Download Appendix E -‐ Elements of Quantum Mechanics

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Appendix E -­‐ Elements of Quantum Mechanics Quantum mechanics provides a correct description of phenomena on the atomic or sub-­‐atomic scale, where the ideas of classical mechanics are not generally applicable. As we describe nuclear phenomena, we will use many results and concepts from quantum mechanics. While it is our goal not to have the reader, in general, perform detailed quantum mechanical calculation, it is important that the reader understand the basis for many of the descriptive statements made in the text. Therefore, we present, in this Appendix, a brief summary of the essential features of quantum mechanics that we shall use. For more detailed discussion of these features, we refer the reader to the references at the end of this Appendix. E-­‐1 Wave Functions All the knowable information about a physical system (i.e., energy, angular momentum, etc.) is contained in the wave function of the system. We shall restrict our discussion to one-­‐body systems for the present. (We could easily generalize to many body systems). The wave function can be expressed in terms of space coordinates and time or momenta and time. In the former notation we write, ψ (x, y, z, t) or just ψ (E-­‐1) −2−
These wave functions, must be “well-­‐behaved”, i.e., they (and their derivatives with respect to the space coordinates), must be continuous, finite and single-­‐valued. The functions Ψ are solutions to a second order differential equation called the Schrödinger equation (see below). The probability of finding a particle within a volume element dxdydz, W dxdydz, is given by W dx dy dz = ψ* ψ dx dy dz (E-­‐2) where ψ* is the complex conjugate of ψ. (To form the complex conjugate of any complex number, replace all occurrences of i (where i =
) with -­‐i. Real numbers are their own complex conjugates. 6-­‐5i is the complex conjugate of 6+5i. So (a+ib)*(a+ib) = (a-­‐ib)(a+ib) = a2 + b2.) The probability per unit volume (the probability density) is W = ψ*ψ. If we look everywhere in the system, we must find the particle so that ∫ψ* ψ dτ = 1 (E-­‐3) where dτ is a volume element dx dy dz. Wave functions possessing this numerical property are said to be normalized. If the value of some physical quantity P is a function of the position coordinates, the average or expectation value of P is given by < P> = ∫ψ*Pψdτ (E-­‐ 4) −3−
This expectation value represents the average outcome of a large number of measurements. E-­‐2 Operators Often we must compute values of quantities that are not simple functions of the space coordinates, such as the y component of the momentum, py, where equation E-­‐4 is not applicable. To get around this, we say that corresponding to every classical variable, there is a quantum mechanical operator. An operator is a symbol that directs us to do some mathematical operation. For example, the momentum operators are (E-­‐5) while the total energy operator is given as (E-­‐6) Thus, to calculate the expectation value of the x-­‐component of the momentum, px, we write (E-­‐7) −4−
Similarly, the classical expression for the kinetic energy is T=p2/2m (E-­‐8) which, translated to quantum mechanics terms, means the kinetic energy operator, , is, in Cartesian coordinates, (E-­‐9) or, using the Laplacian operator, ∇2 (E-­‐10) where ∇2 is (E-­‐11) E-­‐3 The Schrödinger Equation In 1926, Schrödinger found that behavior on the atomic or subatomic scale was correctly described by a differential equation of the form −5−
(E-­‐12) where V represents the potential energy and ψ the wave function of the system. Substituting from equation (E-­‐6), we can write (E-­‐13) This equation is an example of a general class of equations called eigenvalue equations of the form Ωψ = ωψ where Ω is an operator and ω is the value of an observable corresponding to that operator. (The mathematical expression ψ is referred to as an eigenfunction of the operator Ω). To use the Schrödinger equation to gain information about a physical system, we must perform a set of steps that are as follows: (a) Specify the potential energy function of the system, i.e., specify the forces acting (Section 1.6.1). (b) Find a mathematical function, ψ, which is a solution to the differential equation, the Schrödinger equation. (c) Of the many functions that satisfy the equation, reject those that do not conform to certain physical constraints on the system, known as boundary conditions. −6−
