The electrons near the top of the filled band cannot be accelerated. There are no levels with higher energy available to them. Materials in which this occurs are insulators; that is, they do not carry currents when electric fields are applied. In case b the energy levels are only partly filled.
In this case the application of an electric field accelerates the electrons at the top of the stack of levels. These electrons have empty energy levels to move into, and they would accelerate indefinitely in a perfect lattice, as stated in the previous section. What keeps them from doing that is dissipation. The lattice is not perfect for two reasons: one is the presence of impurities, which destroys the perfect periodicity; the other is the effect of thermal agitation on the position of the ions forming the lattice, which has the same effect of destroying perfect periodicity.
Materials in which the energy levels below the gaps are only partially filled are conductors. The width of the gaps in the energy spectrum depends on the materials. For some in- sulators the gaps are quite narrow. When this happens, then at finite temperatures T, there is a calculable probability that some of the electrons are excited to the bottom of the set of energy levels above the gap. These electrons can be accelerated as in a conductor, so that the application of an electric field will give rise to a current.
The current is augmented by an- other effect: the energy levels that had been occupied by the electrons promoted to the higher energy band called the conduction band are now empty. They provide vacancies into which electrons in the lower band called the valence band can be accelerated into, Conduction band Gap Narrow gap Valence band Holes a b c Figure 4C-2 Occupation of levels in the lowest two energy bands, separated by a gap. Electrons cannot be accelerated into a nearby energy level.
These electrons can conduct electricity. The electrons leave behind them holes that act as positively charged particles and also conduct electricity. These vacancies, called holes, propagate in the direction opposite to that of the electrons and thus add to the electric current.
This is the situation shown in Fig. The technology of making very thin layers of compounds of materials has improved in recent decades to such an extent that it is possible to create the analog of the infinite wells discussed in Chapter 3. The outer one has a larger energy gap than the inner one, as shown in Fig. The midpoints of the gaps must coincide1 for equilibrium reasons. The result is that both electrons and holes in the interior semiconductor cannot move out of the region between the outer semi- conductors, because there are no energy levels that they can move to.
Such confined re- gions may occur in one, two, or three dimensions. In the last case we deal with quantum dots. The study of the behavior of electrons in such confined regions is a very active field of research in the study of materials. In summary, one-dimensional problems give us a very important glimpse into the physics of quantum systems in the real world of three dimensions.
Bernstein, P. Fishbane, and S. Gasiorowicz Prentice Hall, Fishbane, S. Gasiorowicz and S. Thornton Prentice Hall, There are, of course, many textbooks on semiconductors, which discuss the many devices that use bandgap engineering in great quantitative detail.
See in particular L. Solymar and D. Supplement 5-A Uncertainty Relations In our discussion of wave packets in Chapter 2, we noted that there is a relationship be- tween the spread of a function and its Fourier transform. Our result depends entirely on the oper- ator properties of the observables A and B. Supplement 7-A Rotational Invariance In this supplement we show that the assumption of a central potential implies the conser- vation of angular momentum.
We make use of invariance under rotations. The kinetic en- ergy, which involves p2, is independent of the direction in which p points. The central potential V r is also invariant under rotations. We show that this invariance implies the conservation of angular momentum. To identify the operators that commute with H, let us consider an infinitesimal rotation about the z-axis.
This parallels the clas- sical result that central forces imply conservation of the angular momentum. They must therefore each be constant. We call the constant m2, without specifying whether this quantity is real or com- plex. We will label the polynomial as Pl z. These polynomials are known as Legendre polynomials.
Let us first write the solution of 7B as Pml z. Feynman and H. See Problem 10 in Chapter The description is arrived at in the following way. We now note that in the presence of a potential flux is still conserved. Let us now consider the special case of a square well. The above argument shows us that we only need to consider the phase shift, since at large distances from the well the only deviation from free particle behavior is the phase shift.
