# Weak interaction SU(2) gauge fields W^{1,2,3} and charge?

1. Jun 18, 2015

### Anchovy

When we start by postulating local SU(2) gauge invariance for our weak isospin doublets \begin{align} \psi &= \begin{pmatrix} \nu_{e} \\ e^{-} \end{pmatrix}_{L} \end{align} etc., we have to introduce massless gauge fields to preserve the Lagrangian's invariance. For SU(2) this demands 3 bosons referred to as $W_{\mu}^{1}, W_{\mu}^{2}, W_{\mu}^{3}$.

We then relate the gauge bosons $W_{\mu}^{1}, W_{\mu}^{2}$ with the particles we actually observed which are $W^{+}, W^{-}$ (for now ignoring the photon, $Z^{0}$ and the U(1) hypercharge gauge boson $B_{\mu}^{0}$, and the whole $W^{+}, W^{-}, Z^{0}$ Higgs mechanism mass aspect of the story).

In all the texts I'm reading the author just simply defines the $W^{+}, W^{-}$ as the following mixtures of $W_{\mu}^{1}, W_{\mu}^{2}$:

$W^{+} = \frac{1}{\sqrt{2}} (W_{\mu}^{1} - W_{\mu}^{2})$
$W^{-} = \frac{1}{\sqrt{2}} (W_{\mu}^{1} + W_{\mu}^{2})$

However, the texts never give any mention of charges of the $W_{\mu}^{1}, W_{\mu}^{2}, W_{\mu}^{3}$ so I assume them to be neutral --> it's not clear to me why mixtures of them result in charged bosons?

2. Jun 18, 2015

### ChrisVer

Before the symmetry breaking there is no charge symmetry... the electromagnetic symmetry is a remnant of the electroweak symmetry breaking...

3. Jun 18, 2015

### fzero

Before symmetry breaking, we have a $U(1)$ charge corresponding to the diagonal component of weak $SU(2)$ isospin, $T_3=\sigma_3/2$. The charges of the W-bosons are given by

$$[ T_3, \mathbf{W}_\mu] = \begin{pmatrix} 0 & \sqrt{2} W^+_\mu \\ -\sqrt{2} W^-_\mu & 0\end{pmatrix},$$

so $W^3$ has charge zero, while $W^\pm$ have charge $\pm 1$. After symmetry breaking, the conserved electric charge is

$$Q = T_3 + \frac{Y}{2},$$

where $Y$ is the weak hypercharge. Since the W-bosons don't carry weak hypercharge, the $T_3$ value corresponds directly to their electric charge.

4. Jun 19, 2015

### samalkhaiat

I’m afraid this is not the right “mixture”, you are missing the crucial $i$: Real combinations of REAL fields do not form charge eigen-states.

Yes, they are neutral, because they are real fields. Gauging the group $SU(2)$ leads to three MASSLESS, REAL gauge fields (transforming in the adjoint representation of the gauge group).
Only COMPLEX combinations of (mass degenerate) REAL fields can carry a $U(1)$ charge. In general, the conserved Noether current associated with $U(1)$ symmetry is of the form $$J_{\mu} = i \left( \varphi \partial_{\mu} \varphi^{*} - \varphi^{*} \partial_{\mu} \varphi \right) .$$ Clearly this current vanishes for real field. However, if we write $$\varphi = \frac{1}{\sqrt{2}}(\phi_{1} + i \phi_{2}) , \ \ \ \varphi^{*} = \frac{1}{\sqrt{2}} (\phi_{1} - i \phi_{2}) ,$$ we can rewrite the above current as $$J_{\mu} = \phi_{1}\partial_{\mu}\phi_{2} - \phi_{2}\partial_{\mu}\phi_{1} .$$ So, in order to have a non-zero $U(1)$ (or $SO(2)$) current, you need at least one complex field (or 2 degenerate real fields). This general fact has nothing to do with “before” or “after” spontaneous breaking of the symmetry.

Sam

5. Jun 19, 2015

### Anchovy

Hmm I'm not yet understanding... so our "$U(1)$ charge" is weak hypercharge, right? ie. $Y = 2(Q - T_{3})$, so $Y_{\nu_{e}} = 2(0 - 1/2) = -1$ and $Y_{e^{-}} = 2(-1 - -1/2) = -1$.

