# String Theory

So I have a few questions about a string theory course I am taking, although I guess the questions are largely on indices/QFT stuff!

(i) Consider the expression
$\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab}$
we were going to take the transpose of this and it was said that the transpose operation only acts on Dirac spinor indices. I thought Dirac spinor indices were the indices attached to the Dirac spinors, $\psi$. However, we calculated
$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T= - \eta_{ab} \partial_\mu \psi^b^T \gamma^\mu^T C^T \psi^a$

i.e. the index on the $\gamma^\mu$ must also be a spinor index as it has been transposed as well. So my question is, what is the actual definition of a spinor index???

(ii) We then showed that
$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T=-\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$
and that
$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^\dagger=+\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$

i.e. that $(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T = - (\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^\dagger$

why does this allow us to conclude that $\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab}$ is real?

(iii) Why does taking the transpose of two anticommuting objects introduce a minus sign?
i.e. if A,B anticommute then why does $(AB)^T=-BA$?

(iii) Does the Polyakov action describe a bosonic particle?

(iv) Why does teh equation of motion $( - \partial_\tau + \partial_\sigma ) \psi_-^a=0$ imply that

$\psi_-^a=k^a(\sigma + \tau)$ i.e. a left moving wave?

I tried substituting it in and I get $-\dot{h}(\sigma + \tau} + h'(\sigma + \tau)$ where dot is tau derivative and ' is sigma derivative. Why should this thing equal zero though?

(v) Given that $\delta \psi^a = \gamma^\nu \epsilon \partial_\nu X^a$ we showed

$\delta \bar{\psi}^a = \delta ( {\psi^a}^T C ) = \partial_\nu X^a ( \epsilon^T {\gamma^\nu}^T C) = - \partial_\nu X^a ( \epsilon^T C \gamma^\nu C^{-1} C ) = - \partial_\nu X^a \bar{\epsilon} \gamma^\nu$
I agree with all of this except I don't understand why the $\partial_\nu X^a$ at the front isn't also transposed?

(vi) Consider the following manipulations:

$\bar{\psi}^a \gamma^\mu \partial_\mu ( \gamma^\nu \epsilon \partial_\nu X^b ) \eta_{ab}$
where $\epsilon$ is a constant anticommutating Majorana spinor that generates the supersymmetry.

We can write this as

$\bar{\psi}^a \gamma^\mu \partial_\nu ( \gamma^\nu \epsilon \partial_\mu X^b ) \eta_{ab}$
This is using the fact that since gamma and epsilon are constant the only contributing term is the one with teh double derivative and then we can use commutativity of partial derivatives to swap $\mu$ and $\nu$

Then (since this is found in the supersymmetry action, it is going to be getting integrated over the string worldsheet), we integrate by parts to get:

$-\partial_\nu \bar{\psi}^a \gamma^\mu \gamma^\nu \epsilon \partial_\mu X^b \eta_{ab}$
I get where this term comes from but what happened to the surface term?
Then this becomes

$- ( \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T \partial_\nu X^a \eta_{ab}$
I do not follow this last step AT ALL!

Thanks very much for any help!

Last edited:

dextercioby
Homework Helper
(iii) because they're anticommuting. So the transposition of the spinor matrices will invert their order, however, because the objects carry Grassmann parity 1, the minus sign pops up to account for that.

(ii) I think there's something fishy with what you wrote. Why would the spacetime derivative switch between spinors ? Or maybe I'm missing something, or probably to make the calculations to verify whether they're right or not. Anyways, perhaps it helps to know that on a Grassmann algebra, an element with definite parity is real wrt an involution iff it's equal to its involute.

(i) the Gamma matrices carry spinorial indices, of course. A spinor index is used each time we're dealing with matrix elements of the operators of finite dimensional representations of a spinor group (SL(2,C) in this case).

(iii) because they're anticommuting. So the transposition of the spinor matrices will invert their order, however, because the objects carry Grassmann parity 1, the minus sign pops up to account for that.

