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\lhead{Physics 253a}
\chead{October 9}
\rhead{Sidney Coleman's Notes}
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\section*{Lorentz transformations}
Under a Lorentz transformation all vectors transform as
\[a^\mu\rightarrow\Lambda^\mu_\nu a^\mu\]
\noindent where $\Lambda^\mu_\mu$ specifies the Lorentz
transformation. $\Lambda^\mu_\nu$ must preserve the Minkowski space
inner product, that is if
\[ b^\mu\rightarrow \Lambda^\mu_\nu b^\nu\;\;\;\text{ then }\;\;\;
a_\mu b^\mu\rightarrow a_\mu b^\mu\]
This must be true for arbitrary $a$ and $b$. (The equation this
condition gives is $g_{\mu\nu}\Lambda^\mu_\alpha\Lambda^\nu_\beta=
g_{\alpha\beta}$.) We'll be interested in one parameter subgroups of
the group of Lorentz transformations parametrized by $\lambda$. This
could be rotations about some specified axis by an angle $\lambda$ or
a boost in some specified direction by a rapidity $\lambda$. In any
case, the Lorentz transformation is given by a family
\[ a^\mu\rightarrow a^\mu(\lambda)=\Lambda(\lambda)^\mu_\nu a^\nu \]
Under this (active) transformation (we are not thinking of this as a
passive change of coordinates) the fields $\phi^a$ transform as
($\phi^a$ is a scalar),
\[ \phi^a(x)\rightarrow
\phi^a(x,\lambda)=\phi^a(\Lambda(\lambda)^{-1}x) \]
We are restricting ourselves to scalar fields. Even though we only
used scalar fields in our examples of $T^{\mu\nu}$, the derivation of
the conservation of $T^{\mu\nu}$ from space-time translation
invariance applies to tensor, or vector, fields. With Lorentz
transformations, we only consider scalars, because there are extra
factors in the transformation law when the fields are tensorial. For
example a vector field $A^\mu(x)\rightarrow \Lambda^\mu_\nu
A^\nu(\Lambda^{-1}x)$.
We need to get $D\phi=\left.\frac{\partial\phi}{\partial\lambda}
\right|_{\lambda=0}$. We'll define
\[ D\Lambda^\mu_\nu\equiv \epsilon^\mu_\nu\;\;\;\substack{\text{
defines some matrix }\epsilon^\mu_\nu}\]
From the invariance of $a^\mu b_\nu$, we'll derive a condition on
$\epsilon^\mu_\nu$.
\begin{align*}
0&=D(a^\mu b_\nu)=(Da^\mu)b_\nu+a^\mu(Db_\nu)\\
\begin{split}
&=\epsilon^\mu_\nu a^\nu b_\mu+ a^\mu\epsilon_\mu^\nu b^\nu\\
&=\epsilon_{\mu\nu}a^\nu b^\mu+\epsilon_{\nu\mu}a^\nu b^\mu
\end{split}\Downarrow\substack{\text{relabel dummy indices}\\
\text{in the second term,}\\ \mu\rightarrow\nu, \nu\rightarrow\mu.}\\
&=(\epsilon_{\mu\nu}+\epsilon_{\nu\mu})a^\nu b^\mu\Rightarrow
\epsilon_{\mu\nu}=-\epsilon_{\nu\mu}
\end{align*}
\noindent since this has to hold for arbitrary $a$ and $b$. $\mu$ and
$\nu$ range from 0 to 3, so there are $\frac{4\cdot(4-1)}{2}=6$
independent $\epsilon$, which is good since we have to generate 3
rotations (about each axis) and 3 boosts (in each direction).
As a second confidence-boosting check we'll do two examples.
