3
$\begingroup$

I am studying about BRST symmetry from the book of P&S (Peskin's and Schroeder's "An Introduction to QFT", Chapter 16.4). The authors construct a nilpotent charge operator and then they describe how it acts on single particle states. In particular (and this is what my question is about), they state that the BRST charge operator $Q$ converts a single forward-polarized gauge boson to a ghost. From what I understand, this happens in the limit of zero coupling.

  1. I can not see why is it particularly the forward-polarized gauge boson that is turned into a ghost and not any other gauge boson with different polarization... It is not obvious for me from the transformation $Q A_{\mu}^a=D_{\mu}^{ac}c^c$.

  2. I can not see why it is turned into a ghost and only into a ghost, as the covariant derivative also contains a gauge boson field inside (and hence can be expanded into gauge boson creation and annihilation operators when acted upon a state). I assume that this has to do with the zero coupling limit, in which the first term of the covariant derivative is significantly larger than the second and hence it makes sense to neglect the second. $$D_{\mu}^{ac}=\frac{1}{g}\delta^{ac}\partial_{\mu}+f^{abc}A^b_{\mu}.$$ If this reasoning is faulty in any way, please correct me.

Furthermore, I have taken a look at Weinberg's book on QFT (Volume 2, Chapter 15.7) and there the author tries to construct commutation relations between creation/annihilation operators and the BRST charge operator. I would like to understand that part as well, but the text confuses me a bit. In particular, why does Weinberg claim that the state $$|e+\alpha p, \psi\rangle=|e,\psi\rangle+\xi\alpha Q|\psi\rangle',$$ where $|\psi\rangle$ is a state s.t. $Q|\psi\rangle=0$, $p_{\mu}$ and $e_{\mu}$ are the gauge boson momentum and polarization vectors and $\alpha$ and $\xi$ are constants, is physically equivalent to the state $|e,\psi\rangle=e_{\mu}a^*(\vec{p})|\psi\rangle$ (This is in p.34)? Is it because, acting on the above-mentioned states with the BRST charge operator yields the same result?

Any help will be appreciated!

P.S.: Also, is there by any chance that when Weinberg writes down $|e,\psi\rangle=e_{\mu}a^*(\vec{p})|\psi\rangle$ he actually means $|e,\psi\rangle=e^{\mu}a^*_{\mu}(\vec{p})|\psi\rangle$?

$\endgroup$

2 Answers 2

6
$\begingroup$
  1. It isn't obvious from that transformation alone. Remember that in P&S, the forward and backward polarization vectors are defined as:

\begin{equation} \epsilon ^{\pm}_{\mu} = \frac{1}{\sqrt{2} |\mathbf{k}|} (k^{0}, \pm \mathbf{k}) \end{equation}

The backward-polarized gauge boson ($\epsilon^{-}_{\mu}$) is the only mode that does not satisfy $k^{\mu} \epsilon _{\mu} = 0$, which from the quantities introduced in P&S corresponds to the field $B^{a}$. That is because, from the BRST Lagrangian, the classical equation of motion for this field is:

\begin{equation} \xi B^{a} = \partial ^{\mu} A^{a}_{\mu} \end{equation}

In other words, the field $B^{a}$ is the only non-zero element that emerges from the divergence of $A^{a}_{\mu}$, which in momentum-space corresponds to $k^{\mu}A^{a}_{\mu}$, and which should normally vanish.

The other two possible polarizations (the transverse ones) also cannot transform into a ghost under a BRST transformation, since that would then imply that the two transverse gauge bosons are part of the $\mathcal{H}_{1}$ vector subspace rather than $\mathcal{H}_{0}$. They would not vanish under the operation of the BRST charge, while the forward-polarized gauge boson (that would now belong to $\mathcal{H}_{0}$) does in fact vanish. This in turn runs contrary to the intended result that the $S$ matrix is unitary when projected upon strictly physical states. If the physical states are now those of $\mathcal{H}_{1}$, the argument that leads to unitarity no longer holds.

The only viable case is thus for the transformed gauge boson to be forward-polarized.

  1. The covariant derivative is actually defined as:

\begin{equation} D_{\mu}^{ac} = \delta ^{ac} \partial_{\mu} + gf^{abc} A^{b}_{\mu} \end{equation}

so at the limit of $g \rightarrow 0$, the covariant derivative becomes a regular derivative, thus transforming the gauge boson strictly into a ghost under the BRST charge operation.

