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"?


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

  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Apr 7 at 20:53

1 Answer 1


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.

  • $\begingroup$ 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"? $\endgroup$
    – user246795
    Apr 20 at 10:58
  • $\begingroup$ @user246795 see my addition to the answer $\endgroup$ Apr 20 at 13:20
  • $\begingroup$ 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. $\endgroup$ Apr 21 at 0:44
  • $\begingroup$ @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. $\endgroup$
    – user246795
    Apr 25 at 8:40
  • $\begingroup$ @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. $\endgroup$ Apr 25 at 9:44

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