How Do You Calculate the Binding Energy of Deuterium's Nucleons?

[carllacan]

Homework Statement

The atomic nucleus of Deuterium is a bound state of two nucleons.

Through a change of coordinates we can transform the situation into a central force problem with a potential described by -V0 for r > R and 0 for r < R.

In the ground state of this nucleus the angular momentum number is l=0.

Find the binding energy of the nucleons.

Homework Equations

The radial equation: http://en.wikipedia.org/wiki/Partic...c_potential#Derivation_of_the_radial_equation

The Attempt at a Solution

I have two (second degree) radial equations (inside and outside of R) with different E, both with l = 0. I also have two conditions at r = R, where the two solutions and their derivatives have to be equal. I also have the normalization condition.

That makes three constraints for four constants (plus the energy), so I can't solve it.

I've tried adding the condition that ψ(0) = 0, because the nucleons can't be in the same place. This would allow me to find the energy inside R, which I think would be the end of the problem (since I am just asked to find the binding energy), but I'm not sure it is right, especially because doing it like this I get a solution with no dependence of V0.

Can you tell me how to continue from here?

Thank you for your time.

 

[mfb]

Where do you get 4 constants from? Energy and amplitudes of the wave function inside and outside are 3.

I've tried adding the condition that ψ(0) = 0, because the nucleons can't be in the same place.

They can, they are different particle types. You can say something about ψ'(0), however.

 

[carllacan]

Three? Each one is a second-degree differential equation, so each wave function has two constants: Asin(kx) + Bcos(kx) inside and Csin(k'x) + Dcos(k'x) outside, isn't that so?

What can I say about ψ'(0)? I'm thinking ψ'(0) = -ψ'(0), because the radial function has to be "symmetric" at each side of the 0 point, but that gives ψ'(0) = 0. Is this right?

 

[dauto]

Three? Each one is a second-degree differential equation, so each wave function has two constants: Asin(kx) + Bcos(kx) inside and Csin(k'x) + Dcos(k'x) outside, isn't that so?

What can I say about ψ'(0)? I'm thinking ψ'(0) = -ψ'(0), because the radial function has to be "symmetric" at each side of the 0 point, but that gives ψ'(0) = 0. Is this right?

Yes, any other value for ψ' would produce a cusp at the origin.

 

[mfb]

Three? Each one is a second-degree differential equation, so each wave function has two constants: Asin(kx) + Bcos(kx) inside and Csin(k'x) + Dcos(k'x) outside, isn't that so?

Ah yeah, that parametrization is better, the amplitudes would have to be complex with their relative phase as parameter... yeah use sine and cosine (or directly an exponential function for the outside area).

What can I say about ψ'(0)? I'm thinking ψ'(0) = -ψ'(0), because the radial function has to be "symmetric" at each side of the 0 point, but that gives ψ'(0) = 0. Is this right?

Right.