Primary constraints for constrained Hamiltonian systems

I would be most thankful if you could help me clarify the setting of primary constraints for constrained Hamiltonian systems. I am reading Classical and quantum dynamics of constrained Hamiltonian systems by H. Rothe and K. Rothe, and Quantization of gauge system by Henneaux and Teitelboim.

Consider a system with Lagrangian $$L(q,\dot{q})$$ and define the momenta $$p_j = \partial L/\partial{\dot{q}_j}$$, with $$j$$ from 1 to $$n$$ (for $$n$$ degrees of freedom), and the Hessian $$W_{ij}(q,\dot{q}) = \partial^2 L/\partial{\dot{q}_i}\partial{\dot{q}_j}$$. Let $$R_W$$ be the rank of the Hessian $$W_{ij}$$.

Assume that $$L$$ is singular and $$R_W < n$$. In order to ask the question, I'm now quoting how the Rothes describe the setting of primary constraints on pages 26 and 27 of their book.

Let $$W_{ab}$$, ($$a,b = 1,...,R_W$$) be the largest invertible submatrix of $$W_{ij}$$, where a suitable rearrangement of the components has been carried out. We can then solve the eqs. $$p_j = \partial L(q,\dot{q})/\partial{\dot{q}_j} \tag{1}$$ for $$R_W$$ velocities $$\dot{q}_a$$ in terms of the coordinates $$q_i$$, the momenta $$\{p_a\}$$ and the remaining velocities $$\{\dot{q}_\alpha\}$$: $$\dot{q}_a = f_a(q,\{p_b\},\{\dot{q}_\beta\})$$, with $$a,b = 1,...,R_W$$ and $$\beta = R_W + 1,...,n$$.

Inserting this expression into the definition of $$p_j$$, one arrives at a relation of the form $$p_j = h_j(q,\{p_a\},\{\dot{q}_\alpha\})$$. For $$j = a$$ ($$a = 1,...,R_W$$) this relation must reduce to an identity. The remaining equations read $$p_\alpha = h_\alpha (q,\{p_a\},\{\dot{q}_\beta\})$$. But the rhs cannot depend on the velocities $$\dot{q}_\beta$$, since otherwise we could express more velocities from the set $$\{\dot{q}_\alpha\}$$ in terms of the coordinates, the momenta, and the remaining velocities.

This is where the presentation from Rothe stops, and my concern is that equations of the form $$p_\alpha = h_\alpha (q, \{p_a\},\{\dot{q}_\beta\})\tag{2}$$ with all $$\{\dot{q}_\beta\}$$ present can still be possible, and yet one cannot solve for more velocities from the set $$\{\dot{q}_\alpha\}$$ in terms of the coordinates, the momenta and the remaining velocities if the conditions stipulated in the Implicit Function Theorem are not met, for not all equations of the type (2) can be solved implicitly for $$\{\dot{q}_\alpha\}$$. Therefore it is not proven that there are $$(n - R_W)$$ primary constraints of the form $$\phi_\alpha (q,p) = 0$$.

Henneaux and Teteilboim even state that these $$(n - R_W)$$ constraints of the form $$\phi_\alpha (q,p) = 0$$ are functionally independent, but give no justification to this statement.

I would be most thankful if you could help clarify my above concern and also if you could clarify the statement by Henneaux and Teitelboin as to the fact that the constraints are functionally independent.

• – Qmechanic Dec 15 '13 at 10:31

OP wrote

This is where the presentation from Rothe stops, and my concern is that equations of the form $p_\alpha = h_\alpha (q, \{p_a\},\{\dot{q}_\beta\})$ (2) with all $\{\dot{q}_\beta\}$ present can still be possible, and yet one cannot solve for more velocities from the set $\{\dot{q}_\alpha\}$ in terms of the coordinates, the momenta and the remaining velocities if the conditions stipulated in the Implicit Function Theorem are not met, for not all equations of the type (2) can be solved implicitly for $\{\dot{q}_\alpha\}$. Therefore it is not proven that there are $(n - R_W)$ primary constraints of the form $\phi_\alpha (q,p) = 0$

I sensed a misunderstanding of Rothe's statements, ignore me if I'm not understanding OP correctly:

Rothe is arguing at least one of the $\dot{q}_\beta$'s can be expressed as a function of $p_a,p_\alpha$ and remaining $\dot{q}_\alpha$'s. For any particular $\alpha$ in your equation (2), by implicit function theorem applied to one-variable($\dot{q}_\beta$) function, it's always doable unless $\frac{\partial h_\alpha}{\partial \dot{q}_\beta}=0$ for our chosen $\alpha$ and $\beta$, but the latter case simply means $h_\alpha$ does not depend on $\dot{q}_\beta$, so either case Rothe's statement is correct.

