# When starting from Qubit State |0>, which operations when applied (from right to left) will result in the qubit state |->

I found this MCQ in a book that my friend gave me. Not entirely sure about this question.

When starting from Qubit State $$|0\rangle$$, which operations when applied (from right to left) will result in the qubit state $$|-\rangle$$?

• ZH
• XH
• HH
• HZ
• Write down the |0> state in its vector format. Write down the |-> state in its vector format. For each of the 4 gate sequences you provide, write down their unitary matrix representation. Perform the matrix-vector product with each one of the 4 matrices on the |0> state. Compare the output you get to the |-> state. You have answered your question! If any of the steps above is not simple for you, edit your question to let us know which one is blocking you. Commented May 30, 2022 at 8:11

## 2 Answers

You can start by writing everything down in matrix-vector form. The $$|0\rangle$$ state is represented in the $$Z$$-basis by the vector $$\begin{bmatrix} 1\\ 0 \end{bmatrix}$$. Applying the Hadamard-gate to that will result in $$\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix} \begin{bmatrix} 1\\ 0 \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1 \end{bmatrix}$$ which is also known as the $$|+\rangle$$ state. Finally, applying the $$Z$$-gate you get $$\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ -1 \end{bmatrix}$$ which is the $$|-\rangle$$ state.

To answer your question, the correct answer is option $$(i)ZH$$. This is because mathematically, the gates are applied in the right-to-left order and hence you apply the $$H$$ gate first taking us from state $$|0\rangle$$ to state $$|+\rangle$$ followed by $$Z$$ gate taking the $$|+\rangle$$ state to $$|-\rangle$$ state. The $$H$$ gate is the superposition gate that transforms the state $$|0\rangle$$ to state $$|+\rangle$$, state $$|1\rangle$$ to state $$|-\rangle$$ and vice versa. The $$Z$$ gate just changes the phase of state $$|1\rangle$$ i.e state $$|0\rangle$$ remains in state $$|0\rangle$$ and the state $$|1\rangle$$ is transformed to state -$$|1\rangle$$. Thus the $$|+\rangle$$ state i.e ($$|0\rangle$$ + $$|1\rangle$$)/$$\sqrt{2}$$ becomes ($$|0\rangle$$ - $$|1\rangle$$)/$$\sqrt{2}$$ which is the $$|-\rangle$$ state upon applying the $$Z$$ gate. The explanation provided by @Cat Telliber supports the same.

Also to justify why the other options are not correct,

Consider option $$(ii)XH$$. $$|0\rangle$$ on $$H$$ gives $$|+\rangle$$ and $$|+\rangle$$ on applying $$X$$ gate stays in the $$|+\rangle$$ state only. This is because the $$X$$ gate acts as a bit flip gate and changes state $$|0\rangle$$ to state $$|1\rangle$$ and vice versa. When it acts on the $$|+\rangle$$ state which is the state ($$|0\rangle$$ + $$|1\rangle$$)/$$\sqrt{2}$$, it is transformed to ($$|1\rangle$$ + $$|0\rangle$$)/$$\sqrt{2}$$ which is also the $$|+\rangle$$ state.

Option $$(iii)HH$$ is essentially the identity gate as $$H$$ gate is its own inverse, hence we end up going from state $$|0\rangle$$ to state $$|+\rangle$$ and back to state $$|0\rangle$$ upon applying the second hadamard gate.

Option $$(iv)HZ$$. State $$|0\rangle$$ stays in state $$|0\rangle$$ upon applying $$Z$$ gate and $$H$$ on state $$|0\rangle$$ gives us the $$|+\rangle$$ state

Hence only option (i) is correct.

As shown in the image attached below, state $$|0\rangle$$ gives us the state ($$|0\rangle$$ - $$|1\rangle$$)/$$\sqrt{2}$$ on applying $$ZH$$ gates(the states $$|0\rangle$$ and $$|1\rangle$$ have the same probability of being observed and the phase angle is $$\pi$$ which corresponds to the $$|-\rangle$$ state)