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

4

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, $012 · · · n − 1$.

Adam Levine

Posted 2019-11-28T05:29:26.923

Reputation: 413

1Please provide a reference yourself! What is the map $\tau$ doing? – Marsl – 2019-11-28T08:12:10.573

@Marsl I provided the required reference – Adam Levine – 2019-11-28T11:21:34.267

1please 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 – 2019-11-28T12:01:46.623

@AdamLevine $\tau(S)$ here is the parity of the permutation $S$, right?

– glS – 2019-11-28T12:11:34.870

Answers

3

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.

DaftWullie

Posted 2019-11-28T05:29:26.923

Reputation: 35 722

2

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.

Danylo Y

Posted 2019-11-28T05:29:26.923

Reputation: 3 940