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Next: Zero Energy n-p scattering Up: PHYS 560: Lectures During Previous: Scattering in a Central

Low-Energy S-Wave Scattering

Let us familiarize ourselves with these scattering amplitudes by examining low energy scattering in the S-wave. In this scenario we have only one partial wave contributing (obviously) with a phase shift $\delta_0$. The cross section for S-wave scattering is then given in terms of this one phase shift

    $\displaystyle \sigma = {4\pi\over k^2}\sin^2\delta_0
\ \ \ .$ (21)

It is clear that for a finite range potential, such as the strong interaction potential between nucleons, the cross section is finite for all energies, and hence the phase shift $\delta_0$ must vanish at least as fast as $k$ in order, so that $\sin\delta_0 / k$ is finite. Hence for small $k$ the phase shift will behave as $\delta_0\rightarrow - k a$, where $a$ is some constant with dimensions of length (to compensate the dimensions of mass of k). In this limit one can also replace the $\sin\delta_0$ with $\delta_0$, to yield a cross section in terms of $a$
    $\displaystyle \sigma = 4\pi a^2
\ \ .$ (22)

The constant $a$ is called the scattering length.

In the limit of $k\rightarrow 0$ as we have been considering above, the solution to the schrodinger equation in the region where the potential vanishes

    $\displaystyle -{1\over 2\mu}\nabla^2 = 0
\ \ ,$ (23)

is a straight line. We already know this as the solution for any $k$ is
    $\displaystyle u_0 \rightarrow {\sin (kr+\delta_0)\over k}$  
    $\displaystyle \rightarrow r-a
\ \ ,$ (24)

in the low momentum regime. We see that the scattering length $a$ is the radius at which the asymptotic (straight line) solution vanishes. This is, of course, a limit of the incident oscillatory waves.

One of the things we should learn from this is that all we see in the low energy (compared to the inverse radius of the potential) is the asymptotic behaviour of the wavefunction characterized by one constant, the scattering length. This is a nontrivial construction from the potential itself and the wavefunction of the state. We see that for momenta much less than the inverse radius of the potential the scattering length is sufficient to describe all of the interactions. This goes back to our original ideas about short distance physics being encoded in the constants of the low energy theory. It is clear that by measuring the scattering length of a system alone we cannot reconstruct the potential uniquely. There are infinitely many different shapes, depths and ranges of potentials that will reproduce a single scattering length.

Let us consider the system I have shown in the

Figure: The wavefunctions at low momentum for the potential in eq. (25).

which corresponds to a finite square well
    $\displaystyle \begin{array}{cll}
V(r) & = -3 & \ \ {\rm for}\ \ r < 1 \\
& = 0 &\ \ {\rm for}\ \ r > 1
\end{array}\ \ \ ,$ (25)

in which a system of reduced mass $\mu=1/2$ moves. The oscillatory curves correspond to the wavefunctions for particles incident upon the potential of energy $E=0.1 = k^2$ for $\mu=1/2$. One is for the system with the potential given above while the other is that for a vanishing potential. I have also shown the $E=0$ wavefunction for the potential. We see that the oscillatory wavefunction intercept with the x-axis is tending to the scattering length $r=a$ defined by the $E=0$ intercept, shown by the straight line for $r>1$. The phase shift (divided by $k$) corresponds to the difference between the intercepts with the x-axis, i.e. the difference between $a\sim 4$ and $r_1\sim 10$. This would say that the scattering amplitude is given by
    $\displaystyle {\rm amp} \sim {\sin\delta\over k}$  
    $\displaystyle = {\sin (k(r_1-a))\over k}$  
    $\displaystyle = {\sin (\pi-k(r_1-a))\over k}$  
    $\displaystyle = {\sin (ka)\over k}$  
    $\displaystyle \rightarrow a
\ \ ,$ (26)

where we used the fact that $kr_1=\pi$ by construction.

It is important to notice that for $a>0$ the wavefunction inside this potential must have just turned over to have a negative derivative at the boundary. This would indicate that a bound state exists for this potential. In order for us to arrive at this conclusion we see that the existence of a bound state requires that the wavefunction at the boundary have a negative derivative so that we can match onto an exponentially falling wavefunction outside the potential. Such a fall off is characteristic of a state with negative energy $E<0$, and is localized about the centre of the potential. We will return to this later on. Also, we note that if the scattering length were negative $a<0$, then the wavefunction would have a positive derivative at the boundary of the potential. This indicates that there is not a loosely bound ($E<\sim 0$) state for this system.

