# Constructing massive N=1 SUSY Multiplets

• A
This isn't really a homework question because our professor has declined to set any exercises or homework (or publish any notes at all) much to the class's chagrin. He claims that we should go through what we do in the lectures and correct all the mistakes he makes on the board (since he refuses to care about sign errors or indices, despite these issues being of UTMOST importance when dealing with a subject that involves fermions!)

Sorry, had to get that out of my system...

Right, so I'm going through our lecture notes and attempting to construct the N=1 SUSY multiplets. I started with the massless multiplets and had no problems, however when I tried the same tactics with the massive ones I encountered unwanted spin-1 states in the scalar multiplet which worry me greatly, since it doesn't match up with what our professor wrote down on the board. I will sketch out what I did.

MASSLESS CASE:

Since the Poincare algebra is a subgroup of the SuperPoincare algebra (SUSY N=1), we know that representations of the SUSY algebra must also be representations of the Poincare group. The Casimir operators are $P^2 = - m^2$ and $\hat{W}^2$ where $\hat{W} = W^{\mu} + Q\bar{Q}$ is the "Super-Pauli-Lubanwski vector".

The SUSY anticommutator reduces to $\{ Q_{\alpha},\bar{Q}_{\dot{\beta}}\} = 4E \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right)$, since we can always boost/rotate to any frame where $P^{\mu} = (E,0,0,E)$ and $\sigma^{0} = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right), \sigma^{3} = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$

We thus obtain a Clifford Algebra with
$\{ Q_{1},\bar{Q}_{1}\} = 4E$ and all other anticommutators vanishing.

We choose a particular state, labelled by helicity and energy $|E,\lambda\rangle = |0\rangle$ (because its a rep of the Poincare algebra) as the "Clifford Vacuum" and demand that the action of $a_{\alpha}|0\rangle = 0$ where $a_{\alpha}=\frac{Q_{\alpha}}{\sqrt{4E}}$.

We then obtain the rest of the multiplet by acting with $a_{1}^{\dagger}=\frac{\bar{Q}_{\dot{\alpha}}}{\sqrt{4E}}$, noting that we can ignore any zero norm states, such as those generated by $a_{2}^{\dagger}$ (since the anticommutator vanishes). Since the square of any element in the algebra is = 0 , we obtain 2 states in our multiplet.

We can work out what effect the action of the creation operator had on $|E,\lambda\rangle = |0\rangle$ by considering the commutator $[J_{3},\bar{Q}_{1}] = \frac{1}{2}\bar{Q}_{1}$. If we act this on the state $|E,\lambda\rangle$, and extract the $J_3$ eigenvalue, we discover that the effect of the SUSY generator was to raise the helicity of the state by 1/2.

Our massless SUSY multiplet is therefore composed of any state with a given helicity and a partner state with a helicity larger by 1/2. This is consistent with what I know (not much!) about SUSY generally - particles get partner particles which have the same mass but spins differ by 1/2 - SUSY rotates bosons into fermions and vice versa.

MASSIVE CASE:

Things get a little trickier here and I seem to be running into difficulties almost immediately trying to apply the analogous method. We boost to a frame where $P^{\mu} = (m, 0, 0, 0)$. Plugging this into the SUSY algebra gives us a Clifford algebra $\{a_{\alpha},a_{\dot{\beta}}^{\dagger}\}=\delta_{\alpha \dot{\beta}}$, where $a_{\alpha}=\frac{Q_{\alpha}}{\sqrt{2m}}$.

I now have 4 states in my multiplet: For a given vacuum state such that $a_{\alpha}|0\rangle = 0$:

$$(|0\rangle, a_{1}^{\dagger}|0\rangle, a_{2}^{\dagger}|0\rangle, a_{1}^{\dagger}a_{2}^{\dagger}|0\rangle )$$

The problem now comes when I select a particular clifford vacuum and try to understand the action of the SUSY generator on that state via the commutator as before. A general state in this case is labelled by 3 parameters, Energy, Total spin ($J^2$ eigenvalue) and Spin along the z axis ($J_3$ eigenvalue), i.e

$$|m,s,s_3\rangle$$

The action of the creation operators is to either raise or lower the eigenvalue of J3: $s_3 \rightarrow s_3\pm \frac{1}{2}$.

