hbar } +b=c,$ $a \mathrm{e}^{\sqrt{8mB}L/\hbar } -b+c=-c\sqrt{2m/B}{\mathcal{V}/\hbar }.$

The four equations are compatible if $\mathrm{e}^{\sqrt{8mB}L/\hbar }={(1-{2B\hbar ^2\over m\mathcal{V}^2})^{-1}}$, which has a non-trivial solution $B\ne 0$ for any $L>0.$ Setting $y=\sqrt{2B/m}{\hbar /\mathcal{V}}$ the compatibility condition reads $\mathrm{e}^{2m\mathcal{V} Ly/\hbar ^2}={1/(1-y^2)}.$ Using the values of L and $\mathcal{V}$ given in the text one obtains $\mathrm{e}^{2m\mathcal{V} L/\hbar ^2} \gg 1$, hence ${2B\hbar ^2 / (m\mathcal{V}^2)} \simeq 1\$ within a good approximation, i.e. $B \simeq {m \mathcal{V}^2 /(2 \hbar ^2)}$, which coincides with the result obtained in the presence of a single thin well. This approximation is indeed equivalent to the limit of a large distance L (hence $\mathrm{e}^{2m\mathcal{V} L/\hbar ^2} \gg 1$) between the two Dirac delta functions; it can be easily verified that in the same limit one has $b \simeq 1$ and $a \simeq 0$, so that, in practice, the state is localized around the attractive delta function in $x = 0$, which is the binding part of the potential, and does not feel the presence of the other term in the potential which is very far away.

As L is decreased, the binding energy lowers and the wave function amplitude, hence the probability density, gets asymmetrically shifted on the left, until the binding energy vanishes in the limit $L \rightarrow 0$. In practice, the positive delta function in $x = L$ acts as a repulsive term which asymptotically extracts, as $L \rightarrow 0$, the particle from its thin well.

2.29

A particle of mass $M=10^{-26}\ \mathrm{kg}$ moves along the x axis under the influence of an elastic force of constant $k= 10^{-6}\ \mathrm{N/m}$. The particle is in its ground state: compute its wave function and the mean value of $x^2$, given by

$$ \langle x^2 \rangle ={\int _{-\infty }^\infty dx x^2|\psi (x)|^2\over \int _{-\infty }^\infty dx |\psi (x)|^2}\ . $$

Answer:

$$ \psi (x) = \left( \frac{kM}{\pi ^2 \hbar ^2}\right) ^{1/8} \mathrm{e}^{ - {\sqrt{k M}} x^2/ 2 \hbar }; \;\;\;\;\;\;\; \langle x^2 \rangle = \frac{1}{2} \frac{\hbar }{\sqrt{kM}} \simeq 5\ 10^{-19} \mathrm{m}^2\, . $$

2.30

A particle of mass $M=10^{-25}\ \mathrm{kg}$ moves in a 3-dimensional isotropic harmonic potential of elastic constant $k= 10\ \mathrm{N}/\mathrm{m}$. How many states have energy less than $2\times 10^{-2}\ \mathrm{eV}$?

Answer: $E_{n_x.n_y,n_z} = \hbar \sqrt{k/M} (3/2 + n_x + n_y + n_z)$. Therefore $E_{n_x.n_y,n_z} < 2 \times 10^{-2}\ \mathrm{eV}$ is equivalent to $n_x + n_y + n_z < 1.54$, corresponding to 4 possible states, $(n_x, n_y, n_z) =$ (0,0,0), (1,0,0), (0,1,0), (0,0,1).

2.31

A particle of mass $M=10^{-26}\ \mathrm{kg}$ moves in one dimension under the influence of an elastic force of constant $k= 10^{-6}\ \mathrm{N}/\mathrm{m}$ and of a constant force $F=10^{-15}\ \mathrm{N}$ acting in the positive x direction. Compute the wave function of the ground state and the corresponding mean value of the coordinate x, given by

$$ \langle x \rangle ={\int _{-\infty }^\infty dx x|\psi (x)|^2\over \int _{-\infty }^\infty dx |\psi (x)|^2}\ . $$

Answer: As in the analogous classical case, the problem can be brought back to a simple harmonic oscillator with the same mass and elastic constant by a simple change of variable, $y = x - F/K$, which is equivalent to shifting the equilibrium position of the oscillator. Hence the energy levels are spaced as for the harmonic oscillator and the wave function of the ground state is

$$ \psi (x) = \left( \frac{kM}{\pi ^2 \hbar ^2}\right) ^{1/8} \mathrm{e}^{ - \frac{\sqrt{k M}}{2 \hbar } \left( x - {F}/{k} \right) ^2}\, , $$

while $\langle x \rangle = {F}/{k} = 10^{-9}\ \mathrm{m}$.

2.32

A particle of mass $m=10^{-30}\ \mathrm{kg}$ and kinetic energy equal to $50\ \mathrm{eV}$ hits a square potential well of width $L=2\times 10^{-10}\ \mathrm{m}$ and depth $V=1\ \mathrm{eV}$. What is the reflection coefficient computed up to the first non-vanishing order in ${V\over 2E}$?

Answer: Let us choose the square well endpoints in $x = 0$ and $x = L$ and fix the potential to zero outside the well. Let $\psi _s$, $\psi _c$ and $\psi _d$ be respectively the wave functions for $x < 0$, $0 < x < L$ and $x > L$. If the particle comes from the left, then $\psi _s=\mathrm{e}^{\mathrm{i}\,\sqrt{2mE}x/\hbar } + a\ \mathrm{e}^{-\mathrm{i}\,\sqrt{2mE}x/\hbar }$, $\psi _c=b\ \mathrm{e}^{\mathrm{i}\,\sqrt{2m(E+V)}x/\hbar } + c\ \mathrm{e}^{-\mathrm{i}\,\sqrt{2m(E+V)}x/\hbar }$ and $\psi _d = d\ \mathrm{e}^{\mathrm{i}\,\sqrt{2mE}x/\hbar }$ where a and c are necessarily of order V / 2E while b and d are equal to 1 minus corrections of the same order. Indeed, as $V \rightarrow 0$ the solution must tend to a single plane wave. By applying the continuity constraints we obtain:

$$ 1+a=b+c \, , $$

$$ 1-a\simeq b-c +{V\over 2E}\, , $$

$$ b\mathrm{e}^{\mathrm{i}\,\sqrt{2m(E+V)}L\over \hbar }+ c\, \mathrm{e}^{-\mathrm{i}\,\sqrt{2m(E+V)}L/\hbar }\simeq (b+{V\over 2E})\mathrm{e}^{\mathrm{i}\,\sqrt{2m(E+V)}L\hbar } - c\, \mathrm{e}^{-\mathrm{i}\,\sqrt{2m(E+V)}L/\hbar } \, , $$

which are solved by $a \simeq {V\over 4E}(\mathrm{e}^{2 \mathrm{i}\,\sqrt{2m(E+V)}L/\hbar }-1)$ and

$$ R = {V^2\over 4E^2}\sin ^2 \left( {\sqrt{2m(E+V)}L\over \hbar } \right) \simeq 0.96\times 10^{-4} \, . $$

