A.39 Deuteron wave function

This addendum examines the form of the wave function of the deuteron. It assumes that the deuteron can be described as a two particle system; a proton and a neutron. In reality, both the proton and the neutron consist of three quarks. So the deuteron is really a six particle system. That will be ignored here.

Then the deuteron wave function is a function of the positions and spin angular momenta of the proton and neutron. That however can be simplified considerably. First of all, it helps if the center of gravity of the deuteron is taken as origin of the coordinate system. In that coordinate system, the individual positions of proton and neutron are no longer important. The only quantity that is important is the position vector going from neutron to proton, {A.5}:

\begin{displaymath}
{\skew0\vec r}\equiv {\skew0\vec r}_{\rm {p}}-{\skew0\vec r}_{\rm {n}}
\end{displaymath}

That represents the relative position of the proton relative to the neutron.

Consider now the spin angular momenta of proton and neutron. The two have spin angular momenta of the same magnitude. The corresponding quantum number, called the spin for short, equals $s_{\rm {p}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_{\rm {n}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12$. However, the proton and neutron can still have different spin angular momentum along whatever is chosen to be the $z$-​axis. In particular, each can have a spin $S_z$ along the $z$-​axis that is either $\frac12\hbar$ or $-\frac12\hbar$.

All together it means that the deuteron wave function depends nontrivially on both the nucleon spacing and the spin components in the $z$-​direction:

\begin{displaymath}
\psi = \psi\left({\skew0\vec r},S_{z,\rm p},S_{z,\rm n}\right)
\end{displaymath}

The square magnitude of this wave function gives the probability density to find the nucleons at a given spacing ${\skew0\vec r}$ and with given spin values along the $z$-​axis.

It is solidly established by experiments that the wave function of the deuteron has net nuclear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and even parity. The question to be examined now is what that means for the orbital angular momentum and the spins of the proton and neutron. To answer that, the wave function needs to be written in terms of states that have definite combined orbital angular momentum and definite combined spin.

The conditions for a state to have definite orbital angular momentum were discussed in chapter 4.2. The angular dependence of the state must be given by a spherical harmonic $Y_l^m(\theta,\phi)$. Here $\theta$ and $\phi$ are the angles that the vector ${\skew0\vec r}$ makes with the axes of the chosen spherical coordinate system. The azimuthal quantum number $l$ describes the magnitude of the orbital angular momentum. In particular, the magnitude of the orbital momentum is $\sqrt{l(l+1)}\hbar$. The magnetic quantum number $m$ describes the component of the orbital angular momentum along the chosen $z$-​axis. In particular, that component equals $m\hbar$. Both $l$ $\raisebox{-.5pt}{$\geqslant$}$ 0 and $\vert m\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ must be integers.

As far as the combined spin angular momentum is concerned, the possibilities were discussed in chapter 5.5.6 and in more detail in chapter 12. First, the proton and neutron spins can cancel each other perfectly, producing a state of zero net spin. This state is called the singlet state. Zero net spin has a corresponding quantum number $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. And since the component of the angular momentum along any chosen $z$-​axis can only be zero, so is the spin magnetic quantum number $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

The second possibility is that the proton and neutron align their spins in parallel, crudely speaking. More precisely, the combined spin has a magnitude given by quantum number $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12+\frac12$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The combined spin angular momentum along the chosen $z$ direction is $m_s\hbar$ where $m_s$ can be $\vphantom0\raisebox{1.5pt}{$-$}$1, 0, or 1.

The wave function of the deuteron can be written as a combination of the above states of orbital and spin angular momentum. It then takes the generic form:

\begin{displaymath}
\psi = \sum_{nlmsm_s}
c_{nlmsm_s} R_n(\vert{\skew0\vec r}\vert) Y_l^m(\theta,\phi) \big\vert s\,m_s\big\rangle %
\end{displaymath} (A.257)

Here the $c_{nlmsm_s}$ are constants. The functions $R_n$ are not of particular interest here; any complete set of orthonormal radial functions will do. Note that the individual terms in the sum above are not supposed to be energy eigenfunctions. They are merely chosen states of definite orbital and spin angular momentum. The ket $\big\vert s\,m_s\big\rangle $ is a way of indicating the combined spin state of the two nucleons. It is defined in terms of the separate spins of the proton and neutron in chapter 5.5.6 (5.26).

