# Prove that the state $\sum_{S\in P_n}(-)^{\tau(S)}|S\rangle$ is invariant up to a phase when changing the basis

I am trying to prove that the $$|S_{n}\rangle$$ is $$n$$-lateral rotationally invariant, where $$|S_{n}\rangle$$ is defined as

$$|S_{n}\rangle=\sum_{S \in P_{n}^{n}} (-)^{\tau(S)}|S\rangle\equiv\sum_{S \in P_{n}^{n}} (-)^{\tau(S)}|s_{0}s_{1}.....s_{n-1}\rangle.$$

A $$n$$ dimensional qudit state on Hilbert space $$H_{n}$$ is the superposition of its computational basis $$\{|i\rangle\}, i\in\{0,1,2,...,n-1\}$$. Here, $$|S_n\rangle$$ is a state of $$n$$ such particles on $$H_{n}^{\otimes n}$$.

Consider now another basis $$|i'\rangle$$ connected with the computational basis by a unitary transformation $$U$$, where $$|i\rangle=\sum_{j} U_{ji}|i'\rangle.$$

Then, in this new basis, the state $$|S_{n}\rangle$$ takes the same form up to a global phase factor $$\phi$$. That is,

$$|S_{n}\rangle=e^{i\phi}\sum_{M \in P_{n}^{n}} (-)^{\tau(M)}|M'\rangle \equiv e^{i\phi}\sum_{M \in P_{n}^{n}} (-)^{\tau(M)}|m_{0}'m_{1}'...m_{n-1}'\rangle.$$

Here $$P_{n}^{n} = \{x_{0}x_{1}x_{2}...x_{n-1}|x_{0},x_{1},x_{2},...,x_{n-1 }\in Z_{n}, \forall j \neq k, x_{j} \neq x_{k}\}$$ and the phase factor is given by $$e^{i\phi} = \det(U).$$

Proof

$$|S_{n}\rangle=\sum_{S \in P_{n}^{n}} (-)^{\tau(S)}\sum_{m_{0}=0}^{n-1}U_{m_{0},s_{0}}|m_{0}'\rangle\otimes...\otimes\sum_{m_{n-1}=0}^{n-1}U_{m_{n-1},s_{n-1}}|m_{n-1}'\rangle$$

$$=\left[\sum_{M \in P_{n}^{n}} + \sum_{M \not\in P_{n}^{n}}\right]\left[\sum_{S \in P_{n}^{n}} (-)^{\tau(S)}U_{m_{0},s_{0}}U_{m_{1},s_{1}}...U_{m_{n-1},s_{n-1}}\right]|M\rangle$$

$$=\left[\sum_{M \in P_{n}^{n}} + \sum_{M \not\in P_{n}^{n}}\right]det(U_{m_{j}, s_{i}})|M\rangle$$

.... continued

Here how $$\left[\sum_{S \in P_{n}^{n}} (-)^{\tau(S)}U_{m_{0},s_{0}}U_{m_{1},s_{1}}...U_{m_{n-1},s_{n-1}}\right]= det(U_{m_{j}, s_{i}})$$?

Please provide a reference if you have any. I have a basic idea about unitary transformations and their matrix representations.

$$\tau(S)$$, named inverse number, is defined as the number of transpositions of pairs of elements of $$S$$ that must be composed to place the elements in canonical order, $$012 · · · n − 1$$.