2.33

A particle of mass $m=10^{-30}\ \mathrm{kg}$ is confined inside a line segment of length $L= 10^{-9}\ \mathrm{m}$ with reflecting endpoints, which is centered around the origin. In the middle of the line segment a thin repulsive potential barrier, describable as $V(x)=W\delta (x)$, acts on the particle, with $W=2\times 10^{-28} \ \mathrm{J}\ \mathrm{m}$. Compare the ground state of the particle with what found in absence of the barrier.

Answer: Let us consider how the solutions of the Schrödinger equation in a line segment are influenced by the presence of the barrier. Odd solutions, contrary to even ones, do not change since they vanish right in the middle of the segment, so that the particle never feels the presence of the barrier. In order to discuss even solutions, let us notice that they can be written, shifting the origin in the left end of the segment, as $\psi _s\sim \sin {(\sqrt{2mE}x/\hbar )}$ for $x < L/2$ and $\psi _d\sim \sin {(\sqrt{2mE}(L-x)/\hbar )}$ for $x > L/2$. Setting $z\equiv {\sqrt{2mE}L/(2\hbar )}$ the discontinuity in the wave function derivative in the middle of the segment gives $\tan z =-z{2\hbar ^2 /(mL W)}\simeq -10^{-1}\ z$. Hence we obtain, for the ground state, $E\simeq {2\hbar ^2\over mL^2}\pi ^2(1-2\times 10^{-1})\simeq 2.75\times 10^{- 19}\ \mathrm{J}$, slightly below the first excited level.

Notice that, increasing the intensity of the repulsive barrier W from 0 to $\infty$, the ground level grows from ${\pi ^2 \hbar ^2 / (2 mL^2)}$ to ${2 \pi ^2 \hbar ^2 / (mL^2)}$, i.e. it is degenerate with the first excited level in the $W \rightarrow \infty$ limit. There is no contradiction with the expected non-degeneracy since, in that limit, the barrier acts as a perfectly reflecting partition wall which separates the original line segment into two non-communicating segments: the two degenerate lowest states (as well as all the other excited ones) can thus be seen as two different superpositions (symmetric and antisymmetric) of the ground states of each segment.

2.34

An electron beam corresponding to an electric current $I = 10^{-12}\ \mathrm{A}\$ hits, coming from the right, the potential step sketched in the figure. The potential energy diverges for $x < 0$ while it is $-V=-10\ \mathrm{eV}$ for $0<x<L$ and 0 for $x>L$, with $L= 10^{-11}\ \mathrm{m}$. The kinetic energy of the electrons is $E_k =0.01\ \mathrm{eV}$ for $x > L$. Compute the electric charge density as a function of x.

Answer: There is complete reflection in $x = 0$, hence the current density is zero along the whole axis and we can consider a real wave function. In particular we set $\psi (x) = a \sin ({\sqrt{2mE}}(x-L)/\hbar +\phi )$ for $x > L$ and $\psi (x) = b \sin ({\sqrt{2m(E+V)} x / \hbar })\$ for $0 < x < L$. Continuity conditions read $b\sin ({\sqrt{2m(E+V)} L / \hbar })=a\sin \phi \simeq b\sin ({\sqrt{2mV} L / \hbar })$, $b \cos ({\sqrt{2m(E+V)} L / \hbar }) = \sqrt{E/(E+V)}a\cos \phi \simeq \sqrt{E/ V}a\cos \phi \simeq b \cos ({\sqrt{2mV} L / \hbar })$ (notice that ${\sqrt{2mV} L / \hbar }\simeq 0.57\ \mathrm{rad}\$ hence $\cos ({\sqrt{2mV} L / \hbar }) \simeq 0.85$). That fixes $\sqrt{E/V} \tan \left( {\sqrt{2mV} L / \hbar }\right) \simeq \tan \phi \simeq \phi$ and $b=a{\sqrt{E/V}/ \cos ({\sqrt{2mV} L / \hbar }) },$ while the incident current fixes the value of a, $I = e a^2 \sqrt{2E/m}$. Finally we can write, for the charge density,

$$ e \rho = I \sqrt{m\over 2E} \sin ^2\left( {\sqrt{2mE}\over \hbar }(x - L)+\phi \right) $$

for $x > L$ and

$$ e \rho \sim I\sqrt{mE\over 2V^2} \sin ^2 \left( {\sqrt{2mE} \over \hbar } x\right) $$

for $0 < x < L$.

2.35

Referring to the potential energy given in Problem 2.34, determine the values of V for which there is one single bound state.

Answer: It can be easily realized that any possible bound state of the potential well considered in the problem will coincide with one of the odd bound states of the square well having the same depth and extending from $-L$ to L. The condition for the existence of a single bound state is therefore ${\pi /2} <{\sqrt{2mV}L/\hbar }<{3\pi /2}\ .$

2.36

A ball of mass $m=0.05\ \mathrm{kg}$ moves at a speed of $3\ \mathrm{m}/\mathrm{s}$ and without rolling towards a smooth barrier of thickness $D=10\ \mathrm{cm}$ and height $H=1\ \mathrm{m}$. Using the formula for the tunnel effect, give a rough estimate about the probability of the ball getting through the barrier.

Answer: The transmission coefficient is roughly

$$ T \sim \exp \left( {-{2D\over \hbar }\sqrt{2m(mgH-{mv^2\over 2})}}\right) \simeq 10^{-1.3\,\times \,10^{32}}. $$

2.37

What is the quantum of energy for a simple pendulum of length $l = 1\ \mathrm{m}$ making small oscillations?

