Suppose I have an unknown state $|\psi\rangle = \sum_i \alpha_i|{\lambda_i}\rangle$, is it possible that I can transform it into $|\psi\rangle = \frac{1}{\sqrt{\sum_i|\alpha_i|^{2r}}} \sum_i \alpha_i^r|{\lambda_i}\rangle$?

I have an idea for one qubit with a measurement, which would be better without measurements.

Suppose the input state is $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ and can be prepared with two copies. An ancilla qubit is provided with state $|0\rangle$, such that

$ (\alpha|0\rangle+\beta|1\rangle)(\alpha|0\rangle+\beta|1\rangle)|0\rangle= \alpha^2|000\rangle + \alpha\beta|010\rangle+\beta\alpha|100\rangle+\beta^2|110\rangle. $

With two CNOT gates in a row, the ancilla qubit is the target qubit, such that

$ \alpha^2|000\rangle+\alpha\beta|011\rangle+\beta\alpha|101\rangle+\beta^2|110\rangle. $

This is followed by a measurement on ancilla qubit if we happen to measure 0, which the state on the first two qubits will be $ \frac{\alpha^2}{\sqrt{|\alpha|^4+|\beta|^4}}|000\rangle+\frac{\beta^2}{\sqrt{|\alpha|^4+|\beta|^4}}|110\rangle. $

With an CNOT gate on the second qubit, using the first qubit as control, such that

$ \frac{\alpha^2}{\sqrt{|\alpha|^4+|\beta|^4}}|00\rangle+\frac{\beta^2}{\sqrt{|\alpha|^4+|\beta|^4}}|10\rangle= (\frac{\alpha^2}{\sqrt{|\alpha|^4+|\beta|^4}}|0\rangle+\frac{\beta^2}{\sqrt{|\alpha|^4+|\beta|^4}}|1\rangle)|0\rangle $

The state in the first qubit will be

$ \frac{\alpha^2}{\sqrt{|\alpha|^4+|\beta|^4}} |0\rangle+\frac{\beta^2}{\sqrt{|\alpha|^4+|\beta|^4}} |1\rangle $

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However, the measurement on ancilla qubit is a nuisance. Can I obtain the powered amplitude state without measurement on arbitrary number of qubits?

  • 1
    $\begingroup$ This is not possible. This powering transformation you're trying to implement is non-linear, so it can't be done via unitary transformations, which are always linear. The only way to do it is probabilistically via measurements, as you did. It's anyway an interesting question what is the optimal way of doing it. $\endgroup$ Jul 18, 2020 at 9:21
  • $\begingroup$ @MateusAraújo Yeah, you're right. As the power increases, leave alone the feasibility for more qubits, the chance to obtain such a state diminishs exponentially. $\endgroup$
    – M. Chen
    Jul 18, 2020 at 10:06

1 Answer 1


As noted by Mateus in the comments, the transformation you are looking for is non-linear. This cannot be done with any matrix transformation. Thus, you will need more qubits, and your solution shows two (+1 scratch qubit) is sufficient. I guess you might wonder if a two-qubit unitary can do it, though?

The problem is that the transformation you want to implement depends on the input state. You can't do this (unitarily) even with extra qubits. I believe the most general result forbidding such requirements is the No-Programming Theorem.

Also note at, as $r\to\infty$, the transformation becomes a projection onto the subspace spanned by the states with highest modulus. You do are doing something like a weak measurement when $r$ is finite.

Nearly final observation: you mention you want $|\psi\rangle$ to be "unknown". You should be cautious taking your solution (as you generalise requiring more copies of $|\psi\rangle$) farther without thinking about no-cloning or more subtle resource counting.

Last thing. A coherent version of something like what you might be looking for is Amplitude Amplification.

  • $\begingroup$ Yeah, I did want to find the eigenstate with the highest modulus, which the amplitude amplification mightn't work. Thus the strategy I came up with is to amplify the highest and diminish the lowest; That is why I thought power transformation might work. So do you have any suggestions on how to achieve this? In other words, how to make the winner take all? $\endgroup$
    – M. Chen
    Jul 20, 2020 at 2:16
  • $\begingroup$ I suggest maybe ask a new top-level question. It's a good one and a reply to a comment might not be the easiest place to answer it. $\endgroup$ Jul 20, 2020 at 23:22

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