Ψ(x) = c 0 φ 0 (x) + c 1 φ 1 (x) (1) Ĥφ n (x) = E n φ n (x) (2) Ψ = c 0 φ 0 + c 1 φ 1 (7) Ĥ φ n = E n φ n (4)

Transcription

1 1 Problem 1 Ψx = c 0 φ 0 x + c 1 φ 1 x 1 Ĥφ n x = E n φ n x E Ψ is the expectation value of energy of the state 1 taken with respect to the hamiltonian of the system. Thinking in Dirac notation 1 an become Ψ = c 0 φ 0 + c 1 φ 1 3 Ĥ φ n = E n φ n 4 E Ψ = E = Ψ Ĥ Ψ = c 0φ 0 + c 1φ 1 Ĥ c 0 φ 0 + c 1 φ 1 = c 0φ 0 + c 1φ 1 c 0 Ĥ φ 0 + c 1 Ĥ φ 1 = c 0φ 0 + c 1φ 1 c 0 E 0 φ 0 + c 1 E 1 φ 1 = c 0c 0 E 0 φ 0 φ 0 + c 0c 1 E 1 φ 0 φ 1 + c 1c 0 E 0 φ 1 φ 0 + c 1c 1 E 1 φ 1 φ 1 = c 0 E 0 + c 1 E 1 Where in the last line the cross terms like φ 0 φ 1 were set to zero because the eigenkets of Ĥ must be orthogonal, an for simplicity they were assume to be normalize φ 1 φ 1 = 1. Now we know that E 0 < E 1, so we can write: c 0 E 0 + c 1 E 1 c 0 E 0 + c 1 E 0 5 where the equal sign woul be for c 1 = 0. Furthermore we know that Ψ is normalize, so c 0 + c 1 which means c 0 E 0 + c 1 E 0 = E 0 so we can write: an finally Now, a few notes on common mistakes. c 0 E 0 + c 1 E 1 E 0 6 E Ψ E 0 There was extensive improper use of Dirac notation on this problem. A ket in Dirac notation is intene to represent one state. Linear combinations of states can be written as sums of kets, but we shouln t have linear combinations appearing insie the ket. So goo notation woul be: Ψ = c 0 φ 0 + c 1 φ 1 7 1

2 Ba notation might look like: Ψ = c 0 φ 0 + c 1 φ 1 8 Why is this an important istinction? Well, for one we want to be consistent with our use of notation. Dirac notation is useful the way it has been establishe because the kets are themselves eigenstates of operators. We know that the little kets in 7 are eigenstates of the Hamiltonian, so we can operate on them an pull out the eigenvalues. However the thing in 8 is not an eigenstate of the Hamiltonian so we on t know what the action of the Hamiltonian woul be on this messy thing. Also, because the kets in 7 are eigenstates of the Hamiltonian, they are orthogonal. This leas us to the convenient ientity that normalize eigenkets in Dirac notation obey the inner prouct relationship: This kin of relationship woul not follow from 8. φ i φ j = δ ij 9 The inner prouct relationship 9 comes from the fact that we make the ientification: φ i φ j = If φ i an φ j are orthogonal, this integral will be zero. However, φ i xφ j x 10 an φ i φ j φ i xφ j x φ i xφ j x δ ij 11 Realize that the left han sie of 11 is a function of x not a number. Think about this graphically if this isn t obvious. You are taking two functions of x an multiplying them together. You will have a new function of x. To get a number out of it you have to integrate over the x epenance. That s what s happening in 10. O.K. So you ve solve the Schröinger equation an foun the eigenkets an eigenvalues of Ĥ such that: Ĥ φ n = E n φ n 1 like state in the problem. This oes not mean that an arbitrary linear combination of the eigenkets will also satisfy the Schröinger equation. Let s see what happens if we let Ĥ operate on a L.C. of the φ n. Ψ = c 0 φ 0 + c 1 φ 1 13 Ĥ Ψ = Ĥ c 0 φ 0 + c 1 φ 1 14 = c 0 Ĥ φ 0 + c 1 Ĥ φ 1 15 = c 0 E 0 φ 0 + c 1 E 1 φ 1 16