The ratio can be related to the phase shift. The first term on the left-hand side is kr l Al 1, 3, 5,. The coef- ficient of the leading power, zl, can be easily obtained from eq. We shall use the simple abbreviation m1, m2 for Y 1 2 l1m1Y l2m2. We merely state the results. The symbol eijk is defined by the following properties: a It is antisymmetric under the interchange of any two of its indices. Two consequences of this rule are i When any two indices are equal, the value of eijk is zero.
We may write the equation in a very suggestive way by using the Levi-Civita symbol in two contexts. First, the symbol may be used to give a matrix representation of the spin 1 angular momentum S.
We need both equations to obtain separate equations for E and B. Supplement A Conservation of Total Momentum In our discussion of angular momentum in Chapter 8 we found that the assumption of in- variance of the Hamiltonian under rotations led to the appearance of a new constant of motion, the angular momentum. In this supplement we show that the assumption of in- variance under spatial displacement leads to the existence of a constant of the motion, the momentum.
The potential energy will change, unless it has the form V x1, x2, x3,. In quantum mechanics the same conclusion holds. We shall demonstrate it by using the invariance of the Hamiltonian under the transformation 13A The invariance im- plies that both HuE x1, x2,. Let us take a infinitesimal, so that terms of 0 a2 can be neglected.
This is a very deep consequence of what is really a statement about the nature of space. The statement that there is no origin—that is, that the laws of physics are invariant under displacement by a fixed distance—leads to a conservation law. In relativistic quan- tum mechanics there are no potentials of the form that we consider here; nevertheless the invariance principle, as stated earlier, still leads to a conserved total momentum.
For light atoms it is possible to solve such an equation on a computer, but such solutions are only meaningful to the ex- pert. We shall base our discussion of atomic structure on a different approach.
The varia- tional principle discussed at the end of Chapter 14 had the virtue of maintaining the single-particle picture, while at the same time yielding single-particle functions that incor- porate the screening corrections.
A more general approach is that due to Hartree. The total can then be set equal to zero, since the constraints on the fi ri are now taken care of. The equation 14A is a rather complicated integral equation, but it is at least an equation in three dimensions we can replace the variable ri by r , and that makes nu- merical work much easier. The trial wave function 14A-2 does not take into account the exclusion principle. To take the exclusion principle into account, we add to the Ansatz represented by 14A-2 the rule: Every electron must be in a differ- ent state, if the spin states are included in the labeling.
A more sophisticated way of doing this automatically is to replace 14A-2 by a trial wave function that is a Slater determi- nant [cf. The resulting equations differ from 14A by the addition of an exchange term.
The new Hartree-Fock equations have eigenvalues that turn out to differ by 10—20 percent from those obtained using Hartree equations supplemented by the con- dition arising from the exclusion principle. It is a little easier to talk about the physics of atomic structure in terms of the Hartree picture, so we will not discuss the Hartree-Fock equations.
We may expect, however, that for low Z at least, the splitting for different l values for a given n will be smaller than the splitting between different n-values, so that electrons placed in the orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f,. Screening effects will accentuate this: Whereas s orbitals do overlap the small r region significantly, and thus feel the full nuclear attraction, the p-, d-,.
This effect is so strong that the energy of the 3d electrons is very close to that of the 4s electrons, so that the anticipated ordering is sometimes disturbed. The same is true for the 4d and 5s electrons, the 4f and 6s elec- trons, and so on.
The dominance of the l-dependence over the n-dependence becomes more important as we go to larger Z-values, as we shall see in our discussion of the peri- odic table. The ionization energy is The radius of the atom is 0. We denote this configuration by 1s 2. The total binding energy is 79 eV. Thus its binding energy is This expression takes into account shielding in the second term.
This property is shared by all atoms whose electrons form closed shells, but the energy re- quired is particularly large for helium. We can estimate the effective Z from the measured ionization energy of 5.
It takes very little energy to excite the lithium atom. The six 2p electronic states lie just a little above the 2s state, and these 2p states, when oc- cupied, make the atom chemically active see our more extended discussion of carbon. Lithium, like other elements that have one electron outside a closed shell, is a very active element. The difference between the 21 eV and the See Fig.
