# Manipulating the amplitude of state based on the state information

In the past, I thought I have seen quantum circuits/algorithm techniques to change the amplitude of state based on the state?

$$\lvert \psi \rangle = \sum_x \ C_x \lvert x \rangle$$, here $$C_x$$ is just constant (i.e. as in simple superpositions)

and after some $$U$$, $$\lvert \psi \rangle = \sum_x \ C_x (x) \lvert x \rangle$$

The amplitude changes based on the $$\lvert x \rangle$$ state.

Any pointer or reference would be appreciated.

• Is there a difference between $C_x$ and $C_x(x)$ here? And between $|X\rangle$ and $|x\rangle$? In general, we can decompose a unitary matrix as $U=\sum_{x,y}u_{x,y}|x\rangle\langle y|$ with orthonormal $\{|x\rangle\}$ such that $U|\psi\rangle=\sum_x \left(\sum_y u_{x,y}C_y\right)|x\rangle$; that's the same as a transformation $C_x\to \left(\sum_y u_{x,y}C_y\right)$. Jun 14 at 18:25
• @QuantumMechanic, sorry about the confusion about Cx, .. which is just meant to be a constant like the amplitude from the uniform superpositions. Jun 14 at 20:15
• Maybe, you are looking for this article: arxiv.org/abs/quant-ph/0407010 Jun 14 at 20:42
• I'm not sure exactly what you're looking for, so let me suggest a couple of possibilities: could it be amplitude amplification you're thinking of? Or, there's a step in the HHL algorithm that achieves something similar to this. Jun 15 at 6:59
• I still don't understand what $C_x(x)$ represents. It should be the value of the coefficient of the $|x\rangle$ state after the evolution, but should it depend on the full $|\psi\rangle$ or only the index $x$?
– glS
Jun 15 at 9:41

The matrix contains information about the vectors(states). To see this, the matrix form can be written as $$U_{ij}=\langle i\mid U\mid j\rangle$$ or the total form of $$U$$ mentioned by @Quantum Mechanic, i.e., $$U=\sum_{ij}U_{ij}\mid i\rangle\langle j\mid$$. To show it more vividly, the stabilizer code will be a good example.
Another easier example is that: when $$U=\left(\begin{array}{ll} 1 & 0 \\ 0 & -1 \end{array}\right) \\\\$$, $$U\mid 0\rangle$$ will become $$\mid 0\rangle$$ while $$U\mid 1\rangle$$ will become $$-\mid 1\rangle$$.