Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula
CHAPTER The quantum theory of motion The three basic modes of motion—translation (motion through space), vibration, and rotation—all play an important role in chemistry because they are ways in which molecules store energy Gas-phase molecules, for instance, undergo translational motion and their kinetic energy is a contribution to the total internal energy of a sample Molecules can also store energy as rotational kinetic energy and transitions between their rotational energy states can be observed spectroscopically Energy is also stored as molecular vibration, and transitions between vibrational states also give rise to spectroscopic signatures In this Chapter we use the principles of quantum theory to calculate the properties of microscopic particles in motion vibrations We see that the energies of oscillator are quantized The acceptable wavefunctions also show that the oscillator may be found at extensions and compressions that are forbidden by classical physics 8C Rotational motion The energy of a rotating particle is quantized, but in this Topic we see that its angular momentum is also restricted to certain values The quantization of angular momentum is a very important aspect of the quantum theory of electrons in atoms and of rotating molecules 8A Translation In this Topic we see that, according to quantum theory, a particle constrained to move in a finite region of space is described by only certain wavefunctions and their corresponding energies Hence, quantization emerges as a natural consequence of solving the Schrödinger equation and the conditions imposed on it The solutions also bring to light a number of non-classical features of particles, especially their ability to tunnel into and through regions where classical physics would forbid them to be found 8B Vibrational motion This Topic introduces the ‘harmonic oscillator’, a simple but very important model for the description of molecular What is the impact of this material? ‘Nanoscience’ is the study of atomic and molecular assemblies with dimensions ranging from 1 nm to about 100 nm and ‘nanotechnology’ is concerned with the incorporation of such assemblies into devices We encounter several concepts of nanoscience throughout the text In Impact I8.1 we explore quantum mechanical effects that render the properties of a nanometre-sized assembly dependent on its size To read more about the impact of this material, scan the QR code, or go to bcs.whfreeman.com/webpub/chemistry/ pchem10e/impact/pchem-8-1.html 8A Translation Contents 8A.1 Free motion in one dimension Brief illustration 8A.1: The wavefunction of a freely-moving particle 8A.2 Confined motion in one dimension The acceptable solutions Brief illustration 8A.2: The energy of a particle in a box (b) The properties of the wavefunctions Example 8A.1: Determining the probability of finding the particle in a finite region (c) The properties of observables Example 8A.2: Estimating an absorption wavelength (a) 8A.3 ➤➤ What you need to know already? 317 318 318 318 319 320 320 321 321 Confined motion in two or more dimensions 322 Separation of variables Brief illustration 8A.3: The distribution of a particle in a two-dimensional box (b) Degeneracy Brief illustration 8A.4: Degeneracies in a two-dimensional box (a) 8A.4 Tunnelling Brief illustration 8A.5: Transmission probabilities for a rectangular barrier Checklist of concepts Checklist of equations You should know that the wavefunction is the solution of the Schrödinger equation (Topic 7B), and be familiar with the techniques of deriving dynamical properties from the wavefunction by using operators corresponding to the observables (Topic 7C) 322 323 324 324 324 326 327 328 In this Topic we present the essential features of the solutions of the Schrödinger equation for translation, one of the basic types of motion We see that quantization emerges as a natural consequence of the Schrödinger equation and conditions imposed on it The solutions also bring to light a number of non-classical features of particles, especially their ability to tunnel into and through regions where classical physics would forbid them to be found 8A.1 Free The Schrödinger equation for a particle of mass m moving freely in one dimension is (Topic 7B) − ➤➤ Why you need to know this material? The application of quantum theory to translation reveals the origin of quantization