We again have a closed shell and the spectroscopic description is 1S0. As far as the energy is concerned, the situation is very much like that of helium.
The shielding sit- uation is somewhat like that for lithium, and if we make a guess that, as in lithium, the binding energy is increased by about 50 percent, we get approximately 9 eV. The experi- mental value is 9. Although the shell is closed, the excitation of one of the electrons to a 2p orbital does not cost much energy. Thus, in the presence of another element a re- arrangement of electrons may yield enough energy to break up the closed shell. We there- fore expect beryllium not to be as inert as helium.
The latter is lower in energy, and it is the 2p shell that be- gins to fill up, starting with boron. These will be discussed below. The ionization energy is 8. This meets the expectation that the value should be somewhat lower than that for beryllium, since the 2p state energy is somewhat higher than that of the 2s orbital.
The second electron could be in the same p-state as the first electron, with the two of them making an up—down spin pair. It is, however, advantageous for the second to stay out of the way of the first electron, thus lowering the repulsion between the electrons. When two electrons go into orthogonally aligned arms, the overlap is minimized and the repul- sion is reduced. The electrons are in different spatial states, so that their spins do not have to be antiparallel.
One might expect carbon to be divalent. This is not so, because of the subtleties that arise from close-lying energy levels. The reduction in the repulsion leads to a somewhat larger ionization energy than that for boron, The spectroscopic description of the ground state is 3P0.
Of the various states, 1S0, 3 P2,1,0, and 1D2, the state of higher spin has the lower energy cf. The three electrons can all be in nonoverlapping p-states, and thus we expect the increase in ionization energy to be the same as the increase from boron to carbon.
This is in agreement with the measured value of Since there are four electrons, it appears as if the determination of the ground-state spectroscopic state would be very difficult. We can, however, look at the shell in another way. We can thus think of oxygen as having a closed 2p shell with two holes in it. These holes are just like anti- electrons, and we can look at two-hole configurations. When the fourth electron is added to the nitrogen configuration, it must go into an orbital with an m-value already occupied.
Thus two of the electron wave functions overlap, and this raises the energy because of the repulsion. It is therefore not surprising that the ionization energy drops to the value of The monotonic increase in the ioniza- tion energy resumes, with the experimental value of Here, as in helium, the first available state that an electron can be excited into has a higher n value, and thus it takes quite a lot of energy to perturb the atom.
Neon, like helium, is an inert gas. In neon, as in helium, the first available state into which an electron can be ex- cited has a higher n-value, so that it takes quite a lot of energy to perturb the atom.
Neon shares with helium the property of being an inert gas. These elements are chemi- cally very much like the series: lithium,. It might appear a little strange that the period ends with argon, since the 3d shell, accommodating ten elements, remains to be filled. The chemical properties of elements at the beginning and end of this period are similar to those of elements at the beginning and end of other periods.
Thus potassium, with the single 4s electron, is an alkali metal, like sodium with its single 3s electron out- side a closed shell. Bromine, with the configuration 4s 2 3d 10 4p 5, has a single hole in a p-shell and thus is chemically like chlorine and fluorine. The series of elements in which the 3d states are being filled all have rather similar chemical properties. The reason for this again has to do with the details of the self-consistent potential.
It turns out that the radii of these orbits2 are somewhat smaller than those of the 4s electrons, so that when the 4s 2 shell is filled, these electrons tend to shield the 3d electrons, no matter how many there are, from outside influences.
The same effect occurs when the 4f shell is being filled, just after the 6s shell has been filled. The elements here are called the rare earths. The knowledge of S, L, and J for the ground states is important, because selection rules allow us to deter- mine these quantities for the excited states of atoms.
What determines the ground-state quantum numbers is an interplay of spin-orbit coupling and the exchange effect discussed in connection with helium in Chapter This means that it is a fairly good approximation to view L and S as separately good quantum numbers: We add up all the spins to form an S and all the orbital angular mo- menta of the electrons to form an L, and these are then coupled to obtain a total J. The former case is described as Russell-Saunders coupling, the lat- ter as j-j coupling.
