Understanding second quantized Hamiltonian

Question

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 $\hat H_0^{(2)}$?

Answer 1

The entire point of second quantization is that we don't need to account for individual particles when writing the Hamiltonian. The information about the number of particles is encoded in the states, and the Hamiltonian is the same for whatever number of particles there is in the system. So there will never be a summation over different particles.

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 ${\bf r},{\bf r}'$.