(ii) I think there's something fishy with what you wrote. Why would the spacetime derivative switch between spinors ? Or maybe I'm missing something, or probably to make the calculations to verify whether they're right or not. Anyways, perhaps it helps to know that on a Grassmann algebra, an element with definite parity is real wrt an involution iff it's equal to its involute.

(i) the Gamma matrices carry spinorial indices, of course. A spinor index is used each time we're dealing with matrix elements of the operators of finite dimensional representations of a spinor group (SL(2,C) in this case).

Hey bigubau, thanks for your reply! Unfortunately, I don't really understand what you wrote!

(iii) Is there a way to show this by writing it out? I have no idea what Grassmann parity is about!
I think he wrote something like $(\bar{A}B)^T = (A^\alpha C_{\alpha \beta} B^\beta)^T= \dots$ where A and B are anticommuting spinors. Can this be used to show why the minus sign appears?

(i) Can you explain what a spinor index is at a more basic level? My knowledge of groups is fairly limited. Why does the gamma carry a spinor index?

Thanks.

fzero
Homework Helper
Gold Member
This is way too many questions for one post. They're not even related to one another.

(i) A spinor index is the index labeling the vector space of a spinor representation of the Lorentz group. The gamma matrices are intertwiners that take a spinor and antispinor representation to the vector representation of the Lorentz group. You should review this material in Weinberg or any decent QFT book.

(ii)
$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T = - (\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^\dagger$

On the RHS, we have the Hermitian conjugate, which is a transpose + complex conjugation. So the LHS and RHS differ by a complex conjugation, which in the case of Grassman variables, comes with an extra minus sign by convention:

$$(\chi \psi)^* = \psi^* \chi^* = - \chi^* \psi^*$$

So the c.c. of the transpose = transpose, we conclude that the quantity is real.

(iii) Consider

$$\begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2\psi_2,$$

while

$$\begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix} =\psi_1\chi_1+ \psi_2\chi_2= -(\chi_1 \psi_1 + \chi_2\psi_2),$$

so transpose must come with a minus sign for every time we commute odd variables.

(iii) The Polykaov action is 2d, it describes a bosonic string.

dextercioby
Homework Helper
(iii) Is there a way to show this by writing it out? I have no idea what Grassmann parity is about!
I think he wrote something like $(\bar{A}B)^T = (A^\alpha C_{\alpha \beta} B^\beta)^T= \dots$ where A and B are anticommuting spinors. Can this be used to show why the minus sign appears?

You don't need to use index gymnastics to show that. You know that your spinors are finite matrices, so $(AB)^{T} = B^{T}A^{T}$. You need though an extra minus sign, because the spinors are really anticommuting. If you don't have the minus, then by plugging A=B, you would NOT have the A^2 =0 condition anymore.

(i) Can you explain what a spinor index is at a more basic level? My knowledge of groups is fairly limited. Why does the gamma carry a spinor index?

I'm aware of the following explanation which I can give to you.

The matrix representations of the 2 psi's are different, one is a 4x1 matrix (the barred one), the other is a 1x4matrix (the one w/o a bar on top of it). To link them we need a 4x4 matrix which would automatically carry one index for the first spinor and one index from the other. The product of the 3 objects must be a real number, that is a 1x1 matrix. So gamma-s carry spinor indices.

I saw you added several points to the original problem. I can't help you on these, because my knowledge of SUSY and strings is very, very well approximated by the number 0.

fzero
Homework Helper
Gold Member
(iv) Why does teh equation of motion $( - \partial_\tau + \partial_\sigma ) \psi_-^a=0$ imply that

$\psi_-^a=k^a(\sigma + \tau)$ i.e. a left moving wave?

I tried substituting it in and I get $-\dot{h}(\sigma + \tau} + h'(\sigma + \tau)$ where dot is tau derivative and ' is sigma derivative. Why should this thing equal zero though?