I don't follow this part: "$U(1)$ charge corresponding to the diagonal component of weak $SU(2)$ isospin, $T_3=\sigma_3/2$".
The $\sigma_{3}$ you mention is... a Pauli matrix, so that would be...
$$\sigma_{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix} \rightarrow T_{3} = \begin{pmatrix} 1/2 & 0 \\ 0 & -1/2 \end{pmatrix}$$
--> I'm not seeing the connection between $Y_{\nu_{e}} = Y_{e^{-}} = -1$ and the diagonal components of that $T_{3}$ matrix?

6. Jun 19, 2015

### Anchovy

Hmm that is odd, I've double-checked my text that I quoted from and it definitely omits the $i$'s, although another one includes them... anyway this has hit on another thing that I was considering posting a question about. Regarding these so-called 'currents':
$$J_{\mu} = i \left( \varphi \partial_{\mu} \varphi^{*} - \varphi^{*} \partial_{\mu} \varphi \right) .$$
These are strange looking objects to me. If I did not keep reading the term 'current' accompanying these expressions I would not think to myself 'that's a current' ie. 'that's some moving charge or another'. We've got the first term being the product of one field (operator?) and the derivative w.r.t. some spacetime coordinate of the conjugate field, then subtracted from that is the same term but with field and conjugate switched. And we've also got an $i$ in there too. How do I interpret that expression? I'm guessing I'm ignorant of what a 'current' really means in a QFT sense.

7. Jun 19, 2015

### fzero

No, there are 2 $U(1)$s, so let me be clearer.

Before electroweak symmetry breaking (EWSB) we have the global symmetries $SU(2)\times U(1)_Y$. A subgroup of this is $U(1)_I\times U(1)_Y$, where $U(1)_I\subset SU(2)$ and we've put $I=$isospin and $Y=$hypercharge indices to distinguish the two $U(1)$s. The group $U(1)_I$ is generated by $T_3$. The reason to make this particular choice is because, when we choose the hypercharge assignments and Higgs VEV in the usual way, then the Higgs VEV is invariant under the particular linear combination $Q$ given above. So $Q$ generates a $U(1)_Q$ group that survives after EWSB, and we identify the conserved charge corresponding to this global symmetry with the electric charge.

Now you can rearrange your calculations above as computations of $Q$, given hypercharge $Y$ assignments and $T_3$ eigenvalues for the electroweak doublets.

As an aside, in my post above, the 2x2 matrix

$$\mathbf{W}_\mu = \frac{\sigma_i}{2} W^i_\mu,$$

and

$$W^\pm_\mu = \frac{1}{\sqrt{2}} ( W^1_\mu \mp i W^2_\mu).$$

8. Jun 19, 2015

### samalkhaiat

Yes, it is called the symmetry current. You normally cover Noether Theorem when you study QFT. Have you studied Lagrangian field theories? If not, how can you ask questions about local gauge theories? The simplest example is the invariance of Dirac Lagrangian $$\mathcal{L}= i \bar{\psi} ( \gamma^{\mu} \partial_{\mu} ) \psi + C.C. ,$$ under infinitesimal $U(1)$ transformation $\delta \psi = i \epsilon \psi$. The Noether current in this case is given by $$J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)} \delta \bar{\psi} = \epsilon \bar{\psi}\gamma^{\mu} \psi .$$ And the $U(1)$ charge, that you were talking about, is simply the integral $$Q = \int d^3 x \ J^{0} = \int d^3 x \ \psi^{\dagger} \psi .$$ Another example is the complex scalar field Lagrangian $$\mathcal{L} = \partial_{\mu}\varphi^{*} \ \partial^{\mu} \varphi .$$ The $U(1)$ symmetry transformations for this model are $$\delta \varphi = i \epsilon \varphi , \ \delta \varphi^{*} = - i \epsilon \varphi^{*} .$$ You calculate the current from $$J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} \delta \varphi + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi^{*})} \delta \varphi^{*}$$
Or
$$J^{\mu }= i \epsilon \left( \varphi \partial^{\mu}\varphi^{*} - \varphi^{*} \partial^{\mu}\varphi \right) ,$$
and the assotiated charge is again given by the volume integral of the time-component of the current.

9. Jun 19, 2015

### Anchovy

Yes I have studied them a bit but I have a lot of gaps to plug regarding QTFs, although I feel like I'm not too far away from a lot of stuff clicking into place. I'm roughly familiar with that theorem yes, ie. Lagrangian invariance under some transformation --> a conserved quantity emerges. I've been through the basic examples you'd find early on in QFT texts.

