Quantum mechanics
Download
1 / 49

Quantum Mechanics - PowerPoint PPT Presentation


  • 33 Views
  • Uploaded on

Quantum Mechanics. Chapter 8 . Approximate Solutions.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Quantum Mechanics' - bennett-delaney


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Quantum mechanics
Quantum Mechanics

Chapter 8.

Approximate Solutions


  • In the overwhelming majority of situations of practical interest, it is impossible to find an exact solution of the Schreodinger equation as we have done up to now. Even the hydrogen atom require; an approximation if the effect of spin is to be considered.

  • Fortunately, we have approximation methods that permit us to calculate the energy levels of a wide variety of systems. These calculations have been confirmed by experiment to an astonishingly high degree of accuracy in three-body systems such as the helium atom.


8 1 time independent perturbation theory
§8.1 interest, it is impossible to find an exact solution of the Schreodinger equation as we have done up to now. Even the hydrogen atom require; an approximation if the effect of spin is to be considered. Time-Independent Perturbation Theory

  • We are often interested in systems for which we could solve the Schreodinger equation if the potential energy were slightly different.

  • Consider a one-dimensional example for which we can write the actual potential energy as

    Vactual(x) = V(x) + υ(x) (12.1)

  • where υ(x) is a small perturbation added to the unperturbed potential V(x).


  • The Hamiltonian operator of the “unperturbed” (i.e. exactly solvable) system is

  • and the Schreodinger equation of that system therefore is

    H0ψl = Elψl (12.3)

    with a known set of eigenvalues El and eigenfunctions ψl.

    • The Schreodinger equation of the “perturbed” system contains the additional term υ(x) in the Hamiltonian:


  • [ exactly solvable) system isH0 + υ(x)]ψn’ = En’ψn’ (12.4)

  • which may make it impossible to solve directly for the eigenfunctions ψn’ and eigenvalues En’.

  • However, Postulate 3 tells us that any acceptable wave function may be expanded in a series of eigenfunctions of the unperturbed Hamiltonian.

  • Therefore we may write each function ψn’ as a series of the functions ψl with constant coefficients:

  • (The letter l is simply an index, having nothing to do with the angular momentum quantum number; this is a one-dimensional analysis.)


  • Substitution into Eq.(12.4) gives exactly solvable) system is

  • The technique used in finding coefficients in a Fourier series may be employed here.

  • We multiply each side of Eq.(12.6) by ψm*—the complex conjugate of a particular unperturbed eigenfunction ψm. We then integrate both sides over all x, to obtain


  • Integrating each side of Eq.(12.7) term by term and removing the space-independent factors anl and En’ from the integrals. we obtain

  • Because the functions ψ are normalized and orthogonal, the right-hand side of Eq.(12.8) reduces to the single term anmEn’ for which l = m, and we have

  • We can rewrite the left-hand side of Eq.(12.8) by using the fact that H0ψl = Elψl [Eq.(12.3)]; this equation then becomes


  • or the space-independent factors a

  • (12.9)

  • Again, because the functions ψ are normalized and orthogonal, the first term on the left reduces to anmEm.

  • The second term can be written in abbreviated form as ∑∞l=1 anlυml, where υml is an abbreviation for the integral ∫∞-∞ψ*mυ(x)ψl dx, called the matrix element of the perturbing potential υ(x) between the states m and l.


  • ( the space-independent factors aThis term can also be written in Dirac notation as <m|υ(x)|l>.)

  • Substitution into Eq.(12.9) now yields

  • and after rearranging terms,

  • Equation (12.10) is exact. No approximations have been used in deriving it. However, it contains too many unknown quantities to permit an exact solution in most cases.


  • The most important of the unknown quantities is the perturbed energy of the nth level, En’,or more precisely,Δ En = En’ - En . So we let m=n in Eq.(12.10) to obtain

  • (12.10a)

  • To find a first approximation to this energy difference, we make the arbitrary assumption that anl = 1 if l = n and anl = 0 if l ≠ n. In that case, the left-hand side of Eq.(12.10a) collapses to a single term: vnn.

