# Can I switch $\alpha_0$ and $\alpha_1$ conditionally to $\alpha_0>0$ in a state $\alpha_0|0\rangle+\alpha_1|1\rangle$?

I have a single qubit $$a$$ in state

$$|s\rangle = \alpha_0|0\rangle + \alpha_1|1\rangle$$

$$\alpha_0$$ may be 0 whereas $$\alpha_1$$ is always positive and above $$0$$. Almost always $$\alpha_0 << \alpha_1$$

Is there any way to flip $$\alpha_0$$ and $$\alpha_1$$ if and only if $$\alpha_0 > 0$$?

I was thinking of having two ancillary qubits and perform two controlled CNOTs to detect both state $$0$$ and state $$1$$ and then CCNOT with those two ancillary qubits to flip $$a_0$$ and $$a_1$$

But does this make sense? I am not sure if detecting a superposition is possible in this way

• a_0>0 does not even make sense: The amplitudes are only defined up to a global phase. Aug 29 '20 at 22:19

No, that's not possible. For example, it would allow you to implement an operation that sent both $$|0\rangle$$ and $$|1\rangle$$ to $$|1\rangle$$. Two states going to the same state means the operation is irreversible and non-unitary.

• would this be irreversible? with q2 being the qubit and q0 and q1 being the ancilla qubits cnot(q2,q0) -> x(q2) cnot(q2,q1) x(q2) -> toffoli(q0, q1, q2) Aug 29 '20 at 19:19
• @CésarLeonardoClementeLópez Any sequence of reversible operations is reversible. Aug 29 '20 at 20:00
• Of course you can map both |0> and |1> to |1>. Aug 29 '20 at 22:18
• @NorbertSchuch Could you elaborate? I'm implicitly assuming the goal is to remain quantum, i.e. have a unitary channel, i.e. no measurement or dumping information into the environment. If you're allowed to decohere the system then yes it's possible; it's similar to a Reset operation. Aug 29 '20 at 22:39
• @NorbertSchuch The questions is talking about conditionally re-ordering amplitudes. Re-ordering preserves the set of distinct amplitudes present in the state vector, as does doing nothing. Any quantum channel that preserves the set of amplitudes must be unitary. Unitarity is implicit in the question. Aug 29 '20 at 23:04

Elaborating more on the previous answer, here is a proof that it's indeed not possible.

Let us assume that there is a unitary $$U$$ acting on qubit $$a$$ and on $$m$$ ancillary qubits such that $$U\Big(|s \rangle \otimes |0 \rangle^m_{anc}\Big)$$ is the desired output. Then on inputs $$|+ \rangle, |- \rangle$$ the outputs are

• $$U\Big(|+ \rangle \otimes |0 \rangle^m_{anc}\Big) = |+ \rangle \otimes |\phi_+ \rangle_{anc}$$
• $$U\Big(|- \rangle \otimes |0 \rangle^m_{anc}\Big) = |- \rangle \otimes |\phi_- \rangle_{anc}$$

So we may write $$U = |+ \rangle \langle +| \otimes W_+ + |- \rangle \langle -| \otimes W_-$$ where $$W_{\pm} |0 \rangle^m = |\phi_{\pm} \rangle$$.

On input $$|1 \rangle$$ we have the output state $$|\chi \rangle = U\Big(|1 \rangle \otimes |0 \rangle^m_{anc}\Big) = \frac{1}{\sqrt{2}} \Big(|+ \rangle \otimes |\phi_+ \rangle_{anc} - |- \rangle \otimes |\phi_- \rangle_{anc}\Big)$$ and since we want the qubit to be in $$|1 \rangle$$ state, it must hold that $$\langle \chi| \Big( |1 \rangle \langle 1| \otimes \mathbb{1} \Big) |\chi \rangle = 1 \implies \text{Re}\{\langle \phi_+ | \phi_- \rangle\} = 1 \implies |\phi_+ \rangle = |\phi_- \rangle := |\phi \rangle$$ But $$U\Big(|0 \rangle \otimes |0 \rangle^m_{anc}\Big) = \frac{1}{\sqrt{2}} \Big(|+ \rangle \otimes |\phi_+ \rangle_{anc} + |- \rangle \otimes |\phi_- \rangle_{anc}\Big) = |0 \rangle |\phi \rangle_{anc}$$ so finally $$U\Big(|s \rangle \otimes |0 \rangle^m_{anc}\Big) = |s \rangle \otimes |\phi \rangle_{anc}$$ and $$U$$ does not flip $$a_0, a_1$$.

To overcome this impossibility, you may want to obtain an algorithm that has the desired output state only with some success probability. A solution towards this direction is the following: if you can obtain two copies of the input state $$|s \rangle$$, then

1. Apply amplitude amplification on the first copy to amplify the amplitude of $$|0 \rangle$$ state.

2. Apply an X gate on the second copy and then a Controlled-X (control is the first copy). Now the state is $$c_0 |0 \rangle X|s \rangle + c_1 |1 \rangle |s \rangle$$.

3. Measure the first qubit. If it's zero, then the state of the second qubit is in the correct state ($$X|s \rangle$$). If it's one, then there is some error probability but this will be small since from step 1, $$|c_0|^2$$ will be big if $$a_0 \neq 0$$.