Chapter 4. Q. A hydrogen atom starts out in the following linear combination of the stationary. (ψ ψ 21 1 ). (1)

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1 Tor Kjellsson Stockholm University Chapter Q. A hydrogen atom starts out in the following linear combination of the stationary states n, l, m =,, and n, l, m =,, : Ψr, 0 = ψ + ψ. a Q. Construct Ψr, t and simplify it as much as you can. The normalized hydrogen wave functions are: ψ nlm = 3 n l! na n [n + l!] 3 e r na r l [ L l+ n l na r/na] Yl m θ, φ where L l+ m n l x and Yl θ, φ are the Laguerre polynomials and Spherical harmonics. In Griffiths second edition you find the formal definitions in chapter 4. and 4. respectively. A few examples have also been listed on p.39 and p.53 which we will refer the reader to. What we do now is to simply put in our two sets of {n, l, m} into eq. : ψ = 3! a [ +!] 3 e r a ψ = r [ L + a r/a] Y θ, φ a e r r [L a 3 a 0 r/a ] Y θ, φ Now we look up the Laguerre polynomial on p.53 and the Spherical Harmonic on p.39 that we need: 3 L 3 0r/a = 6 and Y θ, φ = 8π sinθeiφ. Thus we have arrived at the following expression: ψ = a 5 64π re r a sinθe iφ. 3

2 The same procedure for n, l, m =,, now gives: ψ = a e r r [L a 3 a 0 r/a ] Y θ, φ. Checking the new Spherical harmonic on p.39 we now obtain: and thus: ψ = a 5 64π re r a sinθe iφ Ψr, 0 = ψ + ψ Ψr, 0 = a 5 64π re r a sinθe iφ + a 5 64π re r a sinθe iφ Ψr, 0 = a 5 64π re r a sinθ e iφ e iφ Ψr, 0 = a 5 3π i re r a sinθ sinφ and to get the full time-dependent wave function we just multiply with the usual exponential: Ψr, t = a 5 3π i re r a sinθ sinφ e ie t/ which contains the time dependence and the energy of the state. b Q. Find the expectation value of the potential energy, V. Does it depend on t? Give both the formula and the actual number, in electron volts. The expectation value is, like always, given by: V = Ψ ˆV Ψ and when the states are functions this is given by the integral evaluated over the space: V = Ψ ˆV Ψ dr = Ψ V Ψ dr 4 we could drop the hat on ˆV because it is just a function. Also, note the bold integration letter - this is a notation for a volume integral, which we use because we are working in R 3.

3 Now we simplify the integrand: Ψ ˆV Ψ = [ ] [ ] a 5 3π i re r a sinθ sinφ e ie t/ V r a 5 3π i re r a sinθ sinφ e ie t/ = a 5 3π r e r a sin θ sin φv r Insert this now into eq. 4 and insert the potential V r = V r = e V = a 5 3π e r e r a sin θ sin φ 4πɛ 0 r dr. 4πɛ 0r : Note that r = x, y, z is given in the cartesian coordinate system while our functions are represented using spherical coordinates. Making the transition to spherical coordinates in the integral we get: dr = dxdydz = r sinθdrdθdφ e V = 8a 5 π ɛ 0 r=0 r 3 e r a dr π θ=0 π sin 3 θ dθ sin φ dφ. φ=0 The last two integrals can be evaluated fairly easily but the first might most of you be unfamiliar with. You can evaluate it with the formula: r=0 r n e r a dr = n! a n+ and you can find it in Physics Handbook as the Gamma function. Anyhow, evaluating the three integrals we get: e V = 8a 5 π [6a 4] [ ] 4 [π] = e. ɛ 0 3 6aπɛ 0 a is the Bohr radius and is given by the expression: a 4πɛ 0 me 5 so inserting this into our expectation value of the potential energy we obtain: V = e 6πɛ 0 me 4πɛ 0 = m e = 4πɛ 0 E = 6.8 ev 3

4 which is, as we see, independent of t. 4.8 Q. The raising and lowering operators change the value of m by one unit: L±f m l = A m l f m± l 6 where A m l is some constant. What is this constant if the eigenfunctions f m l are to be normalized? Hint: First show that L is the hermitian conjugate of L ± since L x and L y are observables you may assume that they are hermitian but prove it if you like. Then use: to get the answer: L = L ± L + L z L z 7 A m l = ll + mm ± = l ml ± m +. 8 Recall that L ± = L x ± il y. Following the hint we first show that L + = L by letting L + work on a test function g: g L + g = g L x + il y g = g L x g + g il y g 9 since L x and L y are observables they must be hermitian. It thus follows that: g L + g = g L x g + g il y g = L x g g + il y g g = L g g. 0 You proved in problem 3.5 that you must complex conjugate any complex constant if you move it from the ket to the bra. Continuing to follow the hint we now use eq.7. Recall that to find a scalar from an operator we must compute some sort of inner product. So we compute fl m L fl m : fl m L fl m = fl m L ± L fl m + fl m L }{{}}{{} zfl m + fl m L z fl m. }{{}}{{} 3 4 On the next page we evaluate the terms. 4