Take $\epsilon_{12}=-\epsilon_{21}=1$, all other components zero.
\begin{align*}
Da^1&=\epsilon^1_2a^2=-\epsilon_{12}a^2=-a^2\\
Da^2&=\epsilon^2_1a^2=-\epsilon_{21}a^1=+a^1
\end{align*}
% inserted figure oct. 9, 2/12, p 66 in 1-214
\begin{center}
\includegraphics[width=6cm]{06-fig1.eps}
\end{center}
This says $a^1$ gets a little negative component proportional to $a^2$
and $a^2$ gets a little component proportional to $a^1$. This is a
rotation, in the standard sense about the positive $z$ axis.
Take $\epsilon_{01}=-\epsilon_{10}=+1$, all other components zero.
\begin{align*}
Da^0&=\epsilon^0_1a^1=-\epsilon_{01}a^1=a^1\\
Da^1&=\epsilon^1_0a^0=-\epsilon_{10}a^0=a^0
\end{align*}
This says $x^1$, which could be the first component of the position of
a particle, gets a little contribution proportional to $x^0$, the
time, which is definitely what a boost in the $x^1$ direction does.
In fact, $Da^0=a^1$, $Da^1=a^0$ is just the infinitesimal version of
\begin{align*}
a^0&\rightarrow \cosh\lambda a^0+\sinh\lambda a^1\\
a^1&\rightarrow \sinh\lambda a^0+\cosh\lambda a^1
\end{align*}
Without even thinking, the great index raising and lowering machine
has given us all the right signs.
Now assuming $\mathcal{L}$ is a scalar, we are all set to get the 6
conserved currents.
From $\Lambda^{-1}(\lambda)\Lambda(\lambda)=1$,
$0=D(\Lambda^{-1}\Lambda)$ and $D\left.\Lambda^{-1\mu}_\nu
\right|_{\Lambda=1}=-\epsilon^\mu_\nu$
\begin{align*}
D\dot{\phi}(x)&=\left.\frac{\partial}{\partial\lambda}\phi^a(\Lambda^{-1}
(\lambda)^\mu_\nu x^\nu)\right|_{\lambda=0}\;\;\;
\substack{\text{There are extra}\\ \text{terms in $D\phi^a$ if}\\
\text{$\phi^a$ are not scalars}}\\
&=\partial_\sigma\phi^a(x)D(\Lambda^{-1}(\lambda)^\sigma_\tau
x^\tau)\\
&=\partial_\sigma\phi^a(x)(-\epsilon^\sigma_\tau)x^\tau=
-\epsilon_{\sigma\tau}x^\tau\partial^\sigma\phi^a(x)\\
\end{align*}
Using the assumption that $\mathcal{L}$ is a scalar depending only on
$x$ through its dependence on $\phi^a$ and $\partial_\mu\phi^a$ we
have
\begin{align*}
D\mathcal{L}&=\epsilon_{\lambda\sigma}x^\lambda \partial^\sigma
\mathcal{L}\\
&=\partial_\mu[\epsilon_{\lambda\sigma}x^\lambda g^{\mu\sigma}
\mathcal{L}]
\end{align*}
The conserved current $J^\mu$ is
\begin{align*}
J^\mu&=\sum_a\pi^\mu_a\epsilon_{\lambda\sigma}x^\lambda\partial^\sigma
\phi^a - \epsilon_{\lambda\sigma}x^\lambda g^{\mu\sigma}\mathcal{L}\\
&=\epsilon_{\lambda\sigma}\left(\sum_a\pi^\mu_ax^\lambda
\partial^\sigma \phi^a - x^\lambda g^{\mu\sigma}\mathcal{L}\right)
\end{align*}
This current must be conserved for all six independent antisymmetric
matrices $\epsilon_{\lambda\sigma}$, so the part of the quantity in
parentheses that is antisymmetric in $\lambda$ and $\sigma$ must be
conserved i.e.~$\lambda_\mu M^{\mu\lambda\sigma}=0$ where
\begin{align*}
M^{\mu\lambda\sigma}&=\left(\sum_a\pi^\mu_ax^\lambda
\partial^\sigma \phi^a - x^\lambda
g^{\mu\sigma}\mathcal{L}\right)-(\lambda\leftrightarrow\sigma)\\
&=x^\lambda\left(\sum_a\pi^\mu_a\partial^\sigma \phi^a - g^{\mu\sigma}
\mathcal{L}\right)-(\lambda\leftrightarrow\sigma)\\
&=x^\lambda T^{\mu\sigma}-x^\sigma T^{\mu\lambda}
\end{align*}
If the $\phi^a$ were not scalars, we would have additional terms,
feeding in from the extra terms in $D\phi^a$. The 6 conserved charges
are
\[J^{\lambda\sigma}=\int d^3x M^{0\lambda\sigma}=\int d^3x(x^\lambda
