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### Fund Quantum Mechanics Lect & HW Solutions 33

Course: PHY 3604, Fall 2011
School: UNF
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Word Count: 295

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HERMITIAN 2.6. OPERATORS 2.6.5 15 Solution herm-e Question: Show that an operator such as x2 , corresponding to multiplying by a real function, is an Hermitian operator. Answer: If the operator corresponds to multiplying by a real function of x, call it r(x), then f |rg = all i f rg dx = all i (rf ) g dx = rf |g since the complex conjugate does not aect a real function. 2.6.6 Solution herm-f Question:...

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HERMITIAN 2.6. OPERATORS 2.6.5 15 Solution herm-e Question: Show that an operator such as x2 , corresponding to multiplying by a real function, is an Hermitian operator. Answer: If the operator corresponds to multiplying by a real function of x, call it r(x), then f |rg = all i f rg dx = all i (rf ) g dx = rf |g since the complex conjugate does not aect a real function. 2.6.6 Solution herm-f Question: Show that the operator d/dx is not a Hermitian operator, but id/dx is, assuming that the functions on which they act vanish at the ends of the interval a x b on which they are dened. (Less restrictively, it is only required that the functions are periodic; they must return to the same value at x = b that had they at x = a.) Answer: You rst need to show that f d g dx is not the same as d fg dx in order for d/dx not to be a Hermitian operator. By denition, b dg d g= dx. f dx dx a You can use integration by parts, [1, p. 64], to move the derivative from g to f : f b d f g = f g dx a b a df g dx = f (b)g (b) f (a)g (a) dx b a df dx g dx (the dierentiation can be moved inside the complex conjugate since it is a real operation.) Since the functions f and g are the same at the end points a and b you have d g = f dx b a df dx g dx = d fg dx This makes d/dx a skew-Hermitian operator, rather than a Hermitian one: ipping over the operator to the other side changes the sign of the inner product.
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UNF - PHY - 3604
16CHAPTER 2. MATHEMATICAL PREREQUISITESTo get rid of the change of sign, you can add a factor i to the operator, since the i adds acompensating minus sign when you bring it inside the complex conjugate:df i g =dxbadfidxg dx = idfgdxThis ma
UNF - PHY - 3604
2.6. HERMITIAN OPERATORS17First check the end points. The graph of the sine function, [1, item 12.22], shows that a sineis zero whenever its argument is a whole multiple of . That makes both sin(k 0) = sin(0) andsin(k ) zero. So sin(kx)/ /2 must be ze
UNF - PHY - 3604
18CHAPTER 2. MATHEMATICAL PREREQUISITESCheck that these functions are indeed periodic, orthonormal, and that they are eigenfunctionsof id/dx with the real eigenvalues. . . , 3, 2, 1, 0, 1, 2, 3, . . .Completeness is a much more dicult thing to prove,
UNF - PHY - 3604
UNF - PHY - 3604
Chapter 3Basic Ideas of Quantum Mechanics3.1The Revised Picture of Nature3.2The Heisenberg Uncertainty Principle3.3The Operators of Quantum Mechanics3.4The Orthodox Statistical Interpretation21
UNF - PHY - 3604
22CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS3.4.1Only eigenvalues3.4.2Statistical selection3.5A Particle Conned Inside a Pipe3.5.1The physical system3.5.2Mathematical notations3.5.3The Hamiltonian3.5.4The Hamiltonian eigenvalue problem
UNF - PHY - 3604
3.5. A PARTICLE CONFINED INSIDE A PIPE3.5.5All solutions of the eigenvalue problem3.5.5.123Solution piped-aQuestion: Write down eigenfunction number 6.Answer: Substituting n = 6 in the generic expression for the eigenfunctions,n =n2sinxxx6
UNF - PHY - 3604
24CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSand see how big it is. Note that h = 1.054,57 1034 J s. Express in units of eV, where one eVequals 1.602,18 1019 J.Answer:E1 =(1.054,57 1034 J s)2 2h2 2== 1.506 1018 J2m22 9.109,38 1031 kg (2 1010 m)
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3.5. A PARTICLE CONFINED INSIDE A PIPE3.5.7Discussion of the eigenfunctions3.5.7.125Solution pipef-aQuestion: So how does, say, the one-dimensional eigenstate 6 look?Answer: As the graph below shows, it has six blobs where the particle is likely to
UNF - PHY - 3604
26CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSAnswer: Almost every word in the above story is a gross misstatement of what nature reallyis like when examined on quantum scales. A particle does not have a position, so phrases likehits an end, reect back
UNF - PHY - 3604
3.5. A PARTICLE CONFINED INSIDE A PIPE27Canceling the terms that both sides have in common:3 2 2h8 2 2h=22my2m2xand canceling the common factors and rearranging:23y=.2x8So when y /x = 3/8 = 0.61 or more, the third lowest energy state i
UNF - PHY - 3604
283.5.93.6CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSQuantum connementThe Harmonic Oscillator3.6.1The Hamiltonian3.6.2Solution using separation of variablesh0 (x) =h1 (x) =h2 (x) =h3 (x) =h4 (x) =1(2 )1/4e22 /2e2 2 1ee2 /2(42 )1/4
UNF - PHY - 3604
3.6. THE HARMONIC OSCILLATOR3.6.2.129Solution harmb-aQuestion: Write out the ground state energy.Answer: Taking the generic expression2nx + 2ny + 2nz + 3h2and substituting the lowest possible value, 0, for each of nx , ny , and nz , you get the g
UNF - PHY - 3604
303.6.2.4CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSSolution harmb-dQuestion: Write out the eigenstate 100 fully.Answer: Taking the generic expressionnx ny nz = hnx (x)hny (y )hnz (z )and substituting nx = 1, ny = nz = 0, you get100 = h1 (x)h0 (y
UNF - PHY - 3604
3.6. THE HARMONIC OSCILLATOR3.6.3.131Solution harmc-aQuestion: Verify that the sets of quantum numbers shown in the spectrum gure 3.3 doindeed produce the indicated energy levels.Answer: The generic expression for the energy isEnx ny nz =2nx + 2ny
UNF - PHY - 3604
32CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSnx , ny , and nz is 1, then the other two must be 0 or their sum N would still be greater than1. That means that precisely one of nx , ny , and nz must be 1 and the other two 0. There arethree possibilities
UNF - PHY - 3604
3.6. THE HARMONIC OSCILLATOR3.6.4.133Solution harmd-aQuestion: Write out the ground state wave function and show that it is indeed sphericallysymmetric.Answer: Repeating an earlier exercise, taking the generic expressionnx ny nz = hnx (x)hny (y )hn
UNF - PHY - 3604
34CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSThe other eigenfunctions do not necessarily have their maximum magnitude at the origin: forexample, the shown states 100 and 010 are zero at the origin.For large negative values of its argument, an exponent
UNF - PHY - 3604
3.6. THE HARMONIC OSCILLATOR35The rst polynomial within square brackets in the expression above is zero at x = / 2and x = / 2, producing the two vertical white lines along which there is zero probabilityof nding the particle. Similarly, the second pol
UNF - PHY - 3604
36CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICSsystem (, y , z ) that is rotated 45 degrees counter-clockwise around the z -axis compared toxthe original (x, y, z ) coordinate system.Answer: Take the rotated coordinates to be x and y as shown:yTsdd
UNF - PHY - 3604
3.6. THE HARMONIC OSCILLATORThe same way, you get010 =372y/(2 )3/4e(x2 +y 2 +z 2 )/22.So, the combination (100 + 010 ) / 2 is(x + y )/ (x2 +y2 +z2 )/22100 + 010=e.2(2 )3/4Now x2 + y 2 + z 2 is according to the Pythagorean theorem the squa