The state resulting from the action of the creation operators will be given by a sum over reps with different possible values of "s", the total spin, all with the same J3 eigenvalue (s3 +/- 1/2). Since the SUSY generators are spin 1/2 objects by construction we realise that the representations in the sum (labelled by "s") will be given by the tensor product decomposition of $s\otimes\frac{1}{2} = (s-\frac{1}{2})\oplus(s+\frac{1}{2})$, with some Clebsch-Gordan coefficients $k_i$ that I can't remember.

Now the problem comes when I try to construct the multiplet for say, $|m,0,0\rangle = |0\rangle$.

$$a_1^{\dagger}|m,0,0\rangle = |m,\frac{1}{2},\frac{1}{2}\rangle$$, because total spin (-1/2) can't exist, and therefore similarly
$$a_2^{\dagger}|m,0,0\rangle = |m,\frac{1}{2},-\frac{1}{2}\rangle$$.

Now when I act on either of these things again, with the other generator i.e $a_1^{\dagger} a_2^{\dagger}|m,0,0\rangle$, I should obtain something that is like $|m,0,0\rangle'$, according to my professor and my notes.

However to me it looks like

$$a_2^{\dagger}|m,\frac{1}{2},\frac{1}{2}\rangle = k_1|m,0,0\rangle + k_2|m,1,0\rangle$$.

and

$$a_1^{\dagger}|m,\frac{1}{2},-\frac{1}{2}\rangle = k_3|m,0,0\rangle + k_4|m,1,0\rangle$$,

which is not a spin - 0 state. I don't understand how to get rid of these extra spin-1 guys floating around, and I'm not sure if I've understood this whole thing correctly at all.

Any help would be greatly appreciated. I would also appreciate it if you could direct me to a really good source on the subject matter - our professor recommended Wikipedia and Wess and Bagger (which I find to be a tad formal).

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• Dilatino

Related Beyond the Standard Model News on Phys.org

Hi guys, I know its a bit cheeky to do this, but I'm getting no response at all on the homework questions thread. Strictly this is not actually homework at all because our professor refuses to set homework. I am having trouble constructing N=1 SUSY massive multiplets as detailed in the other thread. I'm struggling to understand why the final state in the multiplet should be proportional to the vacuum state, as I seem to keep getting extra higher spin states floating around...

Any help would be greatly appreciated.

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haushofer
Without going into detail now, if I remember correctly, the SUSY-notes by Bilal have a detailed account of this.

samalkhaiat

Hi guys, I know its a bit cheeky to do this, but I'm getting no response at all on the homework questions thread. Strictly this is not actually homework at all because our professor refuses to set homework. I am having trouble constructing N=1 SUSY massive multiplets as detailed in the other thread. I'm struggling to understand why the final state in the multiplet should be proportional to the vacuum state, as I seem to keep getting extra higher spin states floating around...

