# Probability of measuring the label of representation in quantum Fourier transformaton

I am going through Normal Subgroup Reconstruction and Quantum Computation Using Group Representations. In Definition 2, the authors start with the following function.

$$f : G \to \mathbb{C}$$

Then the Fourier transform of $f$ at the irrep $\rho$ is defined as the $d_\rho \times d_\rho$ matrix:

$$\hat{f}(\rho) = \sqrt{\frac{d_\rho}{|G|}} \sum_{g \in G} f (g) \rho(g)$$

In the quantum computational settings, the authors identify the superposition $\sum_{g \in G} f_g |g\rangle$ ($|a\rangle$ is a quantum state which is the $a$-th computational basis vector), with the function $f:G\to\mathbb{C}$ defined by $f(g) = f_g$.

In this notation, $\sum_{g \in G} f(g)|g\rangle$ is mapped under the Fourier transform to $\sum_{\rho \in \hat{G}, 1\le i,j \le d_\rho} \hat{f}(\rho)_{i,j} |\rho, i, j\rangle$.

$(\rho, i, j)$ is the label of the $(\rho, i, j)$-th computational basis vector $|\rho, i, j\rangle$. $\hat{f}(\rho)_{i,j}$ is a complex number and the $(i,j)$-th element of the matrix $\hat{f}(\rho)$. $\hat{G}$ is the collection of all irreps of $G$.

The the authors say as follows.

When the first portion of this triple is measured, we observe $\rho \in \hat{G}$ with probability

$$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2\\ = tr ((\hat{f}(\rho))^* \hat{f}(\rho))$$

Here, ($\rho, i, j$) is the triple the authors are referring to. Measuring $\rho$ means measuring the qubits which encode the label $\rho$.

My question:

Why is $\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ true?

• What does "$\rho > \in \hat{G}$" mean? – joriki May 2 '16 at 20:55
• @joriki, that was a typo. I have fixed it. Thanks. – Omar Shehab May 2 '16 at 20:57

## 1 Answer

$\sum_{1\le i,j\le d_\rho} |\hat{f}(\rho)_{i,j}|^2 = ||\hat{f}(\rho)||^2$ is just the definition of the Frobenius norm of the matrix $\hat{f}(\rho)$.