Supersymmetry Question

[latentcorpse]

I haven't done any supersymmetry before and am probably going to need it in the future (hopefully anyway).

Anyway, I've been reading through these online notes:

http://www-thphys.physics.ox.ac.uk/people/JosephConlon/LectureNotes/SUSYLectures.pdf

I have a problem with p7:

(i) What is the symbol in the line "However the map is not an isomorphism as N=.... both map to (symbol i don't know) $$\in SO(1,3)$$

(ii) I am completely lost trying to do the first exercise. Where should I start?

Thanks for the help.

 

[dextercioby]

I haven't seen that symbol either. As for the exercise, it's done through the adjoint representation of SL(2,C) which is an automorphism of H(2,C), the group of hermitean 2x2 matrices. Essentially, you build a chain of mappings from the Minkowski space to H(2,C) then through the adjoint action of SL(2,C) you remain in H(2,C) only to return to Minkowski space through the inverse of the first mapping.

I believe this is discussed in some books which deal with spinor calculus. Urbantke and Sexl discuss it in their book (page 236).

 

[fzero]

(i) What is the symbol in the line "However the map is not an isomorphism as N=.... both map to (symbol i don't know) $$\in SO(1,3)$$

He means the identity matrix. I think it's a font artifact from using some hack to generate a fancy blackboard bold 1 (something like $\mbox{1 \hspace{-1.0mm} {\bf l}}$.)

(ii) I am completely lost trying to do the first exercise. Where should I start?

You want to use (16) and (19) to show that (17) implies (18). A more pedagogical method is to use (16) and (18) to derive (19) from (17).

 

[latentcorpse]

He means the identity matrix. I think it's a font artifact from using some hack to generate a fancy blackboard bold 1 (something like $\mbox{1 \hspace{-1.0mm} {\bf l}}$.)



You want to use (16) and (19) to show that (17) implies (18). A more pedagogical method is to use (16) and (18) to derive (19) from (17).

Thanks for both your replies.

Why is X in Spin(3,1) to start with? I think I've confused myself with what Spin(3,1) actually is?

and then the matrix in (16) is 2x2 and has complex entries suggesting it is in GL(2,C)
Why would it be in SL(2,C)? For this to be the case, we would need the det to be 1.
It tells us below that det $$=x_0^2-x_1^2-x_2^2-x_3^2$$. Is this necessarily equal to 1? If so, why?

As for the overall proof, are you saying that I should evaluate

$$\tilde{X} =N \Lambda^\mu_\nu X_\mu \sigma^\nu N^\dagger$$

or is this completely off base? Even if it's right I'm not really sure what I'm doing - am I checking to see that if we transform X, $$\tilde{X}$$ transforms in the way described in (18)?

Thanks very much.

 

[fzero]

Why is X in Spin(3,1) to start with? I think I've confused myself with what Spin(3,1) actually is?

It isn't. X is a 4-vector and Spin(3,1) is the Lorentz group that acts on these.

As for the overall proof, are you saying that I should evaluate

$$\tilde{X} =N \Lambda^\mu_\nu X_\mu \sigma^\nu N^\dagger$$

or is this completely off base?

It's better to label that as $\tilde{X}'$ since it's the result of a transformation. It's also important to keep track of indices a bit better (though the text isn't careful either), since ${\Lambda^\mu}_\nu = - {\Lambda_\nu}^\mu$.

Even if it's right I'm not really sure what I'm doing - am I checking to see that if we transform X, $$\tilde{X}$$ transforms in the way described in (18)?

As I said, that's one way to do it. Show that the solution for the map is consistent with the transformation rules. Though it seems better to use the 2nd way I pointed out, which has you explicitly derive the form of the map instead of guessing.

 

[latentcorpse]

It isn't. X is a 4-vector and Spin(3,1) is the Lorentz group that acts on these.



It's better to label that as $\tilde{X}'$ since it's the result of a transformation. It's also important to keep track of indices a bit better (though the text isn't careful either), since ${\Lambda^\mu}_\nu = - {\Lambda_\nu}^\mu$.



As I said, that's one way to do it. Show that the solution for the map is consistent with the transformation rules. Though it seems better to use the 2nd way I pointed out, which has you explicitly derive the form of the map instead of guessing.

Thanks for the reply. What is the definition of Spin(3,1)

So this means that $$\Lambda \in Spin(3,1)$$ right?

And I can't see how proof is going to work. 16 and 18 tell us what $$\tilde{X}$$ is and how it transforms. How can we use this to establish $$\Lambda^\mu{}_\nu$$

Why do [itex] tags not work anymore?

Cheers.

 

[fzero]

Thanks for the reply. What is the definition of Spin(3,1)

So this means that $$\Lambda \in Spin(3,1)$$ right?

Spin(3,1) is the double cover of SO(3,1). The projective representations one picks up are the spinor representations. At this point, $\Lambda$ may as well be considered to be in SO(3,1), since we're not dealing with spin. There are any number of QFT or othere texts that would go into more detail.

And I can't see how proof is going to work. 16 and 18 tell us what $$\tilde{X}$$ is and how it transforms. How can we use this to establish $$\Lambda^\mu{}_\nu$$

You should write down some equations for $\tilde{X}'$ and then stare at them a bit. It's really obvious how to read off $\Lambda$.

Why do [itex] tags not work anymore?

The tex plugin was changed some months back.

