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Stephen Gasiorowicz. Universjgl qf Minnesota preface. This book is intended to serve as an introduction to quantum physics. In Writing it, I have kept several. By Stephen Gasiorowicz. The emergence of quantum mechanics ; wave particle duality, probability and Schrödinger equation ; the expansion. SOLUTIONS MANUAL to Quantum Physics [Stephen Gasiorowicz]. 哲延 朱. SOLUTIONS MANUAL CHAPTER 1 1. The energy contained in a volume dV is U (ν,T).

Solution manual to Quantum Physics 3rd edition by: Stephen Gasiorowicz. Here 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. We use this to rewrite the energy conservation equation as follows:

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. A short list follows: Let us first write the solution of 7B as Pml z. Consider eigenvalues and eigenfunctions of a Hamiltonian that depends on some pa- rameter—for example, the mass of the electron, or the charge of the electron, or any other parameter that may appear in more complicated problems. 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.

General Rules for Addition of Angular Momenta, and Implications for Identical Particles These two examples illustrate the general features that are involved in the addition of an- gular momenta: We shall use the simple abbreviation m1, m2 for Y 1 2 l1m1Y l2m2. The multiplicities add up to We merely state the results. The symbol eijk is defined by the following properties: 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 Hartree Approximation W This equation has a straightforward interpretation: 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. This deserves comment: 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: 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: 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 first of these rules is easy to understand: 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. 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. We need to evaluate this for the three cases: In the 5 x 5 format, there is only one entry in the bottom left- most corner, and it is 4. 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? The perturbation acts as in the Stark effect. The effect of H1 is to mix up levels that are degenerate, corresponding to a given ml value with different values of l.

There will be a further breakdown of degeneracy. The second order shift for the upper state involves summing over intermediate states that differ from the initial state. A picture of the levels and their spin-orbit splitting is given below. The latter can be rearranged into states characterized by J2, L2 and Jz.

These energies are split by relativistic effects and spin-orbit coupling, as given in Eq. We ignore reduced mass effects other than in the original Coulomb energies. Under these circumstances one could neglect these and use Eq.

The unperturbed Hamiltonian is given by Eq. These will be exactly like the spin triplet and spin singlet eigenstates. According to Eq. This is not surprising. The ionization potential for sodium is 5. Thus the possible states are S1 , P1, D1. Thus the S and D states have positive parity and the P state has opposite parity.

Given parity conservation, the only possible 3 admixture can be the D1 state. B term, if L is not zero. For the S1 stgate, the last term does not contribute. If the two electrons are in the same spin state, then the spatial wave function must be antisymmetric. The problem is one of two electrons interacting with each other. The form of the interaction is a square well potential. The reduction of the two-body problem to a one-particle system is straightforward. Let us denote the lowest bound state wave function by u0 x and the next lowest one by u1 x.

We know that the lowest state has even parity, that means, it is even under the above interchange, while the next lowest state is odd under the interchange.

Hence, for the two electrons in a spin singlet state, the spatial symmetry must be even, and therefor the state is u0 x , while for the spin triplet states, the spatial wave function is odd, that is, u1 x. The energy spectrum is the sum of the energies of the oscillator describing the motion of the center of mass, and that describing the relative motion.

When the electrons are in the same spin state, the spatial two-electron wave function must be antisymmetric under the interchange of the electrons. The first excited state would normally be the 2,2 state, but this is not antisymmetric, so that we must choose 1,3 for the quantum numbers. Suppose that the particles are bosons. Spin is irrelevant, and the wave function for the two particles is symmetric. The changes are minimal. The calculation is almost unchanged. Since the photons are massless, and there are two photon states per energy state, this problem is identical to problem There is no preferred direction in the problem, so that there cannot be any dependence on the eigenvalue of Lz.

Thus all three m values have the same energy. The simplification of the energy shift integrals reduces to the simplification of the integrals in the second part of Eq. We use Eqs. Because of this, the wave function is pushed out somewhat. There is nevertheless some probability that the electron can get close to the nucleus. To describe the quantum properties of any physical system, a new mathematical language is re-. Zoltan Batiz, Bhag C.

The concept of holography has lured philosophers of science for decades Atomic Physics and Quantum Optics Sep 4, Peskin and R. Ryder, Quantum Field Theory.

Weinberg, Quantum Theory of Fields, Vol. Quantum Anomaly in Molecular Physics and H. Adler and W. Abdoul-Carime, N. Khelifa, and J. Cambridge Phil. Quantum Physics Experimentlly Motivated Development May 3, - that the algebra can be used to describe rigid and non-rigid rotating molecules.

Stephen Gasiorowicz. Universjgl qf Minnesota This book is intended to serve as an introduction to quantum physics. In Writing it, I have kept several Quantum Physics. Read more. Quantum Algebras and Quantum Physics.