# What is the right order of creation operators?

I started to learn some basics of second quantisation and specifically its use in quantum chemistry. Currently I'm reading this book by Péter R. Surján, and here is small excerpt from it.

If one has two electrons numbered as 1 and 2 which occupy the orbitals $\phi_{i}$ and $\phi_{k}$ our "first quantized" wave function is a Slater determinant \begin{equation*} \Phi(1,2) = \frac{1}{\sqrt{2}} \bigg| \begin{matrix} \phi_{i}(1) & \phi_{k}(1) \\ \phi_{i}(2) & \phi_{k}(2) \end{matrix} \bigg| \, . \end{equation*} The same entry is denoted in the second quantized notation as: $$\lvert ik \rangle = a_{i}^{+} a_{k}^{+} \lvert \text{vac} \rangle \, .$$ In words, an electron is created on the vacuum in state $\lvert k \rangle$, then another one in state $\lvert i \rangle$. One has the correspondence: $$\Phi(1,2) \leftrightarrow a_{i}^{+} a_{k}^{+} \lvert \text{vac} \rangle \, .$$

As far as I understand the order of creation operators here is just a convention: since we are dealing with antisymmetric fermionic states here, we have to to agree on a specific order of the creation operators and the author's choice was that the order of creation operators in the second quantized expression for an electronic state is the same as the order of spin-orbitals in the corresponding first quantized Slater determinant.

Obviously, once a convention is chosen we have to stick with it, but I fell like the author failed to do so, since later in the text he writes:

Now. creating an $N$-electron wave function $\Psi$ in which spinorbitals $\psi_{1}, \psi_{2}, \dotsc, \psi_{N}$ are occupied, one writes: $$\Psi(1, 2, \dotsc, N) \leftrightarrow a_{n}^{+} \dotsc a_{2}^{+} a_{1}^{+} \lvert \text{vac} \rangle \, .$$

And here the order of creation operators is the reverse of the order spin-orbitals in Slater determinant (although Slater determinant is not written explicitly, I think we can safely assume that it is $| \begin{matrix} \psi_{1} & \psi_{2} & \cdots & \psi_{n} \end{matrix} |$, i.e. that orbitals follow in natural order). But with this convention $\Phi(1,2)$ mentioned above should be equivalent to $a_{k}^{+} a_{i}^{+} \lvert \text{vac} \rangle$ rather than $a_{i}^{+} a_{k}^{+} \lvert \text{vac} \rangle$.

Am I right or totally confused?

Since the states $|\psi\rangle$ and $-|\psi\rangle$ differ only by a phase factor, they really describe the same electron configuration. So, while it is a bit sloppy to have inconsistent sign conventions in different places in a text, it's not surprising to see it, and it's not really a big deal. (Unless you have inconsistent sign conventions within a single equation of course!)

In other words, I agree with what you wrote in your question.

As OP pointed out, one should stick to a single convention. Hence the order of the creation operators should be reversed in one of the two cited equations.

Reversing the order of operators in which of the equations is purely a matter of choice. However, to obtain the simplest relation between the Slater determinant and the occupation number representations, I find it more desirable to use the convention defined by the first cited equation. In this convention, the creation operator connects two normalized Slater determinants in the following way: $$c_{s}^{\dagger} \Psi(s_{1}, s_{2}, \ldots, s_{N}) = \Psi(s, s_{1}, s_{2}, \ldots, s_{N}).$$ Starting from this relation, one can work out the anticommutation relations. The occupation number representation can be defined, e.g., by $$\Psi(1,3,5) = c_{1}^{\dagger}c_{3}^{\dagger}c_{5}^{\dagger}|\mathrm{vac}\rangle = |1,0,1,0,1,0,0,\ldots\rangle.$$ From this definition, one can deduce $$c_{4}^{\dagger}|1,0,1,0,1,0,0,\ldots\rangle =c_{4}^{\dagger}c_{1}^{\dagger}c_{3}^{\dagger}c_{5}^{\dagger}|\mathrm{vac}\rangle = (-1)^2 c_{1}^{\dagger}c_{3}^{\dagger}c_{4}^{\dagger}c_{5}^{\dagger}|\mathrm{vac}\rangle = (-1)^{2}|1,0,1,1,1,0,0,\ldots\rangle,$$ where the factor $(-1)^2$ is introduced as $c_{4}^{\dagger}$ passes through other creation operators twice. More generally, we have $$c_{r}^{\dagger}|\ldots,n_{r}=0,\ldots\rangle = (-1)^{\sum_{s<r}n_{s}} |\ldots,n_{r}=1,\ldots\rangle,$$ which is a widely cited relation.

