# In amplitude amplification, how are the amplitudes of qubits changed?

I am relatively new to quantum computing and I feel like I don't fully understand the power of quantum computing due to a lack of understanding of how amplitude amplification works.

What is confusing me is that the amplitude of the qubit is the square root of the probability and so in amplitude amplification, how is the amplitude of the "correct answer" changed to negative (before then being reflected across the average)?

I understand that it is a gate and the gate is a series of complex operations but at the same time I fail to understand physically how do the gates change the amplitude of the qubit. I am referencing this in response to the webpage: IBM Q Experience Documentation - Grover's algorithm. Any help is greatly appreciated. Thank you!

• can you clarify what you are asking for exactly? why doesn't the current answer address your confusion (or does it)? how is the amplitude of the "correct answer" changed to negative: by applying an appropriate gate. Are you asking how this is done in practice? Or what kind of gate does this? Or something else? – glS Jul 19 '18 at 21:28

EDIT: I completely misunderstood your question and thought that you were confused about what a negative amplitude means, and not about physical mechanisms. I'm leaving this up in case that actually was what you meant. whoops. For the implementation question, how a reflection is implemented physically depends on the qubit implementation you are using.

I think the main confusion you're having is from thinking about the relationship between amplitudes and probabilities in the wrong direction.

You say that the amplitudes are the square root of the probability, but a safer way of thinking, which might help in building intuition, is to say that the amplitude norm squared is the probability.

$$\vert A\vert^2 = P$$

if this is inverted in its most general form, you get

$$A = \sqrt{P} e^{i\theta}$$ for some $\theta$. This additional phase is where your confusion is coming from, as it allows for both negative amplitudes as you are encountering in your example, as well as in many other very important states like $$H\vert 1 \rangle = \vert -\rangle = \frac{1}{\sqrt{2}}\vert 0\rangle - \frac{1}{\sqrt{2}}\vert 1\rangle$$

Without this possibility for negative states, quantum mechanical phenomena like constructive and destructive interference of wave functions would be impossible.

Hope that helps!

• To avoid future confusion, in the last displayed formula, the |+> should be a |->. :) – arriopolis Jul 10 '18 at 10:33