As for the functional independence part, if I'm not mistaken, is quite trivial. It's because in your equation (2), all the $p_\alpha$'s are on LHS and RHS doesn't contain any $p_\alpha$ thus it's impossible to find a inter-relation among these equations(plural in the sense that $\alpha$ can take many values, from $1$ to $M'$). Or in differential calculus language, the constraint functions from (2) will be $\phi_\alpha(q,p)=p_\alpha-h_\alpha(q,\{p_a\})$, thus the Jacobian of these functions will simply be

$\frac{\partial \phi_\beta}{\partial \{p_\alpha,p_a, q\}}= \begin{bmatrix} 1 & 0 & \cdots &0&\cdots&\frac{\partial h_1}{\partial p_a}&\cdots&\frac{\partial h_1}{\partial q_i}&\cdots\\ 0 & 1 & \cdots&0&\cdots&\frac{\partial h_2}{\partial p_a}&\cdots&\frac{\partial h_2}{\partial q_i}&\cdots\\ \vdots & \vdots & \ddots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0 &0 &\cdots&1&\cdots&\frac{\partial h_{M'}}{\partial p_a}&\cdots&\frac{\partial h_{M'}}{\partial q_i}&\cdots\end{bmatrix}$

And this Jacobian is of maximal rank because of the identity submatrix on the left, and this is the same as saying these contraint functions are functionally independent.

P.S.: Now I'm not quite sure what situation H&T were referring to when they said $M<M'$.

• You are absolutely right, and your argument is flawless. Thank you very much! Can you, please, also indicate as to why the (n - R_W) constraints are functionally independent? Can you prove that? Thanks! – user22208 Apr 6 '13 at 7:54
• Where did Henneaux and Teteilboim say that? I actually read a contrary statement on page 5, somewhere below equation (1.6) – Jia Yiyang Apr 6 '13 at 14:10
• On page 5 HT write: "If the rank of $\partial^2 L/\partial{\dot{q}_n}\partial{\dot{q}_{n^\prime}}$ is equal to $N - M^\prime$ there are $M^\prime$ independent equations among (1.6)." In our case the rank of the Hessian is $R_W$ and there should be $(n - R_W)$ functionally independent constraints, but I don't know how to show that. – user22208 Apr 6 '13 at 15:14
• But they said M' can be strictly smaller than M. In other words, M' can be strictly smaller than n-R_w in Rothe's notation. – Jia Yiyang Apr 6 '13 at 16:24
• In Rothe's notation $R_W = N - M^\prime$ and $n = N$. Therefore $n - R_W = N - (N - M^{\prime}) = M^{\prime}.$ – user22208 Apr 6 '13 at 16:45

I) Let us suppress position dependence $$q^i$$ and explicit time dependence $$t$$ in the following, and also assume that the Lagrangian $$L=L(v)$$ is a smooth function of the velocities $$v^i$$, where $$i=1, \ldots, n$$. The Hessian matrix is defined as

$$H_{ij}~:=~\frac{\partial^2 L}{\partial v^i \partial v^j}.\tag{1}$$

Let us consider an open neighborhood$$^1$$ $$V$$ around a fixed point $$v_{(0)}$$. Now a very important assumption is the so-called regularity condition, cf. Refs. 1 and 2. This means that the rank of the Hessian $$H_{ij}$$ should not depend on the point $$v$$. In other words, the Hessian $$H_{ij}$$ should have constant rank $$r$$. (Ref. 3 implicitly assumes this crucial point without stressing its importance.)

