In some texts (see [1],[2],[3]) the two particle interaction operator is defined as: $$ V_{int.} =\frac 1 2\int d\mathbf{r}d\mathbf{r'} V(\mathbf{r},\mathbf{r')} \psi^\dagger(\mathbf{r})\psi^\dagger(\mathbf{r'}) \psi(\mathbf{r})\psi(\mathbf{r'}). $$
Whereas other texts (see [3], [4]) explicitly order the terms (suggesting they don't commute?) like:
$$ V_{int.} = \frac 1 2\int d\mathbf{r}d\mathbf{r'} \psi^\dagger(\mathbf{r})\psi^\dagger(\mathbf{r'}) V(\mathbf{r},\mathbf{r')} \psi(\mathbf{r})\psi(\mathbf{r'}). $$
Do both orderings mean the same thing? Is $V$ an operator or a complex number? And is $\mathbf{r}$ a coordinate or actually an operator $\mathbf{\hat{r}}$, because for first quantized standard representation $\hat{V} = \sum_{i<j} V(\mathbf{r}_i-\mathbf{r}_j)$ I assume the $\mathbf{r}_i$ are formally operators ($\mathbf{\hat{r}}_i$) just like in $\hat{V}(\mathbf{r}) $ for the standard single particle case?
Also, how should one think about operators like this: is it better to think of the operator integrals as the limit of the discrete sum in the sense that "we dont evaluate the integral until it acts on a Hilbert space ket", or do the integrals get 'evaluated' first hence removing the integration dummy variables $\mathbf{r}$ and $\mathbf{r'}$ from the expression? Surely, it cannot be this since then we have "no operators remaining"?
References:
- [1] - equation (1.5.6c)
- [2] - equation (1.71) in INTRODUCTION TO THE MANY-BODY PROBLEM, University Of Fribourg, SPRING TERM 2010
- [3] - equation (51) in Second Quantization, Jörg Schmalian
- [4] - page 43 in Lecture Notes for Quantum Matter, Steven H. Simon
- [5] - equation (101) in Manybody wave function and Second quantization, Ming-Che Chang
I believe this question is related Does $T(x)$ represent a $c$ number or an operator in the second quantization?.