For Russell-Saunders coupling, F. Hund summarized the results of 2 It is understood that this is just a way of talking about the peaking tendencies of the charge distribution.
The Building-Up Principle W various calculations by a set of rules that give the overall quantum numbers of the low- est states. The rules are: 1. The state with largest S lies lowest. For a given value of S, the state with maximum L lies lowest.
In applying these rules we must be careful not to violate the Pauli principle. The second rule emerges qualitatively from the fact that the higher the L-value, the more lobes the wave function has, as shown in Fig. This allows the electrons to stay away from one another, and reduce the effect of the Coulomb repulsion. The third rule follows from the form of the spin-orbit coupling.
Once we get to the point of having a shell that is more than half filled, it is clearer to look at the atom as consisting of a filled shell with a number of holes, as we discussed in our description of oxygen. Thus the multiplet is inverted and it is the largest value of J that gives the lowest lying state.
Let us illustrate the application of the Hund rules to some atoms, and the need to keep track of the Pauli principle. We shall consider the quantum numbers of carbon 2p 2, oxygen 2p 4, and manganese 3d 5. The electrons are placed, as far as is possible, on different shelves, to minimize the repulsion. The largest possible value of Lz gives the L value, which is 1. There are five electrons, and thus each of the spaces is filled by one.
Limitations of space prevent us from a more detailed discussion of the periodic table. A few additional comments are, however, in order. If new, superheavy metastable nuclei are ever discovered, there will presumably exist corresponding atoms, and it is expected that their structure will conform to the prediction of the building-up approach outlined in this supplement.
Consequently, the wave functions of the outermost electrons do not extend much further than that of the electron in the hydrogen atom. Atoms are more or less the same size! The spectroscopy of atoms, once we get beyond hydrogen and helium, is very complicated. Consider, as a relatively simple ex- ample, the first few states of carbon, which are formed from different configura- tions of the two electrons that lie outside the closed shell in the 2p 2 orbitals.
As already pointed out, the possible states are 1S0, 3P2,1,0, and 1D2. The 3P0 state lies lowest, but the other states are still there. The first excited states may be de- scribed by the orbitals 2p 3s. Even with the restrictions provided by the selection rules, there are numerous transitions. Needless to say, the order- ing of these levels represents a delicate balance between various competing ef- fects, and the prediction of the more complex spectra is very difficult.
That task is not really of interest to us, since the main point that we want to make is that quantum mechanics provides a qualitative, and quantitative, detailed explanation of the chemical properties of atoms and of their spectra, without assuming an in- teraction other than the electromagnetic interaction between charged particles.
We shall have occasion to return to the topic of spectra. Supplement C A Brief Discussion of Molecules The purpose of this supplement is to outline the basic approach to the study of simple molecules. We discuss the H2 molecule in some detail, so as to provide an understanding of terms like molecular orbitals and valence bonds. Quantum chemistry has become a field well served by massive computers. Our discussion does not really provide an entry into this field.
It is extremely simple-minded, and its only justification is that it provides an insight into the basic mechanisms that lead to molecular binding. Anything more de- pends on an understanding of electron—electron correlations, and these are way beyond the scope of this book. It is based on the fact that the nuclei are much more massive than the elec- trons, and that therefore a good first approximation treats the nuclei as frozen, with their location determined by the electronic distribution.
We discuss the H2 molecule as a proto- type of other simple diatomic molecules. The H2 Molecule The H2 molecule is a more complicated system, because there are two electrons present, and the exclusion principle therefore plays a role.
The nuclei protons here will be labeled A and B, and the two electrons 1 and 2, re- spectively Fig. We will again compute an upper bound to E RAB by constructing the expectation value of H with a trial wave function.
The electron spin state is a singlet, since the spatial part of the wave function is taken to be symmetric. In this trial wave function, each electron is associated with both protons; that is, the trial wave function is said to be a product of molecular orbitals. The descrip- tion in terms of molecular orbitals is sometimes called the MO method.