T^{0\sigma}-x^\sigma T^{0\lambda})\]
For example, $J^{12}$, the conserved quantity coming from invariance
under rotations about the 3 axis, often called, $J^3$ is
($J^i=\frac{1}{2}\epsilon_{ijk}J^{jk}$),
\[ J^3=J^{12}=\int d^3x(x^1T^{02}-x^2T^{01})\]
If we had point particles with
\[T^{0i}(\vec{x},t)=\sum_ap_a^i\delta^{(3)}(\vec{x}-\vec{r}^a(t))\]
$J^3$ would be
\[\sum_a(x^{a1}p_a^2-x^{a2}p_a^1)=\sum_a(\vec{r}^a\times\vec{p}_a)_3
\]
We have found the field theory analog of the angular momentum. The
particles themselves could have some intrinsic angular momentum.
Those contributions to the angular momentum are not in the $J^i$. We
only have the orbital contribution. Particles that have intrinsic
angular momentum, spin, will be described by fields of tensorial
character, and that will be reflected in extra terms in the $J^{ij}$.
So far we have just found the continuum field theory generalization of
three conserved quantities we learn about in freshman physics. But we
have three other conserved quantities, the $J^{0i}$. What are they?
Consider
\[ J^{0i}=\int d^3x[x^0T^{0i}-x^iT^{00}] \]
This has an explicit reference to $x^0$, the time, something we've not
seen in a conservation law before, but there is nothing a priori wrong
with that. We can pull the $x^0$ out of the integral over space. The
conservation law is $\frac{dJ^{0i}}{dt}=0$ so we have
\begin{align*}
0&=\frac{d}{dt}J^{0i}=\frac{d}{dt}\left[ t\int d^3xT^{0i}-\int d^3x
x^iT^{00} \right]\\
&=t\frac{d}{dt}\underbrace{\cancel{\int d^3x T^{0i}}}_{p^i}+
\underbrace{\int d^3x T^{0i}}_{p^i}-\frac{d}{dt}\int d^3x x^i T^{00}
\end{align*}
Dividing through by $p^0$ gives
\[1=\text{constant}=\frac{p^i}{p^o}=\frac{\frac{d}{dt}\int d^3x
x^iT^{00}}{p^0}=\frac{\frac{d}{dt}(\text{``center of
energy''}^i)}{\text{total energy}} \]
This says that the ``center of energy'' moves steadily. $T^{00}$
is the relativistic generalization of mass. This is the relativistic
generalization of the statement that the center of mass moves
steadily. You aren't used to calling this a conservation law, but it
is, and in fact it is the Lorentz partner of the angular momentum
conservation law.
\section*{Internal Symmetries}
There are other conservation laws, like conservation of electric
charge, conservation of baryon number, and conservation of lepton
number that we have not found yet. We have already found all the
conservation laws that are in a general Lorentz invariant theory.
These additional conservation laws will only occur in specific
theories whose Lagrange densities have special properties.
Conservation laws are the best guide for looking for theories that
actually describe the world, because the existence of a conservation
law is a qualitative fact that greatly restricts the form of the
Lagrange density. All these additional charges are scalars, and we
expect the symmetries they come from to commute with Lorentz
transformations. The transformations will turn fields at the same
spacetime point into one another. They will not relate fields at
different spacetime points. Internal symmetries are non-geometrical
symmetries. Historically, the name comes from the idea that what an
internal symmetry did was transform internal characteristics of a
particle. For us internal just means non-geometrical. We'll study
internal symmetries with two examples.
