# Why spin 1/2 nucleus has no quadrupole?

• xfshi2000
In summary: MeV, which is of same order as mc^2 (Uquark, Ucoul and Ekin are roughly equal at the energy minimum). Has to be considered relativistic I think...
xfshi2000
Hi:
I am confused that we talk in general about only nuclear spin larger than 1/2 which has electric quadrupole. For spin half, there is no quadrupole because of its spherical shape. Anyone can comment on why they are sphere? what is physical significance behind? I read some research articles. They mentioned the recent research shows proton shape maybe is sphere, peanut or bagel shape. How can we conclude spin 1/2 nucleus is sphere?
thanks

xfshi

Please see Slitcher "Principles of magnetic resonance". on the chapter which talks about quadrupole, in the first or second page of that chapter( I don't remember clearly), he mentioned that only nucleus with spin angular moment with more than 1/2 has electric quadrupole. that is source why I question this problem. why don't spin half nucleus has electric quadrupole?
thanks

oh, but that is just basic quantum mechanics, just sandwich the quadropole moment operator between general spherical harmonics with equal L, only the one with L = 0 will contribute. (if i remember this correctly)

The electric quadrupole operator is a Legendre polynomial P_2. The matrix element
<1/2|P_2|1/2> vanishes for any spin 1/2 state. This says nothing about the shape of the nucleus, just that this matrix element vanishes

clem said:
The electric quadrupole operator is a Legendre polynomial P_2. The matrix element
<1/2|P_2|1/2> vanishes for any spin 1/2 state. This says nothing about the shape of the nucleus, just that this matrix element vanishes

Thanks everyone's answer. Hi clem: Could you explain more detail? As I know, commutation [S_z z]=0, we can only derive <1/2,m'|z|1/2,m>=0 unless m'=m. For individual spin 1/2 nucleus, there is no orbital angular momentum(L=0). Intrinsic parity of spin half particle is +1.
If we want to get <1/2,1/2|Q|1/2.1/2>=0, how do you derive <1/2,1/2|z|1/2,1/2>=0?
thanks
xf

Hmm, does it have spherical shape? It does not make sense to me. Proton is uud, right? but charge of u=+2/3e and therefore two u-quarks cannot be at the same place because of Coulomb repulsion. I modeled this once assuming a linear or quadratic quark-quark attractive force to find the equilibrium structure, and i found that it take the shape of a line/rod "u-d-u". I guess that the quadropole term will be very small anyway. You may excuse me here because of my very small knowledge on quark-quark interaction.

clem said:
The electric quadrupole operator is a Legendre polynomial P_2. The matrix element
<1/2|P_2|1/2> vanishes for any spin 1/2 state. This says nothing about the shape of the nucleus, just that this matrix element vanishes

This must have a very restricted validity, assuming some old-fashioned shell-filling model? Assume three electrons in a weak parabolic potential. Then the structure would be a Wigner crystal, i.e., three sharp Gaussian like packages arranged in a triangle for Stot=1/2 (You can carry out this with Hartree-Fock. Shell filling approach using hydrogen orbitals would fail to describe a structure like this). Clearly the quadropole moment is not zero in this case.

per.sundqvist said:
Hmm, does it have spherical shape? It does not make sense to me. Proton is uud, right? but charge of u=+2/3e and therefore two u-quarks cannot be at the same place because of Coulomb repulsion. I modeled this once assuming a linear or quadratic quark-quark attractive force to find the equilibrium structure, and i found that it take the shape of a line/rod "u-d-u". I guess that the quadropole term will be very small anyway. You may excuse me here because of my very small knowledge on quark-quark interaction.

Sorry, I checked my claim (I did a fast estimation many years ago) with the line/rod structure u-d-u and its wrong! (sorry for the mess). It is most likely that the orbitals are bound together in the way that each quark has a spherical symmetric shape, like spherical Gaussians for example.

The pure Coulomb problem collapses the quarks (u-d-u) to a point, but quantum kinetic energy broadens it to typical hydrogen-like structure. Adding the quark-quark attraction U=k*r, with k chosen so that the size of the wave functions are roughly a=1e-15m, gives a positive total binding energy of about 200MeV, which is of same order as mc^2 (Uquark, Ucoul and Ekin are roughly equal at the energy minimum). Has to be considered relativistic I think...

per.sundqvist said:
Sorry, I checked my claim (I did a fast estimation many years ago) with the line/rod structure u-d-u and its wrong! (sorry for the mess). It is most likely that the orbitals are bound together in the way that each quark has a spherical symmetric shape, like spherical Gaussians for example.

