I Can you tell the difference between two neutrons in an alpha particle?

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In an alpha particle, the two neutrons are indistinguishable fermions, and their spins sum to zero, making it impossible to distinguish between a neutron with upspin and one with downspin due to quantum superposition. The discussion highlights that while neutrons in a bound state may be modeled within a potential well, this approach is inadequate for understanding alpha particles, where all nucleons occupy the 1S state. The total wavefunction for the system must remain antisymmetric for neutrons and protons, but no antisymmetry is required between neutrons and protons. The conversation suggests that further clarification from a nuclear physics expert is needed for a deeper understanding. Overall, the complexities of neutron behavior in alpha particles challenge simplistic models.
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Can you tell the difference between two neutrons in an alpha particle?
Can you tell the difference between two neutrons in an alpha particle? In one alpha particle, we know that the sum of the spins of two neutrons is zero. Can a neutron with upspin and a neutron with downspin be distinguished from each other? Or can't you tell because it's superimposed?
 
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Neutrons are indistinguishable fermions. All you can say is that when measuring the spins of the two neutrons you find with some probability each of the 4 possible outcomes, given by the prepared state, in this case a bound state of 2 protons and 2 neutrons within a 4He nucleus.

Where did you get the information about the sum of the spins of the two neutrons being zero?
 
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vanhees71 said:
Where did you get the information about the sum of the spins of the two neutrons being zero?
My knowledge of nuclear physics is almost zero, but I guess that if the two neutrons are understood to be in the ground state of a potential well, then Pauli's principle demands that they are in a singlet state.
 
But it's a bound state of four particles not two particles in a potential well.
 
vanhees71 said:
But it's a bound state of four particles not two particles in a potential well.
That's how nuclei are modeled, particles within a potential well, like Wood-Saxon. Besides, the total wavefunction is antisymmetric for the proton variable and antisymmetric for the neutron variables, but there are no antisymmetry demands between one neutron and one proton.

http://www.personal.soton.ac.uk/ab1u06/teaching/phys3002/course/05_shell.pdf

Anyways, a better answer requires someone that actually knows nuclear physics. I'm curious to know if my guess is correct or not.
 
andresB said:
I'm curious to know if my guess is correct or not.
Not really. Alphas are not ordinary nuclei, since every nucleon is in the 1S state. (Indeed, by some measures, an alpha is smaller than a proton).

The model of one particle in the potential well created by the other three is not very good for alphas. Two and two is even worse.

And guessing tends to confuse the OP more than clarifying. It's better to ask on another thread.
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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