UNF - PHY - 3604
Chapter 4Single-Particle Systems4.1Angular Momentum4.1.1Denition of angular momentum4.1.2Angular momentum in an arbitrary direction4.1.2.1Solution angub-aQuestion: If the angular momentum in a given direction is a multiple of h = 1.054,57 1034J
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404.1.2.2CHAPTER 4. SINGLE-PARTICLE SYSTEMSSolution angub-bQuestion: What is the magnetic quantum number of a macroscopic, 1 kg, particle that isencircling the z -axis at a distance of 1 m at a speed of 1 m/s? Write out as an integer, andshow digits
UNF - PHY - 3604
4.1. ANGULAR MOMENTUM4.1.3.141Solution anguc-aQuestion: The general wave function of a state with azimuthal quantum number l andmagnetic quantum number m is = R(r)Ylm (, ), where R(r) is some further arbitraryfunction of r. Show that the condition f
UNF - PHY - 3604
42CHAPTER 4. SINGLE-PARTICLE SYSTEMSis at leastl(l + 1) 2 l2 h2 = lh2 .h4.1.44.2Angular momentum uncertaintyThe Hydrogen Atom4.2.1The Hamiltonian4.2.2Solution using separation of variables4.2.2.1Solution hydb-aQuestion: Use the tables for t
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4.2. THE HYDROGEN ATOM43The total probability of nding the particle integrated over all possible positions is, using thetechniques of volume integration in spherical coordinates:2r =0|100 |2 d3 r ==0=0or rearranging12r/a0r/a0 =0egivingr2 r
UNF - PHY - 3604
44CHAPTER 4. SINGLE-PARTICLE SYSTEMSwhere a0 = 40 h2 /me e2 , to nd the ground state energy. Express in eV, where 1 eV equals1.602,2 1019 J. Values for the physical constants can be found at the start of this section andin the notations section.Answe
UNF - PHY - 3604
4.2. THE HYDROGEN ATOM4.2.3.345Solution hydc-cQuestion: Based on the results of the previous question, what is the color of the lightemitted in a Balmer transition from energy E3 to E2 ? The Planck-Einstein relation says thatthe angular frequency of
UNF - PHY - 3604
46CHAPTER 4. SINGLE-PARTICLE SYSTEMSAnswer: The square wave function is|100 (r)|2 =1 2r/a0ea30and the value at the nucleus r = 0 is then|100 (0)|2 =1a30For the value at r above to be one percent of this, you must havee2r/a0 = 0.01or taking
UNF - PHY - 3604
4.3. EXPECTATION VALUE AND STANDARD DEVIATION47or multiplying out2px |2px =1( 211 |211 + 211 |211 + 211 | 211 + 211 |211 )2or using the orthonormality of 211 and 211 .2px |2px =1(1 + 0 + 0 + 1) = 1.2For the state 2py , remember that i comes ou
UNF - PHY - 3604
484.3.1.3CHAPTER 4. SINGLE-PARTICLE SYSTEMSSolution esda-cQuestion: Continuing this example, what will be the standard deviation?Answer: The average square deviation from 1.5 is:111(1 1.5)2 + (2 1.5)2 =224Taking a square root, the standard de
UNF - PHY - 3604
4.3. EXPECTATION VALUE AND STANDARD DEVIATION49What are the expectation values of energy, square angular momentum, and z -angular momentum for this state?Answer: Note that the square coecients of the eigenfunctions 211 and 211 are each 1 ,21so each
UNF - PHY - 3604
50CHAPTER 4. SINGLE-PARTICLE SYSTEMSFor the z -angular momentum, the expectation value is zero but the two states have eigenvaluesh and h, so11( 0)2 + (h 0)2 = h.hLz =22Whether h or h is measured, the deviation from zero has magnitude h.4.3.3
UNF - PHY - 3604
4.3. EXPECTATION VALUE AND STANDARD DEVIATION51or multiplying outL2 =1211 + 211 | 2 2 211 + 2 2 211 .hh2multiplying out further to L2 = 2 2 .hFor the z -angular momentum,Lz =or multiplying outLz =multiplying out further to Lz4.3.3.21211
UNF - PHY - 3604
524.4CHAPTER 4. SINGLE-PARTICLE SYSTEMSThe Commutator4.4.1Commuting operators4.4.1.1Solution commutea-aQuestion: The pointer state12px = (211 + 211 ) .2is one of the eigenstates that H , L2 , and Lx have in common. Check that it is not an eige