Yes I believe I've seen an example of the same sort of Dirac Lagrangian U(1) transformation (although not infinitesimal) when you first start reading about QED --> require $\psi(x)$ invariance under local transformation $\psi(x) \rightarrow \psi'(x) = e^{i\alpha(x)}\psi(x)$ and you have to introduce a gauge field $A_{\mu}$ to preserve the Lagrangian's invariance by modifiying the partial derivative to a covariant derivative $\partial_{\mu} \rightarrow D_{\mu} = \partial_{\mu} + ieA_{\mu}$. And $A_{\mu}$ is identified with the photon.

Actually my overall aim in asking these questions is that I'm trying to understand the same sort of approach for electroweak rather than QED and see how the $W^{\pm}, Z^{0}$ come about. So at this moment I'm roughly OK with how the 3 $W_{\mu}$ fields get introduced, similar approach to how $A_{\mu}$ is needed in QED - but now I need to get from there to the $W^{\pm}, Z^{0}$ by way of a $B^{0}$ boson and some U(1) hypercharge... and then there's a Higgs mechanism to stop these bosons being massless that I'll need to tackle... I've got a rough roadmap in my head but filling in the details is proving tough.

This stuff gets hard for me to picture what's going on. I get uncomfortable through not knowing what these objects $\psi, \overline{\psi}$ explicitly look like. So a Lagrangian term like $i\overline{\psi}(\gamma^{\mu}\partial_{\mu})\psi$ means little to me. Looking at its individual parts I know what each gamma matrix looks like, I know what $\partial_{\mu}$ does, I know there's an Einstein summation over the index $\mu$ going on... but I don't have an appreciation of what that term means in the same way that I could instantly look at something classical like $\frac{1}{2}m\partial_{0}\textbf{x}$ and immediately recognise it as the kinetic energy T in L = T - V.
I mean, I keep reading that it's incorrect to think of these objects $\psi$ as wavefunctions from normal quantum mechanics, rather they're operators for each spacetime point instead of states to be operated on, but these Lagrangians I see never show what these field operators are actually operating on... I go through these texts computing what mathematics I can but I'm missing some fundamental stuff here. I realise this has deviated quite far from where the thread started but it's provided an opportunity to ask so...

10. Jun 19, 2015

### samalkhaiat

The question you asked in this thread IS DIRECTLY related to the Lagrangian formalism of field theories, and it can only be answered properly using the tools of relativistic field theory. So, my advice to you is to start learning about the Lagrangian formalism of relativistic fields. And, by the way which textbook that defines the $W^{\pm}$ by the real combinations of $W_{1}$ and $W_{2}$?

11. Jun 19, 2015

### Anchovy

It's not a textbook actually it's just from some lecture course notes.

12. Jun 19, 2015

### samalkhaiat

Then you should correct those lecture notes by sticking $\pm i$ in the definitions.

13. Jun 19, 2015

### fzero

There's a sense in which we can relate the field operator to the wavefunction. Imagine that $\phi(x)$ is expanded in terms of creation/annihilation operators for states with momentum $\vec{k}$

$$\phi(x) \sim \int d^4k \left( e^{i k_\mu x^\mu} a^\dagger(\vec{k}) + e^{-i k_\mu x^\mu} a(\vec{k}) \right),$$

where I choose a real field for simplicity and leave out conventional normalizations, etc that can be filled in by consulting your favorite QFT text. If we act on the state $|0\rangle$ that has zero particles, then

$$\phi(x) |0\rangle \sim \int d^4k e^{i k_\mu x^\mu} |\vec{k}\rangle$$

where $|\vec{k}\rangle$ is the state containing a single particle of momentum $\vec{k}$ and $e^{i k_\mu x^\mu}$ is the free-particle time-dependent wavefunction for the particle. In fact, if we were to compute

$$\langle \vec{k}' |\phi(x) |0\rangle \sim e^{i k_\mu' x^\mu},$$

we are getting something proportional to the single particle wavefunction. Similarly acting on $|0\rangle$ with products of field operators will be connected to multiparticle states and multiparticle wavefunctions.

A better treatment would more carefully extract the nonrelativistic limit, etc. to better connect QFT to QM. In a more general case, the vacuum state might not be the zero-particle state (like in the conducting band of a metal, etc), but it is clear that the quantum field creates particle states from the vacuum and carries information about their wavefunction.

