# How does a Lagrangian with delta potential transform to a Hamiltonian?

Suppose the Lagrangian was given as:

$$L = \frac{1}{2}\int_{-\infty}^{\infty} \underbrace{\left(\dot{A(}z)^2-A(z)^2\right)+\left(\dot{Q(}z)^2\delta(z)-Q(z)^2\delta(z)\right)+2\dot{Q(}z)\cdot A(z) \delta(z)}_{= \mathcal{L}} \,\,\,dz\tag{1}$$

where Q is localized at $$0$$. The above are basically two harmonic oscillators that are coupled to each other in the last term. Now, if I wanted to Legendre-transform this in order to get the Hamiltonian, I wouldn't know what to do with the Dirac-delta function.

The conjugate momentum for A(z) is fairly straight forward:

$$P_A = \frac{\partial \mathcal{L}}{\partial \dot{A\,}(z)} = \dot{A\,}(z)\tag{2}$$

but how should I deal with the other conjugate momentum $$P_Q$$? Would it be

$$P_Q = \frac{\partial \mathcal{L}}{\partial \dot{Q\,}(z)} = \dot{Q\,}(z)\delta(z)+A(z)\delta(z)\tag{3}$$

or

$$P_Q = \frac{\partial \mathcal{L}}{\partial (\dot{Q\,}(z)\delta(z))} = \dot{Q\,}(z)+A(z).\tag{4}$$

I run into problems using the former, but the latter does not seem correct.

• 3 is basically correct, and you are running into trouble because for z=0 PQ(z) is zero, and the velocity cannot be expressed in terms of the momenta. This is called a first class constraint. You have to check if the constraint is compatible with thr hamiltonian – lalala Jul 26 at 18:37

1. Lagrangian formulation. The trick is to appreciate that OP's Lagrangian $$L ~=~ \frac{1}{2}\int_{\mathbb{R}} \! dz ( \dot{A}(z)^2 -A(z)^2) ~+~ \frac{1}{2}( \dot{Q}(0)^2 -Q(0)^2)~+~ \dot{Q}(0) A(0) \tag{A}$$ is a combination of a bulk theory in the $$A$$-sector and a boundary theory in the $$Q$$-sector (say, living on a brane located at $$z=0$$).
2. Euler-Lagrange (EL) equations. The bulk EL eq. is $$\ddot{A}(z)+A(z)~\approx~+\delta(z)\dot{Q}(0), \tag{B}$$ while the boundary EL eq. is $$\ddot{Q}(0)+Q(0)~\approx~-\dot{A}(0). \tag{C}$$ If we assume that the bulk field $$A(z)$$ is continuous without $$\delta(z)$$-contributions, then eq. (B) implies that $$\dot{Q}(0)\approx 0$$. Eq. (C) then implies that $$\ddot{A}(0)\approx 0$$. Eq. (B) then implies that $$A(0)\approx 0$$. Eq. (C) then implies that $$Q(0)\approx 0$$. In conclusion: The two harmonic oscillators $$A$$ and $$Q$$ decouple, and the amplitudes vanish at $$z=0$$.
3. Momenta. The bulk momentum is $$P_A(z)~=~\frac{\delta L}{\delta \dot{A}(z)}~=~\dot{A}(z), \tag{D}$$ while the boundary momentum is $$P_Q(0)~=~\frac{\partial L}{\partial \dot{Q}(0)}~=~ \dot{Q}(0)+A(0).\tag{E}$$
4. Hamiltonian formulation. The Hamiltonian gains a bulk and a boundary part \begin{align} H ~&=~ \int_{\mathbb{R}} \! dz ~ P_{A}(z)\dot{A}(z)~+~P_Q(0)\dot{Q}(0)~-~L\cr~&=~\frac{1}{2}\int_{\mathbb{R}} \! dz ( P_{A}(z)^2 +A(z)^2) ~+~ \frac{1}{2}( (P_Q(0)-A(0))^2 +Q(0)^2). \end{align}\tag{F}
• The problem that I am working on explicitly asks to diagonalize the above Hamiltonian. The idea is to expand the $(P_Q(0)-A(0))^2$ term and get two harm. oscill. and one coupling term: $$\frac{1}{2}\int_{\mathbb{R}} \! dz \left[P_{A}(z)^2 +A(z)^2+A(z)^2\delta(z)\right] ~+~ \frac{1}{2}( P_Q(0)^2 +Q(0)^2) ~+~ P_Q(0)\cdot A(0)$$ I know how to diagonalize the two h.o. and the coupling term. The problem, though, is $A(z)^2+A(z)^2\delta(z)$. I would like to somehow add them up and redefine A(z). Could I do this by defining A(z) with something that cancels the delta-function from the beginning? – xabdax Jul 26 at 19:45