    • Thus Eq.(12.10) is reduced to the approximation


  • This is a first-order approximation to the perturbed energy En’. You can recognize this integral from the formula for the expectation value of an operator.

  • This formula is not exact because the integral contains the wave function of the unperturbed system rather than the actual wave function.

  • Second-order Perturbation Calculation

  • A more accurate value for the energy can be achieved by using a more accurate wave function, for which we again resort to an approximation.


  • We find this approximation in the values of the coefficients anl. We now assume that each coefficient anl is quite small (although not zero) if l ≠ n and that ann ≌ 1.

  • On these basis we can solve Eq.(12.10) for each particular coefficient anm in turn, by assuming that each other coefficient, except anm, may be neglected.

  • With this assumption, Eq.(12.10) becomes

    vmn + anmvmm = (En’ – Em)anm (12.12)

  • or

    (12.13)

    • From Eq.(12.11) we deduce that vmm = Em’ – Em; substitution into (12.13) yields


  • Having found the values of the coefficients a anm, we can substitute these values into Eq.(12.5) to find a better approximation to the perturbed eigenfunctions ψn’.

  • We can then insert these eigenfunctions (instead of the eigenfunctions of the unperturbed system) into Eq.(12.10) to find a second-order approximation to the energy.

  • This second-order value, En”, is


  • This second-order result is particularly important in cases where the matrix elements vnn are all zero, making the first-order perturbation in the energy levels also zero. The following example illustrates this point.

  • Example Problem 12.1 A one-dimensional harmonic oscillator of charge q is perturbed by the application of an electric field E in the positive x direction, making the potential energy

    V(x) = mω2x2/2 - qEx. Show that the first-order perturbation in each energy level is zero and that the second-order perturbation in the energy of the lowest (n = 0) level is -q2E2/2mω2.


  • Solution where the matrix elements v. We insert the perturbing potential v(x) = -eEx into Eq.(12.11). Inserting the harmonic oscillator wave functions [Eqs.(4.27)], we obtain

  • which is proportional to

  • The integral reduces to


  • Regardless of the value of n, the integrand is an odd function of x. Therefore the integral must be zero, and En’ = En for all values of n.

  • The second-order result for level l is found from Eq.(12.15) with n = 0:

  • The first term in the series is

  • which may be written, from Eqs.(4.30),


  • Because ψ is normalized, this becomes –e function of x. Therefore the integral must be zero, and E2E2/2aћω. or -e2E2/2mω2 (since a = mω/ ћ). Each term in the series (12.19) contains the factor |∫ψs*ψ1 dx|2 [analogous to that in Eq.(12.20)].

  • When s ≠ 1, this factor is zero because of the orthogonality of the wave functions). Thus the total energy shift is simply E0” – E0’ = -e2E2/2mω2.

  • Example Problem 12.2 From first-order perturbation theory. find the lowest energy level. and the wave function for this level, in the “stepped” potential well given by (see Figure 12.1):


  • V(x) = +∞ (|x|>a) function of x. Therefore the integral must be zero, and E

  • V(x) = 0 (a/3 < |x| < a)

  • V(x) = δ (|x| < a/3)

  • where δ is small relative to the energy of the lowest level.

  • Solution. The unperturbed wave functions are simply those of the infinitely deep square well (Section 3.2), which are (omitting the time factor), for |x|≤a,

  • ψ(x) = N cos(nπx/2a) (for odd n);

  • ψ(x) = N sin(nπx/2a) (for even n)



  • where N = 1/(a) function of x. Therefore the integral must be zero, and E1/2. The perturbed energy levels are found from Eq.(12.11). For n = l we have

  • because v(x) equals δ only for -a/3 < x < +a/3, and is zero everywhere else.

  • Not surprisingly, this result is simply the product of δ and the probability of finding the particle in the region where v(x) = δ. The numerical result is


  • In other words, because the probability of finding the particle between x = +a/3 and x = +a/3 is 0.61. the energy is increased by 0.61 times the amount of the additional potential energy in that region.

