Say I have a Fock space $H$ with basis $K = \{ | k \rangle \big| k \in \mathbb{N} \}$. Then I consider the following single particle states:

$$ | A \rangle = \sum_{k \in K} a_k | k \rangle, \tag{1}$$ $$ | B \rangle = \sum_{k \in K} b_k | k \rangle. \tag{2}$$

I know that $| k_1 k_2 \rangle = \frac{1}{\sqrt{2}} (| k_1 \rangle \otimes | k_2 \rangle - | k_2 \rangle \otimes | k_1 \rangle)$ is a valid fermionic two-particle state. I expected I could calculate the two particle state which contains particles $A$ and $B$ as

$$ | AB \rangle \overset{?}{=} \frac{1}{\sqrt{2}} (| A \rangle \otimes | B \rangle - | B \rangle \otimes | A \rangle). \tag{3}$$

But it turns out that

$$ \frac{1}{\sqrt{2}} (| A \rangle \otimes | B \rangle - | B \rangle \otimes | A \rangle) = \frac{1}{\sqrt{2}} \sum_{k_1, k_2 \in K} (a_{k_1} b_{k_2}| k_1 \rangle \otimes | k_2 \rangle - a_{k_2} b_{k_1}| k_2 \rangle \otimes | k_1 \rangle) = 0. \tag{5}$$

So how do I write this two particle state $| AB \rangle$? It should be expressible as

$$ | AB \rangle = \sum_{\substack{k_1, k_2 \in K \\ k_1 < k_2}} c_{k_1 k_2} | k_1 k_2 \rangle, \tag{6}$$

but what is $c_{k_1 k_2}$? Is it $c_{k_1 k_2} = a_{k_1} b_{k_2}$? Why?


Why should

$$\frac{1}{\sqrt{2}} \sum_{k_1, k_2 \in K} (a_{k_1} b_{k_2}| k_1 \rangle \otimes | k_2 \rangle - a_{k_1} b_{k_2}| k_2 \rangle \otimes | k_1 \rangle) = 0$$

be true?

By switching the indices you get

$$\begin{align}&\frac{1}{\sqrt{2}} \sum_{k_1, k_2 \in K} (a_{k_1} b_{k_2}| k_1 \rangle \otimes | k_2 \rangle - a_{k_1} b_{k_2}| k_2 \rangle \otimes | k_1 \rangle) = \\ &\frac{1}{\sqrt{2}} \sum_{k_1, k_2 \in K} (a_{k_1} b_{k_2} - a_{k_2} b_{k_1}) | k_1 \rangle \otimes | k_2 \rangle \, .\end{align}$$

As you can see, the symmetric terms vanish, but the antisymmetric ones remain. From this equation, it is easy to see that

$$c_{k1,k2} = \frac{1}{\sqrt{2}} (a_{k_1} b_{k_2} - a_{k_2} b_{k_1}).$$

Further information is given here.

  • $\begingroup$ The antisymmetric terms cancel, as well: $(a_{k_1} b_{k_2}| k_1 \rangle \otimes | k_2 \rangle - a_{k_2} b_{k_1}| k_2 \rangle \otimes | k_1 \rangle)$ cancels with $(a_{k_2} b_{k_1}| k_2 \rangle \otimes | k_1 \rangle - a_{k_1} b_{k_2}| k_1 \rangle \otimes | k_2 \rangle)$. This is precisely my question, why can leave only half of this sum? Moreover, tensor product is not commutative: $| k_1 \rangle \otimes | k_2 \rangle \neq | k_2 \rangle \otimes | k_1 \rangle$. $\endgroup$ – Minethlos Oct 14 '16 at 19:36
  • $\begingroup$ I'm aware that tensor products don't commute, i just switched the summation indices $k_1 \leftrightarrow k_2$. In fact eq. (3) couldn't be true in general because tensor products don't commute. I just saw that you made a little mistake in eq. (5): In the second term you either have to switch the indices of the coefficients or the vectors, otherwise the terms would be the same and it would indeed yield $0$. $\endgroup$ – schlunma Oct 14 '16 at 21:09
  • $\begingroup$ Ah, that's right. Now I get what you mean :) $\endgroup$ – Minethlos Oct 14 '16 at 23:20

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