Before illustrating this procedure for several cases of interest to nuclear chemists, we can point out another important property of the Schrödinger equation. If the potential energy V is independent of time, we can separate the space and time variables in the Schrödinger equation by setting Ψ(x,y,z,t) = ψ(x,y,z) τ(t) (E-­‐13b) Substituting this expression into equation E-­‐13, and simplifying, we have (E-­‐14) The only way this equation can be true is for both sides to equal a constant. If we call this “separation constant” E, we can write (E-­‐15) and (E-­‐16) Equation E-­‐15 is the time independent Schrodinger equation. The solution to equation E-­‐
16 is (E-­‐17) −7−
Using the Euler relation (eiθ = cos θ + i sin θ), we can write τ(t) = cos ωt -­‐ i sin ωt (E-­‐18) where τ(t) is a periodic function with angular frequency ω = E/h . The separation constant E can be shown to be the total energy, i.e. the sum of the kinetic and potential energies, T + V. E-­‐4 The Free Particle To illustrate how the Schrödinger equation might be applied to a familiar situation, consider the case of a “free” particle, i.e., a particle moving along at a constant velocity with no force acting on the particle (V=0). (Figure E-­‐1) For simplicity, let us consider motion in one dimension, the x-­‐direction. For the time independent Schrödinger equation, we have (E-­‐19) or (E-­‐20) −8−
where the constant k is given by (E-­‐21) The allowed values of the energy, E, are (Equation E-­‐21) (E-­‐22) where k can assume any value (E is not quantized). Since V=0, E is the kinetic energy of a particle with momentum p = hk. From de Broglie, we know that (E-­‐23) so that we can make the association that The solution for the Schrödinger equation, including the time-­‐dependent part is (E-­‐24) −9−
where k and ω are given (E-­‐21) as (E-­‐25) (E-­‐26) This solution is the equation for a wave traveling to the right (+x direction, the first term) and to the left (-­‐x direction, second term). We can impose a boundary condition, namely, we can specify the particle is traveling in the +x direction. Then we have (E-­‐27) We can now calculate the values of any observable. For example, to calculate the value of the momentum p, we write (see equation E-­‐7) which agrees, of course, with the classical result. E-­‐5 Particle in a Box (One Dimension) (E-­‐28) − 10 −
Continuing our survey of some simple applications of wave mechanics to problems of interest to the nuclear chemist, let us consider the problem of a particle confined to a one-­‐dimensional box (Figure E-­‐2). This potential is flat across the bottom of the box and then rises at the walls. This can be expressed as: V(x) = 0 0≤ x ≤ L (E-­‐29) V(x) = ∞ x < 0, x > L The particle moves freely between 0 and L but is excluded from x < 0 and x> L. Inside the box, the Schrödinger equation has the form of equation E-­‐19 (the free particle). The time independent solution can be written ψ(x) = A sin kx + B cos kx (E-­‐30) But we know that ψ(x) = 0 at x = 0 and x = L. Thus B must be 0 and A sin kL = 0 (E-­‐31) To have sin kL = 0, we must have and, using the result (E-­‐22), we have kL = nπ n = 1, 2, 3 (E-­‐32) − 11 −
(E-­‐33) In this case, the energy is quantized. Only certain values of the energy are allowed. One can show the normalization condition is satisfied if (E-­‐34) The allowed energy levels, the probability densities and the wave functions are shown for the first few levels of this potential in Figure E-­‐3. Sample Problem: Suppose a neutron is confined to a box that is the size of a nucleus, 10-­‐14m. (a) What is the energy of the first excited state? (b) What is the probability of finding the neutron within a region corresponding to 20% of the width of the box, i.e., between 0.4 x 10-­‐14m and 0.6 x 10-­‐14m in the fourth excited state? Solution: (a) Eo (the energy of the ground state) =