  • To find the perturbed wave function ψ1’, we use the series of Eq.(12.5):

  • ψ1’ = ψ1 + a12ψ2 + a13ψ3 + ┅

  • and we apply Eq.(12.14) to find the coefficients:


  • With N particle between x = +a/3 and x = +a/3 is 0.61. the energy is increased by 0.61 times the amount of the additional potential energy in that region.2 = 1/a, E1’ ≌ E1 = ћ2/32ma2 and E3’ ≌ E3 = (h2/32ma2)/9 (Section 3.2), and integration, Eq.(12.21) becomes

  • Calculation of the other nonzero coefficients, a15, a17, etc., is straightforward. They become progressively smaller, because of the increasing energy difference in the denominator. Therefore the perturbed (unnormalized) wave function is given by


  • Figure 12.2 Wave function for lowest energy in the well of Figure 12.1. Dashed line shows the unperturbed wave function for the infinitely deep well.

    • Figure 12.2 compares ψ1’ and ψ1. Notice several features of this figure:

    • When x = 0, ψ1 = 1, and ψ1’ < 1 because of the negative coefficient a13.


Figure 12.1. Dashed line shows the unperturbed wave function for the infinitely deep well.Near x = ±d, ψ1’ > ψ1, and the second derivative of ψ1’ has a greater magnitude than the second derivative of ψ1, because of the greater value of the perturbed kinetic energy in this region: E’ = E + 0.61δ, and V(x) = 0 in both cases.

• Near the center of the well, the second derivative of ψ1’ is greater than the second derivative of ψ1, because the perturbed kinetic energy E’ - δ is smaller than the unperturbed kinetic energy: E’ - δ = E - 0.39δ.

  • Perturbation of Degenerate Eigenfunctions

  • The preceding discussion covers only the special situations in which no eigenstate is degenerate


  • and the perturbation energy is much larger than the separation between eigenvalues.

  • In such situations it is reasonable to assume that for each perturbed eigenfunction a single coefficient in the expansion (12.5) is nearly equal to 1, and other coefficients are quite small.

  • But in many cases, the unperturbed levels are degenerate.

  • In that case we can only assume that any given perturbed eigenfunction must, in the first approximation, be a linear combination of the eigenfunctions for one of the unperturbed levels.


  • Therefore we must allow as many nonzero coefficients a separation between eigenvalues.mn as there are independent eigenfunctions for that unperturbed level. Then we need to use Eq.(12.10) to set up as many simultaneous equations as are necessary to find these coefficients and to find the perturbed energy levels.

  • In some cases we can find these coefficients without much difficulty. For example, consider the functions that are simultaneous eigentunctions of J2 and Jz for the hydrogen atom (Section 6.1).

  • A typical pair of orthogonal degenerate eigenfunctions could be written as Rnl(r)|l,l>|->s and Rnl(r)|l,l-1>|+>s, respectively.


  • If there were no spin-orbit energy, these eigenfunctions would correspond to the same energy level: that is, the level would have a degeneracy of 2.

  • The perturbation that removes this degeneracy is the spin-orbit energy, which is proportional to L·S, as we saw in Section 6.1 [Eq.(10.11)].

  • We also saw there that

    S·L = (J2 - S2 - L2)/2 (10.17)

    and we saw in Example Problem 10.2 how to construct possible linear combinations of eigenfunctions for the perturbed potential in the case l = 1. The perturbation removes the degeneracy.


8 2 the three body system the helium atom
§8.2 would correspond to the same energy level: that is, the level would have a degeneracy of 2.The Three Body system: The Helium atom

  • In quantum mechanics as in classical mechanics, there is, in general, no exact solution for the motion of three interacting particles. However, there are perturbation techniques that are extremely successful.

  • For example, the helium atom can be solved with great accuracy by taking the “unperturbed” state to be one in which each electron interacts with the proton but not with the other electron.