Answer: In the limit of small oscillations the pendulum can be described as a harmonic oscillator of frequency $\nu = 2 \pi \omega = 2 \pi \sqrt{g/ l}$, where $g \simeq 9.81\ \mathrm{m}/\mathrm{s}$ is the gravitational acceleration on the Earth surface. The energy quantum is therefore $h\nu =3.1\times 10^{-34}\ \mathrm{J}$.

2.38

Compute the mean value of $x^2$ in the first excited state of a harmonic oscillator of elastic constant k and mass m.

Answer: The wave function of the first excited state is $\psi _1 \propto x\ \mathrm{e}^{-x^2\sqrt{km/(4\hbar ^2)}}$, hence

$$ \langle x^2 \rangle ={\int _{-\infty }^{\infty }x^4\mathrm{e}^{-x^2\sqrt{km/ \hbar ^2}}dx\over \int _{-\infty }^{\infty }x^2\mathrm{e}^{-x^2\sqrt{km/ \hbar ^2}}dx }={3\over 2}\sqrt{\hbar ^2\over km}\, . $$

2.39

A particle of mass M moves in a line segment with reflecting endpoints placed at distance L. If the particle is in the first excited state ($n=2$), what is the mean quadratic deviation of the particle position from its average value, i.e. $\sqrt{\langle x^2 \rangle - \langle x \rangle ^2}\$?

Answer: Setting the origin in the middle of the segment, the wave function is $\psi (x)=\sqrt{2/L} \sin ({2\pi x/ L}) \$ inside the segment and vanishes outside. Obviously $\langle x \rangle =0$ by symmetry, while

$$ \langle x^2 \rangle ={2\over L}\int _{-{L\over 2}}^{L\over 2}x^2 \sin ^2\left( {2\pi x\over L}\right) dx=L^2\left( {1\over 12}-{1\over 8\pi ^2}\right) $$

whose square root gives the requested mean quadratic deviation.

2.40

An electron beam of energy E hits, coming from the left, the following potential barrier: $V(x)= \mathcal{V}\delta (x)$ where $\mathcal{V}$ is tuned to $\hbar \sqrt{2E/ m}$. Compute the probability density on both sides of the barrier.

Answer: The wave function can be set to $\mathrm{e}^{\mathrm{i}\,kx}+a\ \mathrm{e}^{-\mathrm{i}\,kx}$ for $x < 0$ and to $b\ \mathrm{e}^{\mathrm{i}\,kx}$ for $x > 0$, where $k={\sqrt{2mE}/\hbar }$. Continuity and discontinuity constraints for $\psi$ and $\psi '$ in $x = 0$ lead to

$$ a={1\over {\mathrm{i}\,k\hbar ^2\over m \mathcal{V}}-1}={1\over \mathrm{i}\,-1} \, , \quad \,\,\,\, b={{ik\hbar ^2\over m \mathcal{V}}\over {\mathrm{i}\,k\hbar ^2\over m \mathcal{V}}-1}={1\over \mathrm{i}\,+1}\ . $$

The probability density is therefore $\rho = {1/2}$ for $x > 0$, while for $x > 0$ it is $\rho = {3/2}\,-\,\sqrt{2}\sin (2kx\,+\,{\pi /4})$.

2.41

A particle moves in one dimension under the influence of the potential given in Problem 2.34. Assuming that

$$ \sqrt{2mV}{L\over \hbar }={\pi \over 2}+\delta \, , \quad \mathrm{with} \quad \delta \ll 1\ , $$

show that, at the first non-vanishing order in $\delta ,$ one has $B\simeq V\delta ^2$, where $B = -E$ and E is the energy of the bound state. Compute the ratio of the probability of the particle being inside the well to that of being outside.

Answer: The depth of the potential is slightly above the minimum for having at least one bound state (see the solution of Problem 2.36), therefore we expect a small binding energy. In particular the equation for the bound state energy, which can be derived by imposing the continuity constraints, is $\cot \left( {\sqrt{2m(V-B)}L/\hbar }\right) =-\sqrt{B/( V-B)}$. The particular choice of parameters implies $B \ll V$, so that $\cot \left( {\sqrt{2m(V-B)}L/\hbar }\right) \simeq \cot ({\pi /2}(1-{B/(2V)}) +\delta )\simeq - \delta +{\pi B/ (4V)}\simeq - \sqrt{B/ V}$, hence $B\simeq V\delta ^2$. Therefore the wave function is well approximated by $k\sin {(\pi x/2L)}$ inside the well and by $k \mathrm{e}^{-{\sqrt{2mB} (x -L)/\hbar }}\simeq k \mathrm{e}^{-{\pi \delta (x - L)/2L}}\$ outside, where k is a normalization constant. The ratio of probabilities is $\pi \delta / 2$: the very small binding energy is reflected in the large probability of finding the particle outside the well.

2.42

A particle of mass $m = 10^{-30}\ \mathrm{kg}$ and kinetic energy $E = 13.9\ \mathrm{eV}$ hits a square potential barrier of width $L = 10^{-10}\ \mathrm{m}$ and height $V = E$. Compute the reflection coefficient R.

Answer: Let us fix in $x = 0$ and in $x = L$ the edges of the square potential barrier, and suppose the particle comes from the left. The wave function is $\psi (x) = \mathrm{e}^{\mathrm{i}\,k x} + {a}\mathrm{e}^{-\mathrm{i}\,k x}$ for $x < 0$ and $\psi (x) = d\ \mathrm{e}^{\mathrm{i}\,k x}$ for $x > L$, where $k = \sqrt{2m E}/\hbar \simeq 2\times 10^{10}\ \mathrm{m^{-1}}$. Instead for $0 \le x \le L$ the wave function satisfies the differential equation $\psi '' = 0$, which has the general integral $\psi (x) = b x + c$. The continuity conditions for $\psi$ and $\psi '$ in $x = 0$ and $x = L$ read

$$ 1 + a = c; \,\,\,\,\,\,\,\,\, \mathrm{i}\,k\ (1 - a) = b; \,\,\,\,\,\,\,\,\, b L + c = d\ \mathrm{e}^{\mathrm{i}\,k L}; \,\,\,\,\,\,\,\,\, b = \mathrm{i}\,k\ d\ \mathrm{e}^{\mathrm{i}\,k L} \, . $$

Dividing last two equations and substituting the first two we get

$$ a = \frac{\mathrm{i}\,k L}{\mathrm{i}\,k L - 2}; \,\,\,\,\,\,\,\,\,\, R = |a|^2 = \frac{k^2 L^2}{4 + k^2L^2} \simeq \frac{1}{2} \, . $$

It is interesting to verify that the same result can be obtained by taking carefully the limit $E \rightarrow V$ in (2.80).

