Measuring Probability of Mixed States

Question

I am a little stuck on understanding the measurement probabilities of a 3 qubit system (QCQI q 4.41).

1)H gates are applied to both $q_1$ and $q_2$

2) $C^{(1,2)}_3(X)$, a Toffoli, controlled by $q_1$ and $q_2$ is then applied to $q_3$

3) A Unitary (S gate) is then applied to $q_3$

4) $C^{(1,2)}_3(X)$, a Toffoli, controlled by $q_1$ and $q_2$ is then applied to $q_3$

5) H gates are applied to both $q_1$ and $q_2$

The probability of measuring $|q_1\rangle = |q_2\rangle = 0$ should be $\frac{5}{8}$, however I can only seem to derive $\frac{4}{8}$, by expanding out the tensors and then cancelling.

After step 4 the state I think is:

$(|00\rangle + |01\rangle + |10\rangle)\otimes S|q_3\rangle + |11\rangle \otimes XSX|q_3\rangle$

Then after applying step 5, expanding out and cancelling I am left with:

$(|00\rangle + |00\rangle + |00\rangle - |11\rangle)\otimes S|q_3\rangle + (|00\rangle - |01\rangle - |10\rangle + |11\rangle)\otimes XSX|q_3\rangle $

however I can't seem to find the missing $|00\rangle$, and also in this result the measurement of $|00\rangle$ corresponds to two different states of $q_3$. I think the error in my understanding is somewhere here:

Applying Hs (step 5) and expanding $|11\rangle \otimes XSX|q_3\rangle$

$(H|1\rangle \otimes H|1\rangle) \otimes IXSX|q_3\rangle$ = $(|00\rangle - |01\rangle - |10\rangle + |11\rangle)\otimes XSX|q_3\rangle $

Could it also be my misunderstanding that the state of $|q_1q_2\rangle$ going into both Toffoli gates can be different for step 2 & 4? I was assuming that if that state was $|11\rangle$ in the first Toffoli then it must be $|11\rangle$ into the second as well.

Answer 1

I'm not sure that I agree with all of your calculation. I agree up to step 4 (but let's be careful and keep normalisation factors in there), $$ \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle)\otimes S|q_3\rangle+|11\rangle\otimes XSX|q_3\rangle, $$ which I would even simplify to $$ \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle)\otimes S|q_3\rangle+|11\rangle\otimes iS^\dagger|q_3\rangle. $$ Now we need to apply the Hadamards $$ \rightarrow\frac{1}{4}(|00\rangle+|01\rangle+|10\rangle+|11\rangle+ |00\rangle-|01\rangle+|10\rangle-|11\rangle+ |00\rangle+|01\rangle-|10\rangle-|11\rangle)\otimes S|q_3\rangle+i\frac{1}{4}(|00\rangle-|01\rangle-|10\rangle+|11\rangle)\otimes S^\dagger|q_3\rangle. $$ Let's simplify the first bracket $$ =\frac{1}{4}(3|00\rangle+|01\rangle+|10\rangle-|11\rangle)\otimes S|q_3\rangle+i\frac{1}{4}(|00\rangle-|01\rangle-|10\rangle+|11\rangle)\otimes S^\dagger|q_3\rangle. $$ So, we can now regroup terms as $$ \frac{1}{4}|00\rangle\otimes(3S+iS^\dagger)|q_3\rangle+\frac{1}{4}(|01\rangle+|10\rangle-|11\rangle)\otimes(S-iS^\dagger)|q_3\rangle. $$ At this point, you might get ahead of yourself and try to read off the amplitude for the $|00\rangle$ term, and determine the measurement probability from that. However, you need to be careful to take into account the normalisation of the state of the third qubit. For example, $$ (S-iS^\dagger)=(1-i)Z=\sqrt{2}e^{-i\pi/4}Z. $$ From this, we conclude that each of the other terms appears with probability $|\sqrt{2}e^{-i\pi/4}/4|^2=1/8$, and applies $Z$ on qubit 3. So, it is clear that the $|00\rangle$ answer must arise with probability 5/8. We just need to check what the rotation is. Let's maipulate it a bit $$ 3S+iS^\dagger=\left(\begin{array}{cc} 3+i & 0 \\ 0 & 3i+1 \end{array}\right). $$ If we write $3+i=\sqrt{10}e^{i\phi}$, then this is $$ \sqrt{10}\left(\begin{array}{cc} e^{i\phi} & 0 \\ 0 & ie^{-i\phi}\end{array}\right)=\sqrt{10}e^{i\pi/4}\left(\begin{array}{cc} e^{i(\phi-\pi/4)} & 0 \\ 0 & ie^{-i(\phi-\pi/4)}\end{array}\right). $$ The $\sqrt{10}$ contributes to the amplitude of the overall state, so we get $|00\rangle$ with probability $|\sqrt{10}e^{i\pi/4}/4|^2$, as required. The unitary is of the form $R_z(\theta)$ with $\theta/2=\pi/4-\phi$. Hence, $$ \cos\theta=\cos\left(\frac{\pi}{2}-2\phi\right)=\sin(2\phi)=\frac{3}{5}. $$