In David Tong's lecture notes on Gauge Theory, in the section on 'Quantising the Colour Degree of Freedom', the following action is discussed,

$$ S_{w}=\int d \tau i w^{\dagger} \frac{d w}{d t}+\lambda\left(w^{\dagger} w-\kappa\right)+w^{\dagger} A(x(\tau)) w.\tag{2.17} $$

here, the complex vector $w$ is the internal, colour degree of freedom, $A= A_{\mu} d x^{\mu} / d \tau$ is a fixed background gauge field $A_\mu(x)$, and $\lambda$ is a Lagrange multiplier to impose the following constraint:

\begin{align} w^{\dagger} w=\kappa.\tag{2.16} \end{align}

Tong claims (halfway down page 36) that the above constraint

is analogous to Gauss' law when quantising Maxwell theory, and we should impose it as a constraint that defines the physical Hilbert space.

I am confused because I don't see what Gauss' law has to do with defining the physical Hilbert space in QED. In Tong's QFT notes, the physical Hilbert space is defined via the Gupta-Bleuler condition, which is an application of gauge choice, the Lorenz gauge: $\partial_\mu A^\mu=0$. I think that I'm failing to see the connection between the gauge choice, and Gauss' law, in the notes on QED. Thus I'm unable to understand the connection to this context in the Gauge Theory notes.


In David Tong: Lectures on Quantum Field Theory, section 6.2 he discussed 2 ways of Maxwell theory quantisation in different gauge:

  1. Coulomb Gauge

  2. Lorenz gauge.

The Lorenz gauge is a way of Lorentz invariant quantisation, so he gave more details on such a quantisation. But he gave all sufficient information about Coulomb Gauge quantisation.

The essential step is to impose constraints on physical Hilbert space. And such constraints have Gauss law form:

$$ \nabla \cdot \vec{E} = \nabla \cdot \vec{A} = 0 $$

The first equation is universal and must be impossed in all gauges and come from:

Second equation is gauge condition.

Similar pictures occur in YM theory. But Coulomb equation is nonlinear in YM. See section 2.2.1 "Canonical Quantisation of Yang-Mills" from Tong's notes . He give more information, related to YM quantisation.

In 'Quantising the Colour Degree of Freedom' Tong consider classical action for color d.o.f. with fixed lenght $\omega^\dagger \omega = \kappa$. $\kappa$ is fixed real number.

$$S_{w}=\int d \tau\; i w^{\dagger} \frac{d w}{d t}+\lambda\left(w^{\dagger} w-\kappa\right)+w^{\dagger} A(x(\tau)) w.\tag{2.17}$$

So, in full analogy with Maxwell theory one must impose restriction on physical Hilpert space: $$ (\omega_i^\dagger \omega_i - \kappa) |phys \rangle =0 $$

Canonical quantization of ssuch action immideatly lead to:

$$ [\omega_i, \omega_j^\dagger] = \delta_{ij} $$

This equation leads to $\kappa \in Z^+$.

Wilson loop as path integral of parallel transport action is also related to your question (this is discussion of following statements from Tong's notes) .

  • 1
    $\begingroup$ A follow-up question, immediately after this point Tong, gives the normal-ordered constraint, $(\omega_i^\dagger\omega_i - \kappa)|\text{phys}\rangle=0$. He says, "This tells us that the physical spectrum of the theory has precisely $\kappa$ excitations. I don't see how this follows. Is $\kappa$ an operator here? I'm not sure I even understand how the objects in this expression work. $\endgroup$ – Adam Trask Sep 18 '20 at 17:06
  • $\begingroup$ @AdamTrask I have edited the answer. $\endgroup$ – Nikita Sep 18 '20 at 20:54
  • $\begingroup$ Thanks. After working with it a bit more, I am specifically confused by the case of $\kappa = 2$, on the next page of Tong's notes. Why would it yield a space of $\frac{1}{2}N(N+1)$ states? Shouldn't it be $\frac{1}{2}N(N-1)$? You pick one of $w_i^\dagger$ with $i = 1...N$, and then the second pick can't be the same $i$, so a factor $(N-1)$. The factor of $1/2$ comes from $[w_i^\dagger,w_j^\dagger] = 0$. $\endgroup$ – Adam Trask Sep 19 '20 at 17:22
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    $\begingroup$ @AdamTrask, you can choose first $\omega_i^\dagger$ in N ways, second $\omega_j^\dagger$ with $j\neq i$ in $N-1$ ways and with $j=i$ only one way. So we have $N(N-1)/2 + N = N(N+1)/2$. $\endgroup$ – Nikita Sep 19 '20 at 19:08

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