Lecture 4 Povh Krane Enge Williams

Transcription

1 Lectue 4 Povh Kane Enge Williams the Deuteon 6. Ch. 4 Ch. Ch 3 d-wave admixtue tenso foce missing S state isospin Poblems on Lectue 4 What is the minimum photon enegy equied to dissociate the deuteon? Take the binding enegy to be.4589 MeV. If the binding enegy of the deuteon was 0 MeV instead of.3 MeV, what would be the appoximate depth of the assumed squae-well potential? Estimate the value of d. 3 Assume the deuteon wave function in the egion of the well, is given by u( ) ii = Asin K( c) whee K = m( Eb V0) h and by u( ) iii = Be h -K ' whee K' = me in egion iii. b What faction of the time do the poton and neuton spend outside the ange of thei foces?

2 Lectue 3 Review Semi-empiical mass fomula Undestand the oigin of the tems Do a calculation to find the most stable isoba Othe popeties a Angula momentum and paity I p of GS and excited states Fo simple nuclei I is esult of j of component nucleons Fo odd A I is half integal (one unpaied nucleon) Fo even A I is integal b Magnetic dipole moment of nucleus µ = glµ N whee µ N = eh m is nucleon mass and l is the AM QNo. m NDM due to spin µ = g s sµ N significance of g s fo poton and neuton egading substuctue of nucleon c Nuclea shape (actually coveed in lectue 4) Quadopole moment (measue of defomation of nucleus) Q = / e ( 3z - ) dv ] and has units of aea usually in ban (0-8 m ). Q +ve polate (football shape) Q ve oblate (cushion)

3 Lectue 4 I think we have spent enough time discussing the geneal popeties of nuclei to give us the basic infomation we need to discuss the details of the foce between nucleons. Fistly let s summaise what we know about the nuclea foce: Some popeties of the Nuclea Foce obsevation Constant nucleon density Few odd-odd nuclei S n fo Ca isotopes implication Shot-ange foce with a epulsive coe. paiing stability paiing enegy ~3 MeV The actual popeties of the nuclea potential ae quite complicated. We will find it depends not only of, but on the spin s of the nucleons, and thei elative AM l. We can find out about the potential in two main ways. One is scatteing of nucleons of each othe o fom nuclei. (This will be discussed a little late). The othe is to study any stable states that might esult fom this inteaction. Thee ae of couse 3 possibilities: pn, pp, nn. It tuns out that only one of these foms a stable nucleus. We will discuss why, and what this means. In this instance we will study the nucleus H, the deuteon, which is the nucleus fomed when a poton and neuton combine. It is the simplest nucleus and in a sense is the nuclea equivalent of the H atom in atomic studies. Unfotunately, unlike the H atom, it does not have any excited states, so thee is no infomation available fom its spectoscopy. None the less the fact that it is bound at all eveals a lot about the nuclea foce. (of couse if it wee not bound, we wouldn t be hee to study nuclea physics, since this nucleus is the basis fo fusion to all the heavie nuclei in the univese) Lets summaise what is known about the deuteon: Binding Enegy.5 MeV Check this fom tables of masses How do I know? Themal neuton captue H(γ,n)p theshold No e vidence of excited states Spin I= Magnetic dipole moment µ d = n m (µ p = n m) (µ n = n m) Quadupole moment Q = ban (expect zeo) Radius R d =. fm ( fm = 0-5 m) 3

4 Solution We need to solve the Schodinge wave equation fo a potential with the chaacteistics we mentioned above. An attactive shot-ange foce between the p and n, and a epulsive coe. Model potential 0 E - E b Phenomenological potential -V o c b Now the fom of V may be V(θϕ), but we will assume that it is only a function of. We should also conside the possibility that the AM of the n el to p is not necessaily zeo, in which case we would need to include a centifugal component in V, which looks like. In fact since we know that thee ae no excited states we will assume that l=0 V = l( l + ) h m Note that this is the potential that the poton feels due to the n, o vice vesa. The solution of this -body poblem can be esolved down to the solution of a single paticle moving in this potential, using the adial scale as the sepaation of one fom the othe, if we wok in the CM system and eplace the mass of the paticle with the educed mass of the two. Thus we ae to solve mpmn m = = m m + m whee, y ( qf ) = u / Y m ( qf ) l p n - h m l n y ( qj ) = ( E -V ) y ( qf ) Howeve if we assume no angula dependence Y lm (/4π) /, and we can solve the -D WE fo u(). 4