The trial wave function also has some undesirable features for large RAB. W Supplement C A Brief Discussion of Molecules The Valence Bond Method The last difficulty can be avoided with the use of the valence bond also called Heitler- London method, in which linear combinations of atomic orbitals are used. We could, in principle, add a triplet term to our variational trial wave function. However, a triplet wave function must be spatially antisymmetric and yields low proba- bility for the electrons being located in the region between the protons.
Although it is not immediately obvious that the attraction is still largest in this configuration when there are two electrons that repel each other in the system, it is in fact so.
There should be no question about the quantitative successes of quantum mechanics in molecular physics. More sophisticated trial wave functions have to be used; for example, a term trial wave function yields complete agreement with observations for the H2 molecule, but it does not, as the MO and VB functions do, give us something of a qualitative feeling of what goes on between the atoms.
In what follows, we will explore the relevance of these approaches to a qualitative understanding of some aspects of chemistry. If this can be large, there will be binding. The two electrons can only overlap significantly, however, if their spins are antiparallel; this is a consequence of the exclusion principle.
The region of over- lap is between the two nuclei, and there the attraction to the nuclei generally overcomes the electrostatic repulsion between the electrons. In the MO picture, too, it is an overlap term—the result in —that is crucial to bonding, and again, bonding occurs because the electron charge distribution is large be- tween the nuclei.
Thus, although here the orbitals belong to the whole molecule rather than to individual atoms, the physical reason for bonding is the same.
We note that in general there may be several bound states of the nuclei, correspond- ing to different electronic configurations. We are not going to pursue this, except to point out the important fact that the E R is different for each electronic state.
The Importance of Unpaired Valence Electrons An important simplification in the study of electronic charge distributions in molecules occurs because we really do not need to take all electrons into account. In the construction of orbitals, be it valence or molecular, only the outermost electrons, not in closed shells— that is, the so-called valence electrons—have a chance to contribute to the bonding.
The inner electrons, being closer to the nucleus, are less affected by the presence of another atom in the vicinity. To see why this is so, consider what happens when an atom with a single va- lence electron is brought near an atom with two paired electrons.
There are two cases to be considered Fig. This is the case for the rare earths. This reduces the overlap, and it turns out that the exchange integral gives a repul- sive contribution to the energy. The original atomic state will frequently no longer be a possible one, and one of the electrons will have to be promoted into another atomic orbital. Sometimes this may cost very little energy, but usually this is not the case, and again bonding is not achieved.
Chemical activity depends on the presence of unpaired outer electrons. An ex- ample of this is the nonexistence of the H-He molecule. In He we have two elec- trons in the 1s state; promotion of one of them into a 2s state costs a lot of energy.
It is for this reason that the atoms for which the outer shells are closed are inert. Not all unpaired electrons are of equal significance. As noted before, the unpaired d- and f-electrons in the transition elements tend to be close to the nucleus, and hence inactive.
Enter the email address you signed up with and we'll email you a reset link. Need an account? Click here to sign up. Download Free PDF. A short summary of this paper. From this we get Here hc 6. The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the energy of the recoiling proton be E. The photon energy is hc 6. With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon.
So that the angle is 10o. This makes sense if one looks at a picture of the potential in the limit of large k. The presence of V0 is there to provide something with the dimensions of an energy. The integral can be looked up. These are very large numbers. For the Done in text. Consider the Schrodinger equation with V x complex. This does not tell us much about ImV x except that it cannot be positive everywhere.
If it has a fixed sign, it must be negative. The problem just involves simple arithmetic. For odd powers the integral vanishes. This is, in fact, the smallest value possible for the product of the 2 dispersions. The justification of this is that the wave functions are expected to go to zero at infinity faster than any power of x , and this is also true of the momentum space wave functions, in their dependence on p.
The linear operators are a , b , f 2. The longest wavelength corresponds to the lowest frequency. Thus the energy eigenvalues are given by En above with n even. There is a typo in the statement of the problem. The sum should be restricted to odd integers. We work this out by making use of an identity. This however is equivalent to the statement that the sum of the probabilities of finding any energy eigenvalue adds up to 1. Hence P2 will be zero. To confirm this, we need to show that the distribution is strongly peaked for large n.
