How Do Three Spin-1/2 Particles Combine to Form States with Total Spin 3/2?

[BeyondBelief96]

Homework Statement

Determine the four states with $s = \frac{3}{2}$ that can be formed by three spin $\frac{1}{2}$ particles. Suggestion: Start with the state $\ket{\frac{3}{2}, \frac{3}{2}}$ and apply the lowering operator. [/B]

Homework Equations

$$S^{2}\ket{s, m} = \hbar^2 s(s+1)\ket{s, m}$$

$$S_z \ket{s, m} = \hbar m \ket{s, m}$$

The Attempt at a Solution

What we want to do is to find what four combinations of interacting spin $\frac{1}{2}$ particles give us a total angular momentum of spin $\frac{3}{2}$. \newline

Consider three interacting spin $\frac{1}{2}$ particles. A natural basis set is to label the states by the value of $S_z$ for each of the particles:

$$ \ket{+z, +z, +z}, \ket{+z, +z, -z}, \ket{+z, -z, +z},\ket{-z, +z, +z}, \ket{+z, -z, -z}, \ket{-z, +z, -z}
\ket{-z, -z, +z}, \ket{-z, -z, -z} $$

Can we reason out with physical intuition what the $s$ and $m$ values will be for each of these states? Consider the first state:

$$ \ket{+z, +z, +z} = \ket{\frac{1}{2}, \frac{1}{2}, \frac{1}{2}} $$

$s = \frac{1}{2}$ for each particle, and m =It was here I realized that when we worked through this problem in class but for a system of two interacting particles, we were evaluating the values of s and m for the eigenstates of the hamiltonian. Does this mean I need to find what the eigenstates of the Hamiltonian will be for a system of three interacting particles? How do I even go about doing so?

 

[BeyondBelief96]

Also, sorry, I'm new to posting here, and I thought it would be able to read LaTeX since I've seen others use it, but I must be doing something wrong, I hope you can read the equations okay.

 

[George Jones]

Also, sorry, I'm new to posting here, and I thought it would be able to read LaTeX since I've seen others use it, but I must be doing something wrong, I hope you can read the equations okay.

Enclose math in-line math with ## at the beginning and the end; enclose stand-alone math with $$ at the beginning and the end.

 

[George Jones]

$$ \ket{+z, +z, +z} = \ket{\frac{1}{2}, \frac{1}{2}, \frac{1}{2}} $$

LaTeX here doesn't support the Dirac notation package. For kets, try, for example, \left| +z, +z, +z \right> , which gives $\left| +z, +z, +z \right>$.

 

[DrClaude]

It was here I realized that when we worked through this problem in class but for a system of two interacting particles, we were evaluating the values of s and m for the eigenstates of the hamiltonian. Does this mean I need to find what the eigenstates of the Hamiltonian will be for a system of three interacting particles? How do I even go about doing so?

You need to find states that are eigenstates of $\hat{S}^2$ with eigenvalue $\frac{15}{4} \hbar^2$ (or $s=3/2$) for three spin-1/2 particles. As the problem suggests:

Determine the four states with $s = \frac{3}{2}$ that can be formed by three spin $\frac{1}{2}$ particles. Suggestion: Start with the state $\ket{\frac{3}{2}, \frac{3}{2}}$ and apply the lowering operator.

You start with the stretched state $| \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \rangle$, which has necessarily maximum spin $s=3/2$ and maximum $m=3/2$ (it is the same as $| s=\frac{3}{2}, m=\frac{3}{2}\rangle$), and then use the lowering operator $\hat{S}_- = \hat{S}_{1-} + \hat{S}_{2-} + \hat{S}_{3-}$ to find the eigenstates with other values of $m$.

 

[George Jones]

Determine the four states with $s = \frac{3}{2}$ that can be formed by three spin $\frac{1}{2}$ particles. Suggestion: Start with the state $\ket{\frac{3}{2}, \frac{3}{2}}$ and apply the lowering operator.

The question also says "apply the lowering operator as in (5.36)." You should look at (5.36) and the steps that led to (5.36) (for two spin 1/2 particles), and then try to do what the text and @DrClaude suggest for three spin 1/2 particles.