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4.5. THE HYDROGEN MOLECULAR ION53Answer: On second thought, maybe I can relax.According to the uncertainty relationship, the uncertainties could be as small as, for example,0.5 1010 m in position and 1024 kg m/s for linear momentum. I am not going to
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54CHAPTER 4. SINGLE-PARTICLE SYSTEMS4.5.4States that share the electron4.5.5Comparative energies of the states4.5.6Variational approximation of the ground state4.5.6.1Solution hione-aQuestion: The solution for the hydrogen molecular ion requires
UNF - PHY - 3604
4.5. THE HYDROGEN MOLECULAR ION55Now evaluate the expectation energy:E = ax( x)|H |ax( x) = |a|2 x( x) h2 2x( x)2m x2You can substitute in the value of |a|2 from the normalization requirement above and applythe Hamiltonian on the function to its r
UNF - PHY - 3604
Chapter 5Multiple-Particle Systems5.15.1.1Wave Function for Multiple ParticlesSolution complex-aQuestion: A simple form that a six-dimensional wave function can take is a product of twothree-dimensional ones, as in (r1 , r2 ) = 1 (r1 )2 (r2 ). Show
UNF - PHY - 3604
58CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSapproximation. For example, two electrons repel each other. All else being the same, theelectrons would rather be at positions where the other electron is nowhere close. As a result,it really makes a dierence for
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5.2. THE HYDROGEN MOLECULE59general, it is just not worth the trouble for the electrons to stay away from the same position:that would reduce their uncertainty in position, increasing their uncertainty-demanded kineticenergy.Answer: The repulsive pot
UNF - PHY - 3604
605.2.2.2CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSSolution hmolb-aQuestion: When the protons are close to each other, the electrons do aect each other, andthe wave function above is no longer valid. But suppose you were given the true wave function,and y
UNF - PHY - 3604
5.2. THE HYDROGEN MOLECULE5.2.4.161Solution hmold-aQuestion: Obviously, the visual dierence between the various states is minor. It may evenseem counter-intuitive that there is any dierence at all: the states l r and r l are exactlythe same physical
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62CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSvice-versa. But there is still almost no probability of nding both protons in the rst quadrant,both near the right proton. Nor are you likely to nd them in the third quadrant, both nearthe left proton.If you aver
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5.2. THE HYDROGEN MOLECULEz263z2z1n1z2z1n1z2z1n1z1n1n2zn2zn2zn2znznznznznznznznzzzzzFigure 5.2: Probability density functions on the z -axis through the nuclei. From left toright: l r , r l , the symmetric combin
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64CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSof the symmetric and antisymmetric states are quite dierent, though they look qualitativelythe same.5.2.5Variational approximation of the ground state5.2.6Comparison with the exact ground state5.35.3.1Two-St
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5.3. TWO-STATE SYSTEMS65and this can be multiplied out, dropping the common factor 2 and noting that 1 |1 and2 |2 are one, as1 |2 2 2 + 1 |2 = 0for which the smallest root can be written as=1+1 |21 1 |22.That is less than 1 |2 , hence is small
UNF - PHY - 3604
66CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSIf this is plugged into the expression for the solution, c1 1 + c2 2 , it takes the formc1 1 + c2 2wherec1 =c1 + c2(1 2 )c2 =c2 + c1.(1 2 )So, while the constants c1 and c2 are dierent from c1 and c2 , the
