In Watrous' Theory of Quantum Information, Example 7.25 discusses the Werner Twirling Channel:

$$\Xi(X) = \int (U \otimes U) X (U \otimes U)^* \mathrm{d}\eta(U)$$

where $\eta$ denotes the Haar measure and $X$ some density matrix. I understand that by their Theorem 7.15 one can conclude that

$$\Xi(X)\in \mathrm{span}\{1\otimes 1, W\} = \mathrm{span}\{\Pi_0,\Pi_1\}, $$

where $W$ is the SWAP operator, and $\Pi_0 = \frac{1}{2}1\otimes 1 + \frac{1}{2}W$, $\Pi_1 = \frac{1}{2}1\otimes 1 - \frac{1}{2}W$. Thus, the original channel can be written as a linear combination

$$\Xi(X) = \alpha(X) \Pi_0+ \beta(X) \Pi_1$$


$$\Xi(X) = \alpha'(X) 1 \otimes 1 + \beta'(X) W.$$

In the book they follow the first equation and determine the prefactors as

$$\alpha(X)=\frac{1}{\binom{n+1}{2}}\langle \Pi_0, \Xi(X)\rangle,\, \beta(X) = \frac{1}{\binom{n}{2}}\langle \Pi_1,\Xi(X) \rangle. \tag{*}$$

What I do not understand here, is how and why they include these binomial coefficients here? The inner product is clear of course, the problem is the binomial coefficient.

This problem interests me in particular because in arXiv:2207.14734 they use this identity of the Werner Twirling Channel in order to derive Equ. 17 to 18 (I adapted the notation slightly to fit the previous better):

$$\int (U|v\rangle \langle v| U^\dagger)^{\otimes 2} \mathrm{d}\eta(U)= \frac{1}{d(d+1)}(1 \otimes 1 + W). \tag{**}$$

Here, $d$ denotes the dimension of the spaces in which e.g. the unitary matrices (2-design) live.

What surprises me here, is that the prefactor ($\frac{1}{d(d+1)}$) is independent of $X=|v\rangle\langle v|$. How can I arrive at this prefactor? In my notes I got different results.

Long story short, I have two questions:

  1. Where do the binomial coefficients come from in $(*)$?

  2. Is there some trick to think about when I want to apply this result to arrive at $(**)$?


1 Answer 1


The binomial coefficients come from the dimension of the subspaces that $\Pi_0,\Pi_1$ project onto. Remember that these are orthogonal projections. From the definition you can see that $\Pi_0$ projects onto the symmetric subspace, and the symmetric subspace of $\mathbb{C}^n\otimes\mathbb{C}^n$ has dimension $\binom{n+1}{2}$, thus $\operatorname{tr}(\Pi_0)=\binom{n+1}{2}$. Similarly for $\operatorname{tr}(\Pi_1)=\binom{n}{2}$, due to $\Pi_1$ projecting onto the antisymmetric subspace.

More directly from the equations you write, just notice that $\langle \Pi_0,\Pi_0\rangle=\operatorname{tr}(\Pi_0^2)=\operatorname{tr}(\Pi_0)$, $\langle \Pi_0,\Pi_1\rangle=0$, and $\langle\Pi_1,\Pi_1\rangle=\operatorname{tr}(\Pi_1)$.

For the other formula the reasoning is pretty much the same. The LHS is just an equivalent way to write the average of $(U\otimes U) (|v\rangle\!\langle v|\otimes|v\rangle\!\langle v|) (U^\dagger\otimes U^\dagger)$, and the RHS is just $\Pi_0/\binom{d+1}{2}$. I guess there's some changes in the notation because you're now denoting Hermitian conjugate with $\dagger$ rather than $*$, and the dimension of the underlying space became $d$ rather than $n$. The reason you get the normalization factor on the RHS here is that you want traces to match up on LHS and RHS. And taking the trace of LHS you easily see that it equals $1$, hence so it might do on the RHS.

  • $\begingroup$ Perfect, thank you! Now I got it:) $\endgroup$
    – Juri V
    Commented Mar 26, 2023 at 7:34

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