The first example is
\[\mathcal{L} = \frac{1}{2}\sum_{a=1}^2\left(\partial_\mu\phi^a
\partial^\mu\phi^a-\mu^2\phi^a\phi^a\right)-g\left(\sum_a(\phi^a)^2
\right)^2\]
This is a special case of a theory of two scalar fields. Both fields
have the same mass, and the potential only depends on the combination
$(\phi^1)^2+(\phi^2)^2$.
This Lagrangian\footnote{We have left particle mechanics behind and
I'll often use Lagrangian to mean Lagrange density.} is invariant
($D\mathcal{L}=0$) under the transformation
\begin{align*}
\begin{split}
\phi^1&\rightarrow\phi^1\cos\lambda+\phi^2\sin\lambda\\
\phi^2&\rightarrow\phi^1\sin\lambda+\phi^2\cos\lambda
\end{split}\;\;\;\text{$SO(2)$ symmetry}
\end{align*}
The same transformation at every space time point. This is a
rotation\footnote{Clockwise rotation makes the signs come out
conventionally if $b$ and $c$ (which we see a little bit later) are
defined like $a$ and $b$ in Itzykson and Zuber p.~121.} in the
$\phi^1$, $\phi^2$ plane. The Lagrangian is invariant because it only
depends on $(\phi^1)^2+(\phi^2)^2$ and that combination is unchanged
by rotations.
\begin{align*}
D\phi^1&=\phi^2,\;\;\;\;\;\; D\phi^2=-\phi^1 & D\mathcal{L}&=0 &F^\mu&=0\\
J^\mu &=\pi^\mu_1D\phi^1+\pi^\mu_2 D\phi^2=(\partial^\mu\phi^1)\phi^2
-(\partial^\mu\phi^2)\phi^1\\
Q&=\int d^3xj^0=\int d^3x(\partial_0\phi^1\phi^2-\partial_0\phi^2
\phi^1)
\end{align*}
We can get some insight into this quantity by going to the case $g=0$
in which case $\phi^1$ and $\phi^2$ are both free fields and can be
expanded in terms of creation and annihilation ops.
\[\phi^a(x)=\int \frac{d^3k}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}
(a_{\vec{k}}^ae^{-ik\cdot x}+a_{\vec{k}}^{a\dagger}e^{ik\cdot x}) \]
Now we'll compute $Q$. Let's have faith in our formalism and assume
that the terms with two creation ops or two annihilation ops go away.
If they didn't $Q$ wouldn't be time independent since they are
multiplied by $e^{\pm 2i\omega_{\vec{k}}t}$.
\[ Q=\int \frac{d^3k}{2\omega_{\vec{k}}} [a_{\vec{k}}^1
a_{\vec{k}}^{2\dagger}(-i\omega_{\vec{k}}-i\omega_{\vec{k}})+
2i\omega_{\vec{k}}a_{\vec{k}}^2a_{\vec{k}}^{1\dagger}] \]
We don't have to worry about the order because
$a_{\vec{k}}^1,a_{\vec{k}'}^{2\dagger}]=0$
\[Q=i\int d^3k[a_{\vec{k}}^{1\dagger}a_{\vec{k}}^2
-a_{\vec{k}}^{2\dagger}a_{\vec{k}}^1\]
We are within reach of something intuitive. Define
\begin{align*}
b_{\vec{k}}&=\frac{a_{\vec{k}}^1+ia_{\vec{k}}^2}{\sqrt{2}} &
b_{\vec{k}}^\dagger&=\frac{a_{\vec{k}}^{1\dagger}
-ia_{\vec{k}}^{2\dagger}}{\sqrt{2}}
\end{align*}
That's not the end of the definitions; we need the other combination
to reconstruct $a_{\vec{k}}^1$, $a_{\vec{k}}^2$,
$a_{\vec{k}}^{1\dagger}$, $a_{\vec{k}}^{2\dagger}$. So define,
\begin{align*}
c_{\vec{k}}&=\frac{a_{\vec{k}}^1-ia_{\vec{k}}^2}{\sqrt{2}} &
c_{\vec{k}}^\dagger&=\frac{a_{\vec{k}}^{1\dagger}
+ia_{\vec{k}}^{2\dagger}}{\sqrt{2}}
\end{align*}
These linear combinations of operators are allowable
\footnote{Exchanging the role of $b_{\vec{k}}$ and $c_{\vec{k}}$ is
the way of making the signs come out conventionally with
counterclockwise rotation.}. They are operators that create (or
destroy) a superposition of states with particle 1 and particle 2.
