4
$\begingroup$

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, $012 · · · n − 1$.

$\endgroup$
4
  • 1
    $\begingroup$ Please provide a reference yourself! What is the map $\tau$ doing? $\endgroup$
    – Marsl
    Nov 28, 2019 at 8:12
  • $\begingroup$ @Marsl I provided the required reference $\endgroup$ Nov 28, 2019 at 11:21
  • 1
    $\begingroup$ 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. $\endgroup$
    – glS
    Nov 28, 2019 at 12:01
  • $\begingroup$ @AdamLevine $\tau(S)$ here is the parity of the permutation $S$, right? $\endgroup$
    – glS
    Nov 28, 2019 at 12:11

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.