3 Now, we can see that for Ψ to satisfy the Schröinger equation, we woul have to have the right han sie of 16 be equal to a constant times Ψ remember, the time inepenent S.E. is just a really cool eigenvalue equation. That is we woul have to have c 0 E 0 φ 0 + c 1 E 1 φ 1 = C c 0 φ 0 + c 1 φ 1 17 But this won t be the case unless E 0 = E 1 which is never true unless φ 0 an φ 1 are egenerate. What s the point of all this? It means that we can not take 4 an interpret it as To o so woul imply that 17 was true. Instea, we have to use Problem a Recall that ˆL z = i φ. Show that ˆL z Y m l Ĥ Ψ = E Ψ Ψ 18 E Ψ = Ψ Ĥ Ψ 19 θ, φ = my m θ, φ. l Solution The spherical harmonic Yl m θ, φ is separable into a prouct of a function of θ an a function of φ. Yl m θ, φ = ΘθΦφ 0 = Θ m l θ A m e imφ where Θ m l θ is the associate Legenre polynomial an A m = 1/ π. ˆL z Yl m θ, φ = ˆL z Θθ Am e imφ 1 = i Θθ Am e imφ φ = Θ m l θ A m i φ eimφ = Θ m l θ A m i ime imφ = Θ m l θ A m m e imφ = m Y m l θ, φ b The operators ˆL + = ˆL x + iˆl y an ˆL = ˆL x iˆl y are calle the raising an lowering operators, respectively. They are so name because they raise or lower the magnetic quantum number, m, by 1. Use the commutators [ˆLz, ˆL ] + = ˆL + an [ˆLz, ˆL ] = ˆL together with the result from part a to argue that this is the case. 3

4 Solution We re aske to prove that ˆL ± raises or lowers the magnetic quantum numbers in the spherical harmonics Yl m θ, φ, i.e. ˆL ± Y m l θ, φ = const Y m±1 l θ, φ which means that ˆL ± Yl m θ, φ must be a simultaneous eigenfunction of ˆL with eigenvalue ll + 1 an ˆL z with eigenvalue m ± 1. We wish to prove: ˆL ˆL± Yl m θ, φ = ll + 1 ˆL± Yl m θ, φ an ˆL z ˆL± Yl m θ, φ = m ± 1 ˆL± Yl m θ, φ 3 We shall use the commutators that we worke out in the homeworks thus Also we re given [ˆL, ˆL x ] = [ˆL, ˆL y ] = [ˆL, ˆL z ] = 0 [ˆL, ˆL ± ] = [ˆL, ˆL x ± iˆl y ] = [ˆL, ˆL x ] ± i[ˆl, ˆL y ] = 0 4 Equation 4 verifies equation because ˆL ˆL± Yl m θ, φ = ˆL ˆL ± Yl m θ, φ Similarly, equation 5 verifies equation 3 ˆL z ˆL± Yl m θ, φ = ˆLz ˆL± ˆL ± ˆLz Y m [ˆL z, ˆL ± ] = ± ˆL ± 5 = [ˆL z, ˆL ± ]Y m l θ, φ + = ll + 1 ˆL± Yl m θ, φ l θ, φ + ˆL± ˆLz Yl m θ, φ = ± ˆL ± Yl m θ, φ + ˆL ± m Yl m θ, φ = m ± 1 ˆL± Yl m θ, φ ˆL± ˆLz Y m l θ, φ 6 where in the first equality we have ae an subtracte the same term to construct an use the commutator. Thus we have foun that ˆL ± Yl m θ, φ is inee a simultaneous eigenfunction of ˆL with eigenvalue ll + 1 an ˆL z with eigenvalue m ± 1. We know Y m±1 l θ, φ is also such a simultaneous eigenfunction. Hence ˆL ± Yl m θ, φ must be proportional to Y m±1 θ, φ. l 4

