# Homework Help: Spherical well for deuteron

1. Apr 3, 2016

### Incand

1. The problem statement, all variables and given/known data
How much of the time are the proton and neutron in a deuteron outside the range of the strong force? Suppose the strong force can be described by a spherical potential with parameters
$V_0 = 35 MeV$, $R = 2.1fm$. The binding energy for deuteron is $E_b = 2.22 MeV$ and the ground state is mostly an $s$-state.

2. Relevant equations
If the radial part of the wave equation is $\Psi(r) = f(r)/r$ then the radial equation is
$\frac{d^2f}{dr^2}+\frac{2m}{\hbar^2}\left[E-V(r)-\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right]f = 0$.

3. The attempt at a solution
The well is
$V(r) = \begin{cases} -V_0, \; \; \; r<R\\ 0, \; \; \; \; \; \; r>R \end{cases}$
and we have a bound state so $E = -E_b$. As I understand the question to find how much the time the proton and neutron is outside the strong force I should find the probability of finding the nucleus outside the well.
For an $s$-state the radial equation becomes
$\frac{d^2f}{dr^2}+\frac{2m}{\hbar^2} \left(E-V(r)\right) = 0$.
The solutions become
$f_1(r) = A\sin k_1 r + B\cos k_1 r$ for $r<R$ and
$f_ 2(r) = Ce^{-k_2r}+De^{k_2r}$ for $r>R$.
$f(r)/r$ should be finite for all $r$. It follows that $B=D=0$ so we the solutions are
$f_1(r) = A\sin k_1 r$ for $r<R$ and
$f_ 2(r) = Ce^{-k_2r}$ for $r>R$
with $k_1 = \sqrt{\frac{2m(E+V_0)}{\hbar^2}}$ and $k_2 = \sqrt{-2mE/\hbar^2}$.

The boundary conditions at $r=R$ for $f$ and $f'_r$ give us
$\begin{cases}A\sin k_1 R = Ce^{-k_2 R}\\ Ak_1 \cos k_1 R = -Ck_2 e^{-k_2 R}\end{cases} \Longrightarrow k_1 \cot k_1 R = -k_2.$
The above values for $E$ and $V_0$ satisfies this condition.

From the first boundary equation it's possible to calculate $\frac{C}{A}=\frac{\sin k_2 R}{e^{-k_2R}} \approx 1.4645$ (I used the mass of the deuteron $m=2.014u$ and converted everything to SI units). If I could norm the function it would be possible to find the probability. The norming condition becomes
$1 = 4\pi A^2 \left[ \int_0^R r^2 \sin^2k_1 r dr + 1.4645^2 \int_R^\infty r^2e^{-2k_2 r} dr \right]$ however the values seem to be so small at this point that numerical integration give zero for the second integral. Am I going at this the right way?
Perhaps it would be simpler to write this is terms of normed eigenfunctions first but while this is simple for $\sin k_1 r$ I still get the same problem for the exponential function.

2. Apr 3, 2016

### drvrm

i just like to know this info will give us some physical info/parameter or not?
the iind issue is how you will know its a p or n -as half of the time its a p and other half time the same one is a n-
being two states of the same particle.
third- when we were doing the real deutron calculation -we took integration up to say 8F so nuclear forces have a tail....its a very simplistic picture of a well of radius 2-3 F.

3. Apr 3, 2016

### Incand

I'm not entirely sure I understand what you are asking. I'm guessing this exercise is mostly for .. exercise and may not be that accurate. The book I'm using does approximate deuterium as a square well and it seems to give a decent approximation however the probability in the exercise is something only referred to in the exercise.

As I interpreted the question (I could be wrong) I'm looking for the probability the entire core is outside the well so I only see deuteron as a fundamental particle in the exercise I guess. It's not true that the deuteron is solely in an $s$-state either, we showed in class it's also in a $d$-state something like $4\%$ of the time.

