# Completeness Relation With Identical Particles

I came across something that is bugging me regarding the completeness relation. This identity states the following

$$\sum_{\mu} | \mu \rangle \langle \mu | = 1,$$

and with identical particles, this is stated as

$$\frac{1}{2} \sum_{ij} | ij \rangle \langle ij | = 1.$$

However, if we now write $$| \mu \rangle = \frac{1}{\sqrt{2}} \left( | ij \rangle - | ji \rangle \right)$$ (which I think we can for fermions), where $$| ij \rangle = | i \rangle \otimes | j \rangle$$ and input it in the first equation we don't get back the second equation

$$\frac{1}{2} \sum_{ij} \left( | ij \rangle - | ji \rangle \right) \left( \langle ij | - \langle ji | \right) = \frac{1}{2} \sum_{ij} \left( | ij \rangle \langle ij | - | ij \rangle \langle ji | - | ji \rangle \langle ij | + | ji \rangle \langle ji | \right) \neq \frac{1}{2} \sum_{ij} | ij \rangle \langle ij |.$$

Where did I go wrong?

You didn't go wrong. Your operator acts as the identity operator on all antisymmetric vectors but annihilates all symmetric vectors, and is therefore clearly not the identity operator on the entire tensor product space. Rather, it is the projection operator from $$\mathscr H\otimes\mathscr H$$ to $$\mathscr H\wedge \mathscr H$$, where the latter is the antisymmetric tensor product of spaces.
• Sorry, I don't understand. There is an identity for symmetric and antisymmetric states? And what is $\frac{1}{2} \sum_{ij} | ij \rangle \langle ij |$ then? Nov 11, 2022 at 15:06
• @RichHardFineMan The operator you propose is the projection operator onto the antisymmetric subspace of $\mathscr H\otimes \mathscr H$. Any vector in the full space can be split up into a symmetric part and an antisymmetric part; your operator annihilates the former and leaves the latter unchanged. $\frac{1}{2}\sum_{ij}|ij\rangle\langle ij|$ is $\frac{1}{2}$ times the identity operator on the full space. Nov 11, 2022 at 15:09