3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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