I want to know in which cases the CNOT gate can change the state of the first qubit (the control qubit), and why is that ?? I know that the CNOT gate has effect only on target qubit..


2 Answers 2


"controlled-not" is a very classical description for the gate which creates an artificial division between the control and the target. It rather implies that nothing should happen to the control, and it's just the target that gets changed. This is very misleading. controlled-not is an entangling gate. This means that both qubits change (and most of the time, you cannot divide them up as one qubit versus the other). It is only very special cases where one or the other does not change:

  • if the control is in a state $|0\rangle$ or $|1\rangle$, the control does not change
  • if the target is in the state $(|0\rangle\pm|1\rangle)/\sqrt{2}$, the target does not change.

I would like the emphasise that there is a lot of symmetry between the control and target. Indeed, the closely related controlled-phase gate is symmetric under a swap of control/target, which emphasises how the control is really not special. Controlled-not can be converted into controlled phase just by applying Hadamard gates to the target, both before and after the gate.


Suppose we have a 2 qubit circuit as follows:

enter image description here

Before the CNOT gate, the tensored state distribution is

$$(a|0\rangle + b|1\rangle)(c|0\rangle + d|1\rangle) = ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle$$

Suppose we now measure $q_2$, then the remaining uncollapsed state of $q_1$ becomes:

Case $q_2=|0\rangle$:

$$\Psi_1 = ac|0\rangle + bc|1\rangle = a|0\rangle + b|1\rangle$$

Case $q_2=|1\rangle$:

$$\Psi_1 = ad|0\rangle + bd|1\rangle = a|0\rangle + b|1\rangle$$

as it is originally.

After the CNOT gate however, the state distribution becomes:

$$ac|00\rangle + ad|01\rangle + bc|1\color{red}1\rangle + bd|1\color{red}0\rangle$$

Suppose we now measure $q_2$ after applying CNOT, then the remaining uncollapsed state of $q_1$ becomes

Case $q_2=|0\rangle$:

$$\Psi_1 = ac|0\rangle + bd|1\rangle$$

Case $q_2=|1\rangle$:

$$\Psi_1 = ad|0\rangle + bc|1\rangle$$

so $a,b,c,d$ has now become hopelessly entangled into the state of $\Psi_1$, despite $\Psi_1$ acting as merely the control (not target) of the CNOT operation...

Since $c, d$ survive long after $q_2$ has collapsed, they might be measurable.


$$\Psi_1 = ac|0\rangle + bd|1\rangle$$

Tensoring with a new qubit $q_3 = e|0\rangle + f|1\rangle$, the state distribution becomes

$$(ac|0\rangle + bd|1\rangle)(e|0\rangle + f|1\rangle)= ace|00\rangle + acf|01\rangle + bde|10\rangle + bdf|11\rangle$$

Applying CNOT, the states become:

$$ace|00\rangle + acf|01\rangle + bde|11\rangle + bdf|10\rangle$$

Case $q_3=|0\rangle$:

$$\Psi_1 = ace|0\rangle + bdf|1\rangle$$

Case $q_3=|1\rangle$:

$$\Psi_1 = acf|0\rangle + bde|1\rangle$$

So $c,d$ survives still, and will survive forever so long as there is an uncollapsed qubit... It is however diluted to more and more extent, such that it might not be possible to extract the value of $c,d$ no matter how many iterations of CNOTs are used.

If $a=\frac{1}{\sqrt 2}, b=\frac{1}{\sqrt 2}$, then the density distribution after CNOT is

$$c|00\rangle + d|01\rangle + c|11\rangle + d|10\rangle$$

Measuring $q_2$ teleports $c,d$ completely into $\Psi_1$:

Case $q_2=|0\rangle$ $$\Psi_1 = c|0\rangle + d|1\rangle$$

Case $q_2=|1\rangle$ $$\Psi_1 = d|0\rangle + c|1\rangle$$

What is the measurement statistics of repeated $q_2$ measurements?

$$|_{q_2=0}\rangle = c + d$$ $$|_{q_2=1}\rangle = d + c$$

Thus, no information about $c,d$ is gained from any measurement statistics of $q_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.