5 : Recall that f m l are eigenfunctions of L : L f m l = ll + f m l = f m l L f m l = ll + : Here we use L + = L : f m l L ± L f m l = L f m l L f m l = A m l f m l A m l f m = A m l f m l l f m l } {{ } f m l L ± L f m l = A m l 3 3 : Recall that f m l are eigenfunctions of L z : Now we use this to show that: L z f m l = mf m l = f m l L z f m l = m. 4 f m l L zf m l = f m l L z mf m l = f m l m mf m l = m f m l f m l }{{} f m l L zf m l = m 5 4 : From eq.4 get: f m l L z f m l = m 6 Inserting the results from -4 into eq. now gives: fl m L fl m = fl m L ± L fl m + fl m L }{{}}{{} zfl m + fl m L z fl m. }{{}}{{} 3 4 ll + = A m l + m m. 7 A m l = ll + m ± m = ll + mm. 8 The last algebraic step in eq.8 is left out. A more important remark is the fact that eq.8 does not determine the phase of A m l. The phases introduced by the ladder operators have no physical significance so we are free to set them to 0. However, in more advanced problems than we consider in this book you must be careful with choosing the phase. Also note that in eq., term, we have used L rather than L ±. Therefore, eq.8 have inverted signs. 5

6 4.9 Starting with the canonical commutation relations for position and momentum: [r i, p j ] = i δ ij, [r i, r j ] = [p i, p j ] = 0. 9 where r, r, r 3 = x, y, z and p, p, p 3 = p x, p y, p z... a Q.... work out the following commutators: [L z, x] = i y, [L z, y] = i x, [L z, z] = 0 0 [L z, p x ] = i p y, [L z, p y ] = i p x, [L z, p z ] = 0 Insert L z = xp y yp x in each commutator and do the algebra. We will start by doing this in detail for [L z, x]: [L z, x] = [xp y yp x, x] = [xp y, x] [yp x, x]. Focus on one commutator at a the time: where we applied: [AB, C] = A[B, C] + [A, C]B eq.9 for [p y, x] = 0 3 [x, x] = 0 for obvious reasons. Turning now to the other commutator: where we used eq.9. [xp y, x] = x[p y, x] + [x, x]x = [yp x, x] = y[p x, x] + [y, x]p x = i y Inserting the results from eq.3,4 into eq. gives: Similarly we get: [L z, x] = [xp y, x] [yp x, x] = 0 i y = i y. [L z, y] = [xp y yp x, y] = x [p y, y] + [x, y] p y y [p x, y] [y, y] p x = i x }{{}}{{}}{{}}{{} i For the remaining commutators [L z, z], [L z, p x ], [L z, p y ], [L z, p z ] and from now on we only write out non-zero commutators using the underbraces we show do the same steps and mark if anything is not 0. 6

7 [L z, z] = [xp y yp x, z] = x[p y, z] + [x, z]p y y[p x, z] [y, z]p x = 0 [L z, p x ] = [xp y yp x, p x ] = x[p y, p x ] + [x, p x ] p y y[p x, p x ] [y, p x ]p x = i p y }{{} i [L z, p y ] = [xp y yp x, p y ] = x[p y, p y ] + [x, p y ]p y y[p x, p y ] [y, p y ] p x = i p x }{{} i [L z, p z ] = [xp y yp x, p z ] = x[p y, p z ] + [x, p z ]p y y[p x, p z ] [y, p z ]p x = 0 b Q. Use the results from the previous problem to obtain [L z, L x ] = i L y directly from: L x = yp z zp y, L y = zp x xp z, L z = xp y yp x. 5 Insert the definition of L x : [L z, L x ] = [L z, yp z zp y ] = [L z, yp z ] [L z, zp y ] = [yp z, L z ] + [zp y, L z ] = y [p z, L z ] + [y, L z ] }{{}}{{} 0 i x p z + z [p y, L z ] }{{} + [z, L z ] p y = i xp z + i zp x = i L y }{{} i p x 0 c Q. Evaluate the commutators [L z, x + y + z ] and [L z, p x + p y + p z]. = [L z, x + y + z ] = [L z, x ] + [L z, y ] + [L z, z ] = [x, L z ] + [y, L z ] + [z, L z ] x [x, L z ] + [x, L z ] x + y [y, L z ] + [y, L z ] y + z [z, L z ] + [z, L z ] }{{}}{{}}{{}}{{}}{{}}{{} i y i y i x i x 0 0 z = 0 [L z, p x + p y + p z] = [L z, p x] + [L z, p y] + [L z, p z] = [p x, L z ] + [p y, L z ] + [p z, L z ] = p x [p x, L z ] + [p x, L z ] x+p y [p y, L z ] + [p y, L z ] p y +p z [p z, L z ] + [p z, L z ] p z = 0 }{{}}{{}}{{}}{{}}{{}}{{} i p y i p y i p x i p x 0 0 7