= 3.3 x 10-­‐13J= 2.0 MeV − 12 −
The energy of the first excited state, n=2, will be 4Eo and the energy spacing between the first excited state and the ground state will be 3Eo = 6 MeV. (b) Probability = which is the result obtained by inspection of the ψ2 curve in Figure E-­‐3. E-­‐6 The Linear Harmonic Oscillator (One Dimension) One of the classic problems of quantum mechanics that is very important for our study of nuclei is the harmonic oscillator. For a simple harmonic oscillator, the restoring force is proportional to the distance from the center, i.e., F = -­‐kx, so that V(x) = kx2/2. The Schrödinger equation is (E-­‐35) The solution of this equation is mathematically complicated and leads to wave functions of the form where (E-­‐36) − 13 −
(E-­‐37) (the oscillator frequency) with a normalization constant of (E-­‐38) The expression Hn (β) is the nth Hermite polynomial (which can be found in handbooks of mathematical functions). The energy eigenvalues can be shown to be (E-­‐39) where m = 0,1,2,3... Thus the energy levels are equally spaced starting with the zero point energy hυ0. (Figure E-­‐4). Note the solutions have the property that there is some probability of finding the particle in classically forbidden regions, i.e., the particle penetrates into the walls. − 14 −
E-­‐7 Barrier Penetration (One Dimension). Another important quantum mechanical problem of interest to nuclear chemists is the penetration of a one-­‐dimensional potential barrier by a beam of particles. The results of solving this problem (and more complicated variations of the problem) will be used in our study of nuclear α-­‐decay and nuclear reactions. The situation is shown in figure E-­‐5. A beam of particles originating at -­‐∞ is incident on a barrier of thickness L and height Vo that extends from x=0 to x=L. Each particle has a total energy E. (Classically, we would expect if E < Vo, the particles would bounce off the barrier while if E > Vo, the particles would pass by the barrier with no change in their properties. Both conclusions are altered significantly in quantum mechanics). It is conventional to divide the space into three regions I, II, and III, shown in Figure E-­‐5. In regions I and III, we have the “free particle” problem treated in E-­‐4. In region I, we have particles moving to the left (the incident particles) and particles moving to the right (reflected particles). So we expect a wave function of the form E-­‐24, whose time independent part can be written (E-­‐40) where . In region III, we have no particles incident from +∞, so, at best, we can only have particles moving in the +x direction (b=0). Thus (E-­‐41) − 15 −
In region II, the time independent Schrödinger equation is (E-­‐42) where k2 = [2m(Vo -­‐ E)]½ /  , assuming Vo > E. The solution is (E-­‐43) Notice that the wave length λ is the same in regions I and III, but the amplitude of the wave beyond the barrier is much less then in front of the barrier. It can be shown that the probability of transmitting particles through the barrier is (E-­‐44) where V is the particle speed. To determine the value of aIII / aI, we eliminate the other constants bI, aII, bII by applying the conditions that ψ and dψ/dx must be continuous through all space. After much algebra (see, for example, the textbook by Evans), we have (E-­‐45) − 16 −
For nuclear applications, the barriers are quite thick (k2L>>1), in which case,
, thus (E-­‐46) The dominant term in this expression is the exponential. For a 6 MeVα-­‐particle, Vo = 20 MeV, L = 10-­‐14 m, we have ≈ 5.1 x 1015 m-­‐1 Thus and T = 16 x 8/20 x (1-­‐8/20)(5.1 x 10-­‐45) = 1.9 x 10-­‐44 So we ignore the pre-­‐exponential term, and write T ≈ e-­‐2G (E-­‐47) − 17 −
where 2G = 2k2L = 2[2m(Vo -­‐ E)]½ / . For an arbitrarily shaped potential that would be more pertinent to nuclear α-­‐decay, one can show (E-­‐48) where x1 and x2 are the points where E = V(x). What about the case where E > Vo. In regions I and III, the situation is the same. In region II, the wave functions will be given as (E-­‐49) (E-­‐50) where Since the wave length