2.43

A particle whose wave function is, for asymptotically large negative times (that is $-t\gg m/(\hbar k_0\varDelta )$), a Gaussian wave packet

$$ \psi (x,t) ={1\over \sqrt{(2\pi )^{3/2}\varDelta }} \int dk e^{\mathrm{i}\,(kx - \hbar k^2 t/(2 m))} e^{-(k - k_0)^2/(2 \varDelta )^2} $$

with $k_0/\varDelta \gg 1$, interacts in the origin through the potential $V(x) = \mathcal{V} \delta (x)$ and its wave function splits into reflected and transmitted components. Considering values of the time for which the spreading of the packets can be neglected, that is $|t|\ll m/(\hbar \varDelta ^2)$ (see Sect. 2.4), compute the transmitted and reflected components of the wave packet.

Answer: The Gaussian wave packet has been studied in detail in Sect. 2.4, it is therefore straightforward to check that, for large negative times, the solution that we are seeking is a wave packet centered in $x = v t$, with $v = \hbar k_0/m$, i.e. a packet approaching the barrier from the left and hitting it at $t \simeq 0$.

It has been shown in Sect. 2.5.2 (see also Problem 2.24) that the generic solution of the time independent Schrödinger equation, obtained in the case of a single plane wave $\mathrm{e}^{ikx}$ hitting the barrier from the left, is

$$ \psi _k(x)=\varTheta (-x)[\exp (ikx)-\mathrm{i}\,\kappa /(k+\mathrm{i}\,\kappa )\exp (-\mathrm{i}\,kx)]+ \varTheta (x)k/(k+\mathrm{i}\,\kappa )\exp (\mathrm{i}\,kx) $$

where $\varTheta$ is the step function ($\varTheta (x) = 0$ for $x < 0$ and $\varTheta (x) = 1$ for $x \ge 0$) and $\kappa =m\mathcal{V}/\hbar ^2$.

The present problem consists in finding a solution of the time dependent Schrödinger equation which, for asymptotically large negative times and $x < 0$, must be a given superposition of progressive plane waves corresponding to the incoming wave packet. Given the linearity of the Schrödinger equation, the solution must be a linear superposition of the generic solutions given above, with the same coefficients of the incoming packet, i.e.

$$ \psi (x,t) ={1\over \sqrt{(2\pi )^{3/2}\varDelta }} \int dk\ \psi _k(x)e^{-\mathrm{i}\,\hbar k^2 t/(2 m)} e^{-(k - k_0)^2/(2 \varDelta )^2}\ . $$

This decomposes into two components for $x<0$ and a single transmitted component for $x>0$.

The first, ingoing component on the negative semi-axis, which corresponds to $\psi _k(x)=\exp (\mathrm{i}\,kx)$, is a standard Gaussian packet which, as discussed above, crosses the origin for $t \sim 0$ and hence disappears for larger times. On the contrary, as we shall show in a while, the second, reflected component describes a packet moving backward, which crosses the origin for $t \sim 0$, hence appears as a part of the solution for $x < 0$ for positive times (i.e. after reflection of the original packet), when it must be taken into account. In much the same way we shall compute the transmitted component, which is a packet moving forward and which appears on the positive semi-axis for positive times. Now we work out the details.

We represent the transmitted and reflected wave packets by

$$ {1\over \sqrt{(2\pi )^{3/2}\varDelta }} \int _{-\infty }^\infty dt\exp (- F_{T/R}(k,x,t)) $$

$$ F_{T}(k,x,t)=(k - k_0)^2/(2 \varDelta )^2-\mathrm{i}\,(kx - \hbar k^2 t/(2 m))-\ln (k/(k+\mathrm{i}\,\kappa )) $$

$$ F_{R}(k,x,t)=(k - k_0)^2/(2 \varDelta )^2+ \mathrm{i}\,(kx + \hbar k^2 t/(2 m))-\ln (-\mathrm{i}\,\kappa /(k+\mathrm{i}\,\kappa ))\ . $$

Then we have the equations:

$$ \partial _k F_{T}(k,x,t)=(k-k_0)/\varDelta ^2-\mathrm{i}\,(x-\hbar t k/m+\kappa /(k(k+\mathrm{i}\,\kappa )))=0 $$

$$ \partial _k F_{R}(k,x,t)=(k-k_0)/\varDelta ^2+\mathrm{i}\,(x+\hbar t k/m-\mathrm{i}\,/(k+\mathrm{i}\,\kappa ))=0\ . $$

The equation for $F_T$ has three solutions: $k_1\sim k_0$, $k_2\sim 0$ and $k_3\sim -\mathrm{i}\,\kappa$ up to corrections of order $\varDelta ^2$. The first solution has a second derivative of order $1/\varDelta ^2$, to be compared with the second derivatives of the other two solutions, which are of order $1/\varDelta ^4$, hence it is the dominant solution, in the same sense discussed in Sect. 2.4, thus we concentrate on it. Setting again $v=\hbar k_0/m$ we have $k_1=k_0 +\mathrm{i}\,\varDelta ^2 (x-vt+ \kappa /(k_0(k_0+\mathrm{i}\,\kappa )))+O(\varDelta ^4)$ and

$$ F_T(k_1,x,t)=F_T(k_0,x,t)-\partial _k^2F_T(k_0,x,t) (k_1-k_0)^2/2 $$

$$ =-\mathrm{i}\,(k_0x - \hbar k_0^2 t/(2 m))-\ln (k_0/(k_0+\mathrm{i}\,\kappa ))+ [x-vt+\kappa /(k_0(k_0+\mathrm{i}\,\kappa ))]^2\varDelta ^2/2)+O(\varDelta ^4)\ . $$

Therefore we have a wave packet centered in $x=vt- \kappa /(k_0^2+\kappa ^2)$. An analogous analysis on the reflected packet gives:

$$ F_R(k_1,x,t)=F_R(k_0,x,t)-\partial _k^2F_R(k_0,x,t) (k_1-k_0)^2/2 $$

$$ =\mathrm{i}\,(k_0x + \hbar k_0^2 t/(2 m))-\ln (-\mathrm{i}\,\kappa /(k_0+\mathrm{i}\,\kappa ))+ (x+vt-\mathrm{i}\,/(k_0+i\kappa ))^2\varDelta ^2/2)+O(\varDelta ^4)\ . $$

Now the packet is centered in $x=-vt+ \kappa /(k_0^2+\kappa ^2).$

The result is almost as anticipated, apart from the fact that the appearance of the transmitted and reflected wave packets is delayed (advanced) with respect to the time the incoming packet hits the potential barrier (well). For large, positive times the particle is in a superposition of reflected and transmitted state, the probability of finding it in one of the two states after a measurement of its position (i.e. the integral of the probability density over the corresponding packets) is given approximately by the reflection or transmission coefficients computed for $k \sim k_0$.

