0
$\begingroup$

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:

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

$\endgroup$
1
  • $\begingroup$ 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

1
+50
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.