If one state is degenerate with another, and for a given $\vec{k}$,
$|\vec{k},1\rangle$ is degenerate with $|\vec{k},2\rangle$, it is
often convenient to work with linear combinations of these states as
basis states. Because $b_{\vec{k}}^\dagger$ and $c_{\vec{k}}^\dagger$
create orthogonal states, $b_{\vec{k}}$ and $c_{\vec{k}}^\dagger$
commute with each other, as is easily checked. The useful thing about
these linear combinations is that $Q$ has a simple expression.
\begin{align*}
i(a_{\vec{k}}^{1\dagger}a_{\vec{k}}^2-a_{\vec{k}}^{2\dagger}
a_{\vec{k}}^1)&=i\frac{b_{\vec{k}}^\dagger+c_{\vec{k}}^\dagger}
{\sqrt{2}}\cdot\frac{b_{\vec{k}}-c_{\vec{k}}}{\sqrt{2}}
+i\frac{b_{\vec{k}}^\dagger-c_{\vec{k}}^\dagger}
{\sqrt{2}}\cdot\frac{b_{\vec{k}}+c_{\vec{k}}}{\sqrt{2}}\\
&=\frac{1}{2}[2b_{\vec{k}}^\dagger b_{\vec{k}}+0b_{\vec{k}}^\dagger
c_{\vec{k}}+0 c_{\vec{k}}^\dagger b_{\vec{k}}-2 c_{\vec{k}}^\dagger
c_{\vec{k}}] = b_{\vec{k}}^\dagger b_{\vec{k}}-c_{\vec{k}}^\dagger
c_{\vec{k}}
\end{align*}
That is,
\[Q=\int d^3k(b_{\vec{k}}^\dagger b_{\vec{k}}-c_{\vec{k}}^\dagger)=
N_b-N_c \]
The $b$'s carry $Q$ charge +1, the $c$'s carry $Q$ charge -1. It's
like particles and antiparticles; the $b$'s and $c$'s have the same
mass and opposite charge. $Q=N_b-N_c$ is true as an operator
equation. $b$ and $c$ type mesons are eigenstates of $Q$. 1 and 2
type mesons are not. An eigenstate of $N_a$ and $N_b$ is an
eigenstate of $Q$, an eigenstate of $N_1$ and $N_2$ is not. By the
way $H$ and $\vec{P}$ have familiar forms in terms of the $b$'s and
$c$'s.
\begin{align*}
H&=\int d^3k\omega_{\vec{k}}(b_{\vec{k}}^\dagger b_{\vec{k}}
+c_{\vec{k}}^\dagger c_{\vec{k}}) & \vec{P}&= \int d^3k\vec{k}
(b_{\vec{k}}^\dagger b_{\vec{k}} +c_{\vec{k}}^\dagger c_{\vec{k}})
\end{align*}
Taking all these linear combinations suggests that we could have
changed bases earlier in the calculation and simplified things.
\begin{align*}
\psi&=\frac{\phi^1+i\phi^2}{\sqrt{2}}\\
&=\int\frac{d^3k}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}[b_{\vec{k}}
e^{-ik\cdot x}+c_{\vec{k}}^\dagger e^{ik\cdot x}]
\end{align*}
$\psi$ always diminishes $Q$ by 1 either by annihilating a $b$ type
particle or by creating a $c$ type particle.