2.44

A particle of mass m moves in one dimension under the influence of the potential

$$ V(x) = V_0 \varTheta (x) - \mathcal{V} \delta (x)\, . $$

If $\mathcal{V} =3\times 10^{-29}\ \mathrm{J}\ \mathrm{m}$ and $m= 10^{-30}\ \mathrm{kg}$, identify the values of $V_0$ for which the particle has bound states. Assuming the existence of a bound state whose binding energy is $B \ll V_0$, compute the ratio of the probabilities for the particle to be found on the right and on the left-hand side of the origin.

Answer: The wave function of a bound state with energy $-B$ would be

$$ \psi _B(x)=N[\varTheta (-x)\exp (\sqrt{2mB}x/\hbar )+\varTheta (x)\exp (-\sqrt{2m(B+V_0)}x/\hbar )] $$

where $\varTheta$ is the step function, N is the normalization factor and the condition $\sqrt{ B}+\sqrt{ B+V_0 }=\sqrt{2m}\mathcal{V}/\hbar$ must be satisfied. Since $\sqrt{V_0}\le \sqrt{ B}+\sqrt{ B+V_0 }<\infty$ the above condition has a solution provided $\sqrt{2m}\mathcal{V}/\hbar \ge \sqrt{ V_0 }$, hence for $V_0\le 2m \mathcal{V}^2/\hbar ^2\simeq 1.62\times 10^{-19}\ \mathrm{J} \simeq 1\ \mathrm{eV}$. The probabilities for the particle to be found on the right and on the left of the origin are respectively $N^2\hbar /(2\sqrt{2m(B+V_0)})$ and $N^2\hbar /(2\sqrt{2mB})$, their ratio for small B is $\sqrt{B/V_0}\simeq \sqrt{2m/V_0}\mathcal{V}/\hbar \,-\,1$.

2.45

A particle of mass m is bound between two spherical perfectly reflecting walls of radii R and $R+\varDelta$. The potential energy between the walls is $V_0=-\hbar ^2\pi ^2/(2m\varDelta ^2)$. If the total energy of the particle cannot exceed $E_M=6\hbar ^2/(2mR^2)$ compute, in the $\varDelta \rightarrow 0$ limit in which the particle is bound on the sphere of radius R, the maximum possible value of its superficial probability density on the intersection point of the sphere with the positive z axis.

Answer: In the $\varDelta \rightarrow 0$ limit, the radial Schrödinger equation tends to the one-dimensional Schrödinger equation of a particle between two reflecting walls with potential energy between the walls equal to $\bar{V}=-\hbar ^2\pi ^2/(2m\varDelta ^2)\,+\,l(l\,+\,1)\hbar ^2/(2mR^2)$. Therefore the possible energy values are $E_{n,l}=(n^2-1)\hbar ^2\pi ^2/(2m\varDelta ^2)\,+\,l(l\,+\,1)\hbar ^2/(2mR^2)$. Only the energies $E_{1,l}=l(l+1)\hbar ^2/(2mR^2)$ remain finite as $\varDelta \rightarrow 0$. In spherical coordinates, the corresponding wave functions are, in the $\varDelta \rightarrow 0$ limit, $\varPsi _{l,m} = \sqrt{2/(\varDelta R^2)}\sin (\pi (r-R)/\varDelta )Y_{l,m}(\theta ,\phi )$.

The harmonic functions $Y_{l,m}$ with $m\not =0$ are proportional to powers of $x_\pm$, hence they vanish on the z axis, therefore and on account of the energy bound, among the possible solutions, we only consider $\varPsi _{l,0}$ for $0\le l\le 2$. Forgetting the radial dependence which, in the $\varDelta \rightarrow 0$ limit corresponds to a probability density equal to $\delta (r-R)$, these are $\varPsi _{0,0}=1/(R\sqrt{4\pi })$, $\varPsi _{1,0}=\sqrt{3/4\pi }\cos \theta /R$ and $\varPsi _{2,0}=\sqrt{5/16\pi }\,(3\cos ^2\theta -1)/R$. The wave function of the particle with the above constraints is written as the linear combination $a_0\, \varPsi _{0,0}+a_1\, \varPsi _{1,0}+a_2\, \varPsi _{2,0}$, with the normalization condition $|a_0|^2\,+\,|a_1|^2\,+\,|a_2|^2=1.$ On the positive z axis the wave function is $1/\sqrt{4\pi }[a_0 \,+\,\sqrt{3}a_1 \,+\,\sqrt{5}a_2]/R^2.$ It is fairly obvious that the maximum absolute value is reached when $a_0=a_1=0$, hence the maximum superficial probability density of the particle is $5/(4\pi R^2)$. The result can be generalized to the case in which different values of the angular momentum can be reached, indeed it can be proved that $|\varPsi _{l,0} (\theta = 0)|^2 = (2 l + 1)/(4 \pi R^2)$.

2.46

A particle of mass m moves along the x axis under the influence of the double well potential:

$$ V(x)=-\mathcal{V}[\delta (x+L)+\delta (x-L)]\ , $$

with $\mathcal{V}>0.$ Study the solutions of the stationary Schrödinger equation. Since the potential is even under x reflection, the solutions are either even or odd. Show that in the even case there is a single solution for any value of L, discuss the range of values of the binding energy B and, in particular, how B depends on L for small L, i.e. when $\alpha (L) \equiv 2m\mathcal{V}L/\hbar ^2 \ll 1$. Compute the “force” between the two wells in this limit. In the odd solution case, compute the range of $\alpha (L)$ for which there are bound solutions and compare the even with the odd binding energies.