and other non-classical features of physical and chemical phenomena This material is important for the discussion of atoms and molecules that are free to move within a restricted volume, such as a gas in a container ➤➤ What is the key idea? The translational energy levels of a particle confined to a finite region of space are quantized, and under certain conditions particles can pass into and through classically forbidden regions motion in one dimension 2 d 2ψ (x ) = Eψ (x ) 2m dx Free motion in one dimension Schrödinger equation (8A.1) and the solutions are (as in eqn 7B.6) ψ k = Aeikx + Be − ikx Ek = k 2 2m Wave Free motion functions in one and dimension energies (8A.2) with A and B constants Note that we are now labelling both the wavefunctions and the energies with the index k The wavefunctions in eqn 8A.2 are continuous, have continuous slope everywhere, are single-valued, and not go to infinity, and so—in the absence of any other information—are acceptable for all values of k Because the energy of the particle is proportional to k2, all non-negative values, including zero, of the energy are permitted It follows that the translational energy of a free particle is not quantized 318 8 The quantum theory of motion • If the particle is shot in the opposite direction, towards negative x, then its linear momentum is −k and its wavefunction is proportional to e−ikx In this case, A = 0 and B is the normalization factor ∞ ∞ Potential energy, V • If it is shot towards positive x, then its linear momentum is +k (Topic 7C), and its wavefunction is proportional to eikx In this case B = 0 and A is a normalization factor Physical interpretation The values of the constants A and B depend on how the state of motion of the particle is prepared: Location, x L moving particle Figure 8A.1 A particle in a one-dimensional region with impenetrable walls Its potential energy is zero between x = 0 and x = L, and rises abruptly to infinity as soon as it touches the walls An electron at rest that is shot out of an accelerator towards positive x through a potential difference of 1.0 V acquires a kinetic energy of 1.0 eV or 0.16 aJ (1.6 × 10 −19 J) The wavefunction for such a particle is given by eqn 8A.3 with B = 0 and k given by rearranging the expression for the energy in eqn 8A.2 into When the particle is between the walls, the Schrödinger equation is the same as for a free particle (eqn 8A.1), so the general solutions given in eqn 8A.2 are also the same However, it will prove convenient to use e±ikx = cos kx ± i sin kx (Mathematical background 3) to write Brief illustration 8A.1 The wavefunction of a freely- 2m E k = e2 k 1/2 × (9.109 ×10−31 kg ) × (1.6 ×10−19 J) = (1.055 ×10−34 Js)2 1/2 = 5.1 ×109 m −1 or 5.1 nm−1 (with 1 nm = 10−9 m) Therefore the wavefunction is ψ(x) = Ae5.1ix/nm Self-test 8A.1 Write the wavefunction for an electron travel- ling to the left (negative x) after being accelerated through a potential difference of 10 kV Answer: ψ(x) = Be−510ix/nm The probability density |ψ|2 is uniform if the particle is in either of the pure momentum states eikx or e−ikx According to the Born interpretation (Topic 7B), nothing further can be said about the location of the particle That conclusion is consistent with the uncertainty principle, because if the momentum is certain, then the position cannot be specified (the operators corresponding to x and p not commute and thus correspond to complementary observables, Topic 8C) 8A.2 Confined dimension motion in one Consider a particle in a box in which a particle of mass m is confined to a finite region of space between two impenetrable walls The potential energy is zero inside the box but rises abruptly to infinity at the walls at x = 0 and x = L (Fig 8A.1) ψ k (x ) = Aeikx + Be − ikx = A(cos kx + i sin kx ) + B(cos kx − i sin kx ) = ( A + B) cos kx + ( A − B)i sin kx If we write C = (A − B)i and D = A + B the general solutions take the form ψ k (x ) = C sin kx + D cos kx General solution for 0 ≤ x ≤ L (8A.3) Outside the box the wavefunctions must be zero as the particle will not be found in a region where its potential energy is infinite: For x < and x > L, ψ k (x ) = (8A.4) At this point, there are no restrictions on the value of k and all solutions appear to be acceptable (a) The acceptable solutions The requirement of the continuity of the wavefunction (Topic 7B) implies that ψk(x) as given by eqn 8A.3 must be zero at the walls, for