# What happens with first phase factor in QFT?

I'm using Mermin's Quantum Computer Science book to understand Shor's algorithm, but I can't figure out why one of the phase factors drops out of the probability for measuring a certain y.

This is the application of the QFT on the superposition of the first register in Shor's algorithm ($$x_0$$ is the offset and $$r$$ is the period): \begin{align*}U_{FT}\frac{1}{\sqrt{m}}\sum_{k=0}^{m-1}\left|{x_0+kr} \right>_n&=\frac{1}{2^{n/2}}\sum^{2^n-1}_{y=0}\frac{1}{\sqrt{m}}\sum_{k=0}^{m-1}e^{2\pi i(x_0+kr)/2^n}\left |{y} \right>_n \\ &=\sum^{2^n-1}_{y=0}e^{2\pi i x_0 y/2^n}\frac{1}{2^n\sqrt{m}}\left(\sum_{k=0}^{m-1}e^{2\pi ikry/2^n}\right)\left |{y} \right>_n\end{align*}

According to Mermin, the probability of getting the result $$y$$ is $$p(y) = \frac{1}{2^nm}\big |\sum_{k=0}^{m-1}e^{2\pi ikry/2^n}\big |^2$$. Why can we just ignore $$e^{2\pi i x_0 y/2^n}$$?

• It’s a global phase that disappears when you take the mod-square. – DaftWullie Jan 27 '19 at 13:14
• what is the mod-square? – jvdh Jan 27 '19 at 13:38
• The absolute value that you’re using to evaluate the probability. – DaftWullie Jan 27 '19 at 14:33

If you have a quantum state like $$|\Psi\rangle_n = a_0|0\rangle_n+a_1|1\rangle_n+...+a_n|2^n-1\rangle_n$$ and you measure it in the $$\{|0\rangle_n,...,|2^{n-1}\rangle_n\}$$ basis, then the probability $$p(y)$$ of getting the state $$|y\rangle_n$$ is $$|a_y|^2$$ where $$a_y \in \Bbb C$$ (i.e it's a complex number).
In your example, $$a_y = e^{2\pi i x_0 y/2^n} \frac{1}{2^n\sqrt{m}} \sum_{k=0}^{m-1}e^{2\pi ikry/2^n}.$$
Remind yourself that for a complex number $$re^{i\theta}$$, the modulus is $$|re^{i\theta}| = |r||e^{i\theta}|=|r| \cdot 1=|r|$$ (since $$e^{i\theta}$$ has a modulus of $$1$$) and the square of the modulus is $$|r|^2.$$ Note that I've not specified $$r$$ is real; it could still be a complex number itself.
In your case, the $$e^{i\theta}$$ is $$e^{2\pi i x_0 y/2^n}$$ and so you don't see it in $$p(y)$$, as $$p(y)$$ is essentially $$|a_y|^2$$. This $$e^{i\theta}$$ has a special name — it's a global phase for $$a_y|y\rangle_n$$.