5.2: The Deuteron
( \newcommand{\kernel}{\mathrm{null}\,}\)
Reduced Hamiltonian in the center-of-mass frame
We start with the simplest problem, a nucleus formed by just one neutron and one proton: the deuteron. We will at first neglect the spins of these two particles and solve the energy eigenvalue problem (time-independent Schrödinger equation) for a bound p-n system. The Hamiltonian is then given by the kinetic energy of the proton and the neutron and by their mutual interaction.
H=12mnˆp2n+12mpˆp2p+Vnuc(|xp−xn|)
Here we stated that the interaction depends only on the distance between the two particles (and not for example the angle...)
We could try to solve the Schrödinger equation for the wavefunction Ψ=Ψ(→xp,→xn,t). This is a wavefunction that treats the two particles as fundamentally independent (that is, described by independent variables). However, since the two particles are interacting, it might be better to consider them as one single system. Then we can use a different type of variables (position and momentum).
We can make the transformation from {→xp,→xn}→{→R,→r} where →R describes the average position of the two particles (i.e. the position of the total system, to be accurately defined) and →r describes the relative position of one particle wrt the other:
{→R=mp→xp+mn→xnmp+mn center of mass →r=→xp−→xn relative position
We can also invert these equations and define →xp=xp(→R,→r) and →xn=xn(→R,→r). Also, we can define the center of mass momentum and relative momentum (and velocity):
{→pcm=→pp+→pn→pr=(mn→pp−mp→pn)/M
Then the (classical) Hamiltonian, using these variables, reads
H=12Mp2cm+12μp2r+Vnuc(|r|)
where M=mp+mn and μ=mpmnmp+mn is the reduced mass. Now we can just write the quantum version of this classical Hamiltonian, using
ˆpcm=−iℏ∂∂→Rˆpr=−iℏ∂∂→r
in the equation
H=12Mˆp2cm+12μˆp2r+Vnuc(|ˆr|)
Now, since the variables r and R are independent (same as rp and rn) they commute. This is also true for pcm and r (and pr and R). Then, pcm commutes with the whole Hamiltonian, [ˆ→pcm,H]=0. This implies that ˆ→pcm is a constant of the motion. This is also true for Ecm=12Mˆ→p2cm, the energy of the center of mass. If we solve the problem in the center-of-mass frame, then we can set Ecm=0 and this is not ever going to change. In general, it means that we can ignore the first term in the Hamiltonian and just solve
HD=−ℏ22μ∇2r+Vnuc(|→r|)
In practice, this corresponds to having applied separation of variables to the original total Schrödinger equation. The Hamiltonian HD (the deuteron Hamiltonian) is now the Hamiltonian of a single-particle system, describing the motion of a reduced mass particle in a central potential (a potential that only depends on the distance from the origin). This motion is the motion of a neutron and a proton relative to each other. In order to proceed further we need to know the shape of the central potential.
Ground state
What are the most important characteristics of the nuclear potential? It is known to be very strong and short range. These are the only characteristics that are of interest now; also, if we limit ourselves to these characteristics and build a simple, fictitious potential based on those, we can hope to be able to solve exactly the problem.
If we looked at a more complex, albeit more realistic, potential, then most probably we cannot find an exact solution and would have to simplify the problem. Thus, we just take a very simple potential, a nuclear square well of range R0≈2.1fm and of depth −V0=−35 MeV.
We need to write the Hamiltonian in spherical coordinates (for the reduced variables). The kinetic energy term is given by:
−ℏ22μ∇2r=−ℏ22μ1r2∂∂r(r2∂∂r)−ℏ22μr2[1sinϑ∂∂ϑ(sinϑ∂∂ϑ)+1sin2ϑ∂2∂φ2]=−ℏ22μ1r2∂∂r(r2∂∂r)+ˆL22μr2
where we used the angular momentum operator (for the reduced particle) ˆL2.
The Schrödinger equation then reads
[−ℏ22μ1r2∂∂r(r2∂∂r)+ˆL22μr2+Vnuc(r)]Ψn,l,m(r,ϑ,φ)=EnΨn,l,m(r,ϑ,φ)
We can now also check that [ˆL2,H]=0. Then ˆL2 is a constant of the motion and it has common eigenfunctions with the Hamiltonian.
We have already solved the eigenvalue problem for the angular momentum. We know that solutions are the spherical harmonics Yml(ϑ,φ):
ˆL2Yml(ϑ,φ)=ℏ2l(l+1)Yml(ϑ,φ)
Then we can solve the Hamiltonian above with the separation of variables methods, or more simply look for a solution Ψn,l,m=ψn,l(r)Yml(ϑ,φ):
−ℏ22μ1r2∂∂r(r2∂ψn,l(r)∂r)Yml(ϑ,φ)+ψn,l(r)ˆL2[Yml(ϑ,φ)]2μr2=[En−Vnuc(r)]ψn,l(r)Yml(ϑ,φ)
and then we can eliminate Yml to obtain:
−ℏ22μ1r2ddr(r2dψn,l(r)dr)+[Vnuc(r)+ℏ2l(l+1)2μr2]ψn,l(r)=Enψn,l(r)
Now we write ψn,l(r)=un,l(r)/r. Then the radial part of the Schrödinger equation becomes
−ℏ22μd2udr2+[Vnuc(r)+ℏ22μl(l+1)r2]u(r)=Eu(r)
with boundary conditions
unl(0)=0→ψ(0) is finite unl(∞)=0→ bound state
This equation is just a 1D Schrödinger equation in which the potential V(r) is replaced by an effective potential
Veff(r)=Vnuc(r)+ℏ2l(l+1)2μr2
that presents the addition of a centrifugal potential (that causes an outward force).