So far we have considered $k=0$ scattering and found that a single parameter can be identified, the scattering length, and we have discussed what this means in terms of wavefunctions and cross sections. However, when we move away from $k=0$ scattering and ask about the cross section for non-zero momenta but still much smaller than the inverse radius, then we expect to be able to perform a taylor expansion on the scattering amplitude, or some quantity related to the scattering amplitude. We have observed in our previous discussions that we can "encode" (in some complicated way) the potential and wavefunction in this region by the phase shift of the eigenstates of the potential. We expect that the dependence on small momenta (compared with the potential range) can also be encoded by this phase shift. We therefore wish to form a derivative expansion of the phase shift, or some function of the phase shift, if it is more appropriate. Let $\chi(r)$ be the asymptotic form of $u_0 (r)$ as $r\rightarrow\infty$, where the radial wavefunction is $\psi = u/r$. We have already seen that $\chi (r) = A \sin (kr+\delta)$ and we can form the quantity

    $\displaystyle {1\over \chi (0)} \left[ {\partial\chi (r)\over\partial r}
\right]_{r=0} =
\ \ \ .$ (27)

We choose to perform the taylor expansion of this object in $k^2$. It is quadratic in $k$ because we are perturbing the solution to the schrodinger equation about $k=0$, and $k$ appears quadratically. We define the effective range $r_{\rm eff}$ by
    $\displaystyle k\cot\delta = -{1\over a} + {1\over 2} r_{\rm eff} k^2 + ....
\ \ \ ,$ (28)

where the $k\rightarrow 0$ limit coincides with our previous definition of the scattering length.

Before we proceed, lets look at this object a slightly different way. Starting with the scattering amplitude $f(\Omega)$ for an S-wave interaction

    $\displaystyle f(\Omega) \ = \
{1\over k} e^{i\delta}\sin\delta
\ =\ {1\over k} {\sin\delta\over\cos\delta-i\sin\delta}$  
    $\displaystyle \qquad\ =\ {1\over k\cot\delta - i k}
\ \ \ .$ (29)

Therefore, if we understand the object $k\cot\delta$, then we understand the scattering amplitude, and hence the cross section. Many years ago, Schwinger and others examined the properties of $k\cot\delta$. For a finite range potential, range $\sim R$, the quantity $k\cot\delta$ has a power series expansion about $k~=~0$ that converges for $k < 1/R$. We write
    $\displaystyle k\cot\delta \ =\ -{1\over a} + {1\over 2} r_0 k^2
\ +\ r_1 k^4\ +\ ...$ (30)

It is easy to see that this correctly reproduces the small $k$ behaviour of $\delta$. In fact one can show, as may already be obvious from the figure that the scatterinbg length is not bounded by the range of the potential, but the $r_i$ are set by the range of the potential, i.e.
    $\displaystyle -\infty < a < \infty
\ \ \ ,\ \ \ r_i\sim R
\ \ \ .$ (31)

We will spend some time discussing where the following numbers come from, but it is found that from neutron proton scattering
    $\displaystyle a^{({}^1\kern-.14em S_0)}\ =\ -23.714\pm 0.013\ {\rm fm}\ \ ,\ \ r^{({}^1\kern-.14em S_0)}\ =\ 2.73\pm 0.03 {\rm fm}$  
    $\displaystyle a^{({}^3\kern-.14em S_1)}\ =\ 5.425\pm 0.0014\ {\rm fm}\ \ ,\ \ r^{({}^3\kern-.14em S_1)}\ =\ 1.749\pm 0.008 {\rm fm}$ (32)

An interesting feature of $k\cot\delta$, is that when the entire denominator of $f(\Omega)$vanishes, the scattering amplitude becomes infinite, corresponding to the existance of a bound state at that location. Therefore, solving $k\cot\delta-ik~=~0$ determines the location in $k$ space of bound states in either channel. In the ${}^3\kern-.14em S_1$ channel, this corresponds to the deuteron bound state with a binding energy of $B=\sim 2.2\ {\rm MeV}$. In the ${}^1\kern-.14em S_0$ channel the pole has positive energy corresponding to a resonance in the scattering amplitude. In fact, the location of the deuteron pole is one of the inputs into the precise determination of $a^{({}^3\kern-.14em S_1)}$ and $r_0^{({}^3\kern-.14em S_1)}$.

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Next: Zero Energy n-p scattering Up: PHYS 560: Lectures During Previous: Scattering in a Central
Martin Savage