Now the eigenfunctions for a box symmetric about the x axis have a definite parity. Since the integral is over an interval symmetric under this exchange, it is zero. They imply that the matrix Str is unitary, and therefore the matrix S is unitary. We have solve the problem of finding R and T for this potential well in the text. This confirms that the S matrix here is unitary.
A sketch of the potential shows that y2 is very large. The other turning point occurs for y not particularly large, so that we can neglect the middle term under the square root, and the value of y1 is 1.
We estimate the contribution from that point on by neglecting the —1 term in the integrand. Thus the estimated time is longer const. This is to be multiplied by the time of traversal inside the box. The important factor is f-2l. It tells us that the lifetime is proportional to kR0 -2l so that it grows as a power of l for small k. Equivalently we can say that the probability of decay falls as kR0 2l. The argument fails because the electron is not localized inside the potential.
Even if f E varies only slightly over the energy range that overlaps small positive E, we can determine the binding energy in terms of the reflection coefficient. The relevant figure is Fig. What does this tell us about the first bound state? All we know is that y is a solution of Eq. All we can say is that the binding energy f the even state will be larger than that of the odd one. These are just the single well conditions. This can be worked out in more detail, but this becomes an exercise in Taylor expansions with no new physical insights.
Substitution into the above equation yields, on the l. We have shown that X is hermitian. There is actually a neater way to do this. However, H is hermitian, so that the eigenvalues are real.
If H is not hermitian, then all four eigenvalues are acceptable. An examination of the equation involving v2 leads to an identical equation, and we associate the — sign with the b2 eigenvalue. We calculate as above, but we can equally well use Eq.
We now take Eq. We then use Eq. We use the fact that Eq. This follows immediately from problems 5 and 6. The quadratic terms change the values of the eigenvalue integer by 2, so that they do not appear in the desired expressions.
This is known as the Poisson distribution. In this calculation it is only the commutator [p 0 , x 0 ] that plays a role. We follow the procedure outlined in the hint. We can now integrate w. We can use the procedure of problem 17, but a simpler way is to take the hermitian conjugate of the result. Thus the two results agree. Consider the H given. We again expect a cyclical pattern. However, we may well 2 2 2 2 have chosen the n direction as our selected z direction, and the eigenvalues for this are again 2,1,0,-1, By the same argument we can immediately state that the eigenvalues are 7m i.
The problem breaks up into three separate, here identical systems. The ground state energy correspons to the n values all zero. The procedure here is exactly the same. These are the only possibilities, so that we have three eigenvalues all equal to zero. Now the sum of the eigenvalues is the trqice of M which is N see problem Thus there is one eigenvalue N and N —1 eigenvalues 0. The set M1, M2 and M3 give us another representation of angular momentum matrices. Now let U be a unitary matrix that diagonalizes A.
The problem is not really solved, till we learn how to deal with the situation when the eigenvalues of A in problem 13 are not all different. The construction is quite simple. It is. We may use the material in Eq. If the matrix M is to be hermitian, we must require that A and all the components of B be real. What may be relevant for a potential energy is an average, assuming that the two particles have equal probability of being in any one of the three Sz states.
We need to calculate 1 1 2 2 the scalar product of this with the three triplet wave functions of the two-electgron system. It is easier to calculate the probability that the state is found in a singlet state, and then subtract that from unity. The eigenfunction of the rotator are the spherical harmonics. This result should have been anticipated. The eigenstates of L2 are also eigenstates of parity. Since the perturbation represents the interaction with an electric field, our result states that a symmetric rotator does not have a permanent electric dipole moment.
The second order shift is more complicated. This can easily be seen from the table of spherical harmonics. The orthogonality of the spherical harmonics for different values of L takes care of the matter. The problem therefore separates into three different matrices. The kinetic energy does not change since p2 is unchanged under rotations.
The energy is the sum of the two energies. The final state is 5-fold degenerate, and the same splitting occurs, with the same intervals. What will be the effect of a constant electric field parallel to B?
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