Any help would be greatly appreciated.
I think you are confused about the labelling of states. Generally speaking, labelling is a confusing business in the representation theory. Couple this with the ridiculous “it is easy to show”-statement of most textbooks, you get a totally confused student. So, students must be careful because, some times, that textbook statement simply means that the problem is not easy for the author to show. Any way, assuming I have understood what you are confused about, I will try to explain the main points using your notations with (I hope) clear labelling. I will be discussing the massive $N = 1$ representations only. So, every where $P^{\mu} = (m , \vec{0})$.
$$W_{\mu} = \frac{1}{2} \epsilon_{\mu\nu\rho\sigma}P^{\nu}M^{\rho\sigma} , \ \ \ W^{2} = -m^{2}j(j+1) .$$
The SUSY Casmir is the square of the tensor
$$C_{\mu\nu} = B_{\mu}P_{\nu} - B_{\nu}P_{\mu},$$ where $$B_{\mu} = W_{\mu} - \frac{1}{4} \bar{Q}_{\dot{\alpha}} \bar{\sigma}^{\dot{\alpha}\beta}_{\mu}Q_{\beta} .$$ In the rest frame
$$B_{i} = -m S_{i} , \ \ C_{0i} = - m B_{i} = m^{2}S_{i} , \ \ C_{ij} = 0 ,$$
where
$$S_{i} = J_{i} - \frac{1}{4m} \bar{Q}\sigma_{i}Q \ \ \ (1)$$
So, $$C^{2} = 2 C_{0i}C^{0i} = 2m^{4}S_{i}S^{i} .$$
Since both $J_{i}$ and $\sigma_{i}$ satisfy the same $su(2)$ algebra, it follows from (1) that
$[ S_{i} , S_{j}] = i \epsilon_{ijk} S_{k} .$ Thus, we can use $S^{2} = s (s+1)$ to label the states, because for our choice of the Clifford vacuum $Q | \Omega \rangle = 0$, we have $J_{i}| \Omega \rangle = S_{i}| \Omega \rangle$. One last piece of notation is the SUSY algebra
$$\big \{ Q_{\alpha} , \bar{Q}_{\dot{\alpha}} \big \} = 2m I_{2},$$
$$a_{\alpha} = \frac{1}{\sqrt{2m}} Q_{\alpha} , \ \ \ a^{\dagger}_{\alpha} = \frac{1}{\sqrt{2m}} \bar{Q}_{\dot{\alpha}} .$$
Notice that the index on $a^{\dagger}_{\alpha}$ is not dotted (why?).
Okay, let’s begin our story. Being an eigen state of spin, the Clifford vacuum is labelled by $j = s$ with the usual $(2j +1)$-fold degeneracy:
$$| \Omega \rangle = | j , j_{3}\rangle , \ \ \ j_{3} = -j , \cdots , +j ,$$ $$J_{3}| \Omega \rangle = S_{3}| \Omega \rangle = j_{3}| \Omega \rangle .$$
Excitations above the Clifford vacuum are obtained by the action of the fermionic operators $(a^{\dagger}_{1}, a^{\dagger}_{2})$:
$$| \Omega \rangle , \ \ a^{\dagger}_{1}| \Omega \rangle , \ \ a^{\dagger}_{2}| \Omega \rangle , \ \ a^{\dagger}_{1}a^{\dagger}_{2}| \Omega \rangle .$$
Thus, there are $4(2j+1)$ states in the SUSY multiplet. To determine the spins of these states, we act on $| \Omega \rangle$ with the following commutation relations (which I assume you know them)
$$[ J_{3} , a^{\dagger}_{1}] = - \frac{1}{2} a^{\dagger}_{1} ,$$ $$[ J_{3} , a^{\dagger}_{2}] = + \frac{1}{2} a^{\dagger}_{2} .$$
It is “really” easy to show
$$J_{3} \ a^{\dagger}_{1}| \Omega \rangle = (j_{3} - 1/2) \ a^{\dagger}_{1}| \Omega \rangle , \ \ \ \ \ \ \ (2)$$ $$J_{3} \ a^{\dagger}_{2}| \Omega \rangle = (j_{3} + 1/2) \ a^{\dagger}_{2}| \Omega \rangle . \ \ \ \ \ \ \ (3)$$
If you like, you can define the states
$$| \Omega^{-}\rangle \equiv a^{\dagger}_{1}| \Omega \rangle , \ \ \ | \Omega^{+}\rangle \equiv a^{\dagger}_{2}| \Omega \rangle .$$
So, in terms of these states (2) and (3) become
$$J_{3} \ | \Omega^{\mp} \rangle = (j_{3} \mp 1/2 ) \ | \Omega^{\mp} \rangle . \ \ \ \ \ \ \ \ \ \ (4)$$
Now, if we act on (2) with $a^{\dagger}_{2}$ and use the relation
$$[a^{\dagger}_{2} , J_{3} ] a^{\dagger}_{1} = - \frac{1}{2}a^{\dagger}_{2} a^{\dagger}_{1} ,$$ we obtain
$$J_{3} \ a^{\dagger}_{2}a^{\dagger}_{1} | \Omega \rangle = j_{3} \ a^{\dagger}_{2}a^{\dagger}_{1} | \Omega \rangle .$$