8 d Q. Show that the Hamiltonian H = p /m + V commutes with all three components of L, provided that V only depends on r. Thus H, L and L z are mutually compatible observables. [H, L] = m [p, L] + [V, L] In the previous problem we proved that [L z, p x + p y + p z] = 0. This also holds for L x and L y - just make the exchange in the previous problem and you will see it. Since all components of L commute with p it follows that [ L, p x + p y + p ] z = 0. Now we will show that [V, L] = 0: Assume that the potential is only a function of r = x + y + z, as we are asked to do. Since V only depends on r it is useful to look what L looks like in spherical coordinates: L = r p = i r 6 L = i r, 0, 0 r, r θ, r sinθ φ ˆr ˆθ ˆφ L = r 0 0 = ˆφ r θ ˆθ sinθ φ r r θ r sinθ φ. 7 8 Note that all terms in eq.8 commute with any function V r since the only operator dependent on r is the scalar valued /r, which certainly commutes with V r. This concludes that [L, V r] = 0 and thus also that: [H, L] = m [p, L] + [V, L] = 0 9 8

9 4. Q. Derive the following equation: L + L = θ + cotθ θ + cot θ φ + i φ 30 using the definition of the raising/lowering operators: L± = ± e ±iφ θ ± i cotθ. 3 φ We will now work with the explicit form of the operators and whenever yo do that it is a good idea to use a test function. Our testfunction in this case will be named gr and from eq. 3 we get: L + L gr = [ e iφ θ + i cotθ ] [ e iφ φ θ i cotθ ] gr. φ Now we first let L act on the testfunction: L + L gr = and then L + shall act on L gr: [ e iφ θ + i cotθ ] [ ] g e iφ g i cotθ φ θ φ [ ] L + L gr = e iφ g e iφ g i cotθ +i cotθ [ ] g e iφ g i cotθ θ θ φ φ θ φ }{{}}{{} A B where we have named two parts to work on separtely in order to avoid algebraic errors. A: B: A = e iφ g θ + e iφ i cotθ g θ φ [ = e iφ g θ i g cotθ θ [ = e iφ g θ + i sin θ B = e φ iφ g θ + i cotθ e φ φ i cotθ θ g φ i cotθ g θ φ iφ g φ ] g φ ]. 3 9

10 [ = e iφ g φ θ + e iφ φ iφ g = i e θ e iφ ] [ g +i cotθ e iφ g θ φ φ + e iφ g φ θ + e iφ cotθ g [ = e iφ i g θ g g + cotθ φ θ φ + i cotθ g φ φ + i e iφ cotθ g φ φ ] g φ ]. 33 From the expression just above eq. 3: L + L gr = e iφ [A + i cotθb] 34 we see that we need to work out A + i cotθb: [ A = e iφ ] g θ + i g sin θ φ i cotθ g θ φ [ i cotθb = i cotθ e iφ i g ] θ g g + cotθ φ θ φ + i cotθ g φ. A normal assumption about functions in physics is that partial derivatives with respect to different variables commute. Hence g θ φ = g φ θ so when we add the two equations above the terms with mixed partial derivatives cancel each other out: [ { }}{ A+i cotθb = e iφ g θ +i sin θ + cot θ g [ = e iφ g θ + i g ] φ + cotθ g θ + cot θ g φ. Now we are finally! done: φ cotθ g θ cot θ g φ L + L gr = e iφ A + i cotθb [ L + L gr = e iφ e iφ g θ + i g ] φ + cotθ g θ + cot θ g φ [ L + L gr = θ + i φ + cotθ ] θ + cot θ φ gr [ L + L = θ + i φ + cotθ ] θ + cot θ φ. 35 ] 0

11 b Q. Use the following three equations: [ L + L = θ + i φ + cotθ ] θ + cot θ φ L z = i φ to derive: [ L = sinθ θ L = L ± L + L z L z 38 sinθ + θ sin θ φ ]. 39 L gr = Now we let the upper version of eq. 38 act on a test function gr: L gr = L + L gr + L zgr L z gr. 40 We know the first term from the tedious calculation in the previous problem. The last two terms can be calculated by using eq. 37: Now we insert this into eq. 40: L z gr = i φ gr = i L z L z gr = g i φ i φ g φ L gr = L + L gr + L zgr L z gr. [ θ + i φ + cotθ ] θ + cot θ φ gr+ so apparently: L = [ θ + i φ + cotθ θ + cot θ φ + φ L = [ 4 = g φ. 4 [ θ + cotθ θ + cot θ φ + φ φ i ] φ ] ] [ gr i [ L = θ + cotθ θ + cot θ + ] φ [ L = θ + cosθ sinθ θ + ] sin θ φ [ L = sinθ sinθ θ + cosθ ] + θ sin θ φ. φ ] gr

