How do I find the hermitian conjugate of this expression?

[gdumont]

Hi,

I'm trying to do problem 3.5 of Peskin & Schroeder and I don't know where to start.

First of all,

I need to get the hermitian conjugate of the following expression

$$\delta \chi = \epsilon F + \sigma^\mu \partial_\mu \phi \sigma^2 \epsilon^\ast$$

where $\epsilon$ is a 2 component-spinor of grassmann numbers, F a complex scalar field $\sigma^\mu = (I,\sigma^i)$ for $i=1,...,3$ and the $\sigma^i$ are the Pauli matrices, $\phi$ is a complex scalar field.

I think the hermitian conjugate would be something like

$$\delta \chi^\dagger = \epsilon^\dagger F^\ast + \epsilon^T \sigma^2 \sigma^\mu \partial_\mu \phi^\ast$$

Am I right?

Thanks

Guillaume

Moderator note: I took the liberty of editing in your LaTeX tags.

-TM

 

[vanesch]

If you want to, I did this problem and you can find it on my homepage (but that's a spoiler of course).

Also, you can use LaTeX here, you just need to surround your LateX with the instuctions tex (between square brackets) and /tex (also between square brackets). It will be much more readable that way!

 

[gdumont]

I had a look at your solution and there is something I don't understand. At the end of the first page you wrote
$$i \chi^\dagger \bar{\sigma}^\mu \sigma^\nu (\partial_\nu \partial_\mu \phi ) \sigma^2 \epsilon^\ast = i\epsilon^\dagger \sigma^2 \chi^\ast (\partial^\mu \partial_\mu \phi)$$
This seems to imply that $$\bar{\sigma}^\mu \sigma^\nu = g^{\mu \nu}$$!?
I tought that
$$\{ \bar{\sigma}^\mu, \sigma^\nu \} = g^{\mu \nu}$$
where the brackets denote the anticommutator. Am I wrong?

P.S. The command \bar{\sigma} doesn't seem to work in the brackets... sorry about that.

 

[vanesch]

I had a look at your solution and there is something I don't understand. At the end of the first page you wrote
$$i\chi^\dagger \bar{\sigma}^\mu \sigma^\nu (\partial_\nu \partial_\mu \phi ) \sigma^2 \epsilon^\ast = i\epsilon^\dagger \sigma^2 \chi^\ast (\partial^\mu \partial_\mu \phi)$$
This seems to imply that $$\bar{\sigma}^\mu \sigma^\nu = g^{\mu \nu}$$!?
I tought that
$$\{ \bar{\sigma}^\mu, \sigma^\nu \} = g^{\mu \nu}$$
where the brackets denote the anticommutator. Am I wrong?

A quick reply:
I didn't check (did this long ago !) it completely, but as $$\mu,\nu$$ are summed with the partial derivatives, that symmetrises the expression, no ?

 

[dextercioby]

Hi,
I'm trying to do problem 3.5 of Peskin & Schroeder and I don't know where to start.
First of all,
I need to get the hermitian conjugate of the following expression
$$\delta \chi = \epsilon F + \sigma^\mu \partial_\mu \phi \sigma^2 \epsilon^\ast$$
where $\epsilon$ is a 2 component-spinor of grassmann numbers, F a complex scalar field $\sigma^\mu = (I,\sigma^i)$ for $i=1,...,3$ and the $\sigma^i$ are the Pauli matrices, $\phi$ is a complex scalar field.
I think the hermitian conjugate would be something like
$$\delta \chi^\dagger = \epsilon^\dagger F^\ast + \epsilon^T \sigma^2 \sigma^\mu \partial_\mu \phi^\ast$$
Am I right?
Thanks
Guillaume
Moderator note: I took the liberty of editing in your LaTeX tags.
-TM

It looks okay. The Pauli matrices are hermitean (they form a basis in $\mbox{su(2)}$, up to a 1/2 ) and involution on the Grassmann algebra goes, under hermitean conjugation (seen as simultaneous involution and transposing), into transposing.

I don't know how Vanesch came up with a second space-time derivative...

Daniel.

 

[alphaone]

I have a very similar question:
I want to find the hermitian conjugate of
$$\epsilon \sigma^\mu \partial_\mu \psi$$
where psi and epsilon are 2 component spinors of grassmann variables.
In that case I think the hermitian conjugate should be:
$$-\partial_\mu \psi^\dagger \sigma^\mu \epsilon^\dagger$$
My main concern is whether a minus sign arises in a hermitian conjugate when commuting the epsilon past the psi.

 

[alphaone]

I realized that there is a mistake in the last post as the psi inthe first expression should be a (psi)^dagger and a psi in the second however my question remains.

 

[dextercioby]

If the Grassmann parity of the the spinors is 1, then they anticommute, so the "-" sign occurs.

 

[alphaone]

Actually I think that even when they anticommute there should not be a - sign, as when taking the hermitian conjugate we are not commuting them but using the definition of the hermitian conjugate so I would say
$$(\epsilon \psi)^\dagger = \psi^\dagger \epsilon^\dagger$$
but please correct me if I am wrong