For a free-particle $|0\rangle$ is the ground state, since adding any particles to it will increase the energy. So this calculation is suggestive that if we compute the expectation value of a current in the ground state $\langle 0 | j^\mu | 0 \rangle$, the result can be related to an expression involving the wavefunctions like $\psi^*(x) \partial \psi(x)$, which gives the probability distribution for the charge based on the probability distribution for the particles. It's perhaps clearest for the spinor where we don't have the derivative operator appearing, but there we would have even more complications involving the nonrelativistic approximation.

I've given a lazy, schematic argument above, but feel that it gives a reasonable picture for connecting relativistic fields in terms of wavefunctions in nonrelativistic QM. If you want to get a better understanding that doesn't suffer the drawbacks of my approach, it would be best to look in a QM or condensed matter text that discusses many body physics via second quantization in terms of nonrelativistic fields, where the issues are more straightforward.

14. Jun 19, 2015

### Anchovy

This I think was useful. So regarding this:
$$\phi(x) \sim \int d^4k \left( e^{i k_\mu x^\mu} a^\dagger(\vec{k}) + e^{-i k_\mu x^\mu} a(\vec{k}) \right),$$
when I'm looking at some Lagrangian featuring some field $\phi(x)$ or $\psi(x)$ or whatever, I can think of it acting on the vacuum $|0\rangle$ to create one particle because the annihilation term achieves nothing and can be ignored. And the integral tells me I'm creating a superposition of all possible momentum states for this particle. What about if it gets applied more than once though, eg. $\phi(x)(\phi(x)|0\rangle)$? This is saying I might end up with one more particle at this point or one less. Is this saying something about what goes on at Feynman diagram vertices? ie. Something along the lines of act on the same spacetime point with two (possibly) different field operators and get a certain probability of destroying or create a particle of with momentum k?

Last edited: Jun 19, 2015
15. Jun 19, 2015

### fzero

Right, just keep in mind that this is true for an ideal case of free-particles that are isolated from interacting with another system. This is typically the case for how the asymptotic states are defined for some high-energy scattering amplitude, but other common examples from condensed matter are very different.

I was still dealing with free-particles so I only wanted to interpret the expression as creating a two particle state. For that to be the case, we should really put the operators at different points or use time or normal ordering to avoid the case where a creation operator cancels against an annihilation operator.

However, it is true that perturbative analysis of interacting theories also involves computing matrix elements of operator products. For example, in $\phi^4$ theory there is a tree-level term (schematically)

$$\langle 3,4 | \phi^4 | 1,2\rangle.$$

As a scattering process, we have an input state with particles 1 and 2 (and associated momenta,etc) that enter the interaction region, interact according to the lowest order process, and then particles 3 and 4 are the output that are detected in the future. Again, instead of "$\phi^4$" we'd have a time-ordered product to keep this better defined. If we expand in the annihilation and creation operators, we'd also see that only the terms with equal numbers of annihilation and creation operators will appear since we are keeping the number of particles in the initial and final states equal to 2.

16. Jun 19, 2015

### Anchovy

OK I should perhaps leave the interaction stuff alone for a while, seems like a bit of a rabbit hole that I can disappear down another time. Going back to earlier, I've got the SU(2) group generators:
$$\sigma^{1} = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} \hspace{1 cm} \sigma^{2} = \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix} \hspace{1 cm} \sigma^{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$
So if I define
$$\sigma^{+} = \frac{1}{2}(\sigma^{1} + i\sigma^{2}) = \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} + \frac{i}{2} \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{pmatrix} + \begin{pmatrix} 0 & \frac{1}{2}\\ \frac{-1}{2} & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$$
$$\sigma^{-} = \frac{1}{2}(\sigma^{1} - i\sigma^{2}) = \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} - \frac{i}{2} \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0\end{pmatrix} - \begin{pmatrix} 0 & \frac{1}{2}\\ \frac{-1}{2} & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$$
I can have these operate on my weak isospin doublet $\psi_{L}$ to get
$$\sigma^{+} \psi_{L} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} \begin{pmatrix} \nu_{e} \\ e^{-} \end{pmatrix} = \begin{pmatrix} 0 \\ e^{-} \end{pmatrix} = \psi_{e^{-}} \hspace{1 cm} \sigma^{-} \psi_{L} = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} \begin{pmatrix} \nu_{e} \\ e^{-} \end{pmatrix} = \begin{pmatrix} \nu_{e} \\ 0 \end{pmatrix} = \psi_{\nu_{e}}$$

The attached image has an expression for the 'currents' $J^{\mu+} = \overline{\psi}_{e^{-}} \gamma^{\mu} (1 - \gamma^{5}) \psi_{\nu_{e}}$ and
$J^{\mu-} = \overline{\psi}_{\nu_{e}} \gamma^{\mu} (1 - \gamma^{5}) \psi_{e^{-}}$. So let's see if I can interpret these objects any better now...

