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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 $|0\rangle \otimes (|1\rangle+ |2\rangle ) \otimes (|0\rangle + |2\rangle)$.

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  • $\begingroup$ 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. $\endgroup$ Commented Mar 17, 2023 at 13:41

1 Answer 1

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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 $a' \neq a$.

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  • $\begingroup$ Thank you for your very clear response. Could you please inform me whether there is a method to make this operator unitary, taking into account any state except antisymmetric ones? $\endgroup$
    – Baloo
    Commented Mar 20, 2023 at 13:30
  • $\begingroup$ @Baloo: No. See my edit. $\endgroup$ Commented Mar 20, 2023 at 14:20
  • $\begingroup$ Thanks a lot for the explanation ! $\endgroup$
    – Baloo
    Commented Mar 20, 2023 at 14:57

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