The pure Coulomb problem collapses the quarks (u-d-u) to a point, but quantum kinetic energy broadens it to typical hydrogen-like structure. Adding the quark-quark attraction U=k*r, with k chosen so that the size of the wave functions are roughly a=1e-15m, gives a positive total binding energy of about 200MeV, which is of same order as mc^2 (Uquark, Ucoul and Ekin are roughly equal at the energy minimum). Has to be considered relativistic I think...

thanks for your helpful answer. Now I focus on the whole nucleus. That means adding all the quark orbital angular momentum and spin angular momentum, nucleus spin are reached. If nuclear spin is half. Now I don't know how to prove <1/2.1/2|Q|1/2,1/2>=0.

malawi_glenn said:
oh, but that is just basic quantum mechanics, just sandwich the quadropole moment operator between general spherical harmonics with equal L, only the one with L = 0 will contribute. (if i remember this correctly)

Do you think my understanding is correct? In general, we say spin angular momentum refers to total angular momentum(orbital angular momentum plus spin angular momentum). When we sandwich the quadrupole operator, wavefunction is the direct product of spatial wavefunction and spinor wavefunction. <J,J_z|Q|J,J_z>. By parity property, we know, x,y z is odd parity, the first J must be different from the second J. So that is reason why matrix element is zero.
thanks

xfshi2000 said:
Do you think my understanding is correct? In general, we say spin angular momentum refers to total angular momentum(orbital angular momentum plus spin angular momentum). When we sandwich the quadrupole operator, wavefunction is the direct product of spatial wavefunction and spinor wavefunction. <J,J_z|Q|J,J_z>. By parity property, we know, x,y z is odd parity, the first J must be different from the second J. So that is reason why matrix element is zero.
thanks
Q is a parity even operator. The Mel you gave is non zero for all J>1/2.

J=1/2 can come from L=0 or L=1. <0|Q|0> is easily seen to be zero.
For L=1, s=1/2 form the eigenstate |1/2,1/2>. Then <1/2,1/2|P_2|1/2,1/2>=0.

clem said:
J=1/2 can come from L=0 or L=1. <0|Q|0> is easily seen to be zero.
For L=1, s=1/2 form the eigenstate |1/2,1/2>. Then <1/2,1/2|P_2|1/2,1/2>=0.

Thank you clem: let me repeat your solution.
Because J=1/2, J=L+S, L can be either 0 or 1. How about S? For example, If L=0 and S=1/2, J could be 1/2. if L=1, S=3/2, we still can acquire 1/2 from the rules J~|L+S| to |L-S|. Ok now I assume S=1/2.
(1) If L=0, S=1/2, J=1/2 and quadrupole operator is even parity, wave function |L,m> also is even parity due to Y_L^m=(-1)^L. <L1,m1|Q|L2,m2> where L1=0,L2=0,m1=0,m2=0. From selection rules, m1=m2 are satisfied, L1 is same to L2. How do you conclude <0,0|Q|0,0>=0? (where m1 and m2 are magnetic quantum number)

(2) If L=1, S=1/2, we still get J=1/2. <L1,m1|Q|L2,m2> L1=1, L2=1, m1 can be 0, +/-1.
m2 can be 0,+/-1. <1,1|Q|1,1>=<1,0|Q|1,0>=<1,-1|Q|1,-1>=0. Notice Q is only function of coordinate. Why are they equal to zero? Sorry I cannot follow your idea.

It's hard for me to do the details here. Look at a book on angular momentum.
I gave one example, but the rule is more general. The three angular momenta in the matrix element <J_1|J_2|J_3> must obey a triangle rule that no J is greater than the sum of the other two.

## 1. Why is the spin of a nucleus described in terms of half-integer numbers?

The spin of a nucleus, like the spin of an electron, is an intrinsic property that cannot be explained by classical physics. Instead, it is described by quantum mechanics which assigns half-integer numbers (such as 1/2) to particles with spin.

## 2. What does it mean for a nucleus to have a quadrupole moment?

A quadrupole moment describes the distribution of charge within a nucleus. A nucleus with a non-zero quadrupole moment has an elongated or flattened shape, while a nucleus with a zero quadrupole moment is more spherical.

## 3. How is the spin of a nucleus related to its quadrupole moment?

The spin of a nucleus is related to its quadrupole moment through the nuclear spin-parity selection rule. This rule states that nuclei with integer spins (such as 0, 1, 2, etc.) have zero quadrupole moments, while nuclei with half-integer spins (such as 1/2, 3/2, 5/2, etc.) have non-zero quadrupole moments.

## 4. Why do spin 1/2 nuclei have no quadrupole moments?

Spin 1/2 nuclei, such as hydrogen and helium-3, have an odd number of nucleons and therefore have a non-zero spin. According to the spin-parity selection rule, this means they also have a non-zero quadrupole moment. However, due to their small size and spherical shape, the quadrupole moment is extremely small and is considered negligible for these nuclei.

## 5. Are there any exceptions to the spin-parity selection rule for quadrupole moments?

While the spin-parity selection rule is generally observed, there are some exceptions. Certain isotopes, such as helium-4 and beryllium-8, have integer spins but also have non-zero quadrupole moments. This is due to their unique nuclear structures and the strong nuclear forces between their nucleons.

• High Energy, Nuclear, Particle Physics
Replies
8
Views
673
• High Energy, Nuclear, Particle Physics
Replies
7
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
11
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
12
Views
3K
• High Energy, Nuclear, Particle Physics
Replies
15
Views
3K
• Quantum Physics
Replies
24
Views
920
• Quantum Physics
Replies
20
Views
4K
• High Energy, Nuclear, Particle Physics
Replies
5
Views
1K
• High Energy, Nuclear, Particle Physics
Replies
4
Views
5K
• High Energy, Nuclear, Particle Physics
Replies
16
Views
2K