It's horribly misleading to use dot and ' notation for a function that is the sum of one variable. Since we use the notation

$$f'(x) = \frac{df(x)}{dx}$$

then

$$\psi'(\sigma+\tau) = \frac{d\psi(\sigma+\tau)}{d(\sigma+\tau)}.$$

You can relate these derivatives to $$\partial_{\tau,\sigma} \psi$$ by using the chain rule.

(v) Given that $\delta \psi^a = \gamma^\nu \epsilon \partial_\nu X^a$ we showed

$\delta \bar{\psi}^a = \delta ( {\psi^a}^T C ) = \partial_\nu X^a ( \epsilon^T {\gamma^\nu}^T C) = - \partial_\nu X^a ( \epsilon^T C \gamma^\nu C^{-1} C ) = - \partial_\nu X^a \bar{\epsilon} \gamma^\nu$
I agree with all of this except I don't understand why the $\partial_\nu X^a$ at the front isn't also transposed?

$\partial_\nu X^a$ has no spinor indices.

(vi) Consider the following manipulations:

$\bar{\psi}^a \gamma^\mu \partial_\mu ( \gamma^\nu \epsilon \partial_\nu X^b ) \eta_{ab}$
where $\epsilon$ is a constant anticommutating Majorana spinor that generates the supersymmetry.

We can write this as

$\bar{\psi}^a \gamma^\mu \partial_\nu ( \gamma^\nu \epsilon \partial_\mu X^b ) \eta_{ab}$
This is using the fact that since gamma and epsilon are constant the only contributing term is the one with teh double derivative and then we can use commutativity of partial derivatives to swap $\mu$ and $\nu$

Then (since this is found in the supersymmetry action, it is going to be getting integrated over the string worldsheet), we integrate by parts to get:

$-\partial_\nu \bar{\psi}^a \gamma^\mu \gamma^\nu \epsilon \partial_\mu X^b \eta_{ab}$
I get where this term comes from but what happened to the surface term?

If you're dealing with a closed string, there is no boundary to support the surface term.

Then this becomes

$- ( \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T \partial_\nu X^a \eta_{ab}$
I do not follow this last step AT ALL!

Thanks very much for any help!

I have no idea where the transpose came from, but ignoring that, it's just a relabeling of $$\mu,\nu$$. I don't think the transpose should be there.

the transpose should be there since you need an $$\bar{\epsilon}$$ term to make the action invariant under super-symmetry transformations and you can transpose it freely since it is a scalar

This is way too many questions for one post. They're not even related to one another.
(ii)
$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T = - (\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^\dagger$

On the RHS, we have the Hermitian conjugate, which is a transpose + complex conjugation. So the LHS and RHS differ by a complex conjugation, which in the case of Grassman variables, comes with an extra minus sign by convention:

$$(\chi \psi)^* = \psi^* \chi^* = - \chi^* \psi^*$$

So the c.c. of the transpose = transpose, we conclude that the quantity is real.

But even if i accept that the cc introduces a minus sign, our equation tells us that dagger=-transpose i.e. that c.c. of transpose = - transpose.
but you wrote that c.c. of transpose = transpose?

(iii) Consider

$$\begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2\psi_2,$$

while

$$\begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix} =\psi_1\chi_1+ \psi_2\chi_2= -(\chi_1 \psi_1 + \chi_2\psi_2),$$

so transpose must come with a minus sign for every time we commute odd variables.

So let's just say $\chi = \begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix}$ and $\psi= \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}$

Then $\chi \psi = \begin{pmatrix} \chi_1 \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2 \psi_2$

Now $- \psi^T \chi^T = - \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix} = - (\psi_1 \chi_1 + \psi_2 \chi_2 ) = - (- (\chi_1 \psi_1 + \chi_2 \psi_2)) = \chi_1 \psi_1 + \chi_2 \psi_2$ where i used the anticommutativity in the 2nd to last equality

but then this means that $\chi \psi = - \psi^T \chi^T$ and we wanted to show $( \chi \psi)^T = - \psi^T \chi^T$. Where did I go wrong?