12 Note now that you can re-write the first parenthesis: sinθ θ + cosθ θ = sinθ + θ θ θ sinθ θ = sinθ θ θ where the last manipulation follows from the differrentiation rule for a product. Hence we have arrived at the desired result: [ L = sinθ θ sinθ + θ sin θ φ ]. 4.4 Q. Two particles of mass m are attached to the ends of a massless rigid rod of length a. The system is free to rotate in three dimensions about the center but the center point itself is fixed. a Q. Show that the allowed energies of this rigid rotor are: E n = nn + ma for n = 0,,, Hint: first express the classical energy in terms of the total angular momentum. Recall that the Hamiltonian operator system: You are used to the appearance: Ĥ always tells you the energy of the Ĥ α n = E n α n. 44 Ĥ = ˆp + V x 45 m but this is just the operators for the kinetic and potential energies of the system! So what do these two operators look like in our current system? If the particles are not moving at relativistic speed each has the kinetic energy mv /. Since there are two of them the total kinetic energy of the system is mv. Since there is no potential energy we thus get our Hamiltonian to be: Ĥ = mv. 46

13 Following the hint given we can try to express this in terms of the angular momentum of the system: ˆL = mrv = m a ˆL v = mav Ĥ = ma 47 but we know both the eigenvalues and eigenfunctions of ˆL! Given an eigenfunction fl m we have: ˆL f m l = ll + f m l 48 so the allowed energies for this system is calling n = l: E n = nn + ma for n = 0,,, b Q. What are the normalized eigenfunctions for this system and what is the degeneracy of the n:th energy level? The normalized eigenfunctions of ˆL are the Spherical Harmonics: Yl m θ, φ. for each l there are l+ values that m can acquire so the degeneracy of the n:th energy level remember, we called n = l in the previous problem is n Q. An electron is in the spin state: χ = A a Q. Determine the normalization constant A. 3i The normalization condition is χ χ =, so: χ χ = A 3i 4 3i A = A which gives: = A 5 A = 5 eiφ 5 3

14 where the last exponential is called a phase. Now, like we discussed in the solution to problem 4.8 the phase carries no physical significance, so you might as well put φ = 0 from now on, we will always do this. Hence A = /5. b Q. Find the expectation values of S x, S y and S z. S x = χ S x χ = 5 3i 4 0 / / 0 3i 4 = 0 53 S y = χ S y χ = 5 0 i / 3i 3i 4 i / 0 4 = 54 5 S z = χ S z χ = / 0 3i 3i / 4 = c Q. Find the uncertainties σ Sx, σ Sy and σ Sz. Note, these are not the Pauli spin matrices! Recall that a variance can be written as: σ a = a a 56 so we see that we need to work out Si for i {x, y, z}. S 0 / 0 / x = = 0 / 0 / S 0 i / 0 i / y = = 0 i / 0 i / S / 0 / 0 z = = 0 0 / 0 / 4 0 which means that: S x = S y = S z = 5 3i i 4 = 4 60 this result will always be true since we are operating on normalized states using the identity matrix multiplied by 4. 4

15 Inserting this result and our previous results into the equation about the variance we get: σ S x = S x S x = 4 6 σ S y = Sy S y = 4 5 = so: σ S x = S z S z = = σ Sx =, σ S y = 7 50, σ S z = d Q. Confirm that your results are consistent with all three uncertainty principles: σ Sx σ Sy Sz, σsz σ Sx Sy, σsy σ Sz Sx 65 Start by computing the left hand sides: σ Sx σ Sy = 7 00, σ Sz σ Sx = 6 5, σ Sy σ Sz = and then look at the respective right hand sides: Sz 7 = 00, Sy 6 = 5, Sx = 0 66 and in each case we see that the uncertainty principles are fulfilled. Note that the two that are not trivial where the right hand side is 0 hit the minimum values. 5

16 4.9 a Q. Find the eigenvalues and eigenspinors of S y. We know the matrix representation of this operator: 0 i / S y = i / 0 67 so we can directly solve for the eigenvalues and eigenspinors by solving the eigenvalue equation: a a S y = λ b b 0 i a a = λ i 0 b b which gives the characteristic equation: dets y λi = 0 λ λ i i = 0 λ = ±. Inserting the eigenvalues into the eigenvalue equation we can then solve for the eigenspinors: 0 i a = ± a 68 i 0 b b which gives the linear system of equations: ib = ±a ia = ±b = b = ±ia. note that the two equations are linearily dependent which comes from the characteristic equation.. Also, note that it is important to keep track of the order of the plus and minus signs because they come from inserting two different eigenvalues. Our normalized eigenspinors are thus: χ y ± =. 69 ±i 6

17 b Q. If you measured S y on a particle in the general normalized state: χ = a χ z + + b χ z 70 what values might you get and with what probabilities? Check that the probabilities add up to. Remember that a and b need not be real! The possible values are the eigenvalues of S y and we know that they are ±. To get the probabilities we must express χ in the eigenvectors of S y. This is done by using the inner product of χ with each of the eigenvectors χ y ± : χ = χ y + χ χ y + + χ y + χ χ y + 7 Working in the usual S z -basis we have that: χ z + = 0 χ z = 0 7 From the previous problem we know then know that the eigenvectors of S y are: χ y + =, χ y i = 73 i which gives us: χ y + χ = a, i = b a ib. 74 χ y χ = a, +i = b a + ib. 75 which gives us the state expressed in the y-basis: χ = a ib χ y + + a + ib χ y 76 Now that we have expressed the state in the right basis we get the probabilities by computing the magnitude squared of the coefficients: P P P = = a ib = a iba + iab + b a + ib = a + iba iab + b + P = a + b = a + b = 77 where the last equality comes from the fact that the original state was normalized. 7