• Please provide a reference yourself! What is the map $\tau$ doing? – Marsl Nov 28 '19 at 8:12
• @Marsl I provided the required reference – Adam Levine Nov 28 '19 at 11:21
• please try to use sensible titles for your questions. I tried to edit in a title actually related to the question being asked. Please check that I interpreted your question correctly. – glS Nov 28 '19 at 12:01
• @AdamLevine $\tau(S)$ here is the parity of the permutation $S$, right? – glS Nov 28 '19 at 12:11

Let $$\rho$$ and $$\sigma$$ be two permutations, i.e. lists of length $$d$$ containing some ordering of the elements 0 to $$d-1$$.
What we want to calculate is $$\langle \rho|U^{\otimes d}|S\rangle$$. If we can show that this is $$(-1)^{\tau(\rho)}$$ up to a global phase, then we know not only that those elements come through the calculation correctly, but by normalisation, it must be that all states not of the form $$|\rho\rangle$$ have 0 amplitude.
Now, $$\langle \rho|U^{\otimes d}|S\rangle=\sum_{\sigma}(-1)^{\tau(\sigma)}\prod_{i=0}^{d-1}U_{\rho_i,\sigma_i}$$ Let us change the order that we do the product, so the first element we take as $$\rho_i=0$$, then $$\rho_i=1$$ etc. So, this expression becomes $$\sum_{\sigma}(-1)^{\tau(\sigma)}\prod_{i=0}^{d-1}U_{i,(\rho^{-1}\sigma)_i}.$$ Next, we change the order that we take the sum over permutations (since the set of objects we're summing over is unchanged by the permutation $$\rho^{-1}$$). $$\sum_{\sigma}(-1)^{\tau(\rho\sigma)}\prod_{i=0}^{d-1}U_{i,\sigma_i}.$$ Next, we use $$\tau(\rho\sigma)=\tau(\rho)\tau(\sigma)$$, so we have $$(-1)^{\tau(\rho)}\left(\sum_{\sigma}(-1)^{\tau(\sigma)}\prod_{i=0}^{d-1}U_{i,\sigma_i}\right).$$ The term in brackets is a standard way of writing $$\text{det}(U)$$. See, for example, Wikipedia. Just to complete the argument in terms of what we had to prove, $$U$$ is unitary, so its determinant has modulus 1, i.e. $$\text{det}(U)$$ just contributes a global phase.
Let unitary operator $$U$$ acts as a permutation $$\boldsymbol{\sigma} = (\sigma_1, \sigma_2,..,\sigma_n)$$ on the standard basis. That is $$U |i \rangle = | \sigma_i \rangle$$ for every $$i \in \{0,1,..,n-1\}$$.
Now if $${\boldsymbol{\rho}} = (\rho_1, \rho_2,..,\rho_n)$$ is some permutation, then $$U^{\otimes n} | \boldsymbol{\rho}\rangle = U^{\otimes n} | \rho_1 \rho_2..\rho_n\rangle = | \sigma_{\rho_1} \sigma_{\rho_2}..\sigma_{\rho_n}\rangle = | (\sigma\rho)_1 (\sigma\rho)_2 ..(\sigma\rho)_n \rangle = | \boldsymbol{\sigma\rho}\rangle$$ Here $$\boldsymbol{\sigma\rho}$$ is the composition of permutations $$\boldsymbol{\sigma}$$ and $$\boldsymbol{\rho}$$.
Now note that $$(-1)^{\tau(\boldsymbol{\sigma\rho})} = (-1)^{\tau(\boldsymbol{\sigma})}(-1)^{\tau(\boldsymbol{\rho})}$$ Finally $$U^{\otimes n} | S_n \rangle = U^{\otimes n} \sum_{\rho \in Sym_n}(-1)^{\tau(\boldsymbol{\rho})}| \boldsymbol{\rho}\rangle = \sum_{\rho \in Sym_n}(-1)^{\tau(\boldsymbol{\rho})}| \boldsymbol{\sigma\rho} \rangle =$$ $$=(-1)^{\tau(\boldsymbol{\sigma})} \sum_{\rho \in Sym_n}(-1)^{\tau(\boldsymbol{\sigma\rho})}| \boldsymbol{\sigma\rho} \rangle = (-1)^{\tau(\boldsymbol{\sigma})} | S_n \rangle = \text{det}(U)| S_n \rangle$$
For the proof for every unitary $$U$$ see the DaftWullie's answer.