It's horribly misleading to use dot and ' notation for a function that is the sum of one variable. Since we use the notation

$$f'(x) = \frac{df(x)}{dx}$$

then

$$\psi'(\sigma+\tau) = \frac{d\psi(\sigma+\tau)}{d(\sigma+\tau)}.$$

You can relate these derivatives to $$\partial_{\tau,\sigma} \psi$$ by using the chain rule.
Ok so we have

$(-\partial_\tau + \partial_\sigma ) \psi_-^a = -\partial_\tau k^a(\sigma + \tau) + \partial_\sigma k^a(\sigma + \tau) = - \frac{\partial k^a(\sigma + \tau)}{\partial \tau} \frac{\partial \tau}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial \sigma} \frac{\partial \sigma}{\partial \sigma} = - \frac{\partial k^a(\sigma + \tau)}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial \sigma}$

So how do I show this is equal to zero now?

$\partial_\nu X^a$ has no spinor indices.
Well we know that transpose only acts on spinor indices. So it won't be transposed by that logic but it will still be moved to the front of the expression because we have transposed the whole thing, right?

If you're dealing with a closed string, there is no boundary to support the surface term.
This kind of makes sense I guess. What do you mean there is no boundary though? A closed string worldsheet is essentially the surface of a cylinder - doesn't this have a boundary?

Thanks again!

fzero
Homework Helper
Gold Member
But even if i accept that the cc introduces a minus sign, our equation tells us that dagger=-transpose i.e. that c.c. of transpose = - transpose.
but you wrote that c.c. of transpose = transpose?

It's hard to be more precise without knowing your conventions. For example are the gamma matrices Hermitian or anti-Hermitian? It might also be the case that

$$i\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab}$$

is the quantity that is real.

So let's just say $\chi = \begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix}$ and $\psi= \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}$

Then $\chi \psi = \begin{pmatrix} \chi_1 \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2 \psi_2$

Now $- \psi^T \chi^T = - \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix} = - (\psi_1 \chi_1 + \psi_2 \chi_2 ) = - (- (\chi_1 \psi_1 + \chi_2 \psi_2)) = \chi_1 \psi_1 + \chi_2 \psi_2$ where i used the anticommutativity in the 2nd to last equality

but then this means that $\chi \psi = - \psi^T \chi^T$ and we wanted to show $( \chi \psi)^T = - \psi^T \chi^T$. Where did I go wrong?

$$\chi\psi$$ is a 1x1 matrix, so it's equal to its transpose. I picked a simple example.

fzero
Homework Helper
Gold Member
Ok so we have

$(-\partial_\tau + \partial_\sigma ) \psi_-^a = -\partial_\tau k^a(\sigma + \tau) + \partial_\sigma k^a(\sigma + \tau) = - \frac{\partial k^a(\sigma + \tau)}{\partial \tau} \frac{\partial \tau}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial \sigma} \frac{\partial \sigma}{\partial \sigma} = - \frac{\partial k^a(\sigma + \tau)}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial \sigma}$

So how do I show this is equal to zero now?

You're not applying the chain rule correctly:

$(-\partial_\tau + \partial_\sigma ) \psi_-^a = -\partial_\tau k^a(\sigma + \tau) + \partial_\sigma k^a(\sigma + \tau) = - \frac{\partial k^a(\sigma + \tau)}{\partial (\sigma +\tau)} \frac{\partial (\sigma +\tau)}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial (\sigma +\tau)} \frac{\partial (\sigma +\tau)}{\partial \sigma}$

Well we know that transpose only acts on spinor indices. So it won't be transposed by that logic but it will still be moved to the front of the expression because we have transposed the whole thing, right?

$$\partial_\nu X^a$$ commutes with everything.