  • In that case each electron state is a state of the helium ion, in which there is no “other electron” with which to interact. The state of the system then must be a superposition like one of the eigenfunctions of Eqs.(11.4), for example,

  • where |a> and |b>, respectively, are eigenstates of the one-electron system, the helium ion (Table 9.1, with Z = 2).

  • For the lowest energy level, each of these states has n = 1; that is, |a> = |b> = |1,0,0>, and if there were no electron-electron interaction the state would have a total energy of -108.8 eV


  • (or -54.4 eV for each electron). ion, in which there is no “other electron” with which to interact. The state of the system then must be a superposition like one of the eigenfunctions of Eqs.(11.4), for example,

  • Obviously, it is a gross oversimplification to overlook the electron-electron interaction energy.

  • The measured ground-state energy of the helium atom is -79.0 eV, or 29.8 eV higher than it would be without the electron-electron interaction. [The value of 79.0 eV is the sum of the first and second ionization energies of the helium atom (Table 12.1).]

  • Can such a large effect be calculated by perturbation methods?

  • Let us compute the first-order approximation by use of Eq.(12.11).


  • TABLE 12.1 Ionization Energies of Light Elements, in Electron Volts

  • Z Element Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ

  • 1 Hydrogen 13.595

  • 2 Helium 24.580 54.400

  • 3 Lithium 5.390 75.619 122.42

  • 4 Beryllium 9.320 18.206 153.85 217.657

  • 5 Boron 8.296 25.149 37.920 259.298 340.127

  • 6 Carbon 11.264 24.376 47.864 64.476 391.986 489.84


  • The “perturbing” potential v is the electrostatic potential of two electrons at positions r1 and r2, or

  • The energy shift produced by this perturbation is therefore, to first order,

  • where u100(1) and u100(2) are simply the hydrogen atom wave functions of Table 9.1 written in terms of the coordinates of particle 1 and particle 2, respectively, and dτ1 and dτ2 are volume elements of the respective coordinates.


  • Written in terms of the six spherical coordinates ρ potential of two electrons at positions r1, θ1, φ1 ,ρ 2,θ2 and φ2 for electrons 1 and 2, respectively, I becomes

  • Except for the factor Ze2/π2a’0, I is identical to the mutual interaction energy of two spherically symmetric charge distributions of charge density

    e-2 ρ 1 and e-2 ρ 2 respectively.

  • The evaluation of this energy is a standard exercise in electrostatic theory. The result, with Z = 2, is


  • I = (5Z/8)m potential of two electrons at positions rrc2α2 = (5/2)×13.60 eV = 34.0 eV (12.26)

  • Recall that the energy of the helium ion is -Z2(mrc2α2/2), and the unperturbed energy E1 is just twice that. or

  • E1 = -2Z2(mrc2α2/2) (12.27)

  • Therefore the total perturbed energy is the sum

  • E1’ = E1 + I = -2Z2(mrc2α2/2) + (5Z/8)mrc2α2

  • = (-Z2 + 5Z/8)(mrc2α2) (12.28)

  • It should be mentioned that the value of mr, depends on the nuclear mass, but mr varies only slightly, from 0.995me for H1 to 0.99996me for C12.


  • The value for Z = 2 is -74.8 eV, which differs by only about 6% from the experimental value of -79.0 obtained from Table 12.1.

  • This is a remarkably good result for a perturbation that could hardly be called small, but we can do much better, as shown in the following.

  • Atoms with larger Values of Z

  • Just as the results for the hydrogen atom can be applied to any one-electron ion (see text above Table 9.1), Eq.(12.28) can be applied to any two-electron ion to yield the approximate total energy that is needed to remove the two electrons.


  • This means that Eq.(12.28) yields the sum of the second and third ionization energies of lithium (Z = 3), the sum of the third and fourth ionization energies of beryllium (Z = 4), etc.