5 fo <c u() = 0 since V = fo c< <c+b u( ) ii = A sin K ( - c ) whee K = m( Eb -V0 ) h fo > c+b u( ) iii = Be - K' wheek' = h me b The complete continuous solution equies continuity of both = c+b Asin Kb = Be and AK cos Kb = - K' Be dividingthese gives K cot Kb = -K' K'( c+ b ) -K'(c+ b ) u( du( ) )and d at K depends on V and E b and K depends on E b V o = 73 MeV b =.337 fm Fo C = 0.4 fm This tanscendental equation gives a elationship between b and V that is consistent with E b. That is we can a ange of values of V and b that will give the coect BE fo the deuteon. We need moe infomation to define eithe the depth V o the width b. This can be got fom the measued adius of the deuteon, and on the vugaph I have witten out the calculation of ms adius of the deuteon V o MeV K cotkb = -K 4 6 b femi Consistent with measued adius 8 5

6 fom ou solution. < > / <d > / < p > / We can calculate the deuteon adius fom ou solution. st need to find A and B by nomalizing j d t = (see Enge page 4) neuton CM poton K' This gives A = + + K'b K'(sin Kb)e B = + K'b K'(c + b ) R d is the measued adius of deuteon =. fm R d = < d > + < p > Whee < d > / = ½< > / This gives anothe elation between E b and b, which leads to a well depth of 73 MeV. Rd < p > / = 0.8 fm < d > = j dt 0 = 0 u( ) whee j = Ylm u( ) = 4p u d since dt = 4p d ( c + b)( + K'b) c K < d >= K' 8K K' 4 4( Known, so this povides anothe elation between Vo and b C = 0.4 fm Vo = 73 MeV B =.337 fm Note that in ode to solve the equation we need to set a value fo the width of the epulsive coe. The value chosen is c = 0.4 fm. So that the deuteon is bound with the coect BE if c = 0.4 fm V o = 73 MeV and the width of the well is b =.337 fm Howeve you should note that the nuclea potential is not a squae well, and that this calculation is only a fist appoximation to the eal situation. A vey impotant chaacteistic of the nuclea foce is evident fom the deuteon. Fistly thee is only one bound state of the deuteon. It has I=. We assumed that l=0 so it must be that the n and p have thei spins aligned, giving S=. This is known as the tiplet S state (thee ae 3 pojections of S=). We assumed that the n and p had a elative AM =0 The AM of the deuteon is I = + s I = = 0 s = o I = 0 = 0 s = 0 Tiplet s Singlet s But if the nuclea foce depended only on the spatial co-odinate, it should make no diffeence if the spins wee paallel o anti-paallel. Thus we would expect at the same enegy an S=0 state (ie the GS should be degeneate). Thee is no S=0 state (singlet S) Conclusion: The nuclea foce is stonge fo paallel spins 6