This kind of makes sense I guess. What do you mean there is no boundary though? A closed string worldsheet is essentially the surface of a cylinder - doesn't this have a boundary?

Thanks again!

But $$-\infty < \tau < \infty$$, so there's no boundary at finite times.

Lol instead of paying cambridge you should pay fzero to teach you

Lol instead of paying cambridge you should pay fzero to teach you

Would probably be more effective.

You're not applying the chain rule correctly:

$(-\partial_\tau + \partial_\sigma ) \psi_-^a = -\partial_\tau k^a(\sigma + \tau) + \partial_\sigma k^a(\sigma + \tau) = - \frac{\partial k^a(\sigma + \tau)}{\partial (\sigma +\tau)} \frac{\partial (\sigma +\tau)}{\partial \tau} + \frac{\partial k^a(\sigma + \tau)}{\partial (\sigma +\tau)} \frac{\partial (\sigma +\tau)}{\partial \sigma}$
So this obviously breaks up into four terms:

$-\frac{\partial k^a}{\partial \tau} \frac[\partial \sigma}[\partial \tau} - \frac[\partial k^a}{\partial \tau} + \partial k^a}{\partial \sigma} + \frac{\partial k^a}{\partial \sigma} \frac{\partial \tau}{\partial \sigma}$

But $\sigma$ can label any point on our string - why should it have any dependence on $\tau$? i.e. why should $\frac{\partial \sigma}{\partial \tau} \neq 0$?

$$\partial_\nu X^a$$ commutes with everything.
So I agree that this term won't have spinor indices and therefore doesn't get transposed but how do we know that it commutes with everything?

Thanks.

It's hard to be more precise without knowing your conventions. For example are the gamma matrices Hermitian or anti-Hermitian? It might also be the case that

$$i\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab}$$

is the quantity that is real.

$$\chi\psi$$ is a 1x1 matrix, so it's equal to its transpose. I picked a simple example.

Our gamma matrices are $\gamma_0 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \gamma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

So I find that when we take the hermitian conjugate (i.e. c.c. and transpose) that
$\gamma_0^\dagger=-\gamma_0 \quad \gamma_1^\dagger = \gamma^1$. Does that help?

the transpose should be there since you need an $$\bar{\epsilon}$$ term to make the action invariant under super-symmetry transformations and you can transpose it freely since it is a scalar

So can you explain (all the steps involved) in going from
$\partial_\nu \bar{\psi}^a \gamma^\mu \gamma^\nu \epsilon \partial_\mu X^b \eta_{ab}$
to
$- ( \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T \partial_\nu X^a \eta_{ab}$

So it looks as if we relabelled $\mu \leftrightarrow \nu , a \leftrightarrow b$
I understand that we can transpose scalars freely, so are you saying that the term in the brackets is a scalar? I don't see how that can be?

Thanks!

fzero
Homework Helper
Gold Member
So this obviously breaks up into four terms:

$-\frac{\partial k^a}{\partial \tau} \frac[\partial \sigma}[\partial \tau} - \frac[\partial k^a}{\partial \tau} + \partial k^a}{\partial \sigma} + \frac{\partial k^a}{\partial \sigma} \frac{\partial \tau}{\partial \sigma}$

But $\sigma$ can label any point on our string - why should it have any dependence on $\tau$? i.e. why should $\frac{\partial \sigma}{\partial \tau} \neq 0$?

I never said that $\frac{\partial \sigma}{\partial \tau} \neq 0$, only that you had applied the chain rule in way that was not useful. Go back and finish the calculation and you'll see what I mean.

So I agree that this term won't have spinor indices and therefore doesn't get transposed but how do we know that it commutes with everything?

Thanks.

It's a bosonic field and you're dealing with classical expressions, so ordering doesn't matter. You'd have to be a bit more careful if you were computing quantum correlation functions or something, but for now those details don't matter.

