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Let's say I have the Lagrangian: $$L=T-V.$$ Along with the constraint that $$f\equiv f(\vec q,t)=0.$$ We can then write: $$L'=T-V+\lambda f. $$ What is my Hamiltonian now? Is it $$H'=\dot q_i p_i -L'~?$$ Or something different? I have found at least one example where using the above formula gives a different answer then the Hamiltonian found by decreasing the degrees of freedom by one rather then using Lagrange multipliers.

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Comments to the question (v2):

To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint

$$g ~:=~ \{f,H^{\prime}\}_{PB} +\frac{\partial f}{\partial t}~\approx~\frac{d f}{d t}~\approx~0.$$

[Here the $\approx$ symbol means equality modulo eqs. of motion or constraints.]

References:

  1. P.A.M. Dirac, Lectures on QM, (1964).

  2. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

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The Hamiltonian is defined by $$ H = \sum_{i=1}^n \left( \frac{\partial L}{\partial \dot q_i} \dot q_i \right) - L $$ So in your case: $ H' = H - \lambda f $

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