# How to normalize the symmetrizer of a qudit state?

Notations :

1. $$\mathcal{H}= \mathbb{C}^d \otimes \cdots \otimes \mathbb{C}^d$$ where $$\mathbb{C}^d$$ appears $$N$$ times.

2. $$S(\mathcal{H})$$ is the symmetric subspace of $$\mathcal{H}$$.

3. $$(|\epsilon_1\rangle,\ldots,|\epsilon_d\rangle)$$ is a basis of $$\mathbb{C}^d$$.

4. $$\mathbb{N}^N_d=$${$$k \in \mathbb{N}^N ~|~ k_i \leq d ~\forall i$$}.

Definition :

1. Let $$\widehat{S} : \mathcal{H} \rightarrow S(\mathcal{H})$$ such that $$\forall |\psi\rangle=|\epsilon_{i_1}\rangle \otimes \cdots \otimes |\epsilon_{i_N}\rangle\in V$$, with $$i_j\in$${$$1,\ldots,d$$} $$\forall j$$, $$\widehat{S}(|\psi\rangle)=\frac{1}{N!} \sum_{\nu \in S_N} |\epsilon_{\nu^{-1}(i_1)}\rangle \otimes \cdots \otimes |\epsilon_{\nu^{-1}(N)}\rangle.$$ $$\widehat{S}$$ is the symmetrizer of a state.

2. Let $$|\phi\rangle= \sum_{k\in \mathbb{N}^N_d} c_k |\epsilon_{k_1}\rangle \otimes \cdots \otimes |\epsilon_{k_N}\rangle$$ with $$c_k \in \mathbb{C}~ \forall k\in \mathbb{N}^N_d$$. $$|\phi\rangle$$ is a normalized state if $$\sum_{k\in \mathbb{N}^N_d} | c_k|^2 =1.$$

Question : $$\widehat{S}(|\psi\rangle)$$ is not a normalized tensor $$\forall |\psi\rangle\in \mathcal{H}$$. How should I modify the definition of $$\widehat{S}$$ to systematically obtain a normalized symmetric state ?

Nota Bene : If one multiplies $$\widehat{S}$$ with $$\sqrt{N!}$$, it solves the problem if $$|\psi\rangle= c_k |\epsilon_{k_1}\rangle \otimes \cdots \otimes |\epsilon_{k_N}\rangle$$, but not if $$|\psi\rangle= \sum_{k\in \mathbb{N}^N_d} c_k |\epsilon_{k_1}\rangle \otimes \cdots \otimes |\epsilon_{k_N}\rangle$$.

For example, you could take the state $$|0\rangle \otimes (|1\rangle+ |2\rangle ) \otimes (|0\rangle + |2\rangle)$$.

• Maybe I'm missing something, but doesn't the state $|\psi \rangle = |0\rangle \otimes |0\rangle \otimes |0\rangle$ get mapped to $\sqrt{6} | \psi \rangle$ if you replace the $N!$ by $\sqrt{N!}$? There will still be six terms in the sum as defined. Commented Mar 17, 2023 at 13:41

A symmetrizer $$\hat{S}$$ can't be a unitary operator (which is what you're asking for) because a symmetrizer is a projection operator onto a proper subspace $$\mathcal{H}_S \subset \mathcal{H}$$. This means that $$\hat{S}$$ will automatically map any vectors orthogonal to $$\mathcal{H}_S$$ to the zero vector, regardless of their normalization.
For example, the antisymmetric state $$|\psi\rangle = \frac{1}{\sqrt{2}} (|0\rangle \otimes |1 \rangle - |1\rangle \otimes |0\rangle),$$ will be mapped to the zero vector by $$\hat{S}$$ or any multiple of $$\hat{S}$$. No normalization convention can make the norm of the zero vector non-zero, and so the norm of $$|\psi\rangle$$ can't be preserved under the action of a symmetrizer.
More generally, any vector $$|\psi\rangle$$ can be decomposed as the linear combination of a normalized vector $$|\psi_S\rangle \in \mathcal{H}_S$$ and a normalized vector $$|\phi\rangle$$ which is orthogonal to $$\mathcal{H}_S$$: $$|\psi\rangle = a |\psi_S\rangle + b |\phi\rangle$$ The norm of this vector will be $$|a|^2 + |b|^2$$. Applying $$\hat{S}$$ to this vector will then yield $$\hat{S} |\psi\rangle = \lambda a |\psi_S\rangle$$ where $$\lambda$$ depends on the chosen normalization for $$\hat{S}$$.
You might think that we could just choose $$\lambda = 1/a$$. But the problem is that $$a$$ differs from vector to vector, whereas the normalization of $$\hat{S}$$ is fixed by the definition of the operator above. So, for example, if we pick the normalization of $$\hat{S}$$ to "work" for the vector $$|\psi\rangle$$ (i.e., $$\lambda = 1/a$$), then that normalization won't work for another vector $$|\psi'\rangle = a' |\psi_S\rangle + b' |\phi\rangle$$ with $$a' \neq a$$.