7 This must mean that the nuclea potential is weake fo two nucleons aligned anti-paallel that aligned paallel. O in othe wods the potential well is less deep fo anti-paallel nucleons. How much shallowe? On the basis of this model, it can be shown that the minimum depth of the well fo the given value of b, that will just bind the p and n is about 60 MeV. So it seems that thee is a significant diffeence in the nuclea potential. Fom scatteing studies, which we will look at shotly we know that the singlet state is only just unbound (by about 60 kev). We should emembe this infomation when we discuss the vaious tems that contibute to the nuclea potential. Thee is anothe woy about ou assumption (calculation) that the deuteon wavefunction is l=0. It comes fom examination of the magnetic dipole moment of the deuteon. Because the l =0 wave function is spheically symmetic, the poton cannot contibute to the mag. moment by vitue of its obital motion (l =0 in semi classical tems means a degeneate ellipse). So the MM of the deuteon should be simply the sum of the intinsic MM of the poton and neuton (this is due to thei spin). Magnetic dipole moment of deuteon mag. mom. (nuclea magnetons) Poton.7975 neuton deuteon expected deuteon obseved discepancy (~.5%) If = 0 the poton cannot poduce a magnetic field The = 0 wave function is spheically symmetic (o the = 0 semi-classical obit is a staight line) Conclusion: The wave function must have a component with >0 the p and n togethe invokes, as we shall see, the exchange of π mesons. These indeed may have chage (π +, π - ) and contibute to the MM. Howeve the likely answe is that the WF is not puely l=0., thee is pobably some ADMIXTURE of l =. (a poton in an l = obit This discepancy is significantly geate than the eos in measuement. Thee may be seveal easons fo this discepancy, including the possibility that the poton may be affected by the close poximity of the neuton within the nucleus of deuteium. In fact of binding Y = a s Y s + a = 0 = Why no =? paity is p =(-) d Y All the tems in y must have same paity y has = and hence negative p, y s and y d have positive p. 4% = contibution d 7

8 of couse can be seen as a cuent sweeping out an aea, and poducing a MM). In which case we should see the WF of the deuteon as: ψ = a s ψ s + a d ψ d fo deuteon I =, Y = a Y s s l= + a d Y d S = But why have we chosen the admixtue as l=; a d-state impuity? What about l=; a p-state impuity? Well, the paity of the GS WF is even, and the paity of an l= WF is neg (Paity = (-) π.) So in ode to maintain the paity of the WF the impuity has to be of even l. S = I = I = This means that the deuteon spends most if the time in an S state and a small faction of the time in a D state. This faction specifies (a d ). When it changes fom S to D the spin of one nucleon has to flip ove in ode to conseve AM. To change S fom to equies an extenal toque. So the nuclea foce has a tenso component, i.e. one that depends NOT just on. T = x F T = F q = - dv/dq F T is a function of S. and q, whee q is the angle between and S. So fo example it will be diffeent fo q = p/ o 0 (if S = 0 thee is no tenso foce) F T epulsive F T attactive The exact size of a d that gives the obseved MM is about 4%. Howeve thee is clea evidence that ou assumption that the deuteon is a pue S state is incoect. This comes fom the measued electic quadupole moment. Fo an S state, Q=0, since the wave function is spheically symmetic. Yet the measued Quadupole moment is Q = ban. So thee must be a WF component that is not spheically symmetic. Well if we allow the obital (but not the total since the spins will flip ove) AM to blithely change, thee must be a toque acting. T= x F = F θ = -dv/dθ. This implies that the potential involved must be a fn of θ,. But the cental potential is a fn of only. This component of the nuclea potential is a non-cental o tenso foce. Since the only diection in space that is elevant to the deuteon is the diection of the spin S, the θ on which this foce depends must be measued fom this diection. In vecto notation the foce must be a function of S.. It will be diffeent fo q = π / o 0. (epulsive) and 8

9 (attactive) Conside the analogue of two ba magnets paallel o in line. They have diffeent enegies and it equies a tenso foce to move them. When the p and n ae in an S=0 state thee is no tenso foce Why is thee no singlet state of the deuteon? Howeve it is wothwhile asking why thee is no S= (tiplet) bound state of the n-n o p-p. Does this imply that the nuclea foce is diffeent between n-n (the di-neuton) o p-p (the dipoton)? We might well imagine that since the deuteon is only just bound, that the coulomb epulsion of potons sepaated by a femi o so might destabilise the system and be unbound: but not neutal neutons. The shot answe is that the Pauli Pinciple fobids it.no two identical paticles can have the same quantum state (o the same set of quantum numbes). The long answe (which is eally the explanation of the Pauli pinciple) is that the total wave function fo identical paticles must be anti-symmetic. The wave function that we obtained fo the p-n system looked like: ψ ( space) = ul ( ) P lm ( θ ) e im l φ We made the assumption that l =0 (that is the wave function is s-state) so the WF became spheically symmetic and the angula pat became (/4π ) -/. The symmety of the WF is whethe it changes fom + to - on eflection though the oigin. Mathematically this symmety is elated to the paity which is (-) l. So this wavefunction is NOT anti-symmetic. This is because this is not the total WF. The spin WF is pat of it so that the moe complete WF is φ ψ = u l ( ) im Plm( θ ) e l χ( spin) χ(spin) is symmetic if the spins ae paallel (S= o tiplet state), and antisymmetic if they ae antipaallel (S=0 o singlet state). Now let s look again at the WF we deived fo the deuteon it has l=0 symmetic S= symmetic so it is NOT a valid WF. 9

