# Nuclear force tensor operator expectation value.

1. Mar 3, 2014

### AntiElephant

1. The problem statement, all variables and given/known data

I have a question asking me to find the expectation value of $S_{12}$ for a system of two nucleons in a state with total spin $S = 1$ and $M_s = +1$, where $S_{12}$ is the tensor operator inside the one-pion exchange nuclear potential operator, equal to

$S_{12} = \frac{3}{r^2}(\sigma^{(1)}\cdot r)(\sigma^{(2)}\cdot r) - \sigma^{(1)} \cdot \sigma^{(2)}$

Where the sigma are the pauli spin matrices.

2. Relevant equations

3. The attempt at a solution

What I personally would've done would note that the system is in a state $\Psi = \psi_{space}\chi_{spin}$, where $\chi_{spin} = \alpha(1)\alpha(2)$ and then just have done

$\int \Psi^* S_{12} \Psi dV$ = $\int |\psi_{space}|^2 \chi_{spin}^* S_{12} \chi_{spin} dV$

So my problem would lie in calculating $\chi_{spin}^* S_{12} \chi_{spin}$ and then evaluating the integral. However, my answer says that the expectation value is simply given by

$\chi_{spin}^* S_{12} \chi_{spin}$

Why is there no integral over the whole of space? The answer doesn't make sense to me because it comes out as;

$\chi_{spin}^* S_{12} \chi_{spin} \propto Y_{20}(\theta,\phi)$

So my expectation value depends on the angles between the two nucleons? How can that make any sense? Has my answer forgotten to neglect the then integral over all space?

Lastly, my answer says that because the expectation value has a $Y_{20}$ term that the operator can transfer 2 units of angular momentum to the orbital motion of the particles. My quantum mechanics knowledge must be lacking because I don't exactly see why this is the case, can anyone link my to a place where I can read up more about what expectation value results mean?

Last edited: Mar 3, 2014
2. Mar 3, 2014

### TSny

Hello, AntiElephant.

I wonder if the first part of the question is just asking you to evaluate the spin part of the matrix element. That is, they just want you to evaluate $\chi_{spin}^* S_{12} \chi_{spin}$.

Now suppose you consider the matrix element $\int \Psi_f^* S_{12} \Psi_i dV$ between two different states $\Psi_i$ and $\Psi_f$. Suppose each state has the same total nucleon spin of S = 1, Ms = +1. But, suppose the spatial part of $\Psi_i$ has orbital angular momentum quantum number $l_i$ and $\Psi_f$ has orbital angular momentum number $l_f$.

How should $l_i$ and $l_f$ be related in order for the spatial part of the integration to be nonzero (taking into account that the spin matrix element is going to throw in a factor of Y20)?

It might help to look at equations (13) through (17) here:http://mathworld.wolfram.com/SphericalHarmonic.html

3. Mar 3, 2014

### Silversonic

Thanks for the reply. Each spatial term has a spherical harmonic, for instance inside $\Psi_f$ there is $Y_{L_f, m_f}$. I can see looking at (13) through (17) any change in angular momentum outside of $-2 \leq \Delta L \leq 2$ gives that integral equal to zero. I feel like it should make the $\Delta L = -1, 1$ states go to zero but I can't see anything from the Wigner 3j-symbols which imply that is the case.

But, forgive me for asking, what is the significance of $\int \Psi_f^* S_{12} \Psi_i dV$? If I were to hazard a guess it's related to Fermi's Golden rule. The nuclear potential $V(r)$ has $S_{12}$ inside it and the probability per second of the potential transitioning the 2-nucleon system from a initial state $\Psi_i$ to a final state $\Psi_f$ is given by

$T \propto \int \Psi_f^* V(r) \Psi_i dV \propto \int \Psi_f^* S_{12} \Psi_i dV$

Hence any transition with $|\Delta L| \geq 3$ has a zero-chance of happening. But $|\Delta L| \leq 2$ has a (possibly) non-zero chance of occurring. Is this correct?

4. Mar 3, 2014

### TSny

According to equation 13 at http://mathworld.wolfram.com/SphericalHarmonic.html, you will have a factor of the 3j symbol

$\left( \begin{array}{ccc} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{array} \right)$

If you look at symmetry property (6) at this link: http://mathworld.wolfram.com/Wigner3j-Symbol.html I think you can see why the above symbol yields zero for $|\Delta L | = 1$. (Here, $\Delta L = l_3-l_1$; and $l_2$ = 2 corresponds to $Y_{2,0}$.)

Yes, I think that's right. If the Hamiltonian has nonzero matrix elements between two states, then there can generally be transitions between those states.

5. Mar 4, 2014

### Silversonic

Thanks for the help.

(Swapped accounts)