, we can note by comparing equations that λ2 > λ1, and the momentum (p (= (2mk2)½)) becomes less. In other words, the particle is scattered. E-­‐8 The Schrödinger Equation in Spherical Coordinates − 18 −
Many problems in nuclear physics and chemistry involve potentials, such as the Coulomb potential, that are spherically symmetric. In these cases, it is advantageous to express the time-­‐independent Schrödinger equation in spherical coordinates (Figure E-­‐6). The familiar transformations from a Cartesian coordinate system (x, y, z) to spherical coordinates (r, θ, φ) are (Figure E-­‐6) x = r sinθcosφ y = r sinθsinφ z = r cosθ (E-­‐51) The time independent Schrödinger equation becomes (E-­‐52) When the potential is spherically symmetric, v=v(r), then the wave function can be written as ψ(r, θ, φ) = R(r) Yℓm (θ, φ) where Yℓm are the spherical harmonic functions. − 19 −
If we substitute this wave function in equation (E-­‐52) and collect terms, we find that all function of r can be separated from the functions of θ and φ. (E-­‐53) Setting both sides of the equation equal to a separation constant, ℓ (ℓ+ 1), where ℓ= 0, 1, 2..., we have (E-­‐54) and (E-­‐55) Working on the equation E-­‐54, it is convenient to change variables (E-­‐56) (E-­‐57) − 20 −
This is called the radial wave equation. Apart from the term involving ℓ, it is the same as the one-­‐dimensional time independent Schrödinger equation, a fact that will be useful in its solution. The last term is referred to as the centrifugal potential, i.e., a potential whose first derivative with respect to r gives the centrifugal force. It is important to note that equation E-­‐55 does not contain the potential energy term, and thus once we have solved it, the solutions will supply to all cases where V does not depend on Θ and φ, i.e., all so-­‐called “central potentials”. The wave functions Yℓm (θ, φ) are known as the spherical harmonic functions and are tabulated. The indices ℓ and m are related to the orbital angular momentum, L, of the particle relative to the origin. The magnitude of L is [ℓ (ℓ+1)]½ h and its 2ℓ+1 possible projections on the z axis are equal to m (m = 0, ±1, ±2...±l)*. ℓ is called the orbital angular momentum quantum number while m is the magnetic quantum number, in reference to the different energies of the m states in a magnetic field (the Zeeman effect). It follows, therefore, that the specification of a particular spherical harmonic function (as a solution to the angular equation) uniquely specifies the particle’s orbital angular momentum and its z-­‐component. *In more formal language, 〈ℓ2〉 =  2ℓ (ℓ + 1) 〈ℓz〉 = m  €
€
− 21 −
E-­‐9 The Infinite Spherical Well As an application of the Schrödinger equation, expressed in spherical coordinates, to a problem of interest in nuclear chemistry, let us consider the problem of a particle in an infinite spherical well (Figure E-­‐7). This potential can be defined as V(r) = 0 r < a (E-­‐58) V(r) = ∞ r > a Following our discussion in section E-­‐8, we expect the solution of the Schrödinger equation to be (E-­‐59) where the radial wave function Rl(r) is a solution to the equation (E-­‐60) inside the well. The solutions of this equation are the spherical Bessel functions (E-­‐61) − 22 −
where
. The boundary conditions require ψ = 0 at r = 0, and r = a. This will happen for values of ka that make the Bessel functions have a value of 0 (the “zeros” of these functions). (Each ℓ value will have its own set of zeros). These resulting values of k can be used to calculate the allowed energy levels (Figure E-­‐8). Each level is labeled with a number (1, 2, 3...) and a letter (s, p, d, e, etc.). The letter follows the usual spectroscopic notation of ℓ (ℓ = 0, s; ℓ= 1, p, etc.) while the number designates how many times that letter has occurred (the first d level is 1d; the second 2d, etc.). E-­‐10 Angular Momentum Classically the angular momentum of a particle can be written as = x . (Section 1.6.2). From this classical expression, we can write down the classical components of the vectorl_; ℓx = ypz -­‐ zpx ℓy = zpx -­‐ xpz ℓz = xpy -­‐ ypx (E-­‐62) − 23 −