Our gamma matrices are $\gamma_0 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \gamma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

So I find that when we take the hermitian conjugate (i.e. c.c. and transpose) that
$\gamma_0^\dagger=-\gamma_0 \quad \gamma_1^\dagger = \gamma^1$. Does that help?

The other important piece of information you want to do those computations is that the fermions in 2d can be taken to be Majorana-Weyl. In particular this means that the components are real.

You really should go through all the steps of showing that

$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T=-\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$

yourself. One thing that's useful is that the MW condition means that $$\bar{\psi}^b = (\psi^b)^T \gamma_0$$. The other is that the rule for taking complex conjugates of Grassman numbers means that

$$(\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a)^* = - \eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$$

This is the proper reality condition for a product of two Grassman numbers.

It's a bosonic field and you're dealing with classical expressions, so ordering doesn't matter. You'd have to be a bit more careful if you were computing quantum correlation functions or something, but for now those details don't matter.
So bosonic fields will commute with everything then? We only need to worry about commutativity of $\psi$'s and $\gamma$'s i.e. objects with spinor indices then?

The other important piece of information you want to do those computations is that the fermions in 2d can be taken to be Majorana-Weyl. In particular this means that the components are real.

You really should go through all the steps of showing that

$(\bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab})^T=-\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$

yourself.
Yep, I can do this.

One thing that's useful is that the MW condition means that $$\bar{\psi}^b = (\psi^b)^T \gamma_0$$. The other is that the rule for taking complex conjugates of Grassman numbers means that

$$(\eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a)^* = - \eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$$

This is the proper reality condition for a product of two Grassman numbers.

So is there a way (like for the transpose earlier) that you can convince me of this minus sign appearing when the cc is taken?

I tried writing something out along the lines of

$\chi \psi = \begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2 \psi_2$

and

$( \chi \psi )^* = \begin{pmatrix} \chi_1^* & \chi_2^* \end{pmatrix} \begin{pmatrix} \psi_1^* \\ \psi_2^* \end{pmatrix} = \chi_1^* \psi_1^* + \chi_2^* \psi_2^*$

but couldn't relate them in any way to get a minus sign?

fzero
Homework Helper
Gold Member
So bosonic fields will commute with everything then? We only need to worry about commutativity of $\psi$'s and $\gamma$'s i.e. objects with spinor indices then?

When you get to quantization (i.e. Virasoro algebra) there are nontrivial commutation relations for bosonic operators. However here we're purely dealing with classical Lagrangians.

So is there a way (like for the transpose earlier) that you can convince me of this minus sign appearing when the cc is taken?

I tried writing something out along the lines of

$\chi \psi = \begin{pmatrix} \chi_1 & \chi_2 \end{pmatrix} \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \chi_1 \psi_1 + \chi_2 \psi_2$

and

$( \chi \psi )^* = \begin{pmatrix} \chi_1^* & \chi_2^* \end{pmatrix} \begin{pmatrix} \psi_1^* \\ \psi_2^* \end{pmatrix} = \chi_1^* \psi_1^* + \chi_2^* \psi_2^*$

but couldn't relate them in any way to get a minus sign?

As I said in post #4. it's a common convention to choose

$$(\chi \psi)^* = \psi^* \chi^* = - \chi^* \psi^*$$

If we were to choose

$$(\chi \psi)^* = \chi^* \psi^*$$

then we would prove that $$i \eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$$ was real instead.

When you get to quantization (i.e. Virasoro algebra) there are nontrivial commutation relations for bosonic operators. However here we're purely dealing with classical Lagrangians.

As I said in post #4. it's a common convention to choose

$$(\chi \psi)^* = \psi^* \chi^* = - \chi^* \psi^*$$

If we were to choose

$$(\chi \psi)^* = \chi^* \psi^*$$

then we would prove that $$i \eta_{ab} \partial_\mu \bar{\psi}^b \gamma^\mu \psi^a$$ was real instead.

Awesome. Thanks a lot. Do you have any ideas for that transpose thing that I posted about at the bottom of the first page in this thread?

