# Understanding second quantized Hamiltonian

Consider an $$N$$-particle system, for which the Hamiltonian is written in first quantization as $$\hat H = \hat H_0 +\hat H_I,$$ where $$\hat H_0 = \sum\limits_{i=1}^{N}\left[-\frac{\hbar^2}{2M}\nabla_i^2 + U(\mathbf r_i)\right]$$ and $$\hat H_I = \frac{1}{2}\sum\limits_{\substack{i,j\\i\neq j}}^N V(\mathbf r_i, \mathbf r_j).$$ When translating this to second quantization, we use the field operators: $$\hat\Psi^{\dagger}(\mathbf r) = \sum\limits_{i=1}^{N} \varphi_i^*(\mathbf r)\hat a_i^{\dagger} \\ \hat\Psi(\mathbf r) = \sum\limits_{i=1}^{N} \varphi_i(\mathbf r)\hat a_i$$ And the formula $$\hat O^{(2)} = \int d^3r \hat\Psi^{\dagger}(\mathbf r) \hat O \hat\Psi(\mathbf r),$$ for single particle operators and $$\hat O^{(2)} = \int d^3rd^3r' \hat\Psi^{\dagger}(\mathbf r) \hat\Psi^{\dagger}(\mathbf r') \hat O \hat\Psi(\mathbf r)\hat\Psi(\mathbf r')$$ for two-particle operators.

Using the formula and substituting $$\hat H_0$$: \begin{align*} \hat H_0^{(2)} & = \int d^3r \hat\Psi^{\dagger}(\mathbf r) \hat H_0 \hat\Psi(\mathbf r)\\ & = \sum\limits_{i=1}^{N} \int d^3r \hat\Psi^{\dagger}(\mathbf r)\left[-\frac{\hbar^2}{2M}\nabla_i^2 + U(\mathbf r_i)\right] \hat\Psi(\mathbf r) \end{align*} However, my book neglects the sum over $$i$$ and writes simply $$\hat H_0^{(2)} = \int d^3r \hat\Psi^{\dagger}(\mathbf r)\left[-\frac{\hbar^2}{2M}\nabla^2 + U(\mathbf r)\right] \hat\Psi(\mathbf r)$$ For the interaction part after substituting $$\hat H_I$$, I get \begin{align*} \hat H_I^{(2)} & = \int d^3rd^3r' \hat\Psi^{\dagger}(\mathbf r)\hat\Psi^{\dagger}(\mathbf r') \hat H_I \hat\Psi(\mathbf r) \hat\Psi(\mathbf r') \\ & = \frac{1}{2}\sum\limits_{\substack{i,j\\i\neq j}}^N \int d^3rd^3r' \hat\Psi^{\dagger}(\mathbf r) \hat\Psi^{\dagger}(\mathbf r') V(\mathbf r_i, \mathbf r_j) \hat\Psi(\mathbf r)\hat\Psi(\mathbf r') \end{align*}

However, my book writes only $$\hat H_I^{(2)} = \frac{1}{2}\int d^3rd^3r' \hat\Psi^{\dagger}(\mathbf r) \hat\Psi^{\dagger}(\mathbf r') V(1,2) \hat\Psi(\mathbf r)\hat\Psi(\mathbf r')$$

Why do we ignore the sums in second quantization? How to interpret $$V(1,2)$$ instead $$V(\mathbf r_i, \mathbf r_j)$$? Why do we write $$\nabla^2$$ and $$U(\mathbf r)$$ instead $$\nabla_i^2$$ and $$U(\mathbf r_i)$$ in the expression of $$\hat H_0^{(2)}$$?

• What you do looks correct. How sure are you the authors didn't make a typo? You could check subsequent equations for this. Or maybe they are (implicitly) computing $\hat H_0^{(2)}$ for a single particle (and maybe taking a sum over all these single-particle $\hat H_0^{(2)}$ later)? – Georg Oct 31 '19 at 12:26

The field operators take care of that for us. An operator like $$\Psi^{\dagger}({\bf r})\Psi({\bf r})$$ will be a delta-function if there is a particle at position $${\bf r}$$, then $$\int d^dr \Psi^{\dagger}({\bf r})\Psi({\bf r})$$ will count the total number of particles.
Similarly, $$\int d^dr\Psi^{\dagger}({\bf r})\left[-\nabla^2+U({\bf r})\right]\Psi({\bf r})$$ will act on each of the particles, without summing over them explicitly. The field-operator will 'pick' on whether there is a particle or not at $${\bf r}$$.
The two-particle operator is similar, only now we have to check whether two particles exist, at positions $${\bf r},{\bf r}'$$.