10 Well thee is clealy something wong hee, since this is what a deuteon looks like: one p and one n in l=0 S=. If we wee dealing with electons (which ae also femions and confom to the same ules as nucleons) this would indeed be an invalid WF. We would have to flip one of the electons ove to make the WF antisymmetic. Fo electons this would be the total WF. Howeve fo nucleons we have not yet specified the WF completely. The total WF is: φ ψ = u l ( ) im Plm( θ ) e l χ ( spin) τ ( T) whee τ(t) is the isospin pat of the WF. This pat of the WF is symmetic if the two nucleons ae the same and antisymmetic if they ae diffeent (i.e. one p and one n). Now we can see that ou WF fo the GS of the deuteon is indeed valid. We can also see that unde these ules the WF fo a di-neuton (nn) o di-poton (pp) is NOT valid. Howeve we must take this a little bit deepe and fomalise this concept of isospin. In essence we need to think of the poton and neuton as diffeent states of the same paticle; the nucleon. To specify the diffeent states we could use coloued pens: ed fo potons and geen fo neutons, and then these colous become the QN of the p and n. In QM we assign a new quantum numbe (the isospin QN) to the nucleon. t= ½. This is in exact analogy to the intinsic spin s= ½, and just as the intinsic spin can have pojections s z = ±½, so can t. t z = + ½ coesponds to a neuton t z = - ½ coesponds to a poton (note this convention is the opposite to that used in paticle physics) t z=+/ neuton t z=-/ poton t s z=+/ s z=-/ s The isospin quantum state of a nucleon pai is the vecto sum of thei isospin. That is, the state can have T=0 o T=, ie a singlet o tiplet isospin state. 30

11 s=/ t=/ S T s=/ t=/ S= S=0 T= T=0 Just as the singlet S state (S=0) can have only pojection (S z =0) and the tiplet S state (S=) have 3 pojections (S z = -,0,+); so thee is only one state associated with the T=0 (singlet) isospin state T z =0. The T= (tiplet) state has 3 pojections T z = -,0,+. What does T z mean? T z = t z. So T z = 0 means a p and a n. T z = + means neutons T z = - means potons T z=+ dineuton T z=0 deuteon T= S z=+ S z=0 S= T z=- dipoton S z=- Now let s look again at the -nucleon wavefunctions possible fo the deuteon. 3

12 φ ψ = u l ( ) im Plm ( θ ) e l χ( spin) τ ( T) =0 S= T=0 deuteon symmetic symmetic anti-sym ANTI =0 S=0 T= (T z =0) deuteon symmetic anti-sym symmetic ANTI 3 =0 S=0 T= (T z =+) dineuton symmetic anti-sym symmetic ANTI 4 =0 S=0 T= (T z = -) dipoton symmetic anti-sym symmetic ANTI So we see that thee ae 4 possible combinations that ae allowed fo a -nucleon system.. is the bound state of the deuteon 3. is a legal state of the deuteon, but we found that this is unbound by about 60 kev. This led us to the impotant discovey that the nuclea cental foce depends on the spin oientation; the potential is shallowe fo anti-paallel spins. 3 and 4 ae legal states, but clealy if is unbound so ae they. 3