Answer: Starting from the even case, we write the solution between the wells as $\psi _I(x)=\cosh kx$, with $B=\hbar ^2k^2/(2m)$, and the external solution as $\psi _E(x)=a\exp (-k(|x|-L))$. The continuity conditions on the wells give: $a=\cosh (kL)$ and $k(\sinh (kL)+a)=\alpha (L)\cosh (kL)/L$. Setting $kL=y$ we get the transcendental equation $\tanh y=\alpha (L)/y-1$. This equation has a single solution $y( \alpha (L))$, corresponding to a single bound state, for any positive value of $\alpha (L)$. In particular for small $\alpha (L)$ also $y( \alpha (L))$ is small and the equation is approximated by $y-y^3/3=\alpha (L)/y-1$ which, up to the second order in $\alpha (L)$, has the positive y solution $y( \alpha (L))=\alpha (L)-\alpha ^2(L)$. The corresponding binding energy is computed noticing that $B=\hbar ^2y^2/(2mL^2)$, from which we have $B=2m\mathcal{V} ^2/\hbar ^2-8m^2\mathcal{V} ^3L/\hbar ^4+O(\alpha ^4(L))$. This implies that there is an attractive force between the two wells which, in the small $\alpha (L)$ limit, is equal to $F=8m^2\mathcal{V} ^3/\hbar ^4$, furthermore $B\le B_{max}=2m\mathcal{V} ^2/\hbar ^2\,$; notice that $B_{max}$ is the binding energy for a single well $-2 \mathcal{V} \delta (x)$, which is indeed the limit of V(x) as $L \rightarrow 0$. For large $\alpha (L)$ also $y( \alpha (L))$ is large and the transcendental equation is well approximated by $1=\alpha (L)/y-1$, which gives $y( \alpha (L))=\alpha (L)/2$, so that the binding energy reaches its minimum value $B_{min}= m\mathcal{V} ^2/(2\hbar ^2)$, which coincides with the binding energy of a single well.

In the odd case the solution between the wells becomes $\psi _I(x)=\sinh kx$ while the external one does not change, therefore the transcendental equation becomes $\tanh y=y/(\alpha (L)-y)$. Here the right-hand side is concave downward and positive, while the left-hand side is concave upward and positive for $0<y<\alpha (L)$, it has a singularity in $\alpha (L)$ and it is negative beyond the singularity. Therefore the equation has a solution for $0<y<\alpha$ if, and only if, the left-hand side is steeper in the origin than the right-hand side, that is if $\alpha (L)>1$. For large values of $\alpha (L)$, $y( \alpha (L))$ tends to $\alpha (L)/2$ from below; notice that in the even case the same limit is reached from above.

In conclusion, for any value of the distance between the two wells, there is an even solution, whose binding energy is larger than that of a single well; on the contrary an odd solution exists only if $L>\hbar ^2/(2m\mathcal{V})$, with a binding energy lower than that of a single well. In the limit of large separation between the two wells, both the odd and the even level approach the energy of a single well, one from above and the other from below, i.e. we get asymptotically two degenerate levels. The presence of two slightly splitted levels (the even ground state and the odd first excited state) is a phenomenon common to other symmetric double well potentials; an example is given by the Ammonia molecule ($\mathrm{N H_3}$), in which the Nitrogen atom has two symmetric equilibrium positions on both sides of the plane formed by the three Hydrogen atoms.

2.47

A particle of mass m is constrained to move on a plane surface where it is subject to an isotropic harmonic potential of angular frequency $\omega$. Which are the stationary states which are found, for the first excited level, by separation of variables in Cartesian coordinates? Show that the probability current density for such states vanishes. Are there any stationary states belonging to the same level having a non-zero current density? Find those having the maximum possible current density and give a physical interpretation for them.

Answer: For the first excited level, $E_1 = 2 \hbar \omega$, the two following stationary states are found in Cartesian coordinates (see Eqs. (2.136) and (2.133)):

$$ \psi _{1,0} = {\sqrt{2}\, \alpha ^2 x \over \sqrt{\pi }} \mathrm{e}^{-\alpha ^2 (x^2 + y^2)/2}; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \psi _{0,1} = {\sqrt{2}\, \alpha ^2 y \over \sqrt{\pi }} \mathrm{e}^{-\alpha ^2 (x^2 + y^2)/2} \,\,\, $$

where $\alpha = \sqrt{m \omega /\hbar }$. In both cases the current density

$$ {\varvec{J}}= -{\mathrm{i}\,\hbar \over 2m} \left( \psi ^*{\varvec{\nabla }}\psi -\psi {\varvec{\nabla }}\psi ^*\right) = {\hbar \over m} \mathrm{Im} \left( \psi ^*{\varvec{\nabla }}\psi \right) $$

vanishes, since the wave functions are real. However it is possible to find different stationary states, corresponding to linear combinations of the two states above, having a non-zero current. Indeed for the most general state, which up to an overall irrelevant phase factor can be written as

$$ \psi = a\, \psi _{1,0} + \sqrt{1 - a^2} \mathrm{e}^{\mathrm{i}\,\phi } \psi _{0,1} $$

where $a \in [0,1]$ is a real parameter, the probability current density is

$$ {\varvec{J}} = {2 \hbar \alpha ^4 \over m \pi } \mathrm{e}^{-\alpha ^2 (x^2 + y^2)}\, a \sqrt{1 - a^2} \sin \phi \ {\varvec{j}} $$

where $(j_x,j_y) = (-y,x)$. The current field is independent, up to an overall factor, of the particular state chosen and describes a circular flow around the origin, hence in general a state with a non-zero average angular momentum; moreover ${\varvec{\nabla }}\cdot {\varvec{J}} =0$, as expected for a stationary state. The current vanishes for $a = 0,1$ or $\phi = 0,\pi$ and is maximum for $a = 1/\sqrt{2}$ and $\phi = \pm \pi /2$, corresponding to the states:

$$ \psi _\pm = \frac{1}{\sqrt{2}} \left( \psi _{1,0} \pm \mathrm{i}\,\psi _{0,1}\right) = {\alpha ^2 \over \sqrt{\pi }} \mathrm{e}^{-\alpha ^2 (x^2 + y^2)/2} (x \pm \mathrm{i}\,y) = {\alpha ^2 \over \sqrt{\pi }} \mathrm{e}^{-\alpha ^2 r^2/2} x_\pm $$

where $r^2 = x^2 + y^2$, which are easily recognized as the states having a well defined angular momentum $L = \pm \hbar$.

