# UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet

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1 UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Solution for take home exam: FYS311, Oct. 7, The Hamiltonian of a charged particle in a weak magnetic field is Ĥ = P /m q mc P A where we have omitted the A term because the field is weak. The vector potential A = B ( y, x, ) gives a uniform field in the z-direction ( B = A). Inserting this into the P A -term we get: qb mc (xp y yp x ) = qb mc ˆL z. Writing out the Hamiltonian in polar coordinates we find Ĥ = h 1 m r r ( r ) L + r mr qb mc ˆL z The particle is constrained to move on a sphere, thus r is constant, and is therefore not a degree of freedom. This reduces the Hamiltonian to the quoted expression Ĥ = ᾱ h L + β ˆL z with α/ h = 1/(mr ) and β = (qb)/(mc). (q < for an electron). Eigenstates of Ĥ are the l, m with energy eigenvalues: E l,m = ᾱ h h l(l + 1) + β hm = hα (l(l + 1) + βα ) m where l {, 1,,...} and m { l, l + 1,..., l 1, l}. The L s represent a spatial degree of freedom(in contrast to spin), thus only integer l s are allowed E[h _ α β/α Figure 1: Energy levels. Shown are the 1 lowest (at β/α = ) levels. 1

2 1. The energy eigenstates are the l, m. Which of these have the lowest energy? Inspecting E l,m we see that the value of m giving the lowest energy for a given l is the smallest possible value of m, m = l. In this state l, l ˆL z l, l = hl. The value of l in the ground state is found by inspecting which l minimizes E l, l for a given value of β/α. From the plot in 1.1 it is clear that for small β/α, E, is the lowest energy. Then as β/α increases higher values of l takes over successively. The ground state will change from l to l + 1 discontinuously when E l+1, (l+1) = E l, l which is equivalent to hα [(l + 1)(l + ) βα (l + 1) = hα [l(l + 1) βα l = β α = l + Thus the ground state will have l = for β/α <. For < β/α < the ground state will have l = 1, etc... This can also be seen from the plot in 1.1. At β/α = {,, 6,...} 1 <L z >[h _ β/α Figure : Plot of ψ ˆL z ψ the ground state manifold is two-dimensional. It is spanned by the states l, l and l + 1, l 1. At these points the value of L z will depend on a further specification of the ground state. Since there are actually two states a further specification would be to specify the probabilities for each of these states. This can be achieved using the framework of density operators which is not part of this course. Without a further specification of the ground state at these points, all we can conclude is that the observed value must lie somewhere in between the values l h and (l + 1) h. In the following we consider the slightly more general state than in the problem set: ψ = c ( l = 1, m = 1 + b l = 1, m = l = 1, m = 1 )

3 To obtain the state in the exam, set c = 1 and b = i. 1.3 ψ(t) = c ( e ie1, 1t/ h 1, 1 + be ie1,t/ h 1, e ie1,1t/ h 1, 1 ) = ce ( iαt e iβt 1, 1 + b 1, e iβt 1, 1 ) = 1 (e e iαt iβt 1, 1 + i ) 1, e iβt 1, 1 The corresponding bra is ψ(t) = c e ( iαt e iβt 1, 1 + b 1, e iβt 1, 1 ) = 1 (e eiαt iβt 1, 1 i ) 1, e iβt 1, 1 The operator ˆL z acting on ψ(t) gives ˆL z ψ(t) = hce iαt ( e iβt 1, 1 e iβt 1, 1 ). Thus the expectation value of ˆL z becomes ψ(t) ˆL z ψ(t) = h c ( 1 + 1) = h ( 1 + 1) = 1. The state ψ(t) has total momentum l = 1. Thus the possible values of the z-axis component of the angular momentum are ± h,. The probabilities for the different values are gotten by projections onto the corresponding eigenfunctions: P ( h) = 1, 1 ψ(t) = ce iαt+iβt = c = 1 P () = 1, ψ(t) = ce iαt b = cb = 1 P (+ h) = 1, 1 ψ(t) = ce iαt iβt = c = 1 These probabilities give the expected value ˆL z = ( h) c + cb + h c = h c ( 1+1) = which agrees with problem 1.3. Note that it is not enough here to show that the most probable value of m is. The value of L z does not depend on how probable it is to get m =. Rather the crucial point here is that it is equally likely to get h as + h. 3

