# Show that the effect of a Controlled Unitary on qubits followed by a measurement is unchanged depending on when measurement is taken

I understand that a Controlled Unitary is just a generalization of a control gate, such as CNOT etc, and that it is given in state representation as $$\hat C_U = |0\rangle\langle0|\otimes\mathbb{I} + |1\rangle\langle1|\otimes \hat U$$or matrix representation of $$\begin{pmatrix} \mathbb{I} & \textbf{0}\\ \textbf{0} & \textbf{U} \end{pmatrix}$$ The question is asking to show that the time of measurement does not impact the effect of a controlled unitary. I'm unsure how to show this.

• Hi Mark, the question was posed by my tutor in university, so I assume it's a true statement!
– Dwye
Commented Dec 9, 2021 at 18:29
• Compute the density matrix of the result in both cases, and show that they are equal. Commented Dec 9, 2021 at 19:07

Suppose $$U$$ acts on $$n$$ qubits and let $$|\psi\rangle$$ be an arbitrary normalized state of $$n+1$$ qubits

$$|\psi\rangle=|0\rangle|\psi_0\rangle+|1\rangle|\psi_1\rangle\tag1$$

where $$|\psi_0\rangle$$ and $$|\psi_1\rangle$$ are not necessarily normalized states of $$n$$ qubits.

## Quantumly controlled gate

Consider first the situation where we begin by applying controlled-$$U$$. After the controlled-$$U$$ the $$n+1$$ qubits are in state

$$CU|\psi\rangle = |0\rangle|\psi_0\rangle+|1\rangle U|\psi_1\rangle.\tag2$$

Let $$P_0=|0\rangle\langle 0|$$ and $$P_1=|1\rangle\langle 1|$$. If measurement yields $$0$$, the unnormalized post-measurement state is

$$(P_0\otimes I_n)CU|\psi\rangle = P_0|0\rangle|\psi_0\rangle+P_0|1\rangle U|\psi_1\rangle = |0\rangle|\psi_0\rangle\tag3$$

where $$I_n$$ denotes the identity on the last $$n$$ qubits. Similarly, if measurement yields $$1$$, the unnormalized post-measurement state is

$$(P_1\otimes I_n)CU|\psi\rangle = P_1|0\rangle|\psi_0\rangle+P_1|1\rangle U|\psi_1\rangle = |1\rangle U|\psi_1\rangle.\tag4$$

## Classically controlled gate

Consider now the situation where we begin by measuring the first qubit. If measurement yields $$0$$, the unnormalized post-measurement state is

$$P_0|\psi\rangle = P_0|0\rangle|\psi_0\rangle+P_0|1\rangle|\psi_1\rangle = |0\rangle|\psi_0\rangle\tag5$$

and if it yields $$1$$, the unnormalized post-measurement state is

$$P_1|\psi\rangle = P_1|0\rangle|\psi_0\rangle+P_1|1\rangle|\psi_1\rangle = |1\rangle|\psi_1\rangle.\tag6$$

Applying classically controlled $$U$$ to the last $$n$$ qubits when measurement result is $$1$$ we get

$$(I_1\otimes U)|1\rangle|\psi_1\rangle=|1\rangle U|\psi_1\rangle.\tag7$$

## Conclusion

Comparing $$(3)$$ with $$(5)$$ and $$(4)$$ with $$(7)$$ we see that the unnormalized post-measurement state is the same in both scenarios regardless of measurement result. Since measurement outcome probability is the squared norm of the unnormalized post-measurement state we see that both measurement outcomes have the same probability in each scenario. Finally, equality of the unnormalized post-measurement states implies equality of the normalized post-measurement states. Thus, measurement outcome probabilities and the post-measurement states of all $$n+1$$ qubits are the same in the two scenarios.

## Summary

The above reasoning can be compressed by noting that

$$CU = P_0\otimes I_n + P_1\otimes U.$$

This way of writing down controlled-$$U$$ makes it clear that

$$(P_i\otimes I_n) \circ CU = CU \circ (P_i\otimes I_n)$$

for $$i=0,1$$ which means that controlled gates and computational basis measurements on the control qubit commute.

• This answer doesn't seem rigorous enough, doesn't the quantum state collapse after the measurement, it should be a probability. Commented Sep 26, 2023 at 7:50