2.48

Starting from the definition of the corresponding harmonic polynomials given in (2.182) and computing the normalization constants, verify Eq. (2.186) for the first spherical harmonics.

Answer: For $l=0$: $\mathcal{Y}_{0,0}$ is a constant, after normalization over the solid angle one finds $Y_{0,0}=\sqrt{1\over 4\pi }$. For $l=1$: $\mathcal{Y}_{1,1}\sim -x_+$. The corresponding normalization constant for $Y_{1,1}=-c_{1,1}\sin \theta e^{i\varphi }$ is given by the equation: $1/|c_{1,1}|^2=2\pi \int _{-1}^1 dx (1-x^2)=8\pi /3$. This, together with (2.180), fixes $Y_{1,\pm 1}$. The normalization constant for $Y_{1,0}=c_{1,0}\cos \theta$ is given by the equation: $1/|c_{1,0}|^2=2\pi \int _{-1}^1 dx x^2=4\pi /3$. The sign is fixed by the condition that $Y_{l,0}(1)>0$. For $l=2$: $\mathcal{Y}_{2,2}\sim x^2_+$. The corresponding normalization constant for $Y_{2,2}=c_{2,2}\sin ^2\theta e^{2i\varphi }$ is given by the equation: $1/|c_{2,2}|^2=2\pi \int _{-1}^1 dx (1-x^2)^2=32\pi /15$. This, together with (2.180), fixes $Y_{2,\pm 2}$. $\mathcal{Y}_{2,1}\sim z x_+$. The corresponding normalization constant for $Y_{2,1}=c_{2,1}\cos \theta \sin \theta e^{i\varphi }$ is given by the equation: $1/|c_{2,1}|^2=2\pi \int _{-1}^1 dx x^2(1-x^2)=8\pi /15$. This, together with (2.180), fixes $Y_{2,\pm 1}$. The normalization constant for $Y_{2,0}=c_{2,0}(3\cos ^2\theta -1)$ is given by the equation: $1/|c_{2,0}|^2=2\pi \int _{-1}^1 dx(3 x^2-1)^2=16\pi /5$. The sign is fixed by the condition that $Y_{l,0}(1)>0$.

2.49

A particle of mass m moves in three dimensions under the influence of the central potential $V(r)=-\hbar ^2\alpha /(2m R)\ \delta (r-R)$, with $\alpha$ positive. Compute the values of $\alpha$ for which the particle has a bound state with non-zero angular momentum.

Answer: For zero angular momentum (S-wave), the solution to the differential equation for the radial wave function $\chi (r)$ defined in (2.187), satisfying the regularity conditions in the origin and at infinity, is equivalent to the odd solution for the one-dimensional double well potential given in Problem 2.46 (setting $L = R$), hence $\chi _{<} \propto \sinh (kr)$ for $r<R$ and $\chi _{>} \propto \exp (k(R-r))$ for $r>R$, the bound state energy being $E = -\hbar ^2k^2/(2m)$ with $k>0.$ We know, from Problem 2.46, that such solution exists only if $\alpha > 1$.

We consider now the P-wave case ($l = 1$). The solution satisfying the correct regularity conditions can be obtained from that written in the S-wave case by applying the recursive equation (2.192). It is, up to an overall normalization factor, $\chi _{<}=\sinh (kr)/(kr)-\cosh (kr)$ for $r<R$ and $\chi _{>} = a\exp (k(R-r))[1/(kr)+1]$ for $r>R$. The continuity conditions at $r=R$, written in terms of $kR=x$, are given by $\sinh x/x-\cosh x=a(1+x)/x$ and $\cosh x/x-\sinh x(1+x^2)/x^2+a(1+1/x+1/x^2)=\alpha (\sinh x/x^2-\cosh x/x),$ from which we have $\tanh x=x(1+x-x^2/\alpha )/(1+x+x^3/\alpha ).$ We know that the graphs of both sides of this equation cross at most once for $x>0$ since we have seen in the one-dimensional case, e.g. in Sect. 2.6, that a thin potential well has at most a single bound state. It remains to be verified if they cross. The graphs are tangent to each other in the origin and, for $x\rightarrow \infty$, the left-hand side tends to $+1$ and the right-hand side to $-1$, therefore if the left-hand side is steeper in the origin the graphs do not cross, otherwise they cross once and there is a bound state. Considering the Taylor expansions of both sides we have $\tanh x\simeq x-x^3/3$ and $x(1+x-x^2/\alpha )/(1+x+x^3/\alpha )\simeq x-x^3/\alpha \, .$ The conclusion is that there is a bound state if $\alpha >3$. It should be clear that if no bound state can be found for $l = 1$ (i.e. $\alpha < 3$), none will be found for $l > 1$ as well, because of the increased centrifugal potential.

2.50

Compute the S-wave scattering length for a particle with mass M in the potential well given in Eq. (2.189).

Answer: Introducing the dimensionless parameters $x=\sqrt{2ME}R/\hbar$ and $y=\sqrt{2MV_0}R/\hbar$, one has, up to an over all normalization factor, the internal solution, for $r<R$, $\chi _<=\sin (\sqrt{x^2+y^2}r/R)$ and, the external solution, for $r>R$, $\chi _>=a\sin (xr/R+\delta _0).$ The continuity conditions at $r=R$ give $x\tan \sqrt{x^2+y^2}=\sqrt{x^2+y^2}\tan (x+\delta _0)$, from which we get

$$ x\cot \delta _0={1+x\tan x\tan \sqrt{x^2+y^2}/\sqrt{x^2+y^2} \over \tan \sqrt{x^2+y^2}/\sqrt{x^2+y^2}-\tan x/x} . $$

The scattering length is apparently equal to $y/(\tan y-y)$. It is positive if $y<\pi /2$, that is in the absence of bound states.

2.51

A particle with mass m moves in three dimensions under the influence of the central potential $V(r)=-\hbar ^2\alpha /(2m R)\delta (r-R)$ with $\alpha$ positive. Compute which are the values of $\alpha$ for which the particle has a bound state in S and P waves.