Again, we can define the state
$$| \bar{\Omega} \rangle \equiv a^{\dagger}_{2}a^{\dagger}_{1} \ | \Omega \rangle ,$$
so that
$$J_{3} \ | \bar{\Omega} \rangle = j_{3} \ | \bar{\Omega} \rangle . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$$
So our irreducible massive multiplet is
$$\mbox{States}: \ \ \ \ |\Omega \rangle , \ \ \ |\Omega^{-} \rangle , \ \ \ |\Omega^{+} \rangle , \ \ \ |\bar{\Omega} \rangle .$$
$$\mbox{Spins}: \ \ \ \ \ \ \ j_{3} , \ \ \ \ j_{3}-1/2 , \ \ \ j_{3}+1/2 , \ \ \ \ \ j_{3} .$$
In general, the states $|\Omega^{\pm} \rangle$ are superposition’s of states with spins $j \pm 1/2$
$$|\Omega^{+} \rangle = C^{++}|m; j + 1/2 , j_{3} + 1/2 \rangle + C^{-+}|m; j - ½ , j_{3}+1/2 \rangle ,$$ $$|\Omega^{-} \rangle = C^{--}|m; j - 1/2 , j_{3} - 1/2 \rangle + C^{+-} |m; j + 1/2 , j_{3} - 1/2 \rangle ,$$
with the Clebsch-Gordan coefficients
$$C^{\eta \eta_{3}} = \langle j + \frac{\eta}{2}, j_{3} + \frac{\eta_{3}}{2} | j , j_{3} , \frac{1}{2}, \frac{\eta_{3}}{2} \rangle .$$
So, for a scalar Clifford vacuum $|\Omega , 0 \rangle = | m ; 0 ,0 \rangle$, each superposition in the definitions of $|\Omega^{\pm}, \pm 1/2 \rangle$ collapse to a single state with spin $j_{3} \pm 1/2 = \pm 1/2$. Thus, the multiplet has two spin states of a massive fermion $|\Omega^{\pm} \rangle = |m; 1/2 , \pm 1/2 \rangle$ and the scalar states $|\Omega \rangle = | m; 0 ,0 \rangle$ and $|\bar{\Omega} \rangle = | m; \bar{0} , 0 \rangle$.
The other massive (vector) multiplets are based the spin one-half Clifford vacua $|\Omega , \pm 1/2 \rangle = | m; 1/2 , \pm 1/2 \rangle$. Starting from $j_{3} = 1/2$ states we obtain multiplet containing 2 inequivalent massive (wino) states with spin projection $+1/2$, a massive vector boson with spin projection $+1$ and a linear combination of a longitudinal vector boson and a scalar:
$$|\Omega , 1/2 \rangle = | m; 1/2 ,1/2 \rangle , \ \ \ | \bar{\Omega}, 1/2 \rangle = | m; \bar{1/2} , 1/2 \rangle$$$$|\Omega^{+},1 \rangle = | m; 1 , 1 \rangle , \ \ \ |\Omega^{-} , 0 \rangle = \frac{1}{\sqrt{2}}\left( | m ; 1 , 0 \rangle + | m ; 0 , 0 \rangle \right) .$$
States with negative spin projections are obtained from the Clifford vacuum with $j_{3}=-1/2$. They contain the orthogonal linear combination of the longitudinal vector boson and the scalar
$$|\Omega , -1/2 \rangle = | m; 1/2 , -1/2 \rangle , \ \ \ |\bar{\Omega}, -1/2 \rangle = | m; \bar{1/2} , -1/2 \rangle$$$$|\Omega^{+} , 0 \rangle = \frac{1}{\sqrt{2}}\left( | m ; 1 , 0 \rangle - | m ; 0 , 0 \rangle \right) , \ \ \ |\Omega^{-} , -1 \rangle = |m; 1 , -1 \rangle .$$
Okay! I hope I didn’t cause more confusion. But, be patient every thing will clear up when you study the representations in terms of finite component super fields.

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• nrqed
Thank you so much for this explanation! Our professor refused to answer when I went to him with this question and instead directed me to a paper from the 1970's with completely different notation to our course's, and then said that it was obvious the higher spin state had to vanish because of Lorentz Invariace!

MathematicalPhysicist
Gold Member
Thank you so much for this explanation! Our professor refused to answer when I went to him with this question and instead directed me to a paper from the 1970's with completely different notation to our course's, and then said that it was obvious the higher spin state had to vanish because of Lorentz Invariace!
So he did give you an answer; just different notation, in graduate courses you are suppoed to know how to search for references by yourself.