 

[latentcorpse]

Spin(3,1) is the double cover of SO(3,1). The projective representations one picks up are the spinor representations. At this point, $\Lambda$ may as well be considered to be in SO(3,1), since we're not dealing with spin. There are any number of QFT or othere texts that would go into more detail.



You should write down some equations for $\tilde{X}'$ and then stare at them a bit. It's really obvious how to read off $\Lambda$.



The tex plugin was changed some months back.

I get $$\tilde{X}' = N \tilde{X} N^\dagger = N X_\mu \sigma^\mu N^\dagger$$

But $$X'_\mu=\Lambda^\nu{}_\mu X_\nu$$ and I can't find an X' term in the line above that would allow me to read off what $$\Lambda^\nu{}_\mu$$ should be.

 

[fzero]

I get $$\tilde{X}' = N \tilde{X} N^\dagger = N X_\mu \sigma^\mu N^\dagger$$

But $$X'_\mu=\Lambda^\nu{}_\mu X_\nu$$ and I can't find an X' term in the line above that would allow me to read off what $$\Lambda^\nu{}_\mu$$ should be.

Your indices are a bit wrong, but we can also write

$\tilde{X}' = {\Lambda^\mu}_\nu X^\nu \sigma_\mu.$

By comparing this with your first line and using an identity for $\sigma$-matrices, we can solve for $\Lambda$.

 

[latentcorpse]

Your indices are a bit wrong, but we can also write

$\tilde{X}' = {\Lambda^\mu}_\nu X^\nu \sigma_\mu.$

By comparing this with your first line and using an identity for $\sigma$-matrices, we can solve for $\Lambda$.

I understand where you got the other expression for $\tilde{X}'$ from so that's fine.

A couple of things:

(i) What was wrong with my indices?

(ii) When we equate our two expressions we get:

$\Lambda^\mu{}_\nu X^\nu \sigma_\mu = N X_\rho \sigma^\rho N^\dagger$

If we right multiply by $\bar{\sigma}^\kappa$ we get

$\Lambda^\mu{}_\nu X^\nu \delta^\kappa{}_\mu = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa$

which simplifies to

$\Lambda^\mu{}_\nu X^\nu = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\mu$

Because $X \in SL(2, \mathbb{C})$, $\text{det } X = X^\alpha X_\alpha = 1$

Therefore, right multiplication by $X_\alpha$ gives

$\Lambda^\mu{}_\nu \delta^\nu{}_\alpha = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\mu X_\alpha$

This of course simplifies to

$\Lambda^\mu{}_\nu = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\mu X_\nu$

Is this on the right lines? How do I get the Pauli matrices next to each other so that I can use an identity for them?

 

[fzero]

I understand where you got the other expression for $\tilde{X}'$ from so that's fine.

A couple of things:

(i) What was wrong with my indices?

You had $\Lambda$ transposed, so you'd be off by a minus sign.

(ii) When we equate our two expressions we get:

$\Lambda^\mu{}_\nu X^\nu \sigma_\mu = N X_\rho \sigma^\rho N^\dagger$

If we right multiply by $\bar{\sigma}^\kappa$ we get

$\Lambda^\mu{}_\nu X^\nu \delta^\kappa{}_\mu = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa$

No, $\sigma^\mu \bar{\sigma}^\kappa \neq \eta^{\mu\kappa}$. The identity that you need is (33) in the notes:

$\mathrm{Tr}\left(\sigma^\mu \bar{\sigma}^\kappa\right) = 2 \eta^{\mu\kappa}.$

which simplifies to

$\Lambda^\mu{}_\nu X^\nu = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\mu$

Because $X \in SL(2, \mathbb{C})$, $\text{det } X = X^\alpha X_\alpha = 1$

This isn't true either, there's no restriction on $\text{det } X$.

You want to write an equation for the coefficient of $X^\nu$. Since $X^\nu$ is arbitrary, the coefficient must vanish. You can solve that equation for $\Lambda$ after tracing over spinor indices.

 

[latentcorpse]

You had $\Lambda$ transposed, so you'd be off by a minus sign.



No, $\sigma^\mu \bar{\sigma}^\kappa \neq \eta^{\mu\kappa}$. The identity that you need is (33) in the notes:

$\mathrm{Tr}\left(\sigma^\mu \bar{\sigma}^\kappa\right) = 2 \eta^{\mu\kappa}.$



This isn't true either, there's no restriction on $\text{det } X$.

You want to write an equation for the coefficient of $X^\nu$. Since $X^\nu$ is arbitrary, the coefficient must vanish. You can solve that equation for $\Lambda$ after tracing over spinor indices.

$\Lambda^\mu{}_\nu X^\nu \sigma_\mu = N X_\rho \sigma^\rho N^\dagger$
$\Lambda^\mu{}_\nu X^\nu \sigma_\mu \bar{\sigma}^\kappa = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa$
$\Lambda_{\mu \nu} X^\nu \sigma^\mu \bar{\sigma}^\kappa = N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa$
$\text{Tr } \left( \Lambda_{\mu \nu} X^\nu \sigma^\mu \bar{\sigma}^\kappa \right) = \text{Tr } \left( N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa \right)$
$2 \Lambda_{\mu \nu} X^\nu \eta^{\mu \kappa} = \text{Tr } \left( N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\kappa \right)$
$\Lambda^\mu{}_\nu X^\nu = \frac{1}{2} \text{Tr } \left( N X_\rho \sigma^\rho N^\dagger \bar{\sigma}^\mu \right)$

How do I move the $\bar{\sigma}^\mu$ to the front on the RHS though?

Cheers.