Are you a genius fzero? Hah! Do you work in String Theory or QFT? Your explanations are awesome and thorough, feynmanesque.

fzero
Homework Helper
Gold Member
Awesome. Thanks a lot. Do you have any ideas for that transpose thing that I posted about at the bottom of the first page in this thread?

I think sgd37 is correct in that

$-\partial_\nu \bar{\psi}^a \gamma^\mu \gamma^\nu \epsilon = (- \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T$

up to a sign because it's a 1x1 matrix.

It's also possible that they were doing something different. The SUSY variation of the fermion kinetic term involves two terms:

$$\delta( \bar{\psi}^a \gamma^\mu \partial_\mu \psi^b \eta_{ab} ) = (\delta\bar{\psi}^a) \gamma^\mu \partial_\mu \psi^b \eta_{ab} - \bar{\psi}^a \gamma^\mu \partial_\mu ( \delta \psi^b) \eta_{ab}.$$

The second term gives

$\bar{\psi}^a \gamma^\mu \partial_\mu ( \gamma^\nu \epsilon \partial_\nu X^b ) \eta_{ab}$

while I believe that the first term can be manipulated into

$( \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T \partial_\nu X^a \eta_{ab},$

again up to signs. These terms cancel against the variation of the bosonic action.

I think sgd37 is correct in that

$-\partial_\nu \bar{\psi}^a \gamma^\mu \gamma^\nu \epsilon = (- \partial_\mu \bar{\psi}^b \gamma^\nu \gamma^\mu \epsilon)^T$

up to a sign because it's a 1x1 matrix.
How do we recognise it as 1x1 though since not all the indices are contracted?

I suppose if $\psi^a$ is 2x1 then $\partial_\nu \bar{\psi}^a$ will be 1x2.
The product of the gammas will still be 2x2.
The epsilon is a Majorana spinor so that will also be 2x1

So this term is really (1x2)x(2x2)x(2x1)=1x1. Is that how you did it? Or is there an easier way to see this?

Why does the fermionic contribution cancel the bosonic contribution? This basically means the action will be invariant under SUSY variations - why do we expect or want this to be the case?

And finally, the SUSY variation in question for all the above work was
$\delta X^a = \bar{\epsilon} \psi^a$
$\delta \psi^a = \gamma^\nu \epsilon \partial_\nu X^a$
Where did this come from? Is this just an example that the lecturer must have picked at random? That's what I thought but then it seems a bit of a coincidence that you started talking about exactly the same variations without me telling you what they were? Are they a particularly common choice or something? Presumably, once you pick $\delta X^a$, that fixes $\delta \psi^a$ since otherwise you could pick any kind of transformation for $\psi^a$ that wouldn't necessarily cancel the bosonic contribution?

fzero
Homework Helper
Gold Member
How do we recognise it as 1x1 though since not all the indices are contracted?

I suppose if $\psi^a$ is 2x1 then $\partial_\nu \bar{\psi}^a$ will be 1x2.
The product of the gammas will still be 2x2.
The epsilon is a Majorana spinor so that will also be 2x1

So this term is really (1x2)x(2x2)x(2x1)=1x1. Is that how you did it? Or is there an easier way to see this?

Essentially that counting is the way that you see it. You could also see it by explicitly putting spinor indices on everything. There's another fundamental issue at play here, which is one of physics. This term was supposed to be a candidate for a kinetic energy term in a Lagrangian. If we had not contracted all of the spinor indices, this would not be a Lorentz invariant, and could never have been in the Lagrangian in the first place.

Why does the fermionic contribution cancel the bosonic contribution? This basically means the action will be invariant under SUSY variations - why do we expect or want this to be the case?

There are a couple of reasons why we want worldsheet supersymmetry. The first has to do with a pathology of the bosonic string. If you work out the states that propagate in spacetime for the bosonic string, the lightest state actually has $$m^2<0$$, which means it's a tachyon. These particles travel faster than light and would violate causality. Furthermore, they signal an instability of the vacuum, since they behave like they are at a local maximum in a potential.