4 1.5 The operator ˆL x = (ˆL + + ˆL )/, and ˆL ± l, m = h l(l + 1) m(m ± 1) l, m ± 1. In particular this gives ˆL x 1, 1 = h 1, ˆL x 1, = h ( 1, 1 + 1, 1 ) ˆL x 1, 1 = h 1, Using these relations we find Thus ˆL x ψ(t) = h ce iαt ( b 1, 1 + (e iβt e iβt ) 1, + b 1, 1 ) = h ( e iαt i 1, 1 + (e iβt e iβt ) 1, + i ) 1, 1 ψ(t) ˆL x ψ(t) = h c ( be iβt + b (e iβt e iβt ) be iβt) = h c ( e iβt (b b) e ibt (b b) ) = h c sin(βt)i(b b) = h sin(βt) Applying ˆL x once more gives which results in ˆL x ψ(t) = h ce iαt ( (e iβt e iβt ) 1, 1 + b 1, + (e iβt e iβt ) 1, 1 ) ψ(t) ˆL x ψ(t) = h c ( (1 e iβt ) + b (e iβt 1) ) = h c ( (1 e iβt ) + b (e iβt 1) ) = h c ( + b cos(βt) ) = h c ( + b cos(βt) ) = h = h (3 cos(βt)) ( 1 + sin (βt) )

5 1.6 The state ψ(t) has l = 1, thus the possible measurement values of ˆL x is ± h,. In order to find the probabilities of each we need the eigenstates of ˆL x. From the action of ˆL x l, m on the l = 1 states listed above (1.5) it is not difficult to see that ˆL x ( 1, 1 1, 1 ) = ˆL x ( 1, 1 + ) ( 1, + 1, 1 = h 1, 1 + ) 1, + 1, 1 ˆL x ( 1, 1 ) ( 1, + 1, 1 = h 1, 1 ) 1, + 1, 1 Normalizing these, the eigenstates of L x are And the probabilities 1 are Added together we find = 1 ( 1, 1 ) 1, + 1, 1 = 1 ( 1, 1 1, 1 ) + = 1 ( 1, 1 + ) 1, + 1, 1 P ( h) = ψ(t) = c eiβt b e iβt = 1 (1 sin(βt)) P () = ψ(t) = c eiβt + e iβt = 1 cos (βt) P (+ h) = + ψ(t) = c eiβt + b e iβt = 1 (1 + sin(βt)) P ( h) + P () + P (+ h) = 1 (1 sin(βt)) + 1 cos (βt) + 1 (1 + sin(βt)) = sin(βt) + 1 cos (βt) = = In the position representation the state ψ(t) is represented as ψ(θ, φ, t) = ce ( iαt e iβt Y1 1 (θ, φ) + by1 (θ, φ) + e iβt Y1 1 (θ, φ) ) = 1 ( e iαt e iβt Y1 1 (θ, φ) + i ) Y1 (θ, φ) + e iβt Y1 1 (θ, φ) 1 Alternative (equal) expressions are P (± h) = 3/8 ± sin(βt)/ cos(βt)/8 and P () = 1/ + cos(βt)/. 5

6 where Yl m (θ, φ) are the spherical harmonic functions. Specifically 3 3 Y 1 ±1 (θ, φ) = 8π sin θe±iφ, Y1 (θ, φ) = π cos θ Alternatively we can project ψ(t) on the position eigenkets θ, φ, ψ(θ, φ, t) = θ, φ ψ(t), and use θ, φ l, m = Yl m (θ, φ). The probability per area of finding the particle with angular coordinates θ and φ at time t is ψ(θ, φ, t) which is normalized such that π π dθ sin θ dφ ψ(θ, φ) = 1. We are interested in the most likely value of θ, specific values of φ do not matter, thus we integrate over φ in order to obtain a probability distribution over θ alone P (θ, t) = sin θ π dφ ψ(θ, φ, t) which is normalized as π dθp (θ, t) = 1 The extra factor sin θ comes from the measure of the integral in polar coordinates. One can see that it must be included in the probability density of finding specific θ values from the following argument. If the particle is equally likely to be anywhere on the sphere, the most probable value of θ will θ = π/ simply because there is more area on the sphere which has θ = π/ (the equator) than any other values of the polar angle. Compare this for instance to θ = the north pole. The sin θ factor expresses this fact. P (θ, t) = sin θ π dφ ψ(θ, φ, t) = sin θ c π ( Y1 1 + b Y1 + Y1 1 ) = sin θ c 3 ( sin (θ) + b cos (θ) ) = sin θ c 3 ( (1 b ) sin (θ) + b ) = sin θ 3 ( sin (θ) ) 8 Where we have used the fact that the integral over φ is zero for products of spherical harmonics that differ in their m-value. Therefore all cross-terms vanish and with it the 6