4. Apr 3, 2016

### drvrm

now i understand that you are looking for a particle 'deuteron outside the potential well - in that case one will have to take a energy of the deuteron E <0 ,which may not be the E-binding energy- and problem reduces to finding out the probability of finding it outside the potential well so it reduces to normal well calculation and transmission coefficient of this 'particle'- one can proceed on that line taking current of transmission and reflection from the wall and applying conservation theorem like jincident= jrefl. +jtransmitted - and the probabilty of foinding it outside gives you percentage of time its outside-like alpha particle tunneling.(if i am right in picturising)

5. Apr 3, 2016

### Incand

Well I'm not entirely sure that's what I'm looking for it's just my guess at what the exercise meant. I'm guessing this shouldn't be to complicated since it's an introductory question.

I think from what you're saying I understand what I did wrong. I should use the positive energy $E$ and then calculate the transmission coefficients.
$k_1 = \frac{2m(V_0+E)}{\hbar^2} \approx 1.8925e+15$ and $k_2 =\sqrt{2mE/\hbar^2} = 4.6220e+14$.
The transmission coefficient is then
$T = \frac{4k_1k_2}{(k_1+k_2)^2} \approx 0.63$. Is this what you meant?

The answer says $62\%$ which is close enough I guess

6. Apr 3, 2016

### drvrm

well i have not checked the math but i hope the picture is now clear just like tunnelling of the well by a deuteron.
however one could use the regular deuteron wavefunction with normalized amplitudes but surprisingly for a good part of time its found outside- that is perhaps due to well depth which is very shallow and its just sufficient to give bound state.

7. Apr 3, 2016

### Incand

The actual value I get isn't really important, more if I'm going about it the correct way. The one thing I'm wondering is how I can consider this as a particle scattering against a step potential. I mean there's nothing saying I have a deuteron starting at zero moving outwards, the wave function is already "everywhere". And does the transmission coefficient really tell me how much time the particle is outside the well? Even if the particle is reflected instead it may transmit at a later time and if It's outside the well it may enter again (with the same transmission coefficient).

As I understand it deuteron is very weakly bound compared to other nuclei which would fit with the picture that it's outside the well a large part of the time.

8. Apr 3, 2016

### drvrm

actually the question you posedis about the probability of finding the particle in a certain region 'say outside the wll'- it should be the probability calculation -if you have the amplitude of the wave function in the outside region (normalized in the sense that total probability is unity of inside and outside taken together-then exact calculation can be made.
reflection and transmission are part of the same particle and R+T=1 or incident currennt= refl.part + trans. part , so in a way it gives an estimate of time spent if incident beam intensity per unit time is taken as 1.

9. Apr 3, 2016

### Incand

I believe I understand now, thanks for all the help! I also seem to be able to calculate the integrals in my first post now, doing it analytically instead seem to give me a value in the right region. Still doesn't give me the correct value but perhaps I made some errors so I'm going go over that.

10. Apr 3, 2016

### TSny

Incand,

I think your approach in the first post is the way to go. But, note a couple of things. The mass m in the differential equation for $f$ is not the mass of the deuteron. $f$ describes the part of the wave function dealing with the relative motion of the two particles in the system. Do you recall what mass is relevant here?

Also, when normalizing the wavefunction, don't forget that $\Psi = f(r)/r$. You want $\Psi$ to be normalized when integrating over all of 3D space.

(I noticed a little typo in your expression for $\frac{C}{A}$. You have $k_2$ in both the numerator and denominator.)

11. Apr 4, 2016

### Incand

Cheers! I believe I got it right now! Didn't think of the the $\Psi(r)=f(r)/r$ part nor to use the reduced mass.

I used matlab to do the numerical calculation to avoid errors, posting the code if anyone is interested
Code (Text):

h  = 1.0546e-34; %hbar
mp =1.00727647; %u
mn = 1.00866501;
m = mp*mn/(mp+mn)*1.660566e-27; %reduced mass in kg
V0 = 5.6e-12; %converted to joule
E = -3.552e-13;
R=2.1e-15;
k1 = sqrt(2*m*(V0+E)/h^2)
k2 = sqrt(-2*m*E/h^2)
c = sin(k1*R)/exp(-k2*R); %from boundary conditions
f1 = @(r) sin(k1*r).^2;
c1 = integral(f1,0,R)
c2 = exp(-2*k2*R)/(2*k2)
P = c^2*c2/(c1+c^2*c2) %easier to divide by the total than calculating the norming coefficient

which outputs $P = 0.6213$ which is in agreement with the answer!