We now permute/rename the velocities $$(v^1,\ldots, v^n)$$ such that the $$r\times r$$ minor $$A_{ab}$$ is invertible in the top left corner of the Hessian

$$H~=~ \begin{bmatrix} A & B \\ C & D \end{bmatrix} ~=~ \underbrace{\begin{bmatrix} 1 & 0 \\ CA^{-1} & 1 \end{bmatrix}}_{\text{invertible}} \begin{bmatrix} A & 0 \\ 0 & D-CA^{-1}B \end{bmatrix} \underbrace{\begin{bmatrix} 1 & A^{-1}B \\ 0 & 1 \end{bmatrix}}_{\text{invertible}}. \tag{2}$$

The renaming is assumed to be done in the same way in the whole neighborhood $$V$$. This is possibly by going to a smaller open neighborhood (which we also call $$V$$) if necessary. (Later when we apply inverse function theorem below we might have to implicitly restrict $$V$$ further.) It follows from the constant rank condition that

$$D~=~CA^{-1}B. \tag{3}$$

II) We next perform the singular Legendre transformation $$v\leadsto p$$. Define functions

$$g_i(v)~:=~\frac{\partial L(v)}{\partial v^i}, \qquad i=1, \ldots, n. \tag{4}$$

The momenta are defined in the Lagrangian theory as

$$p_i~:=~g_i(v), \qquad i=1, \ldots, n. \tag{5}$$

The velocities

$$v^i~\longrightarrow~ (u^a,w^{\alpha}) \tag{6}$$

split into two sets of velocity coordinates

$$u^a, \quad a=1, \ldots, r, \qquad\text{and}\qquad w^{\alpha}, \quad \alpha=1, \ldots, n-r, \tag{7}$$

which we shall call primary expressible (unexpressible) velocities, respectively. Similarly, the momenta

$$p_i~\longrightarrow~ (\pi_a,\rho_{\alpha}) \tag{8}$$

split into two sets of momentum coordinates

$$\pi_a, \quad a=1, \ldots, r, \qquad\text{and}\qquad \rho_{\alpha}, \quad \alpha=1, \ldots, n-r. \tag{9}$$

The primary expressible velocities

$$u^a~=f^a(\pi,w), \qquad a=1, \ldots, r. \tag{10}$$

are extracted from the $$r$$ first momentum relations (5) via the inverse function theorem with the $$w$$-variables as passive spectator parameters.

III) Next define composite functions

$$h_i(\pi,w) ~:=~ g_i(f(\pi,w),w), \qquad i=1, \ldots, n. \tag{11}$$

It follows immediately that

$$h_a(\pi,w) ~=~ \pi_a, \qquad a=1, \ldots, r. \tag{12}$$

because the functions $$g$$ and $$f$$ are each other's inverse for fixed $$w$$. Differentiation of (12) wrt. $$w^{\alpha}$$ leads to

\begin{align} 0~=~&\frac{\partial h_a(\pi,w)}{\partial w^{\alpha}} \cr ~\stackrel{(11)}{=}~&\left.\frac{\partial g_a(u,w)}{\partial w^{\alpha}}\right|_{u=f(\pi,w)} +\left. \frac{\partial g_a(u,w)}{\partial u^b} \right|_{u=f(\pi,w)} \frac{\partial f^b(\pi,w)}{\partial w^{\alpha}} \cr ~\stackrel{(1)+(2)+(4)}{=}&\left.B_{a\alpha}\right|_{u=f(\pi,w)} +\left. A_{ab} \right|_{u=f(\pi,w)} \frac{\partial f^b(\pi,w)}{\partial w^{\alpha}} . \end{align}\tag{13}

Theorem 1. The $$h_i$$-functions (11) do not depend on the $$w$$-variables.

Proof of theorem 1:

\begin{align} \frac{\partial h_{\alpha}(\pi,w)}{\partial w^{\beta}} ~\stackrel{(11)}{=}~&\left.\frac{\partial g_{\alpha}(u,w)}{\partial w^{\beta}}\right|_{u=f(\pi,w)} +\left. \frac{\partial g_{\alpha}(u,w)}{\partial u^a} \right|_{u=f(\pi,w)} \frac{\partial f^a(\pi,w)}{\partial w^{\beta}} \cr ~\stackrel{(1)+(2)+(4)}{=}&\left.D_{\alpha\beta}\right|_{u=f(\pi,w)}+\left. C_{\alpha a} \right|_{u=f(\pi,w)} \frac{\partial f^a(\pi,w)}{\partial w^{\beta}} \cr ~\stackrel{(13)}{=}~&\left. \left(D_{\alpha\beta}-C_{\alpha a}(A^{-1})^{ab}B_{b\beta} \right)\right|_{u=f(\pi,w)}\cr ~\stackrel{(3)}{=}~&0.\end{align}\tag{14}

End of proof. The $$n-r$$ last momentum relations (5) now become $$n-r$$ functionally independent primary constraints

$$\phi_{\alpha}(\pi,\rho)~:=~\rho_{\alpha}- h_{\alpha}(\pi)~\approx~0,\tag{15}$$

where the $$\approx$$ symbol means equality modulo constraints. The primary constraints (15) are clearly functionally independent as each of them depend on different $$\rho$$ momenta.