it must match the wavefunction inside the material of the walls where the functions meet That is, the wavefunction must satisfy the following two boundary conditions, or constraints on the function at certain locations: ψ k (0) = and ψ k (L) = Particle in a onedimensional box Boundary conditions (8A.5) As we show in the following Justification, the requirement that the wavefunction satisfy these boundary conditions implies that only certain wavefunctions are acceptable and 319 8A Translation that the only permitted wavefunctions and energies of the particle are En = nh 8mL2 2 n = 1, 2, n = 1, 2, 1/2 (8A.6a) (8A.6b) where C is an as yet undetermined constant Note that the wavefunctions and energy are now labelled with the dimensionless integer n instead of the quantity k Justification 8A.1 The energy levels and wavefunctions of a particle in a one-dimensional box From the boundary condition ψk(0) = 0 and the fact that, from eqn 8A.3, ψ k(0) = D (because sin 0 = 0 and cos 0 = 1), we can conclude that D = 0 It follows that the wavefunction must be of the form ψk(x) = C sin kx From the second boundary condition, ψk(L) = 0, we know that ψk(L) = C sin kL = 0 We could take C = 0, but doing so would give ψ k(x) = 0 for all x, which would conflict with the Born interpretation (the particle must be somewhere) The alternative is to require that kL be chosen so that sin kL = 0 This condition is satisfied if kL = nπ n = 1, 2,… The value n = 0 is ruled out, because it implies k = 0 and ψ k(x) = 0 everywhere (because sin 0 = 0), which is unacceptable Negative values of n merely change the sign of sin kL (because sin(−x) = −sin x) and not result in new solutions The wavefunctions are therefore ψ n (x ) = C sin (nπx / L) n =1, 2,… as in eqn 8A.6a At this stage we have begun to label the solutions with the index n instead of k Because k and Ek are related by eqn 8A.2, and k and n are related by kL = nπ, it follows that the energy of the particle is limited to En = n2h2/8mL2, as in eqn 8A.6b We conclude that the energy of the particle in a one-dimensional box is quantized and that this quantization arises from the boundary conditions that ψ must satisfy This is a general conclusion: the need to satisfy boundary conditions implies that only certain wavefunctions are acceptable, and hence restricts observables to discrete values So far, only energy has been quantized; shortly we shall see that other physical observables may also be quantized We need to determine the constant C in eqn 8A.6a To so, we normalize the wavefunction to by using a standard integral from the Resource section Because the wavefunction is zero outside the range 0 ≤ x ≤ L, we use L ∫ ψ dx = C ∫ 2 L nπx sin dx L Integral T.2 = L 2 C × = 1, so C = L 1/2 2 nπx for ≤ x ≤ L ψ n (x ) = sin L L for x < and x > L ψ n (x )= En = n2 h 8mL2 n = 1, 2,… Onedimen Wave sional functions (8A.7a) box One-dimensional box Energy levels (8A.7b) where the energies and wavefunctions are labelled with the quantum number n A quantum number is an integer (in some cases, as we see in Topic 9B, a half-integer) that labels the state of the system For a particle in a one-dimensional box there is an infinite number of acceptable solutions, and the quantum number n specifies the one of interest (Fig 8A.2) As well as acting as a label, a quantum number can often be used to calculate the energy corresponding to the state and to write down the wavefunction explicitly (in the present example, by using the relations in eqn 8A.7) 100 Energy, E/E1; E1 = h2/8mL2 nπx ψ n ( x ) = C sin L for all n Therefore, the complete solution for the particle in a box is n 10 Classically allowed energies 81 64 49 36 25 16 41 Figure 8A.2 The allowed energy levels for a particle in a box Note that the energy levels increase as n2, and that their separation increases as the quantum number increases Classically, the particle is allowed to have any value of the energy in the continuum shown as a shaded area Brief illustration 8A.2 The energy of a particle in a box A long carbon nanotube can be modelled as a one-dimensional structure and its electrons described by particle-in-a-box wavefunctions The lowest energy of an electron in a carbon nanotube of length 100 nm is given by eqn 8A.7b with n = 1: kg m2 s−2 (1)2 × 6.626 ×10−34 J s E1 = = 6.02 × 10−24 J × (9.109 × 10−31 kg ) × (100 ×10−9 m)2 or 0.00602 zJ and its wavefunction is ψ1(x) = (2/L)1/2sin(πx/L) Self-test 