Equivalent Representations of Clifford Algebra I'm reviewing David Tong's excellent QFT lecture notes here and am a little confused by something he writes on page 94.
We've considered the standard chiral representation of the Clifford Algebra, and he is now generalising to a different representation. He writes that this will involve
$$\gamma^{\mu}\rightarrow U\gamma^{\mu}U^{-1} \ \textrm{and} \ \psi \rightarrow U\psi$$
What does the second transformation mean? I don't see how it has anything to do with a representation of the Clifford algebra (which is an assignment of a matrix to every element of the algebra, as far as I know). Does he mean that in the resulting projective representation of the Lorentz group we should transform
$$\psi\rightarrow S[\Lambda]U\psi$$
or am I barking up the wrong tree?
This is all very reminiscent of changing pictures in QM, but I've never talked about representation theory in that context! Is there a rigorous link? 
Many thanks!
 A: I was really confused by something like this too (it was a statement that $Y_{lm}$ was 'a representation' of a group of rotations). The problem is that in physics textbooks the distinction between a group and an action of this group is usually not stressed enough.  
In your case we are considering two representations (lets call them $A$ and $B$). For an element $g_\Lambda$ of the Lorentz group the corresponding linear operators would act on a vector space of Dirac spinors:
$$ A(g_\Lambda)\cdot\psi = S[\Lambda]^\alpha_\beta \psi^\beta,\quad B(g_\Lambda)\cdot\psi = S'[\Lambda]^\alpha_\beta \psi^\beta$$
(I'm dropping coordinate dependence, and I won't write indices any more.)
Now we are saying that the two representations $A$ and $B$ are (unitary) equivalent if there is a unitary transformation $U$ which allows you to 'compensate' for the switch of those representations. So let us take two spinors $\psi$ and $S[\Lambda]\psi$ and transform them like:
$$\psi\to U\psi,\quad S[\Lambda]\psi\to US[\Lambda]\psi$$
But also we have:
$$A\to B \quad\Rightarrow\quad S[\Lambda]\to S'[\Lambda]$$
So, for consistency, we must have:
$$ US[\Lambda]\psi = S'[\Lambda]U\psi \quad\Rightarrow\quad S'[\Lambda] = US[\Lambda]U^{-1}$$
Finally, gathering stuff up:
$$\psi\to U\psi$$
$$S[\Lambda]\to US[\Lambda]U^{-1}$$