IV) The Lagrangian energy function is defined as

$$h(v)~\stackrel{(4)}{:=}~g_i(v) v^i -L(v).\tag{16}$$

Define the Hamiltonian as a composite function

\begin{align} H(\pi,w)~:=~&h(f(\pi,w),w)~\cr \stackrel{(10)+(11)+(16)}{=}&h_a(\pi) f^a(\pi,w) +h_{\alpha}(\pi) w^{\alpha} - L(f(\pi,w),w).\end{align}\tag{17}

Theorem 2. The Hamiltonian (17) does not depend on the $$w$$-variables.

Proof of theorem 2:

\begin{align} \frac{\partial H}{\partial w^{\alpha}} ~\stackrel{(17)}{=}~&\left(h_a(\pi) -\left. \frac{\partial L(u,w)}{\partial u^a} \right|_{u=f(\pi,w)} \right) \frac{\partial f^a(\pi,w)}{\partial w^{\alpha}} + h_{\alpha}(\pi) -\left. \frac{\partial L(u,w)}{\partial w^{\alpha}} \right|_{u=f(\pi,w)} \cr ~\stackrel{(4)+(11)}{=}~0.\end{align} \tag{18}

End of proof.

V) Example with $$n=2$$ and $$r=1$$. Let the Lagrangian be

$$L ~=~\frac{1}{2}\frac{u^2}{1-w}~=~\frac{u^2}{2}\sum_{n=0}^{\infty}w^n, \qquad |w|~<~1. \tag{19}$$

The Hessian

$$H_{ij}~=~\begin{bmatrix} \frac{1}{1-w} & \frac{u}{(1-w)^2} \\ \frac{u}{(1-w)^2} & \frac{u^2}{(1-w)^3}\end{bmatrix} \tag{20}$$

has two eigenvalues $$\frac{1}{1-w}+ \frac{u^2}{(1-w)^3}>0$$ and $$0$$, i.e. it has constant rank $$r=1$$ when $$|w|< 1$$.

The first momentum relation

$$\pi~=~\frac{\partial L}{\partial u}~=~ \frac{u}{1-w} \tag{21}$$

can be inverted to yield

$$u~=~ (1-w)\pi. \tag{22}$$

The second momentum relation

$$\rho~=~\frac{\partial L}{\partial w}~=~ \frac{u^2}{2(1-w)^2} ~=~\frac{1}{2}\pi^2\tag{23}$$

$$\phi~:=~ \rho -\frac{1}{2}\pi^2~\approx~0. \tag{24}$$

The Hamiltonian (17) becomes

$$H~=~ \pi u + \rho w -L~=~\frac{1}{2}\pi^2. \tag{25}$$

It is easy to check that there is no secondary constraint. End of example.

References:

1. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, (1994), p. 5-7.

2. D.M. Gitman and I.V. Tyutin, Quantization of fields with constraints, (1990), p. 13-16.

3. H. Rothe and K. Rothe, Classical and quantum dynamics of constrained Hamiltonian systems, (2010), p. 24-27.

--

$$^1$$ We will only make a local argument, i.e. ignore global issues.

• I asked almost this same question here over a year ago, and it's been in the back of my mind ever since. Thank you so much for finally clearing this up for me. I only have one small question still. How are we sure that we can rename the coordinates locally and not just pointwise? – Charles Hudgins Mar 19 '20 at 23:59
• We use the assumption that the rank of the Hessian doesn't jump. – Qmechanic Mar 20 '20 at 5:45
• Looking at it again, it seems clear. 1. Did you draw on a particular reference for this answer, or did you come up with it on your own? Asking since I think I would prefer that reference's exposition to Rothe's on this topic. 2. Also, out of curiosity, does the Lagrangian in that post have some significance, or is it just a good example? – Charles Hudgins Mar 21 '20 at 2:25
• 1. I more or less constructed the proof from scratch. 2. It was just the simplest example I could find. – Qmechanic Mar 22 '20 at 8:07