How does Poisson bracket vanish in this equation?

Question

I am studying the book "Lectures on Quantum Field Theory", by Ashok Das. I am stuck in the last step of equation (10.35) as I will explain below.

Under section 10.2 (Dirac method and Dirac bracket), we are trying to go between the Lagrangian formulation and the Hamiltonian formulation. In Lagrangian formulation, the coordinates and the velocities are $q^i$ and $\dot{q}^i$. In Hamiltonian formulation, the coordinates and momenta are $q^i$ and $p_i$.

It is assumed that we can't determine all $\dot{q}^i$ from $p_i$. Namely, we are dealing with a constrained system. Let's say we can find $R$ (out of $N $) velocities in terms of momenta (or vice versa): $$ \dot{q}_a = f^a ( q^i, p_b ), \hspace{2cm} a,b=1,2,...,R, \tag{10.25}$$

We can also write: $$ p_a = g_a( q^i, \dot{q}^b), \hspace{2cm} a,b=1,2,...,R, $$ $$ p_\alpha = g_\alpha( q^i, p_\alpha), \hspace{2cm} \alpha=R+1,R+2,...,N. \tag{10.26} $$

The remaining $N-R$ equations actually define constraints which we can denote as: $$ \phi_\alpha = p_\alpha - g_\alpha(q^i,p_a) = 0. \tag{10.27}$$

Now there is discussion about writing primary Hamiltonian in terms of canonical Hamiltonian with the (primary) constraints as: $$ H_p = H_{can} + \lambda^\alpha \phi_\alpha .\tag{10.31} $$

These $\lambda$ are lagrange multipliers which are undetermined at this stage. It can be shown however that on constrained hypersurface ${\mathrm{T}}$ in the phase space, we can identify these lagrange multipliers with the velocities: $$ \lambda^\alpha \approx \dot{q}^\alpha. \tag{10.34} $$

Now, we want that the constraints be satisfied after time evolution also. This gives:

$$ \dot{ \phi }_\alpha \approx \{ \phi _ \alpha ,H _ {can} + \lambda ^ \beta \phi _ \beta \} $$ $$ \hspace{4cm} = \{ \phi _ \alpha ,H _ {can} \} + \lambda ^ \beta \{ \phi _ \alpha , \phi _ \beta \} + \{ \phi _ \alpha , \lambda ^ \beta \} \phi _ \beta $$ $$ \hspace{3cm} \approx \{ \phi _ \alpha ,H _ {can} \} + \lambda ^ \beta \{ \phi _ \alpha , \phi _ \beta \} \approx 0. \tag{10.35} $$

It seems to me that in last step they have made the substitution: $$ \{ \phi_\alpha , \lambda^\beta \} \phi_\beta \approx 0. $$

Can someone explain why it has to be zero (on the constrained hypersurface)? They say they are using (10.27) in the last step.

Answer 1

The quantity $\{\phi_\alpha, \lambda^\beta\}$ is zero because $\phi_\beta=0$ on the constraint surface. There are derivatives only within the Poisson bracket.

Answer 2

Ref. 1 means that the term $\{\phi_{\alpha},\lambda^{\beta}\}\phi_{\beta}\approx 0$ vanishes on the primary constrained surface $\phi_{\beta}\approx 0$.

Interestingly, this is true even though the canonical Poisson bracket $\{\phi_{\alpha},\lambda^{\beta}\}$ is not defined at this stage of the Dirac-Bergmann analysis! To quote p. 12 in Ref. 2 (using the notation of Ref. 1)

The Poisson bracket $\{\phi_{\alpha},\lambda^{\beta}\}$ is not defined, but it is multiplied by something that vanishes, $\phi_{\beta}$.

The issue is that because of the later consistency condition $\dot{\phi}_{\alpha}\approx 0$, the solution to the $\lambda^{\beta}$ variables might depend on the canonical variables $q^i$ and $p_j$, and we might want to eliminate (some of) the $\lambda^{\beta}$ variables from the onset if possible, cf. Refs. 2 & 3.

References:

1. A. Das, Lectures on QFT, 2008; eq. (10.35).

2. P.A.M. Dirac, Lectures on QM, 1964; eq. (1-31).

3. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; eq. (1.25).