UNF - PHY - 3604
5.5. MULTIPLE-PARTICLE SYSTEMS INCLUDING SPIN67Answer:s(s + 1) 2 =h5.515 2h.4Multiple-Particle Systems Including Spin5.5.1Wave function for a single particle with spin5.5.1.1Solution complexsa-aQuestion: What is the normalization requirement
UNF - PHY - 3604
68CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSof nding it somewhere with spin down is | |2 d3 r. The sum of the two integrals must beone to express the fact that the probability of nding the particle somewhere, either with spinup or spin down, must be one, ce
UNF - PHY - 3604
5.5. MULTIPLE-PARTICLE SYSTEMS INCLUDING SPIN695.5.4Wave function for multiple particles with spin5.5.4.1Solution complexsb-aQuestion: As an example of the orthonormality of the two-particle spin states, verify that| is zero, so that and are indeed
UNF - PHY - 3604
70CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSThis is similar to the observation in calculus that integrals of products can be factored intoseparate integrals:all r 1all r 2all r 1f (r1 )g (r2 ) d3 r1 d3 r2 =f (r1 ) d3 r1all r 2g ( r 2 ) d3 r 2Answer:
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5.6. IDENTICAL PARTICLES71For gs to be normalized, its square norm must be one:gs |gs = 1.According to the previous subsection, this inner product evaluates as the sum of the innerproducts of the matching spin components:a+ gs,0 |a+ gs,0 + a+ gs,0 |
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725.6.1CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSSolution ident-aQuestion: Check that indeed any linear combination of the triplet states is unchanged underparticle exchange.Answer: In the notations of the previous section, the most general linear combina
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5.7. WAYS TO SYMMETRIZE THE WAVE FUNCTION5.75.7.173Ways to Symmetrize the Wave FunctionSolution symways-aQuestion: How many single-particle states would a basic Hartree-Fock approximation useto compute the electron structure of an arsenic atom? How
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74CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSall r1all r 2l (r1 )r (r2 ) r (r1 )l (r2 ) d3 r1 d3 r2can according to the rules of calculus be factored into three-dimensional integrals asSS1 |2=all r 1l (r1 ) r (r1 ) d3 r1all r 2r (r2 ) l (r2 ) d3 r2=
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5.8. MATRIX FORMULATION75So there are two energy eigenvalues:E1 = J LandE2 = J + L.In the rst case, since according to the equations above(J E1 ) a1 La2 = 0=La1 La2 = 0it follows that a1 and a2 must be equal, producing the eigenfunctionSSa1 1
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76CHAPTER 5. MULTIPLE-PARTICLE SYSTEMSelectrons around the same nucleus, the remaining C = 4 Slater determinants can be writtenout explicitly to give the two-particle statesS1 =l r r l 2S2 =l r r l 2S3 =l r r l 2S4 =l r r l 2Note that
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5.9. HEAVIER ATOMS77Seigenfunction 1 = 1 has, according to its denitionabove, both electrons spin-up and inthe excited antisymmetric spatial state (l r r l ) / 2.A similar guess that works is(a1 , a2 , a3 , a4 ) = (0, 0, 0, 1).SThis corresponds t
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78CHAPTER 5. MULTIPLE-PARTICLE SYSTEMS5.9.4Lithium to neon5.9.5Sodium to argon5.9.6Potassium to krypton5.9.7Full periodic table5.10Pauli Repulsion5.11Chemical Bonds5.11.1Covalent sigma bonds
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5.11. CHEMICAL BONDS5.11.2Covalent pi bonds5.11.3Polar covalent bonds and hydrogen bonds5.11.4Promotion and hybridization5.11.5Ionic bonds5.11.6Limitations of valence bond theory79