18 c Q. If you measured S y, what values might you get and with what probabilities? Recall from eq. 58 what Sy looks like in matrix form: S y = We don t need to find its eigenstates because they are just: 0 and 0 78 and have the same eigenvalue the matrix is diagonal and the entries are the same. Since there is only one possible value to get, the answer is: 4 with probability. 4.3 Q. Construct the spin matrices S x, S y and S z for a particle of spin. Hint: How many eigenstates of S z are there? Determine the action of S z, S + and S on each of these states. Follow the procedure used in the text for spin /. Also, study pages carefully before attempting this problem. First we need to determine the number of eigenstates of Ŝz. This is given by the total number of different m s we can have: m s { s, s +,, 0,, s}. For a spin particle this is just m s {, 0, }. Let us call these states: s m = 0 - By knowing the action of the operators on these states we can determine their matrix representation - precisely as was done in the book in section From eqns in the text book we read: Ŝ m = m Ŝ z m = m m This is of course the same number as for Ŝy, Ŝx. 8

19 Ŝ ± m = mm ± m ±. Since we are asked to find the operators on matrix form we make the natural choice: = 0 0 = = 0 0 With this choice, we easily obtain S z : 0 0 S z = and now we proceed with the ladder operators. To find the matrix version of an operator Â, recall that the elements are given by A mn = e m  e n : S ± : Note: m Ŝ± n = n nn ± n ± = nn ± m n ± = m Ŝ+ n = nn + δ m,n+ m Ŝ n = nn δ m,n. Now we need to check the different combinations of m and n. Start with the ones for Ŝ+: For m = we will see that only n = 0 will give a contribution: e Ŝ+ e = Ŝ+ = 0 = 0 e Ŝ+ e = Ŝ+ 0 = = = }{{} e Ŝ+ e 3 = Ŝ+ - = 0 = 0 = 0 }{{} 0 so thus far we know that our matrix must have the form: If you do not see this, pause for a second and breathe. Now look at what Ŝz does to these kets. 9

20 S + = 0 0 Continuing with the next row we obtain: e Ŝ+ e = 0 Ŝ+ = 0 0 = 0 e Ŝ+ e = 0 Ŝ+ 0 = 0 = 0 = 0 }{{} 0 e Ŝ+ e 3 = 0 Ŝ+ - = 0 0 = 0 0 = }{{} and thus we now know the second row in the matrix: S + = Finally, for the last row all matrix elements are 0: e 3 Ŝ+ e = - Ŝ+ = - 0 = 0 e 3 Ŝ+ e = - Ŝ+ 0 = - = - = 0 }{{} 0 e 3 Ŝ+ e 3 = - Ŝ+ - = - 0 = - 0 = 0 }{{} S + = The exact same procedure do it! for the lowering operator gives: S = Now we proceed to construct S x and S y. On p.74 we see that we get these as: so: S x = S + + S S y = i S + S 0

21 S x = S y = i Q. An electron is at rest in an oscillating magnetic field: where B 0 and ω are constants. a Q. Construct the Hamiltonian matrix for this system. B = B 0 cosωtẑ, 79 The Hamiltonian for an electron at rest in a magnetic field B is: H = µ B = γs B 80 where µ is the magnetic dipole moment and γ is the gyromagnetic ratio. Inserting B we obtain: H = γs z B 0 cosωt = γb 0 cosωt since the magnetic field is pointing in the ẑ-direction. b Q. The electron starts out at t = 0 in the spin-up state with respect to the x-axis. that is: χ0 = χ x +. Determine χt at any subsequent time. Beware: this is a time-dependent Hamiltonian, so you cannot get χt in the usual way from stationary states. Fortunately, in this case you can solve the time-dependent Schrödinger equation: directly. i χ t = H χ 8

22 First we express our state as a column vector: at χt = bt and plug it in, together with our Hamiltonian, into the Schrödinger equation: 83 at i χ t = H χ 0 at i t i d dt bt at bt which gives us a system of equations: = γb 0 cosωt 0 bt = γb 0 cosωt at bt 84 i dat dt i dbt dt = γ B 0 cosωt at = γ B 0 cosωt bt. Note that the two equations are almost identical. We can thus focus on one of them say, the upper one and later re-use the derivation. i da dt = γ B 0 cosωt a now we separate the variables: and integrate both sides: i da a = γ B 0 cosωt dt where A = e ic D. da i a = γ B 0 cosωtdt i log a + C = γ B 0 ω sinωt + D log a = iγ B 0 at = Ae γ B 0 ω sinωt sinωt + ic D ω Following the same steps to find bt we obtain: bt = Be iγ B 0 ω sinωt.