  • The accuracy of the calculation improves as Z increase; for Z = 4, Eq.(12.28) yields E1’ = (-16 + 2.5)(27.2 eV) = -367.2 eV, which is only about 1% different in magnitude from the sum of the third and fourth ionization energies (shown in Table 12.1 to be 371.5).

  • The reason for this improvement is simply that the perturbing potential energy becomes a smaller part of the total energy as Z increases. See also Exercise 3 in this chapter.


  • The Variational Technique third ionization energies of lithium (Z = 3), the sum of the third and fourth ionization energies of beryllium (Z = 4), etc.

  • To obtain greater accuracy, we must vary the wave function. Rather than use the method leading to Eq.(12.15), let us try a different method, called the variational technique, to alter the wave function.

  • The basis of this technique is that, because the allowed wave functions form a complete, orthogonal set of functions (Postulate 3), any well-behaved function ψ’ can be expanded in a series of functions belonging to this set.

  • Therefore, if we calculate E' by using the arbitrary function ψ' (which is presumably not one of the eigenfunctions of the perturbed system), the resulting value of E' will be a weighted sum of all of the energy eigenvalues of the perturbed system.


  • We set third ionization energies of lithium (Z = 3), the sum of the third and fourth ionization energies of beryllium (Z = 4), etc.

    ψ' = ∑anψn’ and Hψn’ = En’ψn’

    (12.29)

    where the ψ’n are precisely the normalized perturbed eigenfunctions, the En’ are the corresponding energy eigenvalues, and H is the Hamiltonian operator for the complete system, including perturbation(s).

    • The expectation value of the energy is now

    E' = ∫ψ’*Hψ' dτ (12.30)

  • This becomes, using Eq.(12.29),

    E' = ∫{∑an*ψn’ *}H{∑anψn’} dτ (12.31)


  • Substituting from Eq.(12.29) into Eq.(12.31), expanding both series, and then integrating term-by-term and using the orthogonality of the ψ’n we finally have

  • E' = ∑|an|2 En (12.32)

  • Equation (12.32) shows that the value of E', which is computed from Eq.(12.30), must be greater than or equal to the ground-state energy E1 for the perturbed system. If E' = E1, the “arbitrary” function ψ' is actually the ground-state eigenfunction.

  • [If this is not clear, substitute E2 = n2E1, E3 = n3E1. etc. into Eq.(12.32), where n2, n3, etc. are numbers greater than 1, because En > E1 for n > 1. Then

  • E' = |a1|2E1 + |a2|2n2E1 + |a3|2n3E1 …


  • where each term after the first is of the form |a series, and then integrating term-by-term and using the orthogonality of the ψ’n|2 multiplied by a number greater than E1.

  • Therefore the entire series must have a value greater than or equal to E1 ∑|an|2. and since we know that ∑|an|2 = 1. we see that E'≥E1. (They are equal only if an = 0 for n > 1, or if there is no energy level higher than E1.)]

  • These facts show us how to improve an approximation even when we don't know the actual ground-state energy. If we compute the values of E' for a number of trial wave functions, the smallest value of E' must be the one that is closest to the actual ground-state energy of the perturbed system.


  • Therefore, a good strategy for computing the ground state energy of a “perturbed” system is to write ψ' in terms of a parameter that can be varied systematically to find the minimum value of E'.

  • For example, we can use the eigenfunctions of a two-electron ion as in the first-order calculation above [Eq.(12.25)], except that we substitute a variable Z' in place of the actual atomic number Z in the wave functions (but not in the potential V). The total energy is then

  • E' = ∫ψ’*Hψ’dτ (12.33)

    where the H operator is given by


  • In Eq.(12.34), E state energy of a “perturbed” system is to write ψ' in terms of a parameter that can be varied systematically to find the minimum value of E'.k = -(ћ2/2m)(▽12 + ▽22) is the operator for the total kinetic energy,

    Vp = -(Ze2/4πε0r1) + (Ze2/4πε0r2) is the total potential energy of the interaction between each electron and the nucleus, and Ve = e2/(4πε0|r1 – r2|) is the potential energy of the interaction between the two electrons.