In particular, why does Weinberg claim that the state $$|e+\alpha p, \psi\rangle=|e,\psi\rangle+\xi\alpha Q|\psi\rangle',$$ where $|\psi\rangle$ is a state s.t. $Q|\psi\rangle=0$, $p_{\mu}$ and $e_{\mu}$ are the gauge boson momentum and polarization vectors and $\alpha$ and $\xi$ are constants, is physically equivalent to the state $|e,\psi\rangle=e_{\mu}a^*(\vec{p})|\psi\rangle$ (This is in p.34)?

Because physical modes require to uphold $k^{\mu}\epsilon_{\mu} =0$ as we explained, therefore for a lightlike particle such as the non-Abelian gauge bosons (before the advent of the Higgs mechanism), any arbitrary change:

\begin{equation} \epsilon_{\mu} ' = \epsilon _{\mu} + \alpha k_{\mu} \end{equation}

for constant $\alpha$ maintains the orthogonality relation.

Is it because, acting on the above-mentioned states with the BRST charge operator yields the same result?

No, but the fact $Q$ acting on both of them yields the same result compensates for the fact that, had it not been for this BRST symmetry, the orthogonality relation wouldn't arbitrarily hold (as you probably observed with the existence of non-physical modes without the Fadeev-Popov method). It is essentially ensuring that the trivial equivalency I wrote above (and which Weinberg claims in the text) holds for non-Abelian gauge bosons where there is a polarization vector direction $\lambda$ which gives $k^{\mu}\epsilon ^{\lambda}_{\mu} \neq 0$.

P.S.: Also, is there by any chance that when Weinberg writes down $|e,\psi\rangle=e_{\mu}a^*(\vec{p})|\psi\rangle$ he actually means $|e,\psi\rangle=e^{\mu}a^*_{\mu}(\vec{p})|\psi\rangle$?

The creation and annihilation operators do not have a Lorentz index, so no. The states written are all implied to have Lorentz indices, since you have states of fermions and gauge bosons, the latter of which will have a free Lorentz index. After all, if the creation/annihilation operators for gauge bosons had Lorentz indices, then the gauge boson field operator would suddenly acquire two Lorentz indices (one from these operators, and one from the polarization vector).

$\endgroup$
4
  • 1
    $\begingroup$ if you look at Weinberg's book (Vol. 2), there exist commutation relations in which the Fourier coefficients are vectors and hence they indeed do have Lorentz indices. That does not necessarily imply that the Lorentz indices are assigned to the creation/annihilation operators. It may imply something like $a^{\mu}(\vec{k})=\sum_r\epsilon_r^{\mu}(\vec{k})a_r(\vec{k})$, where $\epsilon_r^{\mu}(\vec{k})$ is the polarization vector. This is standard practice from what I understand. $\endgroup$
    – schris38
    Jul 25, 2022 at 12:58
  • $\begingroup$ Having said that, and looking at the Commutation Rels, I think that he may have missed placing an index at some places, i.e. in the Eq. right below Eq. (15.7.36) or in the first line (in the text) below (15.7.35)... $\endgroup$
    – schris38
    Jul 25, 2022 at 12:59
  • 1
    $\begingroup$ I don't have the book readily available at the moment to check, but in that case you could be right. At a first glance it just seems to me to be a simple state with a fermion and a gauge boson, so having a Lorentz index only for the polarization vector suffices. If some other state is instead implied, your assumption in your question could be correct. $\endgroup$
    – rhomaios
    Jul 25, 2022 at 13:07
  • 2
    $\begingroup$ As far as (1) is concerned, I was looking for a more algebraic way to view why it is the forward-polarized gauge bosons that are turned into ghosts and I think I found one (I will make a complement answer to your already enlightening one). $\endgroup$
    – schris38
    Jul 25, 2022 at 13:13
6
$\begingroup$

I will post an answer because I have understood things in a certain way and I would like to share it. In this way, if it is wrong, I will get to know why it is wrong (if someone is kind enough to correct me) and if it is correct, it will complement @rhomaios's answer in a nice way.

So, after reading both P&S and Weinberg, I came to the conclusion that one can tell whether a transverse, forward-polarized or backward-polarized gauge boson belongs in $\mathcal{H}_0,\mathcal{H}_1$ or $\mathcal{H}_2$ in the following way. Assume $a^{\dagger}_r(\vec{k}),\ r=\pm,1,2$ is the creation operator for a gauge boson, where $+$ denotes forward-polarized gauge bosons, $-$ denotes backward-polarized gauge bosons and $r=\{1,2\}$ will be used to label the transversely polarized gauge bosons. The same notation is (roughly) used in P&S.

