- #1

- 46

- 10

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.

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 [itex] P^2 = - m^2 [/itex] and [itex] \hat{W}^2 [/itex] where [itex] \hat{W} = W^{\mu} + Q\bar{Q} [/itex] is the "Super-Pauli-Lubanwski vector".

The SUSY anticommutator reduces to [itex] \{ Q_{\alpha},\bar{Q}_{\dot{\beta}}\} = 4E

\left( \begin{array}{cc}

1 & 0 \\

0 & 0 \\

\end{array} \right)

[/itex], since we can always boost/rotate to any frame where [itex] P^{\mu} = (E,0,0,E) [/itex] and [itex] \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) [/itex]

We thus obtain a Clifford Algebra with

[itex] \{ Q_{1},\bar{Q}_{1}\} = 4E [/itex] and all other anticommutators vanishing.

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

We then obtain the rest of the multiplet by acting with [itex]a_{1}^{\dagger}=\frac{\bar{Q}_{\dot{\alpha}}}{\sqrt{4E}} [/itex], noting that we can ignore any zero norm states, such as those generated by [itex] a_{2}^{\dagger} [/itex] (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 [itex] |E,\lambda\rangle = |0\rangle[/itex] by considering the commutator [itex] [J_{3},\bar{Q}_{1}] = \frac{1}{2}\bar{Q}_{1} [/itex]. If we act this on the state [itex] |E,\lambda\rangle [/itex], and extract the [itex] J_3 [/itex] 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.

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 [itex] P^{\mu} = (m, 0, 0, 0) [/itex]. Plugging this into the SUSY algebra gives us a Clifford algebra [itex] \{a_{\alpha},a_{\dot{\beta}}^{\dagger}\}=\delta_{\alpha \dot{\beta}} [/itex], where [itex] a_{\alpha}=\frac{Q_{\alpha}}{\sqrt{2m}} [/itex].

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

[tex] (|0\rangle, a_{1}^{\dagger}|0\rangle, a_{2}^{\dagger}|0\rangle, a_{1}^{\dagger}a_{2}^{\dagger}|0\rangle )[/tex]

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 ([itex] J^2 [/itex] eigenvalue) and Spin along the z axis ([itex]J_3[/itex] eigenvalue), i.e

[tex] |m,s,s_3\rangle [/tex]

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

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 [itex]s\otimes\frac{1}{2} = (s-\frac{1}{2})\oplus(s+\frac{1}{2}) [/itex], with some Clebsch-Gordan coefficients [itex]k_i[/itex] that I can't remember.

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

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

[tex] a_2^{\dagger}|m,0,0\rangle = |m,\frac{1}{2},-\frac{1}{2}\rangle [/tex].

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

However to me it looks like

[tex] a_2^{\dagger}|m,\frac{1}{2},\frac{1}{2}\rangle = k_1|m,0,0\rangle + k_2|m,1,0\rangle [/tex].

and

[tex] a_1^{\dagger}|m,\frac{1}{2},-\frac{1}{2}\rangle = k_3|m,0,0\rangle + k_4|m,1,0\rangle [/tex],

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).

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 [itex] P^2 = - m^2 [/itex] and [itex] \hat{W}^2 [/itex] where [itex] \hat{W} = W^{\mu} + Q\bar{Q} [/itex] is the "Super-Pauli-Lubanwski vector".

The SUSY anticommutator reduces to [itex] \{ Q_{\alpha},\bar{Q}_{\dot{\beta}}\} = 4E

\left( \begin{array}{cc}

1 & 0 \\

0 & 0 \\

\end{array} \right)

[/itex], since we can always boost/rotate to any frame where [itex] P^{\mu} = (E,0,0,E) [/itex] and [itex] \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) [/itex]

We thus obtain a Clifford Algebra with

[itex] \{ Q_{1},\bar{Q}_{1}\} = 4E [/itex] and all other anticommutators vanishing.

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

We then obtain the rest of the multiplet by acting with [itex]a_{1}^{\dagger}=\frac{\bar{Q}_{\dot{\alpha}}}{\sqrt{4E}} [/itex], noting that we can ignore any zero norm states, such as those generated by [itex] a_{2}^{\dagger} [/itex] (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 [itex] |E,\lambda\rangle = |0\rangle[/itex] by considering the commutator [itex] [J_{3},\bar{Q}_{1}] = \frac{1}{2}\bar{Q}_{1} [/itex]. If we act this on the state [itex] |E,\lambda\rangle [/itex], and extract the [itex] J_3 [/itex] 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 [itex] P^{\mu} = (m, 0, 0, 0) [/itex]. Plugging this into the SUSY algebra gives us a Clifford algebra [itex] \{a_{\alpha},a_{\dot{\beta}}^{\dagger}\}=\delta_{\alpha \dot{\beta}} [/itex], where [itex] a_{\alpha}=\frac{Q_{\alpha}}{\sqrt{2m}} [/itex].

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

[tex] (|0\rangle, a_{1}^{\dagger}|0\rangle, a_{2}^{\dagger}|0\rangle, a_{1}^{\dagger}a_{2}^{\dagger}|0\rangle )[/tex]

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 ([itex] J^2 [/itex] eigenvalue) and Spin along the z axis ([itex]J_3[/itex] eigenvalue), i.e

[tex] |m,s,s_3\rangle [/tex]

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

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 [itex]s\otimes\frac{1}{2} = (s-\frac{1}{2})\oplus(s+\frac{1}{2}) [/itex], with some Clebsch-Gordan coefficients [itex]k_i[/itex] that I can't remember.

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

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

[tex] a_2^{\dagger}|m,0,0\rangle = |m,\frac{1}{2},-\frac{1}{2}\rangle [/tex].

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

However to me it looks like

[tex] a_2^{\dagger}|m,\frac{1}{2},\frac{1}{2}\rangle = k_1|m,0,0\rangle + k_2|m,1,0\rangle [/tex].

and

[tex] a_1^{\dagger}|m,\frac{1}{2},-\frac{1}{2}\rangle = k_3|m,0,0\rangle + k_4|m,1,0\rangle [/tex],

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).

Last edited: