Nuclear force tensor operator expectation value.

In summary: The significance of the integral is that it's related to Fermi's Golden rule. It states that 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|
  • #1
AntiElephant
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Homework Statement



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

[itex] S_{12} = \frac{3}{r^2}(\sigma^{(1)}\cdot r)(\sigma^{(2)}\cdot r) - \sigma^{(1)} \cdot \sigma^{(2)} [/itex]

Where the sigma are the pauli spin matrices.

Homework Equations


The Attempt at a Solution



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

[itex] \int \Psi^* S_{12} \Psi dV [/itex] = [itex] \int |\psi_{space}|^2 \chi_{spin}^* S_{12} \chi_{spin} dV [/itex]

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

[itex] \chi_{spin}^* S_{12} \chi_{spin} [/itex]

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

[itex] \chi_{spin}^* S_{12} \chi_{spin} \propto Y_{20}(\theta,\phi) [/itex]

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 [itex] Y_{20} [/itex] 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?
 
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  • #2
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 [itex] \int \Psi_f^* S_{12} \Psi_i dV [/itex] 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
TSny said:
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 [itex] \int \Psi_f^* S_{12} \Psi_i dV [/itex] 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

Thanks for the reply. Each spatial term has a spherical harmonic, for instance inside [itex] \Psi_f [/itex] there is [itex] Y_{L_f, m_f} [/itex]. I can see looking at (13) through (17) any change in angular momentum outside of [itex] -2 \leq \Delta L \leq 2 [/itex] gives that integral equal to zero. I feel like it should make the [itex] \Delta L = -1, 1 [/itex] 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 [itex] \int \Psi_f^* S_{12} \Psi_i dV [/itex]? If I were to hazard a guess it's related to Fermi's Golden rule. The nuclear potential [itex] V(r) [/itex] has [itex] S_{12} [/itex] inside it and the probability per second of the potential transitioning the 2-nucleon system from a initial state [itex] \Psi_i [/itex] to a final state [itex] \Psi_f [/itex] is given by

[itex] T \propto \int \Psi_f^* V(r) \Psi_i dV \propto \int \Psi_f^* S_{12} \Psi_i dV [/itex]

Hence any transition with [itex] |\Delta L| \geq 3 [/itex] has a zero-chance of happening. But [itex] |\Delta L| \leq 2 [/itex] has a (possibly) non-zero chance of occurring. Is this correct?
 
  • #4
Silversonic said:
Thanks for the reply. Each spatial term has a spherical harmonic, for instance inside [itex] \Psi_f [/itex] there is [itex] Y_{L_f, m_f} [/itex]. I can see looking at (13) through (17) any change in angular momentum outside of [itex] -2 \leq \Delta L \leq 2 [/itex] gives that integral equal to zero. I feel like it should make the [itex] \Delta L = -1, 1 [/itex] states go to zero but I can't see anything from the Wigner 3j-symbols which imply that is the case.
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}##.)

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

[itex] T \propto \int \Psi_f^* V(r) \Psi_i dV \propto \int \Psi_f^* S_{12} \Psi_i dV [/itex]

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

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.
 
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  • #5
Thanks for the help.

(Swapped accounts)
 

Related to Nuclear force tensor operator expectation value.

1. What is a nuclear force tensor operator expectation value?

A nuclear force tensor operator expectation value is a mathematical quantity that describes the average value of a nuclear force tensor operator in a given quantum state. This operator describes the strength and direction of the nuclear force acting between two particles.

2. Why is the nuclear force tensor operator expectation value important?

The nuclear force tensor operator expectation value is important because it provides crucial information about the nuclear force between particles in a nucleus. This information is essential for understanding the structure and behavior of atomic nuclei and for making predictions about nuclear reactions and processes.

3. How is the nuclear force tensor operator expectation value calculated?

The nuclear force tensor operator expectation value is calculated using quantum mechanics principles and mathematical equations. It involves taking the inner product of the nuclear force tensor operator with the wave function of the quantum state in question. This calculation can be complex and requires specialized knowledge and tools.

4. What factors affect the nuclear force tensor operator expectation value?

The nuclear force tensor operator expectation value is affected by several factors, including the type of particles involved, their distance from each other, and their quantum states. Additionally, the strength of the nuclear force and any external forces or fields acting on the particles can also impact the expectation value.

5. How does the nuclear force tensor operator expectation value relate to nuclear stability?

The nuclear force tensor operator expectation value is closely related to nuclear stability. In stable nuclei, the expectation value is typically high, indicating a strong and balanced nuclear force between particles. In contrast, in unstable nuclei, the expectation value may be low, indicating an imbalanced or weak nuclear force, which can lead to nuclear decay or other reactions.

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