Now, assume that we wish to add to a state $|\psi\rangle$ that belongs in $\mathcal{H}_0$ a transversely polarized gauge boson! The way to do it is the following (normalization factors are completely ignored) $$|r,\vec{k};\psi\rangle=a^{\dagger}_r(\vec{k})|\psi\rangle =-\sum_s\delta_{rs}a^{\dagger}_s(\vec{k})|\psi\rangle =-\epsilon_r^T(\vec{k})\cdot\sum_s \epsilon_s^{*T}(\vec{k})\ a^{\dagger}_s(\vec{k})|\psi\rangle =-\epsilon_r^{\mu}(\vec{k}) a^{\dagger}_{\mu}(\vec{k})|\psi\rangle$$ where $r=1,2$ (such that the gauge boson polarization vector is the transverse one) and in the last step I dropped the $T$ superscript and I defined $a^{\dagger}_{\mu}(\vec{k})=\sum_r\epsilon_{r\mu}^*(\vec{k})a^{\dagger}_r(\vec{k})$. According to Weinberg, the BRST charge operator obeys the commutation relation $[Q, \alpha_{\mu}^{\dagger}(\vec{k})]=k_{\mu}c^{\dagger}(\vec{k})$, where $c^{\dagger}(\vec{k})$ is the creation operator we would get if we had expanded the ghost field into normal modes (Note that all the indices regarding the gauge group are suppressed!). Then, acting on the above-mentioned state with the BRST charge operator yields $$Q|r,\vec{k};\psi\rangle=-\epsilon^{\mu}_r(\vec{k})k_{\mu} c^{\dagger}(\vec{k})|\psi\rangle= -k\cdot\epsilon_r(\vec{k})|g,\vec{k};\psi\rangle$$ where with $|g,\vec{k};\psi\rangle$ I have denoted the state containing an additional ghost which carries momentum $\vec{k}$! Now it can be seen that, since $|r,\vec{k};\psi\rangle$ can not be written as a state $\sim Q|\psi_1\rangle$, where $|\psi_1\rangle\in\mathcal{H}_1$, and since $Q|r,\vec{k};\psi\rangle=0$ due to the polarization of the gauge boson chosen to be transverse, then the $|\psi_1\rangle\in\mathcal{H}_1$ state belongs in $\mathcal{H}_0$!

Similar line of reasoning can be applied for demonstrating that forward-polarized gauge bosons belong in $\mathcal{H}_1$ and that backward-polarized gauge bosons belong in $\mathcal{H}_2$. The only difference is that in the latter case, one needs to show that the respective state can be written as the charge operator times a state belonging in $\mathcal{H}_1$, whereas in the former case, one simply needs to show that upon acting with the BRST charge operator on a forward-polarized gauge boson state, the result is not zero!

As far as the zero coupling limit is concerned, indeed, the latter implies that the covariant derivative reduces to the regular derivative in that limit, and hence this is the reason a gauge boson turns into a ghost that carries some momentum, say $p_{\mu}$. This might be a detail, but the book states that the covariant derivative is given by $$D_{\mu}^{ac}=\frac{1}{g}\delta^{ac}\partial_{\mu}+f^{abc}A^b_{\mu} \xrightarrow[]{g\rightarrow0}D_{\mu}^{ac}= \frac{1}{g}\delta^{ac}\partial_{\mu}$$

P.S.#1: And for anyone reading about BRST symmetry from Weinberg's Volume II book on QFT, please be aware that some Lorentz indices may be missing...

P.S.#2: The states used in Weinberg $|e+\alpha p,\psi\rangle$ and $|e,\psi\rangle$ are equivalent because when the BRST charge operator acts on both of them, it yields states that belong to the same subspace (hence actually the same states up to some constants) $$Q|e,\psi\rangle=e_{\mu}p^{\mu}c^{\dagger}|\psi\rangle$$ $$Q|e+\alpha p,\psi\rangle= (e_{\mu}+\alpha p_{\mu})p^{\mu}c^{\dagger}|\psi\rangle$$ and in this way I think it is obvious why the latter state being equivalent to the former implies invariance under a gauge transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.