\[ [Q,\psi]=-\psi\;\;\;\text{ a $Q$ ``eigenfield''} \]
There is also the hermitian conjugate of $\psi$, $\psi^\dagger$,
\begin{align*}
\psi^\dagger&=\frac{\phi^1-i\phi^2}{\sqrt{2}}\\
&=\int\frac{d^3k}{(2\pi)^{3/2}\sqrt{2\omega_{\vec{k}}}}[c_{\vec{k}}
e^{-ik\cdot x}+b_{\vec{k}}^\dagger e^{ik\cdot x}]
\end{align*}
($\psi^\dagger$ will be denoted $\psi^*$ in the classical limit. A
quantum field $\psi^*$ will be understood to be $\psi^\dagger$.)
$\psi^\dagger$ always increases $Q$ by 1.
$[Q,\psi^\dagger]=+\psi^\dagger$.
$\psi$ and $\psi^\dagger$ have neat commutation relations with $Q$.
$\phi^1$ and $\phi^2$ have messy commutation relations with $Q$.
\begin{align*}
[Q,\phi^1]&=\frac{1}{\sqrt{2}}[Q,\psi+\psi^\dagger]=\frac{1}{\sqrt{2}}
(-\psi+\psi^\dagger)\\
\begin{split}
&=-i\phi^2\\
[Q,\phi^2]&=i\phi^1
\end{split}\;\;\;\substack{\text{In agreement with a formula which}\\
\text{is usually true $[Q,\phi^a]=-iD\phi^a$}}
\end{align*}
Under our transformation $\psi\rightarrow e^{-i\lambda}\psi$,
$\psi^*\rightarrow e^{i\lambda}\psi^*$. This is called a $U(1)$ or
phase transformation. It is equivalent to $SO(2)$.
{\scriptsize Digression; we'll get back to symmetries.}
As quantum fields, $\psi$ and $\psi^\dagger$ are as nice as $\phi^1$
and $\phi^2$. They obey
\begin{align*}
(\square+\mu^2)\psi&=0 & (\square+\mu^2)\psi^\dagger&=0
\end{align*}
They obey simple equal time commutation relations:
\begin{align*}
[\psi(\vec{x},t),\psi(\vec{y},t)]&=0=[\psi^\dagger(\vec{x},t),
\psi^\dagger(\vec{y},t)]=[\psi(\vec{x},t),\psi^\dagger(\vec{y},t)]\\
[\psi(\vec{x},t),\dot{\psi}(\vec{y},t)]&=0=[\psi^\dagger(\vec{x},t),
\dot{\psi}^\dagger(\vec{y},t)]\\
[\psi(\vec{x},t),\dot{\psi}^\dagger(\vec{y},t)]&=i\delta^{(3)}(\vec{x}
-\vec{y})=[\psi^\dagger(\vec{x},t),\dot{\psi}(\vec{y},t)]
\end{align*}
These equations can be obtained by doing something completely idiotic.
Back to the classical Lagrangian which is
\[ \mathcal{L}=\partial_\mu\psi^*\partial^\mu\psi-\mu^2\psi^*\psi\;\;
\; \left(\substack{\text{no }\frac{1}{2}}\right) \]
Imagine a person who once knew a lot of quantum field theory, but has
suffered brain damage, is going to canonically quantize this theory.
He has forgotten that $*$ stands for complex conjugate, and he is
going to treat $\psi$ and $\psi^*$ as if they were independent fields.