These classical expressions can be converted to the operator language of quantum mechanics by substitutions (such as x → x, px →i  (∂/∂x), etc.) €
(E-­‐63) As remarked earlier (Section E-­‐9), the expectation values of 〈ℓz〉 and 〈 ℓ2〉 for a central potential are 〈ℓz〉 = m  m = 0, ±1, ±2...±l (E-­‐64) and € €
€
〈ℓ2〉 = ℓ(ℓ+1)  2 (E-­‐65) €
− 24 −
We can give these results a pictorial interpretation that is worth noting. Consider a state of definite orbital angular momentum ℓ. Then The z component of ℓ may have any value up to ±ℓ  . The possible values of ℓz can be represented as the projection of a vector of length ℓ on the z axis (figure E-­‐9). This €
situation is referred to as spatial quantization. Only certain values of ℓz are allowed. Due to the Uncertainty Principle, the values of ℓx and ℓy are completely uncertain. In the language of Figure E-­‐9, the vector representing ℓ is rotating about the z axis, so that ℓ and ℓz are fixed, but ℓx and ℓy are continuously changing. In chemistry, we found that to describe the complete quantum state of an electron in an atom, we had to introduce another quantum number, the intrinsic angular momentum or spin. This quantum number is designated as s. By analogy to the orbital angular momentum quantum number ℓ, we have 〈s2〉 = s(s + 1)  2 〈s2〉 = ms  ms= ±½ (E-­‐66) €
€
Nucleons also have values of the spin quantum number of s = ½, like electrons. The total angular momentum of a nucleon j can be written as − 25 −
= + (E-­‐67) The usual quantum mechanical rules apply to j, i.e., 〈j〉 = j (j+1)  2 €
〈jz〉 = mj  = 〈ℓz + sz〉 €
where mj = -­‐j, -­‐j + 1...j-­‐1, j Thus we have mj = m + ms = m ± ½. €
€
Since mℓ is always an integer, then mj must have a half integer and j must be a half integer, either j = ℓ -­‐ ½ or j = ℓ + ½. Alternatively, for a given ℓ value, we have two possible values of j, j = ℓ -­‐ ½ or j = ℓ + ½. For example, for ℓ = 1 (p state), we have j = ℓ -­‐1/2 = ½ or j = ℓ + ½ = 3/2. We designate these states as p1/2 and p3/2, respectively. E-­‐11 Parity − 26 −
A wave function has positive (or even) parity if it does not change sign by reflection through the origin. ψ (-­‐x, -­‐y, z) = ψ(x, y, z) positive parity, π = + (E-­‐69) Alternatively if reflection through the origin produces a change of sign, the parity of the wave function is negative (-­‐). ψ (-­‐x, -­‐y, z) = -­‐ ψ(x, y, z) negative parity, π = -­‐ (E-­‐70) When ψ is expressed in spherical coordinates as ψ(r, θ, φ), then “reflection through the origin” is accomplished by replacing θ, and φ by (π-­‐θ) and (π + φ), respectively. (r cannot change sign as it is just a distance). In other words, the parity of the wave function is determined only by its angular part. For spherically symmetric potentials, the value of ℓ uniquely determines the parity as π = (-­‐1)ℓ (E-­‐71) A corollary of this in that for a system of particles, the parity is even if the sum of the individual orbital angular momentum quantum numbers Σℓi is even; the parity is odd if Σℓi is odd. Thus the parity of each level depends on its wave function. An excited state of a nucleus need not have the same parity as the ground state. − 27 −
Parity will be valuable to us in our discussion of nuclei because it is conserved in beta decay which will tell us that a different force, the weak interaction, is acting in beta decay compared to nuclear reactions. Also the rates of the γ-­‐ray transitions between nuclear excited states depend on the changes in parity and can be used to determine the parity of nuclear states. − 28 −
