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Suppose there are $3$ bosons living on a 1-dimensional ring of length $L$. The Hamiltonian is given by $$H=-\sum_{i=1}^3\frac{\partial^2}{\partial x_i^2}+\sum_{1\leq j<k\leq 3}2c\delta(x_j-x_k).$$ Now the exercise wants me to derive conditions which must be fulfilled when 2 particles meet, i.e. there is a discontinuity in the derivative of the wave function because of the $\delta$-potential.

What I tried to do was integrate $H\psi=E\psi$ with respect to the variables $x_1-x_2$, $x_1-x_3$ and $x_2-x_3$ from $-\varepsilon$ to $\varepsilon$ to 'catch' the discontinuity:

$$-\int_{-\varepsilon}^{\varepsilon}d(x_1-x_2)\sum_{i=1}^3\frac{\partial^2\psi}{\partial x_i^2}+2c\int_{-\varepsilon}^{\varepsilon}d(x_1-x_2)\sum_{1\leq j<k\leq 3}\delta(x_j-x_k)\psi\rightarrow 0\text{ if }\varepsilon\rightarrow 0.$$ the second term has only zeros except for $j=1$ and $k=2$ because $x_3$ is fixed in which case it becomes $2c\psi(x_1,x_1,x_3)$. Something similar also happens to the first term: only the integral of $\frac{\partial^2\psi}{\partial x_1^2}$ and $\frac{\partial^2\psi}{\partial x_2^2}$ does not become zero as $\varepsilon\rightarrow 0$ in which case it becomes $$-\lim_{\varepsilon\rightarrow 0}\left(\frac{\partial\psi}{\partial x_1}(x_2-\varepsilon)-\frac{\partial\psi}{\partial x_1}(x_2+\varepsilon)\right)+c\psi(x_1,x_1,x_3)=0.$$ We can do the same thing by integrating with respect to $x1-x_3$ and $x_2-x_3$: $$-\lim_{\varepsilon\rightarrow 0}\left(\frac{\partial\psi}{\partial x_1}(x_3-\varepsilon)-\frac{\partial\psi}{\partial x_1}(x_3+\varepsilon)\right)+c\psi(x_1,x_1,x_3)=0.$$ $$-\lim_{\varepsilon\rightarrow 0}\left(\frac{\partial\psi}{\partial x_2}(x_3-\varepsilon)-\frac{\partial\psi}{\partial x_2}(x_3+\varepsilon)\right)+c\psi(x_1,x_1,x_3)=0.$$

My question is whether I have missed something here or if there is a flaw in my reasoning. I really need help on this, I have been thinking for a long time now.

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