23 This gives us our state: Ae χt = γ B 0 ω sinωt Be iγ B 0 ω sinωt and from the initial value condition we can determine A and B: χ0 = χ x + = = A = B = 85 giving: χt = e iγ B 0 ω sinωt e iγ B ω sinωt c Q. Find the probability of getting / if you measure S x. Note: this is of course at time t. Recall that in order to find the probability of obtaining an eigenvalue in a specific basis you must first express the state in that basis. Currently our state is expressed in the z-basis 3 : χ z t = e iγ B 0 ω sinωt e iγ B 0. ω sinωt Now we change basis by using the inner product of the state in our current basis with the eigenstates of the new basis: χ x t = χ x + χ z t χ x + + χ x χ z t χ x. 87 Note here that to answer the question we only need to compute the inner product χ x χ z t because we are asked about /. In problem 4.9 you found the eigenstates of S y. If you follow the same procedure for S x you will find that: χ x = 88 so: χ x χ z t = e iγ B 0 ω sinωt e iγ B 0 ω sinωt where we in the last step used one of Euler s formulae. = i sin γ B 0 ω sinωt 89 3 Because we got this state from working with S z = it must be expressed in the z-basis Since this matrix is diagonal, 3

24 Computing the square of the absolute value of eq.89 gives the probability: P = sin γ B 0 ω sinωt. 90 d Q. What is the minimum field B 0 required to force a complete flip in S x? The system starts out in χ x + and we want to estimate the minimum value of B 0 so that the system is changed to χ x. This would mean that: P = sin γ B 0 ω sinωt sin γ B 0 ω sinωt = = ± γ B 0 ω sinωt = ±nπ n Z +. Now, since we want the minimum value of B 0 we set n = for larger n then B 0 gets larger.. The term sinωt is of course dependent on time but it reaches maximum magnitude of ±, so the minimum value of B 0 which occurs at sinωt = ± is: B 0 min = πω γ = πω γ. 4

25 4.34 a Q. Apply S to 0 and confirm that you get -. Let us first start to clarify the notation that we will use: i For the composite system we will write S M where S can go from s s up to s + s in integer steps. The M value can then range in integer steps from s to s. ii The composite system can be written as a linear combination of the product of the individual systems: S M = i C i s m s m 9 The coefficient C i is called Clebsch-Gordan coefficient, but in this problem we will not discuss these further. iii When s = s it is customary to only use m, m for brevity: s m = i C i m m. 9 In this problem we consider s = s =. The first thing to note is that the total spin can either be 0 or. The values m i can only be ± and we will simply refer to them as ±. Thus the following is true in our notation: m m m m = + 93 if m = + and m =. From the book eq.77 we know the following: S M m m = = [ ] 94 - = In the composite system of particle and particle, our spin operator is the sum of the respective spin operators: S = S + S 95 5

26 in the sense that they each act on their own particle. From the book eq we know the ladder operators to a general spin state: so in our case: S j ± s j m = ss + mm ± s j m ± S + = +, S + + = 0 96 S = 0, S + =. 97 Now we use eq 94 and these to get: S 0 = S + S [ ] S 0 = [ ] S + S + + S + S + S 0 = [ ] S 0 = [ ] S 0 = which we see from eq 94 is: S 0 =. b Q. Apply S ± to 0 0 and confirm that you get 0. From the book we see: So applying the ladder operators one at a time: S = 0 0 = [ + + ] 98 S + + S + [ + + ] S = [ ] S + + S + + S + + S + + 6

27 S = [ ] Similarly: S = S 0 0 = S + S [ + + ] S 0 0 = [ ] S + S + S + S + S 0 0 = [ ] S ± 0 0 = c Q. Show that and are eigenstates of S, with the appropriate eigenvalues. For a state S M to be an eigenstate with the proper eigenvalue it needs to fulfil the following equation: S S M = SS + S M 0 so we need to show the the right hand side is indeed what we get if we start with the left hand side. Starting with = + + in the individual systems: S = S = S = S = S + S + + = S + S + S S + + S + S + S S + + S + S + S S + + s s s s S S + + S = S S

28 The first two terms might have been easy but what about the last one? Note here that: so: S S = S x S x + S y S y + S z S z 03 S S + + = S x S x S y S y S z S z Recall now that we are using the z-basis so the last term is easy: S z S z + + = For the S x S x S y S y + + we have to do some more work. The fastest way is to use the matrix representation of S x, S y : S x = 0 S 0 y = 0 i i 0 S x + = 0 = 0 0 = 0 S y + = 0 i = i 0 = i i We let the reader verify the following using the information in eq06: S x S x S y S y + + = + i = In total we therefore get: S = s s s s S S S 3 = S = + + =. As an exercise, follow the same steps to show that S - = -. 8

