Controlled U gate with arbitrary U

Question

I'm working through MIT's Quantum Information Science I, Part 1 on OpenCourseWare, and for the life of me I can't figure out the following problem:

Given the circuit above, where the meter sign denotes the measurement with $\{|0⟩,|1⟩\}$ basis:

• Suppose $U$ is a unitary having eigenvalue $e^{i\theta_1}$ with eigenstate $|u_1\rangle$, and eigenvalue $e^{i\theta_2}$ with eigenstate $|u_2\rangle$. Also, suppose that $e^{i\theta_1} \ne e^{i\theta_2}$. Let $|\psi_i=a|u_1\rangle+b|u_2\rangle$ where $a,b$ are real numbers satisfying $a^2+b^2=1$. Find the probability of getting the measurement result 0. Express the probability in terms of $a$, $\cos(\theta_1)$, and $\cos(\theta_2)$.

So far, I think I've managed to work out that, after the second Hadamard gate, we have $a(1+e^{i\theta_1}+b(1+e^{i\theta_2})) \otimes |+\rangle$. However, I'm absolutely confused about where to go next.

Our starting state is $|0\rangle \otimes |\psi_i\rangle$.

Then, we perform a Hadamard operation on the first qubit, and we get $|+\rangle \otimes |\psi_i\rangle$.

Then, we perform the controlled-U operation, and we get $(|0\rangle \otimes (a|u_1\rangle + b|u_2\rangle) + |1\rangle \otimes (e^{i\theta_1}a|u_1\rangle + e^{i\theta_2}b|u_2\rangle))$.

Finally, we perform a Hadamard operation on the first qubit again, getting state $(|0\rangle \otimes (a(1+e^{i\theta_1})|u_1\rangle + b(1+e^{i\theta_2})|u_2\rangle)) + (|1\rangle \otimes (a(1-e^{i\theta_1})|u_1\rangle + b(1-e^{i\theta_2})|u_2\rangle))$.

However, when I try to calculate the probability of measuring |0>, the math explodes and I wind up with crazy terms like $a^2(2i\cos\theta_1+2i\sin\theta_1+2\cos^2\theta_1+2\cos\theta_1)\cos^2\theta_1)+b^2(2i\cos\theta_2+2i\sin\theta_2+2\cos^2\theta_2+2\cos\theta_2)\cos^2\theta_2)$. Am I going wrong conceptually here or mathematically?

Answer 1

Your initial state is $$ |0\rangle\otimes|\psi_i\rangle=|0\rangle\otimes(a|u_1\rangle+b|u_2\rangle). $$ After the first Hadamard, this becomes $$ \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes(a|u_1\rangle+b|u_2\rangle). $$

Now, the key is what to do with controlled-$U$. If the top qubit is $|0\rangle$, do nothing: $$ |0\rangle\otimes(a|u_1\rangle+b|u_2\rangle)\rightarrow |0\rangle\otimes(a|u_1\rangle+b|u_2\rangle) $$ while if the first qubit is $|1\rangle$, apply $U$ to the second register: $$ |1\rangle\otimes(a|u_1\rangle+b|u_2\rangle)\rightarrow |1\rangle\otimes U(a|u_1\rangle+b|u_2\rangle)=|1\rangle\otimes(ae^{i\theta_1}|u_1\rangle+be^{i\theta_2}|u_2\rangle). $$ This means that, after the controlled-$U$, the full state is $$ \frac{1}{\sqrt{2}}\left( |0\rangle\otimes(a|u_1\rangle+b|u_2\rangle)+|1\rangle\otimes(ae^{i\theta_1}|u_1\rangle+be^{i\theta_2}|u_2\rangle) \right). $$ Now we do the final Hadamard: $$ \frac{1}{2}\left( (|0\rangle+|1\rangle)\otimes(a|u_1\rangle+b|u_2\rangle)+(|0\rangle-|1\rangle)\otimes(ae^{i\theta_1}|u_1\rangle+be^{i\theta_2}|u_2\rangle) \right). $$ Collect the terms of $|0\rangle/|1\rangle$ on the first qubit: $$ \frac12|0\rangle(a(1+e^{i\theta_1})|u_1\rangle+b(1+e^{i\theta_2})|u_2\rangle)+\frac12|1\rangle(a(1-e^{i\theta_1})|u_1\rangle+b(1-e^{i\theta_2})|u_2\rangle). $$ Finally, you want to know the probability that the first qubit is measured in state $|0\rangle$. This is just the square of the length of that term: $$ \|\frac12|0\rangle(a(1+e^{i\theta_1})|u_1\rangle+b(1+e^{i\theta_2})|u_2\rangle)\|^2=\frac14\|a(1+e^{i\theta_1})|u_1\rangle+b(1+e^{i\theta_2})|u_2\rangle\|^2. $$ You don't need to know what the two vectors $|u_i\rangle$ look like in the standard basis. All you need to know is that they have length 1, and are orthogonal to each other (because they are eigenvectors). Thus, the probability is $$ \frac14\left(|a(1+e^{i\theta_1})|^2+|b(1+e^{i\theta_2})|^2\right), $$ which is the same as $$ \frac14\left(|a(e^{-i\theta_1/2}+e^{i\theta_1/2})|^2+|b(e^{-i\theta_2/2}+e^{i\theta_2/2})|^2\right), $$ and thus $$ |a|^2\cos^2(\theta_1/2)+|b|^2\cos^2(\theta_2/2) $$