# State of qubit after gate and measurement

Suppose we have a gate with 1 control qubit and 1 target qubit:

Lets say the state of the control qubit initially was $$\begin{pmatrix} a\\ b \end{pmatrix}$$ and the state of the target qubit was initially $$\begin{pmatrix} c\\ d \end{pmatrix}.$$

The gate transforms the state of the target from $$c|1\rangle+d|0\rangle$$ into $$a|00\rangle+b|01\rangle+f(c,d)|10\rangle+g(c,d)|11\rangle$$ . Now after the gate has acted on the target bit if we measure it which means that it will be no longer entangled with the control qubit the probabilities to measure it in state $$|0\rangle$$ will be equal to $$f(c,d)$$ right?

My question is suppose we have this quantum circuit:

Which would be the state of the target qubit at the location pointed out by the arrow

You're not normalizing properly as it's hard to speak of the "state of the target" - you need all four coefficients both before and after application of the control gate. You can speak of the state of the entire system before and after application of the control gate, using all four coefficients collectively. Regardless you are also fundamentally forgetting the Born rule - the amplitudes need to be squared to talk about probabilities.

Let's work through a specific example, for example the controlled Hadamard gate:

$$\begin{pmatrix} 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& \frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\\ 0& 0& \frac{1}{\sqrt 2}& -\frac{1}{\sqrt 2} \end{pmatrix}.$$

Let your control qubit be written as $$a|0\rangle+b|1\rangle$$, while your target qubit is $$c|0\rangle+d|1\rangle$$. Assume throughout that both the control and the target qubits start off as pure states; then we must have $$|a|^2+|b|^2=|c|^2+|d|^2=1$$.

Before application of the controlled Hadamard gate, I would describe the joint system as:

$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle,$$

while after the controlled Hadamard is applied, I would describe the state as:

$$ac|00\rangle+ad|01\rangle+\frac{bc}{\sqrt 2}|10\rangle+\frac{bc}{\sqrt 2}|11\rangle+\frac{bd}{\sqrt 2}|10\rangle-\frac{bd}{\sqrt 2}|11\rangle.$$

The Born rule gives the probability of any state as the squared magnitude of the contributions of the amplitudes to those respective states.

For example, the probability of measuring $$|10\rangle$$ is given as $$\big|\frac{bc+bd}{\sqrt 2}\big|^2$$, and of measuring $$|11\rangle$$ as $$\big|\frac{bc-bd}{\sqrt 2}\big|^2$$.

Editing further, responsive to the comments and to the OP's edit to the question

Immediately after measuring the target qubit (in the standard basis) after application of a controlled-$$A$$ gate, it is correct to say that the target is either $$|0\rangle$$ or $$|1\rangle$$ and hence is in a pure state (in the standard basis).

But the probability that the target qubit is measured as $$|0\rangle$$ or $$|1\rangle$$ may nonetheless be dependent on the probability that the control qubit is $$|1\rangle$$, as well as the particular properties of $$A$$ and of the target qubit before application of the controlled-$$A$$ gate.

For example, if the control qubit is always $$|0\rangle$$ then the $$A$$ gate never gets activated, and hence the target qubit never gets changed.

• But once we measure the target qubit we disentangle it with the control qubit which means it becomes again a pure state no? Commented Dec 16, 2023 at 16:33
• Yes it's pure again after measuring. Commented Dec 16, 2023 at 16:37
• So what would be the probability of finding the target qubit at state |0> after we measure it? Commented Dec 16, 2023 at 16:38
• The probability of measuring $|00\rangle$ plus the probability of measuring $|10\rangle.$ Commented Dec 16, 2023 at 16:43
• @MarkSpinneli check the edit.But since it is pure again it is not related to the control qubit anymore so talking about state $|00>$ or |$10>$ is meaningless I think. Commented Dec 16, 2023 at 16:45