29 4.36 a Q. A particle of spin and a particle of spin are at rest in a configuration such that the total spin is 3, and its z component is. If you measured the z component of the angular momentum of the spin- particle what values might you get and what is the probability of each one? When you combine angular momentum J and J, the possible results are: J tot = J J, J J +,..., J + J and the new M-value the projection on an axis can range from J tot to J tot in integer steps. In our case we are told that J tot = 3 and that M = since the z-component is M =. The system is thus in the total state 3. Recall that this system is built up by eigenstates of the two individual particles. To find the probability of obtaining a certain value after a measurement of one of the particles we can use the Clebsch-Gordan table. From the table labelled we find the column 3 +. Then we see that there are three rows with fractions and to the left of these one can see the individual m-values. The state is thus 4 : 3 = J M = m +m =M C s,s,j m,m,m s m s m From eq.08 we see that the possible m-values of the spin particle are,, 0. The possible results and probabilities are thus:, with P = 5, with P = , 0 with P 0 = You need to practice this by yourself - there are no shortcuts!. 9

30 b Q. An electron with spin down in the state ψ 50 of the hydrogen atom. If you could measure the total angular momentum squared of the electron alone not including the proton spin, what values might you get and what is the probability of each? Note that this problem takes the opposite perspective from the previous problem. In the previous problem you knew the total angular momentum and had to find out information about the individual systems. In this problem you actually have information about the individual systems: L m l s m s = 0 - and want to find information about the total system JM : L m l s m s = J 0 - = J C L,s,J m,m,m J M C L,s,J m,m,m J M. To find the coefficients Cm L,s,J,m,m we go to the Clebsch-Gordan / table. In the previous problem you read off a column - in this case we instead read off a row. We are specifically looking for the row 0 -. Once you find it you will see: 0 - = So now we can answer the question of the possible values we might obtain and with what probabilities when we measure the total angular momentum squared: = 5 4 with 5 P + = 3 4 with 3 P 4 4 = 3 = 3 30

31 4.38 Q. Consider the three-dimensional harmonic oscillator, for which the potential is: V r = mω r. 09 a Q. Show that separation of variables in Cartesian coordinates turns this into three one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. Consider the Schrödinger equation: where Ψx, t i = ĤΨx, t 0 t Ĥ = ˆp + V x = m m x + y + z + mω x + y + z. Assume now a separable solution: which inserted to the Schrödinger equation gives: Ψx, t = ψx, y, zt t i ψx, y, z T [ t = m x + y + z + mω x + y + z ] ψx, y, zt i T T t = [ ψx, y, z m x + y + z + mω x + y + z ] ψx, y, z now, from the assumption about separable variables it is implied that the variables must be independent of each other. Seeing that the left hand side is only a function of T and the right hand side is only a function of X,Y,Z we must deduce that: i T T t = E 3 where E is a constant 5. If it was not a constant, then T would depend on X,Y,Z which contradicts the assumption. 5 Try to work out the dimension on the left hand side and you ll see the appropriateness of the letter E. Hint: T t should give you a quantity measured in a time unit. 3

32 Now we can insert this back to obtain the time-independent Schrödinger equation: Eψx, y, z = [ m x + y + z + mω x + y + z ] ψx, y, z which of course something familiar to us all. Assume further now, in our separation of variables, that ψx, y, z = XxY yzz: E XY Z = [ m x + y + z + mω x + y + z ] XY Z E XY Z = Y Z d X m dx + XZ d Y dy and now divide by XY Z: d X m X dx + d Y Y dy + Z + XY d Z dz + mω x + y + z XY Z d Z dz + mω x + y + z = E. Lastly, group together terms that are dependent on the same variable: m X d X dx + mω x + m Y d Y dy + mω y + m Z d Z dz + mω z = E. Now, each of the terms in the round brackets are only dependent on that respective variable and our assumption is that they are independent of each other. Since the right hand side is constant it follows that each of these terms must be constant! Let us call these constants for E x, E y, E z : d X m X dx + mω x }{{} Ex d Y m Y dy + mω y }{{} Ey d Z m Z dz + mω z = E }{{} Ez where E x + E y + E z = E so effectively we now have one second order differential equation per variable: d X m dx + mω x X = E x X 4 d Y m dy + mω y Y = E y Y 5 d Z m dz + mω z Z = E z Z. 6 3

33 These three differential equations are each equivalent to the one dimensional Harmonic Oscillator from section.3 in the text book! The allowed energies in that section for one dimension are derived to be: E n = n + ω 7 and in our case it applies in each direction i {x, y, z}: E i n i = n i + ω 8 with: E n = E x n x + E y n y + E z n z = n + 3 ω 9 where n = n x + n y + n z and n x, n y, n z goes from 0,,... b Q. Determine the degeneracy dn of E n. Given the criteria on n x, n y, n z, the degeneracy is given by how many ways we can get the same value n. The easiest way is to look for some sort of structure: We start by choosing n x, then look at the possibilities left: If n x = n then n y = 0 and n z = 0, this gives one way. If n x = n then n y = 0, and n z =, 0, this gives two ways. If n x = n then n y = 0,, and n z =,, 0, this gives three ways. Since we can continue with this down to: If n x = 0 then n y = 0,,,... n and n z = n, n, n,..., 0, this gives n+ ways. Summing all these possibilities we obtain: n+ dn = i = i n + n

