Second quantization interaction operator - what does the integration variable $\mathbf{r}$ represent?

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 $$\hat{V} = \sum_{i I assume the $$\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:

I believe this question is related Does $T(x)$ represent a $c$ number or an operator in the second quantization?.

• Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Apr 7 at 20:53

In the right-hand sides of the given formulas the $$\psi^\dagger$$ and $$\psi$$ are operators, and $$V$$ is just a complex number (depending on $$\mathbf{r}$$ and $$\mathbf{r}'$$, which are coordinates, not operators).

So both formulas mean the same thing. Only the ordering of the operators matters.
You may want to write the operators with a $$\hat{}$$ to make it more clearly:

\begin{align} \hat{V}_{int.} &=\frac 1 2\int d^3r\ d^3r'\ V(\mathbf{r},\mathbf{r'}) \hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r'}) \hat{\psi}(\mathbf{r})\hat{\psi}(\mathbf{r'}) \\ &= \frac 1 2\int d^3r\ d^3r'\ \hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r'}) V(\mathbf{r},\mathbf{r')} \hat{\psi}(\mathbf{r})\hat{\psi}(\mathbf{r'}) \end{align}

(And by the way: the differentials in the integrals are volume elements, not vectors. Therefore I wrote $$d^3r$$ and $$d^3r'$$, instead of $$d\mathbf{r}$$ and $$d\mathbf{r}'$$.)

The equations above are operator equations. That means they are true for any state $$\left|u\right>$$ acted on by the operators. Like this:

\begin{align} \hat{V}_{int.}\left|u\right> &=\frac 1 2\int d^3r\ d^3r'\ V(\mathbf{r},\mathbf{r'}) \hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r'}) \hat{\psi}(\mathbf{r})\hat{\psi}(\mathbf{r'}) \left|u\right> \\ &= \frac 1 2\int d^3r\ d^3r'\ \hat{\psi}^\dagger(\mathbf{r})\hat{\psi}^\dagger(\mathbf{r'}) V(\mathbf{r},\mathbf{r')} \hat{\psi}(\mathbf{r})\hat{\psi}(\mathbf{r'}) \left|u\right> \end{align}

So, first the operators act on a state $$\left|u\right>$$, and then the integrals are evaluated.

• What do the integrals of the operators mean (e.g. see second part of question)? Are the integrals evaluated before or after the operator "acts on a state"? Apr 20 at 10:58
• @user246795 see my addition to the answer Apr 20 at 13:20
• It is common to choose to write the volume elements as if they were vectors. Funny that you are specifically making a comment about not doing that. Apr 21 at 0:44
• @ThomasFritsch Are you writing the volume elements to specifically separate them from the $\mathbf{r}$ in the field operators? I am still not 100% sure when the $\mathbf{r}$ changed from operators '$\mathbf{\hat{r}}$' in the discrete case to simply label vectors in the continuous case. Apr 25 at 8:40
• @naturallyInconsistent I'm writing $d^3r$ to make it dimensionally correct. $d^3r$ has dimension of length$^3$, not length$^1$. And it is just a number, not a vector of 3 numbers. Apr 25 at 9:44