How to increase the amplitude of a specific quantum state?

If my quantum circuit produces the state $a|0000\rangle+b|0011\rangle$, how to make $b>a$?

  • 1
    $\begingroup$ how does "how to make" mean? You mean what gate/circuit can be used to change this state into another one with $b>a$? $\endgroup$
    – glS
    Aug 1, 2019 at 11:26
  • $\begingroup$ yes, is there any gate or circuit can do that? $\endgroup$ Aug 1, 2019 at 11:46
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    $\begingroup$ You could invert about the mean. This is tagged grovers-algorithm. What do you know of Grover's algorithm? Are you getting confused about a specific step? $\endgroup$
    – Mark S
    Aug 1, 2019 at 12:21

1 Answer 1


Your question is rather unclear. Amplitude amplification is a specific algorithm that you can run, but it doesn't make any sense to apply it to the example state that you're giving.

You don't specify whether or not the parameters $a,b$ are known. If they are, then the task is trivial. First, apply a controlled-not from qubit 3 to qubit 4, leaving you in the state $$ |00\rangle(a|0\rangle+b|1\rangle)|0\rangle. $$ Now apply a unitary on qubit 3 (e.g. a Y-rotation by some angle of your choosing), so that you change to a state $$ |00\rangle(c|0\rangle+d|1\rangle)|0\rangle. $$ Then you apply controlled-not again to give the final state $$ |00\rangle(c|00\rangle+d|11\rangle). $$ However, I suppose this is not what you had in mind?

Alternatively, I might infer that you're intending $a>b$, perhaps you don't know $a,b$, but all you care about is changing the coefficients so that the amplitude of the 11 component is larger than that of the 00 component. In that case, it's even easier: apply Pauli X on both qubits 3 and 4. It'll convert you to $$|00\rangle(b|00\rangle+a|11\rangle),$$ satisfying those requirements.

  • $\begingroup$ Thank you for your answer. I didn't ask the question correctly. I have two states like that a|100> + b|001> with same amplitude(a=b). My question is how to make the state with |1> in the third qubit has the highest amplitude? $\endgroup$ Aug 4, 2019 at 20:36
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    $\begingroup$ @EbtehalAli So that's like the first case I describe. Since you know exactly what the state is, you can find a unitary to map it to exactly what you want. In fact, in the case you give in the comment, that's just a maximally entangled state. So, you don't even need to do any two-qubit gates to convert it, just use the strategy for deterministic conversion that's associated with Nielsen's majorization criteria (section 12.5.1 of Neilsen & Chuang) $\endgroup$
    – DaftWullie
    Aug 5, 2019 at 8:33

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