8A.2 What are the energy and wavefunction for the next higher energy electron of the system described in this Brief illustration? Answer: E2 = 0.0241zJ, ψ (x) = (2/L)1/2 sin(2πx/L) (b) The properties of the wavefunctions Figure 8A.3 shows some of the wavefunctions of a particle in a one-dimensional box We see that: • Shortening the wavelength results in a sharper average curvature of the wavefunction and therefore an increase in the kinetic energy of the particle (its only source of energy because V = 0 inside the box) • The number of nodes also increases as n increase; the wavefunction ψn has n − 1 nodes • Increasing the number of nodes between walls of a given separation increases the average curvature of the wavefunction and hence the kinetic energy of the particle Physical interpretation • The wavefunctions are all sine functions with the same amplitude but different wavelengths • The probability density for a particle in a onedimensional box is Wavefunction, ψ 54 (8A.8) • and varies with position The non-uniformity in the probability density is pronounced when n is small (Fig. 8A.4) The most probable locations of the particle correspond to the maxima in the probability density Example 8A.1 Determining the probability of finding the particle in a finite region The wavefunctions of an electron in a conjugated polyene can be approximated by particle-in-a-box wavefunctions What is the probability, P, of locating the electron between x = 0 (the left-hand end of a molecule) and x = 0.2 nm in its lowest energy state in a conjugated molecule of length 1.0 nm? Method According to the Born interpretation, ψ (x)2 dx is the probability of finding the particle in the small region dx located at x; therefore, the total probability of finding the electron in the specified region is the integral of ψ(x)2dx over that region, as given in eqn 7B.11 The wavefunction of the electron is given in eqn 8A.7a with n = 1 The integral you need is in the Resource section: Answer The probability of finding the particle in a region between x = 0 and x = l is x L ∫ l P = ψ n2dx = Figure 8A.3 The first five normalized wavefunctions of a particle in a box Each wavefunction is a standing wave; successive functions possess one more half wave and a correspondingly shorter wavelength L l 2 ∫ sin l nπx 2πnl dx = − sin L 2nπ L L Now set n = 1, L = 1.0 nm, and l = 0.2 nm, which gives P = 0.05 The result corresponds to a chance of in 20 of finding the electron in the region As n becomes infinite, the sine term, which is multiplied by 1/n, makes no contribution to P and the classical result for a uniformly distributed particle, P = l/L, is obtained Self-test 8A.3 Calculate the probability that an electron in the state with n = 1 will be found between x = 0.25L and x = 0.75L in a conjugated molecule of length L (with x = 0 at the left-hand end of the molecule) n=1 (a) nπx ψ n2 (x)= sin2 L L Physical interpretation 320 8 The quantum theory of motion Answer: P = 0.82 n=2 n=2 n=1 (b) n=2 n=1 (c) Figure 8A.4 (a) The first two wavefunctions, (b) the corresponding probability densities, and (c) a representation of the probability density in terms of the darkness of shading The probability density ψ n2 (x ) becomes more uniform as n increases provided we ignore the fine detail of the increasingly rapid oscillations (Fig 8A.5) The probability density at high quantum numbers reflects the classical result that a particle bouncing between the walls spends, on the average, equal times at all points That the quantum result corresponds to the classical prediction at high quantum numbers is an illustration of the correspondence principle, which states that classical mechanics emerges from quantum mechanics as high quantum numbers are reached implies that 〈p2〉 ≠ 0, which implies that the particle must always have nonzero kinetic energy |ψ (x)|2 • If the wavefunction is to be zero at the walls, but smooth, continuous, and not zero everywhere, then it must be curved, and curvature in a wavefunction implies the possession of kinetic energy Physical interpretation 321 8A Translation The separation between adjacent energy levels with quantum numbers n and n + 1 is x/L ψ 2(x) Figure 8A.5 The probability density for large quantum number (here n = 50, blue, compared with n = 1, red) Notice that for high n the probability density is nearly uniform, provided we ignore the fine detail of the increasingly rapid oscillations (c) The properties of observables The linear momentum of a particle in a box is not well defined because the wavefunction sin kx is not an eigenfunction of the linear momentum operator However, each wavefunction is a linear combination of the linear momentum eigenfunctions eikx and e−ikx Then, because sin x = (eix − e−ix)/2i, we can write 1/2 1/2 2 nπx ψ n ( x ) = sin (eikx − e − ikx ) = L L 2i L k= nπ L (8A.9) It follows from the discussion in Topic 7C that half the measurements of the linear momentum will give the value +k︀ and −k for the other half This detection of opposite directions of travel with equal probability is the quantum mechanical version of the classical picture that a particle in a one-dimensional box rattles from wall to wall and in any given period spends half its time travelling to the left and half travelling to the right Because n cannot be zero, the lowest energy that the particle may possess is not zero (as would be allowed by classical mechanics, corresponding to a stationary particle) but E1 = h2 8mL2 Particle in a box Zero-point energy (8A.10) • The Heisenberg uncertainty principle requires a particle to possess kinetic energy if it is confined to a finite region: the location of the particle is not completely indefinite (Δx ≠ ∞), so the uncertainty in its momentum cannot be precisely zero (Δp ≠ 0) Because Δp = (〈p2〉 − 〈p〉2)1/2 = 〈p2〉1/2 in this case, Δp ≠ 0 Physical interpretation This lowest, irremovable energy is called the zero-point energy The physical origin of the zero-point energy can be explained in two ways: En+1 − En = (n + 1)2 h2 n2 h2 h2 − = (2n + 1) 2 8mL 8mL 8mL2 (8A.11) This separation decreases as the length of the container increases, and is very small when the container has macroscopic dimensions The separation of adjacent levels becomes zero when the walls are infinitely far apart Atoms and molecules free to move in normal laboratory-sized vessels may therefore be treated as though their translational energy is not quantized Example 8A.2 Estimating an absorption wavelength β-Carotene (1) is a linear polyene in which 10 single and 11 double bonds alternate along a chain of 22 carbon atoms If we take each Ce C bond length to be about 140 pm, then the length L of the molecular box in β-carotene is L = 2.94 nm Estimate the wavelength of the light absorbed by this molecule from its ground state to the next higher excited state β-Carotene Method For reasons that will be familiar from introductory chemistry, each C atom contributes one p electron to the π-orbitals Use eqn 8A.11 to calculate the energy separation between the highest occupied and the lowest unoccupied levels, and convert that energy to a wavelength by using the Bohr frequency relation (eqn 7A.12) Answer There are 22 C atoms in the conjugated chain; each contributes one p electron to the levels, so each level up to n = 11 is occupied by two electrons The separation in energy between the ground state and the state in which one electron is promoted from n = 11 to n = 12 is ∆E = E12 − E11 = (2 ×11 + 1) (6.626 ×10−34 Js)2 × (9.109 ×10−31 kg ) × (2.94 ×10−9 m)2 = 1.60 × 10−19 J 322 8 The quantum theory of motion or 0.160 aJ It follows from the Bohr frequency condition (ΔE = hν) that the frequency of radiation required to cause this transition is = 1.60 × 10−19 J ∆E = = 2.41 × 1014 s −1 h 6.626 × 10−34 Js or 241 THz (1 THz = 1012 Hz), corresponding to a wavelength λ = 1240 nm The experimental value is 603 THz (λ = 497 nm), corresponding to radiation in the visible range of the electromagnetic spectrum Considering the crudeness of the model we have adopted here, we should be encouraged that the computed and observed frequencies agree to within a factor of 2.5 Self-test 8A.4 Estimate a typical nuclear excitation energy in electronvolts (1 eV = 1.602 × 10 −19 J; 1 GeV = 109 eV) by calculating the first excitation energy of a proton confined to a one-dimensional box with a length equal to the diameter of a nucleus (approximately 1 × 10−15 m, or 1 fm) Answer: 0.6 GeV 8A.3 Confined dimensions motion in two or more Now consider a rectangular two-dimensional region of a surface with length L1 in the x-direction and L2 in the y-direction; the potential energy is zero everywhere except at the walls, where it is infinite (Fig 8A.6) As a result, the particle is never found at the walls and its wavefunction is zero there and everywhere outside the two-dimensional region Between