5 General avice For this type of problems where you re aske to prove something, start by writing own what you know such as what the commutator means, an try to apply what is given such as [ˆL z, ˆL ± ] = ± ˆL ±. 3 Problem 3 a There are a couple of ways to o this problem. One shortcut that you coul have use, before going to the meat of the long way : the term symbols for a p 5 state are the same as those for a p 1 state, which you may fin easier to calculate, or which you might even just know. Since a p-orbital contributes up to one unit of orbital angular momentum, we know that L = 1, which gives us the P from our term symbol. Since we only have one electron, S + 1 = =, so we have P. Finally, we take J { L + S,..., L S } = {1 + 1,..., 1 1 } = { 3, 1 }. So our term symbols for p 1 are P 3/ an P 1/, an these will also be the term symbols for p 5 states. Now for the long way. The first thing to remember is that fille shells on t contribute to the term symbol. This is because M L = M S = 0 for a fille subshell there s a own spin for every up spin, so M S = 0, an since the values of m l are equally populate from l to l, M L = 0. This means that we only have to look at the unfille subshell, 3p 5. We re going to enumerate all the microstates, but before we o that, we ll see how many microstates we ll have to make sure we on t miss any. The number of microstates is given by, where G is the number of possible electrons in the given kin of orbital G = l+1, G! N!G N! where l = 0 for s, l = 1 for p, etc. an N is the number of electrons. So for p 5 we have G = = 6 an N = 5, which gives us a number of microstates equal to 6! 5!6 5! = 6. Our six microstates are given in table 1. For each microstate, we calculate the total z-projection of the angular momentum, M L = m l an the total z-projection of the spin, where spin up gives +1/, an spin own gives -1/. State Label m l = 1 m l = 0 m l = 1 M L M S 1-1 1/ -1-1/ 3 0 1/ 4 0-1/ 5 1 1/ 6 1-1/ Table 1: Microstates for a p 5 electron configuration Now we make a table of the possible values of M L an M S of the microstates enumerate in table 1. The possible values of M L which we will use as our rows are -1, 0, an 1. The possible values of M S which we will use as our columns are ±1/. Then we can ientify 5

6 where each state belongs on this table, an put the state number 1 for it onto the table. The result is table Table : Microstates organize by M S an M L From table 1, we can see that the highest possible value of M L is 1, which means that our term symbol will be of the P type. There are two such possibilities, which means that the multiplicity will be. So for each column in which we have a M L = 1, we cross out one state from each row. In this example, that leaves our little table empty, so we re one. The only kin of state is a P state. Now we just nee to etermine the values of J. As mentione in the first paragraph, the possible values of J are the numbers from L + S own to L S, with a separation of 1 between each. In this case, there will be only states: L + S = 1 + 1/ = 3/ an L S = 1 1/ = 1/. So our final term symbols are P 3/ an P 1/. b This part is really just a fancy way of asking which state has the lower energy? From Hun s rule, you shoul know that since this subshell is more than half-fille, the term symbol with the higher value of J has the lower energy. So the energies of the states are orere P 3/ < P 1/ Those are the two energy states for Ar +. The neutral argon atom, which has a full shell, must have only one term symbol: 1 S 0. We also know that the neutral argon atom has a lower energy than the argon ion because we re tol that the energy of the photon must be greater than or equal to the ionization energy of the atom, if nothing else. So we get an energy level iagram that looks like figure 1: Ar + P 1/ Ar + P 3/ Ar 1 S 0 Figure 1: Energy level iagram for the photo-ionization of Ar From figure 1, we can see that it takes more energy to get to the P 1/ state than it will to get to the P 3/ state. So when we fill in the spectrum on the exam, we get: Notice that the energy scale grows from right to left in figure. 1 Remember that the value of the state number has no physical significance. It is just a label, so your states coul have ifferent state numbers. However, the your table shoul have the same number of entries in each box, no matter what orer your states were in. 6

7 Figure : Solution to 3b 4 Problem 4 Before aressing the problem itself, it s worth taking a moment to iscuss its title. The virial theorem is an important statement about partitioning of energy in physical systems; it makes a quantitative preiction of how the total energy of a system is ivie on average between kinetic an potential energy. Although the virial theorem appears in essentially ientical form in classical mechanics, some of the mathematical etails are ifferent ; hence the preface, quantum mechanical. We ll have more to say on this topic at the en of the solution. We are aske in part a to show that the quantity given by ] ] ψ [Ĥ, Â] = ψ [Ĥ, Â = ψ x [Ĥ, Â ψx 7 will be equal to 0. We ve been given the following information: The Hamiltonian is Ĥ = ˆK + V x = m + V x. See francisr/quanprob/noe90.html for a walkthrough of the erivation. 7