Answer: We note that the S-wave equation gives a bound state if $\alpha >1$ as we can see comparing with the one dimensional case with a reflecting wall in the origin. Then we consider the solution of the P-wave radial Schrödinger equation. Due to the regularity conditions in the origin and at infinity, the radial wave function defined in Eq. (2.195) is, up to an over all normalization factor, $\chi _{<}=\sinh (kr)/(kr)-\cosh (kr)$ for $r<R$ and $\chi _{>}=a\exp (k(R-r))[1/(kr)+1]$ for $r>R$, the bound state energy being $-\hbar ^2k^2/(2m)$ and $k>0.$ The continuity conditions at $r=R$, written in terms of $kr=x$, are given by $\sinh x/x-\cosh x=a(1+x)/x$ and $\cosh x/x-\sinh x(1+x^2)/x^2+a(1+1/x+1/x^2)=\alpha (\sinh x/x^2-\cosh x/x)\ ,$ from which we have $\tanh x=x(1+x-x^2/\alpha )/(1+x+x^3/\alpha ).$ The graphs of both sides of this equation are tangent to each other in the origin and, for $x\rightarrow \infty$, the left-hand side tends to $+1$ and the right-hand side to $-1$, therefore if the left-hand side is steeper in the origin the graphs do not cross, otherwise they cross once and there is a bound state. Considering the Taylor expansions of both sides we have $\tanh x\simeq x-x^3/3$ and $x(1+x-x^2/\alpha )/(1+x+x^3/\alpha )\simeq x-x^3/\alpha .$ The conclusion is that there is a bound state if $\alpha >3$.

2.52

A particle with mass m moves in three dimensions under the influence of the central potential $V(r)=-\hbar ^2\alpha /(2m R)\delta (r-R)$. Compute the S-wave phase shift comparing the case of $\alpha$ positive with that of $\alpha$ negative.

Answer: Denoting the solutions as in Problem 2.51 one has, up to an over all normalization factor, $\chi _<(r)\sim \sin kr$, $\chi _>(r)\sim a\sin (kr+\delta _0)$. The (dis-)continuity relations give, for $x=kR$, $x\cot \delta _0=(2x^2/\alpha -x\sin (2x))/2\sin ^2x.$ The solution of physical interest corresponds to $\delta _0$ vanishing with x, thus $\delta _0= \alpha x/(1-\alpha )+O(x^2).$ Therefore in the present case the scattering length is equal to $\alpha R/(1-\alpha )$, which is negative either with $\alpha >1$, or with $\alpha$ negative, otherwise the scattering length is positive. If the scattering length is positive the low energy phase shift increases with the energy, otherwise it decreases. It is worth recalling here that the potential has a bound state with $l=0$ for $\alpha >1$. Thus we see that, when there is a bound state, the scattering length is negative.

If the equation $2x/\alpha =\sin 2x$ has solutions, these correspond to energies for which $\delta _0= \pi /2\pm n\pi$ and hence $\sin ^2\delta _0$ reaches its maximum value. If $1/\alpha =1-\epsilon$ with $\epsilon$ and x small, from Eq. (2.233) we have that the S-wave contribution to the total cross section is

$$ \sigma _S\simeq {4\pi R^2\over x^2 +\epsilon ^2}\, . $$

For $1>\alpha >0$ the phase shift starts increasing without reaching $\pi$ and vanishes at high energy. If $\alpha$ is negative the phase shift starts decreasing but for large enough energy goes back to zero.

2.53

A particle with mass m moves in three dimensions under the influence of the central potential $V(r)=-\hbar ^2y^2/(2m R^2)\varTheta (R-r)+\hbar ^2\alpha /(2m R)\delta (r-R)$ with $y=\pi -\epsilon$, $\alpha =\pi /\epsilon -1/2$ and $\epsilon \ll 1$. Compute the S-wave phase shift in the energy range $kR=O(\epsilon )$.

Answer: Denoting the internal and external solutions as in Problems 2.51 and 2.52, and introducing the variable $x=kR$ we have, up to an over all normalization factor, $\chi _<(r)\sim \sin (\sqrt{y^2+x^2}r/R)$, $\chi _>(r)\sim a\sin (xr/R+\delta _0)$. The (dis-)continuity relations give $\sqrt{y^2+x^2}\cot \sqrt{y^2+x^2}=x\cot (x+\delta _0)-\alpha .$ After short calculations we get:

$$ x\cot \delta _0={\sqrt{y^2+x^2}\cot \sqrt{y^2+x^2}+\alpha + x\tan x \over 1-\tan x(\sqrt{y^2+x^2}\cot \sqrt{y^2+x^2}+\alpha )/x}\, , $$

and, with the given choice of the parameters, $\sqrt{y^2\,+\,x^2}\cot \sqrt{y^2\,+\,x^2}+\alpha =(1\,-\,x^2/\epsilon ^2)/2\,+\,O(\epsilon ^2).$ Thus $x\cot \delta _0\simeq (1-x^2/\epsilon ^2)/(1+x^2/\epsilon ^2).$ The corresponding S-wave cross section is

$$ \sigma _0(x)=4\pi R^2{(1+x^2/\epsilon ^2)^2 \over x^2(1+x^2/\epsilon ^2)^2+(1-x^2/\epsilon ^2)^2}\, . $$

For $x=0$ we have $\sigma _0(0)\simeq 4\pi R^2$ and the cross section has a sharp maximum for $x=\epsilon$, indeed $\sigma (\epsilon )=4\pi R^2/\epsilon ^2.$ This is usually called a resonance. The phase shift grows and crosses $\pi /2$, while the cross section reaches its maximum possible value, $\lambda ^2/\pi$.

2.54

A particle with mass m moves in three dimensions under the influence of the central potential $V(r)=-\hbar ^2y^2/(2m R^2)\varTheta (R-r)+\hbar ^2\alpha /(2m R)\delta (r-R)$ with $y=\pi -\epsilon$, $\alpha =\pi /\epsilon -1/2$ and $\epsilon \ll 1$. Discuss the existence of bound states in S wave.

Answer: Denoting the internal and external solutions as in Problems 2.51 and 2.52, and introducing the parameter $x=\sqrt{2m B}R/\hbar$ we have, up to an over all normalization factor, $\chi _<(r)\sim \sin (\sqrt{y^2-x^2}r/R)$, $\chi _>(r)\sim a \exp (-xr/R)$. Then, the (dis-)continuity relations give: $\sqrt{y^2-x^2}\cot \sqrt{y^2-x^2}=-x-\alpha .$ It is not difficult to verify that for large $\alpha$ and $y<\pi$ the above equation has no solution. Once again we find positive scattering length in the absence of bound states.