the walls, because the particle has contributions to its kinetic energy from its motion in both the x and y directions, the Schrödinger equation has two kinetic energy terms, one for each axis For a particle of mass m the equation is ∞ V ∞ Particle confined to surface ∞ ∞ − 2 ∂2ψ ∂2ψ + = Eψ 2m ∂x ∂y (8A.12) This is a partial differential equation (Mathematical background 4), and the resulting wavefunctions are functions of both x and y, denoted ψ(x,y) This dependence means that the wavefunction and the corresponding probability density depend on the location in the plane, with each position specified by the values of the coordinates x and y (a) Separation of variables A partial differential equation of the form of eqn 8A.12 can be simplified by the separation of variables technique (Mathematical background 4), which divides the equation into two or more ordinary differential equations, one for each variable We show in the Justification below using this technique that the wavefunction can be written as a product of functions, one depending only on x and the other only on y: ψ (x , y ) = X (x )Y ( y ) and that the total energy is given by E = E X + EY Justification 8A.2 The separation of variables technique applied to the particle in a two-dimensional box We follow the procedure in Mathematical background and apply it to eqn 8A.12 The first step to confirm that the Schrödinger equation can be separated and the wavefunction can be factored into the product of two functions X and Y is to note that, because X is independent of y and Y is independent of x, we can write x Figure 8A.6 A two-dimensional square well The particle is confined to the plane bounded by impenetrable walls As soon as it touches the walls, its potential energy rises to infinity d 2Y ∂2ψ ∂2 XY = =X 2 dy ∂y ∂y Note the replacement of the partial derivatives by ordinary derivatives in each case Then eqn 8A.12 becomes y L1 (8A.13b) where EX is the energy associated with the motion of the particle parallel to the x-axis, and likewise for EY and motion parallel to the y-axis d2 X ∂2ψ ∂2 XY = =Y 2 dx ∂x ∂x L2 (8A.13a) − d 2Y 2 d X Y X + = EXY 2m dx dy Next, we divide both sides by XY, and rearrange the resulting equation into d X d 2Y 2mE + =− X dx Y dy 8A Translation The first term on the left, (1/X)(d2X/dx2), is independent of y, so if y is varied only the second term on the left, (1/Y)(d2Y/dy2), can change But the sum of these two terms is a constant, 2mE/2, given by the right-hand side of the equation Therefore, if the second term did change, then the right-hand side could not be constant Consequently, even the second term cannot change when y is changed In other words, the second term, (1/Y)(d2Y/dy2), is a constant, which we write −2mEY/2 By a similar argument, the first term, (1/X)(d2X/dx2), is a constant when x changes, and we write it −2mEX/2, with E = EX + EY Therefore, we can write d2 X 2mE =− X X dx + – – – + + + – + (a) d 2Y 2mE =− Y Y dy 323 (b) These expressions rearrange into the two ordinary (that is, single-variable) differential equations 2 d X − = EX X 2m dx 2 d 2Y − = EY Y 2m dy (8A.14) (c) Each of the two ordinary differential equations in eqn 8A.14 is the same as the one-dimensional particle-in-a-box Schrödinger equation (Section 8A.2) The boundary conditions are also the same, apart from the detail of requiring X(x) to be zero at x = 0 and L1, and Y(y) to be zero at y = 0 and L2 We can therefore adapt eqn 8A.7a without further calculation: 1/2 2 n πx Xn (x ) = sin for ≤ x ≤ L1 L1 L1 1/2 2 n πy Yn ( y ) = sin for ≤ y ≤ L2 L2 L2 (d) Figure 8A.7 The wavefunctions for a particle confined to a rectangular surface depicted as contours of equal amplitude (a) n1 = 1, n2 = 1, the state of lowest energy; (b) n1 = 1, n2 = 2; (c) n1 = 2, n2 = 1; (d) n1 = 2, n2 = 2 Some of the wavefunctions are plotted as contours in Fig 8A.7 They are the two-dimensional versions of the wavefunctions shown in Fig 8A.3 Whereas in one dimension the wavefunctions resemble states of a vibrating string with ends fixed, in two dimensions the wavefunctions correspond to vibrations of a rectangular plate with fixed edges Brief illustration 8A.3 The distribution of a particle in a Then, because ψ = XY, two-dimensional box ψ n ,n (x , y ) = × (L1L2 )1/2 n πx n πy sin sin L1 L2 for ≤ x ≤ L1 , ≤ y ≤ L2 ψ n ,n (x , y ) = outside box 1 2 Twodimensional box Wave functions (8A.15a) ψ 12,2 (x , y ) = Similarly, because E = EX + EY, the energy of the particle is limited to the values n2 n2 h En ,n = 12 + 22 L1 L2 8m Consider an electron confined to a square cavity of side L, and in the state with quantum numbers n1 = 1, n 2 = 2 Because the probability density is Two-dimensional Energy levels box (8A.15b) with the two quantum numbers taking the values n1 = 1, 2, … and n2 = 1, 2, … independently The state of lowest energy is (n1 = 1, n2 = 1) and E1,1 is the zero-point energy πx 2πy sin2 sin2 L2 L L the most probable locations correspond to sin 2(πx/L) = 1 and sin 2(2πx/L) = 1, or (x,y) = (L/2, L/4) and (L/2, 3L/4) The least probable locations (the nodes, where the wavefunction passes through zero) correspond to zeroes in the probability density within the box, which occur along the line y = L/2 Self-test 8A.5 Determine the most probable locations of an electron in a square cavity of side L when it is in the state with quantum numbers n1 = 2, n2 = 3 Answer: points (x = L/4 and 3L/4; y = L/6, L/2, and 5L/6) 324 8 The quantum theory of motion We treat a particle in a three-dimensional box in the same way The wavefunctions have another factor (for the z-dependence), and the energy has an additional term in n32 / L23 Solution of the Schrödinger equation by the separation of variables technique then gives – + 1/2 n πx n πy n πz ψ n ,n ,n (x , y , z ) = sin sin sin L1L2 L3 L1 L2 L3 for ≤ x ≤ L1 , ≤ y ≤ L2 , ≤ z ≤ L3 Three-dimensional box n n n h En ,n ,n = + + L L L 8m 2 3 2 2 Wavefunctions (8A.16a) Threedimensional box Energy levels (8A.16b) The quantum numbers n1, n2, and n3 are all positive integers 1, 2, … that can be varied independently The system has a zero-point energy (E1,1,1 = 3h2/8mL2 for a cubic box) (b) Degeneracy A special feature of the solutions arises when a two-dimensional box is not merely rectangular but square, with L1 = L2 = L Then the wavefunctions and their energies are n πx n πy ψ n ,n (x , y ) = sin sin L L L for ≤ x ≤ L, ≤ y ≤ L outside box ψ n ,n (x , y ) = 1 En ,n = (n12 + n22 ) h2 8mL2 (a) (b) Figure 8A.8 The wavefunctions for a particle confined to a square well Note that one wavefunction can be converted into the other by rotation of the box by 90° The two functions correspond to the same energy True degeneracy is a consequence of symmetry ψ2,1 are then not degenerate Similar arguments account for the degeneracy of states in a cubic box Other examples of degener acy occur in quantum mechanical systems (for instance, in the hydrogen atom, Topic 9A), and all of them can be traced to the symmetry properties of the system Brief illustration 8A.4 Degeneracies in a two- dimensional box + – Square Wave box functions (8.17a) The energy of a particle in a two-dimensional square box of side L in the state with n1 = 1, n2 = 7 is E1,7 = (12 + 72 ) Square box Energy levels (8.17b) Consider the cases n1 = 1, n2 = 2 and n1 = 2, n2 = 1: πx 2πy ψ 1,2 = sin sin L L L E1,2 = 5h2 8mL2 2πx πy ψ 2,1 = sin sin L L L E2,1 = 5h2 8mL2 Although the wavefunctions are different, they have the same energy The technical term for different wavefunctions corres ponding to the same energy is degeneracy, and in this case we say that the state with energy 5h2/8mL2 is ‘doubly degenerate’ In general, if N wavefunctions correspond to the same energy, then we say that the state is ‘N-fold degenerate’ The occurrence of degeneracy is related to the symmetry of the system Figure 8A.8 shows contour diagrams of the two degenerate functions ψ1,2 and ψ2,1 Because the box is square, one wavefunction can be converted into the other simply by rotating the plane by 90° Interconversion by rotation through 90° is not possible when the plane is not square, and ψ1,2 and h2 50h2 = 8mL 8mL2 This state is degenerate with the state with n1 = 7 and n 2 = 1 Thus, at first sight the energy level 50h2/8mL2 is doubly degener ate However, in certain systems there may be states that are not apparently related by symmetry but are ‘accidentally’ degenerate Such is the case here, for the state with n1 = 5 and n 2 = 5 also has energy 50h 2/8mL Accidental degeneracy is also encountered in the hydrogen atom (Topic 9A) Self-test 8A.6 Find a state (n1, n2) for a particle in a rectangular box with sides of length L1 = L and L2 = 2L that is accidentally degenerate with the state (4,4) Answer: (n1 = 2, n2 = 8) 8A.4 Tunnelling If the potential energy of a particle does not rise to infinity when it is in the wall of the container, and E