Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a step in the construction of this algorithm which I'm not understanding:

$\displaystyle \left[\sum_x \frac{| x \rangle}{\sqrt{2^n}}\right]\left[\frac{ | 0 \rangle -| 1 \rangle }{\sqrt{2}} \oplus | f(x) \rangle\right]=\sum_x \frac{(-1)^{f(x)} | x \rangle }{\sqrt{2^n}}\left(\frac{| 0 \rangle - |1 \rangle}{\sqrt{2}}\right)$

where $f:\{0,1\}^n \to \{0,1\}$.

I don't see how they are equal, and I believe part of my confusion is on how $\oplus$ work for $2$-qubits. $\oplus$ denotes addition $\!\!\!\!\mod 2$... so does it work in each element separately?

share|cite|improve this question
You can get the right formating for braces and parentheses enclosing big stuff by using \left and \right in front of them. – joriki Aug 13 '12 at 21:16
Also you're still using \mid instead of |, as Michael pointed out in this question. \mid is good for the "divides" bar, but not as part of a bra or ket -- as you can see, it produces too much spacing in that case. – joriki Aug 13 '12 at 21:21
up vote 2 down vote accepted

We can best write the step in question as $$\frac{1}{\sqrt{2^{n+1}}} \sum_x|x\rangle\left(|f(x)\rangle - |1 \oplus f(x) \rangle \right) = \frac{1}{\sqrt{2^{n+1}}} \sum_x(-1)^{f(x)}|x\rangle \left(|0\rangle - |1\rangle \right).$$ $f(x)$ is equal to either $0$ or $1$. If $f(x)=0$, then $$|f(x)\rangle - |1 \oplus f(x) \rangle=|0\rangle - |1\oplus 0 \rangle = |0\rangle - |1 \rangle.$$ If $f(x)=1$, then $$|f(x)\rangle - |1 \oplus f(x) \rangle=|1\rangle - |1\oplus 1 \rangle = |1\rangle - |0 \rangle = (-1)\left(|0\rangle - |1 \rangle\right).$$ So, $(-1)^{f(x)}\left(|0\rangle - |1 \rangle\right)$ is another way of writing $|f(x)\rangle - |1 \oplus f(x) \rangle$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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