# Symmetric and antisymmetric states of spin singlet/triplet

I'm studying the book "Quarks and Leptons: An Introductory Course In Modern Particle Physics". When I reach chapter 2 (page 33), I encounter this exercise problem that I couldn't understand:

EXERCISE 2.1 Justify the decomposition shown in (2.1) by either (1) considering the symmetry of the states under interchange of the nucleons or (2) using the angular momentum "step-down" operator.

Here's the solution in the book:

My questions are: Why do these "justify" the decomposition in (2.1)? Why the state $|S=1,M_S=0>$ is said to be "also" symmetric? What does it mean that the state $|S=0,M_S=0>$ be antisymmetric due to orthogonality? I'm really confused...

I'm really new to the subject, so it'd be great if someone could help me. Thank you!

## 1 Answer

If you interchange the spins, then the states are either symmetric or antisymmetric under permutation: this is more explicit if you write $$\sqrt{2}\vert S=1,M=0\rangle=\vert\uparrow\rangle_1\vert \downarrow\rangle_2 + \vert\downarrow\rangle_1\vert\uparrow\rangle_2\, , \tag{1}$$ where $\vert \uparrow\rangle_1$ denotes particle 1 in the spin-up state etc.

Clearly (1) is symmetric under permutation of the particle indices, i.e. it comes back to $+1$ times itself under permutation. On the other hand $$\sqrt{2}\vert S=0,M=0\rangle=\vert\uparrow\rangle_1\vert \downarrow\rangle_2 - \vert\downarrow\rangle_1\vert\uparrow\rangle_2\, , \tag{2}$$ is clearly antisymmetric under permutation of particles indices, i.e. this states comes back to $-1$ times itself under permutation.

Finally, note that all $S=1$ states are symmetric in the sense above irrespective of the value of $M$.

Since $L_x=L_x^{(1)}+L_x^{(2)}$, i.e. the total $L_x$ is the sum of the projection for particles $1$ and $2$, with $L_x^{(1)}$ action on states for particle 1 etc, is unchanged under permutation of particles, the operator $L_\pm = L_\pm^{(1)}+L_\pm^{(1)}$ will not change the permutation symmetry of the state $\vert S,M\rangle$ when stepping up or down since, if $\vert S,M\rangle$ has a definite permutation symmetry, $$L_\pm \vert S,M\rangle$$ will have the same permutation symmetry as $\vert S,M\rangle$.

• Thank you! But may I ask one more question: What does orthogonality have to do here? Why did he say "Orthogonality then demands $|S=0,M=0>$ to be anti symmetric? – John Smith Aug 18 '18 at 12:05
• If you have a linear combination $\vert a\rangle+\vert b\rangle$, the orthogonal state to that will be $\vert a\rangle - \vert b\rangle$. If you apply this to $\vert a\rangle = \vert \uparrow\rangle\vert\downarrow\rangle$ and $\vert b\rangle = \vert\downarrow\rangle\vert\uparrow\rangle$, you automatically get the antisymmetric combination. Since the $S=1$ states must be orthogonal to the $S=0$ state, the $\perp$ argument holds for orthogonality of the $\vert S,M=0\rangle$ states, albeit only because there are two $M=0$ states. – ZeroTheHero Aug 18 '18 at 13:52