E-­‐12 Quantum Statistics The parity of a system is related to the symmetry properties of the spatial portion of the wave function. Another important quantum mechanical property of a system of two or more identical particles is the effect on the wave function of exchanging the coordinates of two particles. If no change in the wave function occurs when the spatial and spin coordinates are exchanged, we say the wave function is symmetric and the particles obey Bose-­‐Einstein statistics. If, upon exchange of the spatial and spin coordinates of the two particles the wave function changes sign, the wave function is said to be antisymmetric and the particles obey Fermi-­‐Dirac statistics. The “statistics” these particles followed, profoundly affects the property of an assembly of such particles. Particles with half-­‐integer spins, such as neutrons, protons, and electrons, are fermions, and obey Fermi-­‐Dirac statistics, have antisymmetric wave functions, and as a consequence, obey the Pauli principle. (No two particles can have identical values of the quantum numbers, m, ℓ, mℓ, s, and ms). Photons, or other particles with integer spins, such as the π meson, are bosons, obey Bose-­‐Einstein statistics, have symmetric wave functions and do not obey the Pauli principle. This difference between fermions and bosons is reflected in how they occupy a set of states, especially as a function of temperature. Consider the system shown in Figure E-­‐
10. At zero temperature (T = 0), the bosons will try to occupy the lowest energy state (a Bose-­‐Einstein coordinate) while for the fermions, the occupancy will be one per quantum − 29 −
state. At high temperatures the distributions are similar and approach the Maxwell Boltzman distribution. The Fermi-­‐Dirac distribution can be described by the equation (E-­‐72) where fFD is the number of particles per quantum state, k is Boltzman’s constant and EF is the Fermi energy. At T = 0, all energy levels up to EF are occupied (fFD = 1) and all energy levels above EF are empty (fFD = 0). As T increases, some levels above EF become occupied at the expense of levels below EF. References K.S. Krane, Modern Physics (Wiley, New York, 1983) A well written introductory treatment of quantum physics. M. Scharff, Elementary Quantum Mechanics (Wiley, London, 1969) A very lucid, elementary treatment of quantum mechanics, emphasizing physical insight rather than formal theory. − 30 −
L.I. Schiff, Quantum Mechanics (McGraw-­‐Hill, New York, 1955) An old classic treatment that contains several applications of interest. E. Merzbacher, Quantum Mechanics (Wiley, New York, 1961) Another treatment with several nuclear physics applications. C. Cohen-­‐Tannoudji, B. Diu, F. Laloe, Quantum Mechanics (Wiley, New York, 1977) An encyclopedic treatment. R.M. Eisberg, Fundamentals of Modern Physics (Wiley, New York, 1961). A comprehensive treatment of modern physics. − 31 −
E
0
-x
x
Figure E-­‐1. The free particle problem. ∞
∞
v=∞
E
x=0
x=L
v=0
Figure E-­‐2. A schematic diagram of a particle in a one-­‐dimensional box. The particle is free to move between x = 0 and x = L, but not allowed to have x < 0 or x > L. − 32 −
Figure E-­‐3. The allowed energy levels of a particle in a one-­‐dimensional box. The wave function is shown as a solid line for each level while the shaded area gives the probability density. − 33 −
Figure E-­‐4. The low-­‐lying levels and associated probability densities for the harmonic oscillator. − 34 −
V0
L
Figure E-­‐5. A schematic diagram of a particle of energy E incident on a barrier of height V0 and thickness L. The wave function ψ is shown also. − 35 −
Figure E-­‐6. Spherical polar coordinates. − 36 −
∞
∞
v(r)
0
R
r=a
Figure E-­‐7. Schematic diagram of the infinite square well potential. − 37 −
Figure E-­‐8a. Energy levels of an infinitely deep spherical square well. The radical probability density functions are shown for different values of ℓ − 38 −
Figure E-­‐8b. The three-­‐dimensional probability densities, ∫n, ℓ, m (r,θ) for an infinitely deep three-­‐dimensional square well. − 39 −
Figure E-­‐9. The spatial orientation and z components of a vector with ℓ = 2. − 40 −
(a)
# particles/
level
(b)
# particles/
level
(c)
Figure E-­‐10. (a) The Bose-­‐Einstein distribution function. (b) The Fermi-­‐Dirac distribution function. − 41 −
(c) The filling of levels by fermions at T=0 and T=T1 > 0. The dashed line indicates the Fermi energies EF