34 4.40 a Q. Prove the three-dimensional Virial theorem: for stationary states. T = r V In problem 3.3 we used eq.3.7: d dt ˆQ = i [Ĥ, ˆQ] + ˆQ t to show that d xp = T x dv dt dx 3 and since all expectation values are constant in time for stationary states we obtained the one dimensional Virial theorem: T = x dv. dx Now we exchange Q = xp to the three-dimensional analogy Q = rp in eq.: d dt r p = i r p [Ĥ, r p] +. 4 t For a stationary state we know that the left hand side must be 0. Also, the last term on the right hand side is 0 because there is no explicit time dependence in neither r or p. Therefore: 0 = i [Ĥ, r p]. 5 Now we expand commutator in eq.5: [Ĥ, r p] = [Ĥ, r p + r p + r 3 p 3 ] = [r p + r p + r 3 p 3, Ĥ] = [r p, Ĥ] + [r p, Ĥ] + [r 3p 3, Ĥ] = r [p, Ĥ] + [r, Ĥ]p + r [p, Ĥ] + [r, Ĥ]p + r 3 [p 3, Ĥ] + [r 3, Ĥ]p 3 34

35 Inserting Ĥ = p m + V r gives: [Ĥ, r p] = r [p, p m + V r] + [r, p m + V r]p ˆp + r [p, m + V r] + [r, p m + V r]p + r 3 [p 3, p m + V r] + [r 3, p m + V r]p 3. In the equation above we have commutators if we expand each of the six into two commutators. However, a lot of them will actually be 0 and now we hunt those down. Recall problem 4.9 and the canonical commutators: [r j, p k ] = i δ jk, [r j, r k ] = [p j, p k ] = 0. Any component of p will commute with p so: [p j, p m + V r] = [p j, p m ] +[p j, V r] }{{} 0 which simplifies some of the terms in our hideous expression of [Ĥ, r p]: [Ĥ, r p] = r [p, V r] + [r, p m + V r]p + r [p, V r] + [r, p m + V r]p + r 3 [p 3, V r] + [r 3, p m + V r]p 3. Note now that any function fr must commute with the coordinate variables x, y, z r, r, r 3 : Thus: [r j, V r] = r j V r V rr j = r j V r r j V r = 0 6 [r j, p m + V r] = [r j, p m ] + [r j, V r] }{{} 0 which then gives: [Ĥ, r p] = r [p, V r] + [r, p m ]p + r [p, V r] + [r, p m ]p + r 3 [p 3, V r] + [r 3, p m ]p 3. From problem 3.3c we know that: V r [p j, V r] = ih 7 r j 35

36 so inserting that we get: [Ĥ, r p] = r ih [r, p m ]p + r 3 V r + [r, p r ih m ]p + r V r r 3 ih V r r. + [r 3, p m ]p 3 + Now we only have three more commutators and we will evaluate them all at once: [r j, p m ] = [rj, p + p + p m 3] = m 3 [r j, p k] = m k= 3 [p k, r j ] = k= = m 3 k= p k [p k, r j ] + [p k, r j ]p k = m 3 p k i + i p k δ kj = i m which we now insert into our commutator and then simplify the expression: [Ĥ, r p] = V r i r ih + p V r + r ih + r m r i p V r i + r 3 ih + p 3 m r 3 m [Ĥ, r p] = ih V r V r V r r + r + r 3 i p r r r 3 m + p + p 3 k= [Ĥ, r p] = ih r V i p m. Referring back to back into eq. 5 we get: i [Ĥ, r p] = 0 i ih r V i p = 0 m r V p = m and since T = p m we have finally arrived at: r V = T. 36

37 b Q. Apply the Virial theorem to the case of Hydrogen and show that: The Virial theorem states: T = E n ; V = E n. 8 r V = T. For Hydrogen we have the potential V r = e 4πɛ 0r which gives: V r V = ˆr + V r V r ˆθ + r r θ r sinθ φ ˆφ = e 4πɛ 0 r ˆr = r V e e = rˆr 4πɛ 0 r ˆr = = V. 4πɛ 0 r Exchanging the left hand side in the Virial theorem with this gives: V = T. 9 Lastly, recall Ehrenfest s theorem - expectation values obey classical laws. So: Combining eq.9 and eq.30 we obtain: T + V = E n. 30 T = E n V = E n. 3 37

38 c Q. Apply the Virial theorem to the three-dimensional harmonic oscillator Problem 4.38 and show that in this case: The same steps as the previous problem: The Virial theorem states: r V = T. T = V = E n /. 3 For 3D harmonic oscillator we have the potential V r = mω r giving: V = V r ˆr + r r V r ˆθ + θ V r r sinθ φ ˆφ = mω rˆr = r V = rˆr mω rˆr = mω r = V. Exchanging the left hand side in the Virial theorem with this gives: V = T V = T. 33 Lastly, recall Ehrenfest s theorem - expectation values obey classical laws: T + V = E n. 34 Combining eq.33 and eq.34 gives: T = V = E n 35 which is what we wanted to show. 38