It turns out that if you add enough fermions on the worldsheet to make a supersymmetric theory, there is a natural way to project out all tachyonic states. You will no doubt learn about this so called GSO projection soon.

Furthermore, worldsheet SUSY allows us to construct string theories which have SUSY on target space. While SUSY has not yet been found in nature, there are several reasons why it is phenomenologically desirable, such as explaining why the cosmological constant is not absurdly huge and solving the hierarchy problem of the electroweak sector.

And finally, the SUSY variation in question for all the above work was
$\delta X^a = \bar{\epsilon} \psi^a$
$\delta \psi^a = \gamma^\nu \epsilon \partial_\nu X^a$
Where did this come from? Is this just an example that the lecturer must have picked at random? That's what I thought but then it seems a bit of a coincidence that you started talking about exactly the same variations without me telling you what they were? Are they a particularly common choice or something? Presumably, once you pick $\delta X^a$, that fixes $\delta \psi^a$ since otherwise you could pick any kind of transformation for $\psi^a$ that wouldn't necessarily cancel the bosonic contribution?

What you try to pick as a transformation doesn't matter. As long as there are equal numbers of worldsheet bosons and fermions (with appropriate Lagrangians), that transformation is a symmetry, whether someone recognizes it or not.

Essentially that counting is the way that you see it. You could also see it by explicitly putting spinor indices on everything. There's another fundamental issue at play here, which is one of physics. This term was supposed to be a candidate for a kinetic energy term in a Lagrangian. If we had not contracted all of the spinor indices, this would not be a Lorentz invariant, and could never have been in the Lagrangian in the first place.

There are a couple of reasons why we want worldsheet supersymmetry. The first has to do with a pathology of the bosonic string. If you work out the states that propagate in spacetime for the bosonic string, the lightest state actually has $$m^2<0$$, which means it's a tachyon. These particles travel faster than light and would violate causality. Furthermore, they signal an instability of the vacuum, since they behave like they are at a local maximum in a potential.

It turns out that if you add enough fermions on the worldsheet to make a supersymmetric theory, there is a natural way to project out all tachyonic states. You will no doubt learn about this so called GSO projection soon.

Furthermore, worldsheet SUSY allows us to construct string theories which have SUSY on target space. While SUSY has not yet been found in nature, there are several reasons why it is phenomenologically desirable, such as explaining why the cosmological constant is not absurdly huge and solving the hierarchy problem of the electroweak sector.

What you try to pick as a transformation doesn't matter. As long as there are equal numbers of worldsheet bosons and fermions (with appropriate Lagrangians), that transformation is a symmetry, whether someone recognizes it or not.

Today we were discussing the vierbein. This involved the formula

$g_{ab}e^a_\alpha e^b_\beta = \eta_{\alpha \beta}$

Now, I don't really understand what is going on with the indices here. We appear to have mixed and matched abstract indices and basis indices. All we said about them was that $\alpha, \beta$ are tangent space indices. Even that confuses me though because they are "lowered" indices and so surely they should be cotangent space indices. Can you explain what is going on here?

So what is the point of the veirbein? Is it purely a function that relates the curved metric to the flat metric by the above equation?

And then, we said that although under local lorentz transofrmations, the metric is invariant but that $e^a_\alpha \rightarrow e^a_\alpha \Lambda^\alpha_\beta$ I tried to prove this invariance by substitution
Making a Lorentz transformation gives
$g_{ab}e^a_\alpha \Lambda^\alpha_\gamma e^b_\beta \Lambda^\beta_\delta = \eta_{\alpha \beta}$
Now I need to get those $\Lambda$'s over to teh other side where I can use the property $\Lambda^T \eta \Lambda = \eta$. But how do I move them over? Do I just flip the indices on them or do i need to invert the matrices themselves? Either way, I don't see how it is going to give me what I want....

Thanks.