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
  • 2
    $\begingroup$ 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. $\endgroup$ Commented May 30, 2022 at 8:11

2 Answers 2


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)

enter image description here


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