Notice that if l is large, the centrifugal potential is higher. The ground state is then found for l = 0. In that case there is no centrifugal potential and we only have a square well potential (that we already solved).
[−ℏ22μ1r∂2∂r+Vnuc(r)]u0(r)=E0u0(r)
This gives the eigenfunctions
u(r)=Asin(kr)+Bcos(kr),0<r<R0
and
u(r)=Ce−κr+Deκr,r>R0
The allowed eigenfunctions (as determined by the boundary conditions) have eigenvalues found from the odd-parity solutions to the equation
−κ=kcot(kR0)
with
k2=2μℏ2(E0+V0)κ2=−2μℏ2E0
(with E0<0).
Recall that we found that there was a minimum well depth and range in order to have a bound state. To satisfy the continuity condition at r=R0 we need λ/4≤R0 or kR0≥142π=π2. Then R0≥π2k.
In order to find a bound state, we need the potential energy to be higher than the kinetic energy V0>Ekin. If we know R0 we can use k≥π2R0 to find
V0>ℏ2π22μ4R20=π28ℏ2c2μc2R20=π28(191MeVfm)2469MeV(2.1fm)2=23.1MeV
We thus find that indeed a bound state is possible, but the binding energy E0=Ekin−V0 is quite small. Solving numerically the trascendental equation for E0 we find that
E0=−2.2MeV
Notice that in our procedure we started from a model of the potential that includes the range R0 and the strength V0 in order to find the ground state energy (or binding energy). Experimentally instead we have to perform the inverse process. From scattering experiments it is possible to determine the binding energy (such that the neutron and proton get separated) and from that, based on our theoretical model, a value of V0 can be inferred.
Deuteron excited state
Are bound excited states for the deuteron possible?
Consider first l = 0. We saw that the binding energy for the ground state was already small. The next odd solution would have k=3π2R0=3k0. Then the kinetic energy is 9 times the ground state kinetic energy or E1kin=9E0kin=9×32.8MeV=295.2MeV.. The total energy thus becomes positive, the indication that the state is no longer bound (in fact, we then have no longer a discrete set of solutions, but a continuum of solutions).
Consider then l > 0. In this case the potential is increased by an amount ℏ2l(l+1)2μR20≥18.75MeV (for l = 1). The potential thus becomes shallower (and narrower). Thus also in this case the state is no longer bound. The deuteron has only one bound state.
Spin dependence of nuclear force
Until now we neglected the fact that both neutron and proton possess a spin. The question remains how the spin influences the interaction between the two particles.
The total angular momentum for the deuteron (or in general for a nucleus) is usually denoted by I. Here it is given by
ˆ→I=ˆ→L+ˆ→Sp+ˆ→Sn
For the bound deuteron state l = 0 and ˆ→I=ˆ→Sp+ˆ→Sn=ˆ→S. A priori we can have ˆ→S=0 or 1 (recall the rules for addition of angular momentum, here ˆ→Sp,n=12).
There are experimental signatures that the nuclear force depends on the spin. In fact the deuteron is only found with ˆ→S=1 (meaning that this configuration has a lower energy).
The simplest form that a spin-dependent potential could assume is Vspin∝ˆ→Sp⋅ˆ→Sn (since we want the potential to be a scalar). The coefficient of proportionality V1(r)/ℏ2 can have a spatial dependence. Then, we guess the form for the spin-dependent potential to be Vspin=V1(r)/ℏ2ˆ→Sp⋅ˆ→Sn. What is the potential for the two possible configurations of the neutron and proton spins?
The configuration are either ˆ→S=1 or ˆ→S=0. Let us write ˆ→S2=ℏS(S+1) in terms of the two spins:
ˆ→S2=ˆ→S2p+ˆ→S2n+2ˆ→Sp⋅ˆ→Sn
The last term is the one we are looking for:
ˆ→Sp⋅ˆ→Sn=12(ˆ→S2−ˆ→S2p−ˆ→S2n)
Because ˆS2 and ˆ→S2p,ˆ→S2n commute, we can write an equation for the expectation values wrt eigenfunctions of these operators10:
⟨ˆ→Sp⋅ˆ→Sn⟩=⟨S,Sp,Sn,Sz|ˆ→Sp⋅ˆ→Sn|S,Sp,Sn,Sz⟩=ℏ22(S(S+1)−Sp(Sp+1)−Sn(Sn+1))
since Sp,n=12, we obtain
⟨ˆ→Sp⋅ˆ→Sn⟩=ℏ22(S(S+1)−32)=+ℏ24 Triplet State, |S=1,1212,mz⟩−3ℏ24 singlet State, |S=0,12,12,0⟩
If V1(r) is an attractive potential (< 0), the total potential is Vnuc|S=1=VT=V0+14V1 for a triplet state, while its strength is reduced to Vnuc|S=0=VS=V0−34V1 for a singlet state. How large is V1?
We can compute V0 and V1 from knowing the binding energy of the triplet state and the energy of the unbound virtual state of the singlet (since this is very close to zero, it can still be obtained experimentally). We have ET=−2.2MeV (as before, since this is the experimental data) and ES=77keV. Solving the eigenvalue problem for a square well, knowing the binding energy ET and setting ES ≈ 0, we obtain VT = −35MeV and VS = −25MeV (Notice that of course VT is equal to the value we had previously set for the deuteron potential in order to find the correct binding energy of 2.2MeV, we just –wrongly– neglected the spin earlier on). From these values by solving a system of two equations in two variables:
{V0+14V1=VTV0−34V1=VS
we obtain V0 = −32.5MeV V1 = −10MeV. Thus the spin-dependent part of the potential is weaker, but not negligible.
10 Note that of course we use the coupled representation since the properties of the deuteron, and of its spin-dependent energy, are set by the common state of proton and neutron