8 ψx is a stationary state for which Ĥψx = Eψx.  is a linear but not necessarily Hermitian operator. With those facts in han, we may prove that 7 vanishes in several ways. They are all essentially equivalent, but some may make more sense at first than others. Approach 1: Using the Heisenberg equation of motion. An insightful approach is to recall the Heisenberg equation of motion for the time epenent expectation value of A: i ] t A t Ψt = Ψt [Â, Ĥ [ ] which we may immeiately rewrite using the stanar commutator property [Â, ˆB] = ˆB, Â. The result, which looks very much like our expression of interest, is i ] t A t Ψt = Ψt [Ĥ,  On the previous exam, we saw that there are two principal situations in which A t will be constant i.e., in which its time erivative will be equal to 0. 3 The first is when  an Ĥ commute, which we have to exclue since arbitrary pairs of operators on t commute. The secon is when the state for which we re calculating the expectation value is the time-epenent version of a stationary state, Ψx, t = ψxe ie t. We may rewrite the equation of motion in that case as [ ] ] ψ xe ie t Ĥ,  ψxe ie t = ψ x [Ĥ,  ψx A glance at this expression shows that it is ientical to 7, an our reasoning emans that it be equal to 0. Voilà! Approach : Using the integral form of the expression. If the above approach seems too abstract, we may manipulate the integral instea. This also allows us to make real use of the turnaroun rule arising from the hermiticity of the Hamiltonian, [ φ 1xĤφ x = φ xĥφ 1x ] 8 where the φ i s are any function of x. We make our first step in that irection by expaning out the commutator in the right sie of 7 to get ψ xĥâψx ψ xâĥψx We may now use hermiticity in the first term an the action of the Hamiltonian in the secon term to obtain [ ] Âψx Ĥψx E ψ xâψx 3 We won t rehash that explanation here; refer to your lecture notes from September 13 th or the solutions to exam 1 as necessary. 8

9 Note that we have Âψx on the left of Ĥ in the first term. This grouping results from the fact that ψx is not necessarily an eigenfunction of Â. As such, Âψx may be a wholly ifferent function from ψx, as oppose to simply its constant multiple, an nees to be manipulate as a single object. Moving on, we may use the facts that Ĥψx = Eψx an E is real to get [ ] E Âψx ψx E Although it appears we re in a bin here, we re really not. ψ xâψx The ientity operator I or, less peantically, the number 1 is Hermitian. Obviously, we can place a 1 between Âψx an ψx without having altere our expression. Then we can use its hermiticity to get us the rest of the way to our answer: E ψ xâψx E ψ xâψx = 0 Once again, we ve confirme that 7 is 0. Approach 3: Using Dirac notation an complex conjugation. If you prefer to use Dirac notation, we can o that, too. We can write out the Dirac form of the turnaroun rule 8 as φ 1 Ĥ φ = φ Ĥ φ 1 an the expane right han sie of 7 as ψ Ĥ ψ E ψ  ψ 9 where we ve alreay allowe the Hamiltonian to o its business in the secon term. Now we use a trick: if we complex conjugate the first term twice, there will be no overall change. That is, ψ Ĥ ψ = ψ Ĥ ψ We can now apply the turnaroun rule an then manipulate the result to get ψ Ĥ ψ = ψ  Ĥ ψ ψ = E ψ  = E ψ  ψ where we ve starre  in the intermeiate steps because it isn t necessarily Hermitian. This is the same term as on the right of 9, an so, for the thir time, we ve shown that 7 is 0. Approach 4: Using Dirac notation an operators acting left. 9