  • The first two terms are evaluated by comparison with the corresponding values for hydrogen, as follows:


  • Each of the ▽ state energy of a “perturbed” system is to write ψ' in terms of a parameter that can be varied systematically to find the minimum value of E'.2 operators yields a factor of Z’2 in the kinetic energy, which makes Ek the sum of two kinetic energies of Z’2×13.60 eV, because the electron in the hydrogen atom has a kinetic energy of 13.60 eV. Thus Ek = 2Z’2×13.6 eV.

  • The expectation value of Vp must be Z' times twice the potential energy of an electron in a one-electron system of nuclear charge Ze, because the exponential factor e-2ρ1 (where ρ1 = Zr1/a0’) is now e-2ρ1’, where Z has been replaced by Z'.

  • Thus the combined potential energy of the two electrons in the field of the nucleus is Vp = -2ZZ'×27.2 eV. (One factor of Z remains, because this term is proportional to the nuclear charge Ze.)


  • The third term, V state energy of a “perturbed” system is to write ψ' in terms of a parameter that can be varied systematically to find the minimum value of E'.e, is computed as in Eq.(12.25), where the result was an energy of 5Z/4×13.60 eV. When Z is replaced in the wave function by Z', the resulting energy is 5Z'/4×13.60 eV.

  • The three terms yield a total energy of

    E' = Ek + Vp + Ve = 2Z'2×13.6 eV - 2ZZ'×27.2 - 5Z'/4×13.60 (12.35)

  • or

    E' = (2Z'2 - 4ZZ' + 5Z'/4)×13.60 eV (12.36)

  • For helium, with Z = 2, we have

    E' = (2Z'2 - 27Z'/4)×13.60 eV (12.37)


  • The minimum value of E' is -77.46 eV, obtained when Z' = 27/16. This is considerably lower than the -74.8 eV given by Eq.(12.28), and it is less than 2% above the experimental value of -78.98 eV. Improvements in the trial function have reduced the result to within a few parts per million of the experimentally observed value.

  • Higher Energy Levels

  • Although the variational technique computes only the lowest energy level of a system, the general perturbation method can be used for any level.

  • Consider the 1s2s and 1s2p states of a two-electron atom or ion.


  • Without the electron-electron interaction, the energies of these states would differ only because of the tiny spin-orbit energy (Section 10.2).

  • The electron-electron interaction not only raises the energy of both of these states, it also greatly increases the splitting between them.

  • The reason for this is seen in the radial probability amplitudes for hydrogen (Figure 9.2b).

  • The 1s probability amplitude is confined to very small values of r, relative to the 2s or 2p amplitudes, and the 2s amplitude extends to much larger values of r than the 2p amplitude.


  • ( of these states would differ only because of the tiny spin-orbit energy (Section 10.2). The 2s amplitude, with zero angular momentum, can be visualized as representing a more eccentric “orbit” than the 2p amplitude.)

  • Consequently, the mean distance between a 1s electron and a 2s electron is greater than the average distance between a 1s electron and a 2p electron, and therefore the Coulomb repulsion energy is smaller for a 1s2s state than for a 1s2p state, and the energy difference can be computed by the perturbation approach, starting with Eq.(12.24) and using the relevant functions [u100, u210, and u200].

  • Recall that the spin-orbit energy is proportional to L·S.


  • Thus the 1s2p, S = 1 states ( of these states would differ only because of the tiny spin-orbit energy (Section 10.2). 3p, Figure 12.3) become three energy levels (one for each possible orientation of S relative to L), spaced about 10-4 eV apart as in the spin-orbit splitting of hydrogen. The 1s2s level is unsplit, because L = 0 for this state. The other 1s2p states, having S = 0, are not split.

  • Thus, when the electron states are n = 1 and n = 2, there are six distinct energy levels.


  • The End of these states would differ only because of the tiny spin-orbit energy (Section 10.2).

  • Thank Your for Your Attention!


ad