7 time-dependence. to. We get To find the most likely value of θ we differentiate p(θ) and set it equal ( 3(1 b ) sin (θ) + b ) cos(θ) = which has solutions cos(θ) = and sin (θ) = b 3( b 1) = 3 Thus the extrema of the probability distribution is at θ = π/ and at sin θ = /3. The θ = π/ is a minimum while the other corresponds to a maximum. This can be seen by differentiating once more or by plotting the probability density p(θ). Two angles fulfill the..15 p(θ) θ maximum condition: θ = sin 1 ( /3) =.957 rad = 5.7 θ = π.957 rad =.1863 rad = These corresponds to a latitude of roughly 36 north and south of the equator (Gibraltar/Cape Town). 1.8 We use ˆL z = x ˆP y y ˆP x which gives x = [x ˆP y y ˆP x, ˆP [ x = x, ˆP x ˆPy = i h ˆP y y = [x ˆP y y ˆP x, ˆP [ y = y, ˆP y ˆPx = i h ˆP x These commutators imply [ˆL z, [ˆL z, ˆP x = [ˆL z, i h ˆP y = i h( i) h ˆP x = h ˆPx Alternatively we could first have computed ψ(θ, φ, t) = 3 8π (sin θ cos (βt φ) + cos θ) and then performed the integral over φ. From this expression it is clear that when time changes by t the effect on the state is to rotate it an angle β t about the z-axis. Therefore the probabilities of getting specific values of φ are time-dependent, while probabilities for values of θ are not. 7

8 Using this we find h l, m ˆP x lm = l m [ˆL z, [ˆL z, ˆP x l, m = l m ˆL ˆP z x ˆL z ˆPx ˆLz + ˆP x ˆL z l, m = h ( (m ) m m + m ) l m ˆP x l, m = h (m m) l m ˆP x l, m Assembling everything on one side of the equation we get h [(m m) 1 l m ˆP x lm = Which implies that l m ˆP x lm = whenever the parenthesis [ (m m) 1 which is equvivalent to the condition m m ± 1. q.e.d. We might as well have used the commutation relations to establish the double commutator [ˆL z, [ˆL z, ˆP y = [ˆL z, i h ˆP x = i hi h ˆP y = h ˆPy. Thus the same relation holds for l m ˆP y lm. For ˆP z we have z =. Thus = l, m z lm = (m m) l, m ˆP z lm This implies that l, m ˆP z lm = for m m. So this differs from the other components. Note here that although z =, the commutator [L, ˆP z. So although ˆP z and ˆL z have common eigenstates the state l, m is not an eigenstate of ˆPz. By considering the commutator with L one can also show that l m ˆP i lm = for l l ± 1. The relations shown here are very general. In fact they hold for all vector operators, not just the momentum operator treated here. 1.9 Using the defining commutation relations for the angular momentum operators, among them [ˆL y, ˆL z = iˆl x it becomes clear that the extra term can be written H e = γ h i (ˆLy ˆLz ˆL z ˆLy ) = γ ˆL x Thus the Hamiltonian now reads Ĥ = ᾱ h L + (γ,, β) L = ᾱ h L + γ + β n L 8

9 where n is the unit vector (γ,, β)/ γ + β This problem is equivalent to our initial problem if we redefine our z-axis to lie along n and identify γ + β with β. Thus the energy eigenvalues are E l,m = hαl(l + 1) + γ + β hm The physical explanation for the same form of the energy levels is that the extra term corresponds to a change of the magnetic field, it is tilted towards the x-direction and changed in magnitude. As there is nothing special about which direction the magnetic field points in, the form of the energy levels must remain the same, regardless of the direction of the field, and we need only to correct for the changed magnitude. This is achieved by letting β β + γ. 9

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