 

[BeyondBelief96]

You need to find states that are eigenstates of $\hat{S}^2$ with eigenvalue $\frac{15}{4} \hbar^2$ (or $s=3/2$) for three spin-1/2 particles. As the problem suggests:

You start with the stretched state $| \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \rangle$, which has necessarily maximum spin $s=3/2$ and maximum $m=3/2$ (it is the same as $| s=\frac{3}{2}, m=\frac{3}{2}\rangle$), and then use the lowering operator $\hat{S}_- = \hat{S}_{1-} + \hat{S}_{2-} + \hat{S}_{3-}$ to find the eigenstates with other values of $m$.

Okay, so I did so and got $(\hat{S}_1 + \hat{S}_2 + \hat{S}_3)\left | \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right > = \hbar (\left | -z, +z, +z \right > , \left | +z, -z, +z \right >, \left | +z, +z, -z \right >$ following the example from the book, only how do I express this state in terms of eigenstates, given that I don't even know what the eigenstates of 3 interacting particles are?

 

[BeyondBelief96]

I feel as if there's something fundamental connection I'm not seeing. I'm assuming $\left | +z, +z, +z\right>$ is an eigenstate, sort of like how $\left | \pm{z}, \pm{z} \right >$ were eigen states of two interacting particles? So we make the assumption that it would follow the same logic? I'm not sure why we are applying the lowering operator as well.

 

[DrClaude]

Okay, so I did so and got $(\hat{S}_1 + \hat{S}_2 + \hat{S}_3)\left | \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right > = \hbar (\left | -z, +z, +z \right > , \left | +z, -z, +z \right >, \left | +z, +z, -z \right >$ following the example from the book, only how do I express this state in terms of eigenstates, given that I don't even know what the eigenstates of 3 interacting particles are?

Be careful, you are missing a numerical factor when calculating the action of the lowering operator on the state. Also, you forgot the minus sign on the operators, $\hat{S}_{1-}$, and that should be + signs on the right-hand side, not commas.

I feel as if there's something fundamental connection I'm not seeing. I'm assuming $\left | +z, +z, +z\right>$ is an eigenstate, sort of like how $\left | \pm{z}, \pm{z} \right >$ were eigen states of two interacting particles? So we make the assumption that it would follow the same logic? I'm not sure why we are applying the lowering operator as well.

The idea is that you want to express the three-particle states $|s,m\rangle$, which are eigenstates of $\hat{S}^2$ and $\hat{S}_z$, in terms of the single-particle states
$$
| \pm z \rangle_1 \otimes | \pm z \rangle_2 \otimes | \pm z \rangle_3 \equiv | \pm z, \pm z, \pm z \rangle
$$
The starting point is a state that you know how to express in terms of single-particle states, which are the stretched states
$$
|s=\frac{3}{2},m=\frac{3}{2}\rangle = | + z, + z, + z \rangle
$$
or
$$
|s=\frac{3}{2},m=-\frac{3}{2}\rangle = | - z, - z, - z \rangle
$$
since only all particles having spin pointing in the same direction can result in the maximum possible total spin and spin projection.

Then you can find the other states of the same total spin using raising or lowering operators,
$$
\hat{S}_{-} |s=\frac{3}{2},m=\frac{3}{2}\rangle = \left( \hat{S}_{1-} + \hat{S}_{2-} + \hat{S}_{3-} \right) | + z, + z, + z \rangle
$$
or
$$
\hat{S}_{+} |s=\frac{3}{2},m=-\frac{3}{2}\rangle = \left( \hat{S}_{1+} + \hat{S}_{2+} + \hat{S}_{3+} \right) | - z, - z, - z \rangle
$$
On the left-hand side, you will get $\hat{S}_{-} |s=\frac{3}{2},m=\frac{3}{2}\rangle = c |s=\frac{3}{2},m=\frac{1}{2}\rangle$ (I'll let you look up the factor $c$ that appears there), so by applying the operator on the right-hand side you will get a representation of $|s=\frac{3}{2},m=\frac{1}{2}\rangle$ in terms of single-particle states. You can then repeat the procedure to get the other values of $m$.