The above expression for the wave function is quite generally valid for a system of two fermions. But it can be made much more specific based on the mentioned known properties of the deuteron.

The simplest is the fact that the parity of the deuteron is even. Spherical harmonics have odd parity if $l$ is odd, and even if $l$ is even, {D.14}. So there cannot be any odd values of $l$ in the sum above. In other words, the constants $c_{nlmsm_s}$ must be zero for odd $l$.

Physically, that means that the spatial wave function is symmetric with respect to replacing ${\skew0\vec r}$ by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. It may be noted that this spatial symmetry and the corresponding even parity are exactly what is expected theoretically. The reasons were explored earlier in {A.8} and {A.9}. The wave function for any given spin state should not change sign, and odd parity cannot meet that requirement. However, it should be noted that the arguments in {A.8} and {A.9} are not valid if the potential includes terms of second or higher order in the momentum. Some more advanced potentials that have been written down include such terms.

The spatial symmetry also means that the wave function is symmetric with respect to swapping the two nucleons. That is because ${\skew0\vec r}$ is the vector from neutron to proton, so swapping the two inverts the sign of ${\skew0\vec r}$. This does assume that the small difference in mass between the neutron and proton is ignored. Otherwise the swap would change the center of gravity. Recall that the (part of the) wave function considered here is relative to the center of gravity. In any case, the hypothetical wave functions for a bound state of two protons or one of two neutrons would be exactly symmetric under exchange of the positions of the two identical particles.

The condition that the nuclear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is a bit more complex. First a brief review is needed into how angular momenta combine in quantum mechanics. (For a more complete description, see chapter 12.) A state with definite quantum numbers $l$ and $s$ has in general quantum uncertainty in the net nuclear spin $j_{\rm {N}}$. But the values of $j_{\rm {N}}$ cannot be completely arbitrary. The only values that can have nonzero probability are in the range

\begin{displaymath}
\vert l-s\vert \mathrel{\raisebox{-.7pt}{$\leqslant$}}j_{\rm {N}} \mathrel{\raisebox{-.7pt}{$\leqslant$}}l+s
\end{displaymath}

The key is now that unless a state $l,s$ has a nonzero probability for $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, it cannot appear in the deuteron wave function at all. To verify that, take an inner product of the state with the representation (A.257) of the deuteron wave function. In the left hand side, you get zero because the deuteron wave function has $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and states of different $j_{\rm {N}}$ are orthogonal. In the right hand side, all terms except one drop out because the states in the sum are orthonormal. The one remaining term is the coefficient of the considered state. Then this coefficient must be zero since the left hand side is.

Using the above criterion, consider which states cannot appear in the deuteron wave function. First of all, states with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are according to the inequalities above states of nuclear spin $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$. That cannot be 1, since $l$ had to be even because of parity. So states with $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 cannot appear in the deuteron wave function. It follows that the deuteron wave function has a combined nucleon spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 without quantum uncertainty.

Secondly, states with $l$ $\raisebox{-.5pt}{$\geqslant$}$ 4 have $j_{\rm {N}}$ at least equal to 3 according to the above inequalities. So these states cannot appear. That leaves only states with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or 2 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 as possibilities.

Now states with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are states with $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Any such state can appear in the deuteron wave function. To what amount remains unknown. That would only be answerable if an exact solution to the proton-neutron deuteron would be available. But surely, based on arguments like those in {A.8} and {A.9}, it is to be expected that there is a significant $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 component.

States with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are also a possibility. But they cannot appear in arbitrary combinations. Any such state has multiple possible values for $j_{\rm {N}}$ in the range from 1 to 3. That uncertainty must be eliminated before the states are acceptable for the deuteron wave function. It turns out that pure $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 states can be obtained by taking specific combinations of states. In particular, groups of states that vary only in the quantum numbers $m$ and $m_s$ can be combined into states with $j_{\rm {N}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. (For the curious, the specific combinations needed can be read off in figure 12.6).

The bottom line is that the deuteron wave function can have uncertainty in the orbital angular momentum. In particular, both orbital angular momentum numbers $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 can and do have a nonzero probability.