$J^{\mu+}$ is bookended by $\overline{\psi}_{e^{-}}$ and $\psi_{\nu_{e}}$... so I have a $\nu_{e}$ creation operator (for every momentum state) and a $e^{-}$ annihilation operator acting on the vacuum at this $(t, \textbf{x})$? So this expression represents or relates to a charged-current weak interaction with a $W^{-}$ boson emitted?
And there's also $\gamma^{\mu} (1 - \gamma^{5})$ sandwiched in the middle... The $(1 - \gamma^{5})$ ensures we just deal with the left-handed part of the field it acts on I think, ($\psi_{L} = (\frac{1 - \gamma^{5}}{2}) \psi$, not obvious to me how that equation works but for now I'll just believe it)... and that just leaves the $\gamma^{\mu}$ in there, which I'm not sure of the significance of?

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17. Jun 19, 2015

### fzero

These are complex fields so typically $\psi_e^\dagger$ would have a creation operator for the electron and an annihilation operator for the antielectron. $\psi_\nu$ would create antineutrinos and annihilate neutrinos.

Similarly the $W^-$ field annihilates $W^-$ bosons and creates $W^+$. So $J^+ W^-$ describes processes like $e^+\nu_e\rightarrow W^+$ or $W^-\rightarrow e^-\bar{\nu}_e$. The Hermitian conjugate term $J^- W^+$ leads directly to the other combinations.

The $\gamma^\mu$ is forced by Lorentz invariance because of the 4-vector index on the gauge field. It is also tied by the gauge invariance to the $\gamma^\mu$ that appears in the fermion kinetic term.

18. Jun 20, 2015

### Anchovy

OK thanks for that. One other thing I'm wondering is regarding getting to expressions for $W^{\pm}$ from $\sigma^{\pm}$. I'm aware that the gauge field $W_{\mu}$ can be expressed as linear combinations of the generators: $W_{\mu} = \sum^{3}_{i=1} W^{i}_{\mu}\sigma_{i}$ but it's not clear to me how to get to

$W^{+} = \frac{1}{\sqrt{2}} (W_{\mu}^{1} - iW_{\mu}^{2}) \hspace{1 cm} W^{-} = \frac{1}{\sqrt{2}} (W_{\mu}^{1} + iW_{\mu}^{2})$
from
$\sigma^{+} = \frac{1}{2} (\sigma^{1} + i\sigma^{2}) \hspace{1 cm} \sigma^{-} = \frac{1}{2} (\sigma^{1} - i\sigma^{2})$

Also, something I forgot to ask earlier - I notice that in post #7 you stated the $W_{\mu}$ was a 2x2 matrix, yet the gauge bosons are referred to as vector bosons so I would have expected a row or column vector?

Last edited: Jun 20, 2015
19. Jun 20, 2015

### fzero

The map

$$\frac{\sigma_i}{2} \mathbf {V}^i = \frac{1}{2} \begin{pmatrix} V^3 & V^1 - iV^2 \\ V^1 + iV^2 & - V^3 \end{pmatrix} \equiv \mathbf{V}$$

is precisely an isomorphism between real 3-vectors and complex traceless Hermitian 2x2 matrices. The price of this transformation is that the $su(2)$ algebra now acts on $V$ with the adjoint action:

$$[ U , \mathbf{V} ] = \mathbf{V}'.$$

You gave expressions for $\sigma^\pm$ above, so we can see explicitly that

$$\begin{split} \sigma^+ \mathbf{V} & = \frac{1}{2} \begin{pmatrix} V^1 + iV^2 & - V^3 \\ 0 & 0 \end{pmatrix}, \\ \sigma^- \mathbf{V} & = \frac{1}{2} \begin{pmatrix}0 & 0 \\ V^3 & V^1 - iV^2 \end{pmatrix}. \end{split}$$

Therefore

$$V^1 \pm iV^2 = 2 \text{Tr} ( \sigma^\pm \mathbf{V}).$$

You can convert this into an expression for $W^\pm$ depending on your convention for the definition of $W^\pm$ (there's a factor of $\sqrt{2}$ involved in the one that I used before).