I would interpret $1$ and $2$ in the function arguments as shorthands for the coordinates belonging to particles 1 and 2; but the subscripts $1, 2, i, k$ of the single-particle wave functions $\phi$ / $\psi$ and corresponding creation/annihilation operators $a^+$ / $a$ as specifying a single-particle state.

The first part is quite a common convention: instead of $\Phi(r_1, r_2)$ one writes $\Phi(1, 2)$, essentially to make the notation lighter. (If your book follows that convention, it should of course be properly introduced.)

So, $\phi_1(2)$ would mean “particle 2 in state 1”. Conversely, the creation operator $a^+_1$ has only one index; since the particles are indistinguishable and it does not make sense to say which particle is in state 1, just “there is a particle in state 1”. This is the point of second quantization, if you want. (In first quantized notation, the Slater determinant is needed to take the indistinguishability into account.)

To try and answer your actual question, you are right that the order is a matter of convention. Your second quote would establish that convention; since we do not know if $i<k$, we cannot really say if there is a contradiction with the first quote. But the order of elements in the Slater determinant does suggest that $i<k$, which would be inconsistent, like you say.

Warning: This post may contain a really stupid but important error.

There's a bit more to this than the other answers are covering.

As you already noted, in expressions like this

$$(c_1^\dagger c_2^\dagger \ldots) |0\rangle,$$

it is important to keep a consistent ordering. The reason is that fermion operators for different modes anticommute, so you have things like

$$c_1^\dagger c_2^\dagger|0\rangle = -c_2^\dagger c_1^\dagger |0\rangle \, .$$

This is as higgsss already explained. Now, what has not been discussed so far is what happens when you also have annihilation operators. Consider an expression like this

$$|A \rangle = c_1^\dagger c_2^\dagger c_1 c_2 |\Psi \rangle \, .$$

The creation and annihilation operators are both in ascending order, which appears to be consistent. However, this is not the way you should actually do it. To make equations simpler and easier to understand it's actually much better to have the creation and annihilation operators ordered oppositely, e.g.

$$|B \rangle = c_1^\dagger c_2^\dagger c_2 c_1 |\Psi \rangle\,.$$

The ordering used here (and the fact that the creation operators are all to the left of the annihilation operators) is called "normal ordering". The reason normal ordering is good is that once you have a normal ordered string of operators you don't have to worry about any minus signs showing up as you compute matrix elements. For example assuming $|\Psi\rangle$ has an excitation in modes 1 and 2:

$$\langle \Psi | c_1^\dagger c_2^\dagger c_2 c_1 |\Psi\rangle = 1$$

while

$$\langle \Psi | c_1^\dagger c_2^\dagger c_1 c_2 |\Psi \rangle = -1 \, .$$

When you have an expression like $|A\rangle$ or $|B\rangle$ it is often a good idea to rewrite it as a sum of normal ordered strings of operators by using the commutation and anticommutation rules. That, however, can be enormously tedious. Fortunately, there is a theorem called "Wick's theorem" which makes this procedure much easier.

If you have an expression like this: $$\langle 0 | \text{not normally ordered string of fermion operators} | 0 \rangle$$ you use Wick's theorem to break it down into a sum of normally ordered strings of operators. One of the main benefits of Wick's theorem is that it shows you that most of the possible terms in that sum are zero.

• good explanation, but usually one refers to normal order when the creation operators are to the left of the annihilation operators. Then the action on coherent states becomes particularly simple. – ulf Dec 9 '14 at 4:22
• @ulf: Oh jeez. I can't believe I screwed that up. Thanks. – DanielSank Dec 9 '14 at 5:44
• By making the operators normal ordered, you made the equation incorrect. $\langle 0|c_{1}^{\dagger}c_{2}^{\dagger}c_{2}c_{1}|0\rangle$ vanishes because of the annihilation operators applied to the vacuum state. – higgsss Dec 10 '14 at 15:58
• @higgsss: You are right, of course. This is not my day. I must have brain cancer. Will fix soon-ish. – DanielSank Dec 10 '14 at 17:23
• @higgsss: Could take a look at the edited version and see if I am still making dumb errors? If not I'll remove the warning at the top of the post. – DanielSank Dec 11 '14 at 19:12