Here he goes
\begin{align*}
\pi^\mu_\psi&=\frac{\partial\mathcal{L}}{\partial\partial_\mu\psi}
=\partial^\mu\psi^* & \pi^\mu_{\psi^*}&=\partial^\mu\psi
\end{align*}
The Euler-Lagrange equations he gets are
\[ \partial_\mu\pi^\mu_\psi=\frac{\partial\mathcal{L}}{\partial\psi}\;
\;\;\text{ i.e.~}\;\;\;\square\psi^*=-\mu^2\psi^* \]
The idiot got it right. He also gets $\square\psi=-\mu^2\psi$. How
about that!? Now he says ``I'm going to deduce the canonical
commutation relations'':
\[i\delta^{(3)}(\vec{x}-\vec{y})=[\psi(\vec{x},t),\pi^0_\psi(\vec{y},
t)]=[\psi(\vec{x},t),\partial_0\psi^*(\vec{y},t)] \]
By God, he gets that right, too. How can you work with complex
fields, which can't be varied independently, treat them as if they can
be and nevertheless get the right answers?
We'll demonstrate that this works for the equations of motion. It
works very generally. You can demonstrate that it works for the equal
time commutation relations.
Suppose I have some action $S(\psi,\psi^*)$. The equations of motion
come from
\[0=\delta S=\int d^4x(A\delta\psi+A^*\delta\psi^*) \]
{\sc Naive Approach:} Treating the variations in $\delta\psi$ and
$\delta\psi^*$ as independent we get
\[ A=0 \;\;\;\text{ and }\;\;\;A^*=0 \]
{\sc Sharp Approach:} However $\delta\psi$ is not independent of
$\psi^*$. We can use the fact that $\psi$ and $\psi^*$ are allowed to
be complex. We can make purely real variations in $\psi$ so that
$\delta \psi=\delta\psi^*$ and we can make purely imaginary variations
in $\psi$ so that $\delta\psi=-\delta\psi^*$. From the real variation
we deduce
\[ A+A^*=0 \]
\noindent and from the imaginary variation we deduce
\[ A-A^*=0\]
\noindent which implies $A=A^*=0$.
For the ETCR write $\psi=\frac{\psi_r+i\psi_i}{\sqrt{2}}$,
$\psi^*=\frac{\psi_r-i\psi_i}{\sqrt{2}}$. Nothing tricky or slick.
Back to internal symmetries. For our second example, take
\[\mathcal{L}=\frac{1}{2}\sum_{a=1}^n(\partial_\mu\phi^a\partial^\mu
\phi^a-\mu^2\phi^a\phi^a)-g\left(\sum_{a=1}^n(\phi^a)^2\right)^2\]
This is the same as our first example except we have $n$ fields
instead of 2. Just as in the first example the Lagrangian was
invariant under rotations mixing up $\phi^1$ and $\phi^2$, this
Lagrangian is invariant under rotations mixing up $\phi^1$, $\dots$,
$\phi^n$, because it only depends on $(\phi^1)^2+\cdots+(\phi^n)^2$.
The notations are,
\[\phi^a\rightarrow\underset{\substack{n\times n \text{ rotation
matrix}}}{\sum_b\underbrace{R^a_b}\phi^b} \]
There are $\frac{n(n-1)}{2}$ independent planes in $n$ dimensions and
we can rotate in each of them, so there are $\frac{n(n-1)}{2}$
conserved currents and associated charges. This example is quite
different from the first one because the various rotations don't in
general commute. (They all commuted in the first one by virtue of the
fact that there was only one.) We don't expect the various charges to
commute. If they did they would generate symmetries that commute.
Anyway, for any single rotation axis, the symmetry is just like the
one we had in the first example, so we can read off the current,
\[ J_\mu^{[a,b]}=\partial_\mu\phi^a\phi^b-\partial_\mu\phi^b\phi^a \]
You can't find combinations of the fields that have simple commutation
relations with all the $Q^{[a,b]}$'s. For $n=3$ you can choose the
fields to have a simple commutation relation with one charge, say
$Q^{[1,2]}$. This is just like isospin, which is a symmetry generated
by $I_1$, $I_2$ and $I_3$. You can only choose the particle states
$\pi^+$, $\pi^0$ and $\pi^-$ to be eigenstates of one of them, $I_3$.
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