10 There s one last approach, an it s probably the simplest of all. For a Hermitian operator Ô with eigenstates o an real eigenvalues o, we can write own the expression Ô o = o o. An equivalent statement can be obtaine by using complex conjugation: Ô o = o o = o Ô = o o In other wors, Ô can act on its eigenstates to the left. Applying this iea to 9, it becomes E ψ  ψ E ψ  ψ = 0 For the last time, we ve shown that 7 is 0. Now let s see what happens when we set  = xˆp = i x an evaluate its commutator with the Hamiltonian part b. Since we have a one-imensional problem, we ll make the simplifying assumption that the Laplacian is just before proceeing. 4 We begin by simplifying our commutator as much as possible using the commutator rules: [ Ĥ, i x ] = i [ m + V x, x ] { [ = i m, x ] [ + V x, x ]} { [ ] [ = i x m, ] [ +, x + V x, x ]} { = i 0 + [ ] [ ] m, x +, x [ + V x, x ] } { [ ] [ ] = i m, x +, x [ ] } + x V x, + [V x, x] { [ ] [ ] = i m, x +, x [ ] } + x V x, + 0 { [ ] [ ] = i m, x +, x [ ] } +x V x, 4 This isn t necessary, though you coul o the entire problem using the full Laplacian for three-imensional Cartesian coorinates an get the same result. 10

11 We ve simplifie our commutator to the point where we just have to compute two others that are significantly more elementary. They are: [ ], x f = [ ] xf xf = f =, x = 1 [ ] V, f = V f [ ] V f = V f = V x, = V where the x epenence of the potential V x an the arbitrary function fx is implie on the left. Substituting these results back into our expression for the original commutator, we get [ Ĥ, i x ] = i { m + x V } V = i x i m = i x V i ˆK 30 which is just what we were aske to show. We may use this result a little cleverly by substituting it into 7. We have [Ĥ, Â] = i x V i ˆK = 0 Taking an expectation value is a linear operation, though, so this is equivalent to saying that i x V i K = 0 where we ve roppe the hat on K. A little more rearrangement an iviing out i gets us to the result for part c: x V = K What happens if our potential is the harmonic oscillator, V x = 1 kx? Plugging in, we fin that kx = K = V = K In other wors part, the kinetic an potential energies for a quantum mechnical harmonic oscillator will, on average, be equal. Now that we ve seen the virial theorem in action for a specific example, let s consier it more formally. It may be state as follows 5 : for a system whose potential energy is an n th orer homogeneous function of its coorinates, V λq 1, λq,..., λq N = λ n V q 1, q,..., q N 31 5 See muchomas/8.04/lecs/lec TISE/noe8.html for a proof, albeit a highly mathematical one. 11

12 the kinetic an potential energies will be relate by the expression where ˆK is the kinetic energy operator as before. K = n V 3 Note that the harmonic oscillator potential we ve examine in this problem is n orer homogeneous, such that application of 3 immeiately leas to our result from part. We can also apply the virial theorem in this form to other potentials: any potential of the form V x = cx n will be n th orer homogeneous; the Coulomb potential V r = 1 r is 1st orer homogeneous; an the gravitational potential V z = mgz is 1 st orer homogeneous. 6 Of course, most potentials aren t as simple as this, an it won t necessarily always be obvious that the virial theorem applies. That s why the more general form is so useful rather than wasting a lot of time calculating a potentially messy 7 erivative of V with respect to one or more coorinates, we can instea test it for homogeneity using 31. What s more, we can immeiately write own the appropriate version of 3 if it passes. In all of these cases, the virial theorem coul be taken to suggest that we treat the potential energy as funamental an use it to learn everything we want to know about the system s energetics at any given time. It turns out, though, that this approach can lea us astray from important insights. At one time, the conventional wisom in quantum mechanics hel that the chemical bon coul be explaine primarily by potential energy effects like electron-nuclear attraction, electron-electron repulsion, etc. This notion was challenge by some meticulous theoretical work one by Klaus Rueenberg in the early 1960s. In fact, he showe that the formation of the chemical bon is strongly epenent on a ecrease in the electronic kinetic energy. For an overview of these ieas as applie to the H + molecular cation, see the paper by Frank Rioux given in the footnote.8 6 That last example is a little cheeky. Gravitation an quantum mechanics on t really mesh... If you can figure out a way to reconcile the two, the King of Sween will give you at least one Nobel Prize. 7 Pun intene. 8 frioux/6rio897.pf 1