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

César Leonardo Clemente López

Posted 2020-08-29T19:06:35.560

Reputation: 541

a_0>0 does not even make sense: The amplitudes are only defined up to a global phase. – Norbert Schuch – 2020-08-29T22:19:35.370



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.

Craig Gidney

Posted 2020-08-29T19:06:35.560

Reputation: 11 207

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) – César Leonardo Clemente López – 2020-08-29T19:19:12.230

@CésarLeonardoClementeLópez Any sequence of reversible operations is reversible. – Craig Gidney – 2020-08-29T20:00:44.087

Of course you can map both |0> and |1> to |1>. – Norbert Schuch – 2020-08-29T22:18:54.517

1@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. – Craig Gidney – 2020-08-29T22:39:45.077

Why would you exclude the latter? If the desired operation can be implemented with a CP map, it can be implemented. – Norbert Schuch – 2020-08-29T22:40:40.130

@NorbertSchuch When I say you can't send both |0> and |1> to |1>, I am including the environment in my accounting. I would describe operations like the one you are considering as sending |0,env0> to |1,env0> and |1,env0> to |1,env1>. I like to explicitly purify my channels when describing them. Sorry if that created confusion. – Craig Gidney – 2020-08-29T22:42:30.050

Well, then it does not answer the question. – Norbert Schuch – 2020-08-29T22:50:40.533

1@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. – Craig Gidney – 2020-08-29T23:04:45.873

I disagree. This is not implicit in the question. I interpret it as asking "is it physically possible". – Norbert Schuch – 2020-08-29T23:11:33.960


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 $.


Posted 2020-08-29T19:06:35.560

Reputation: 1 151