How do I measure the first qubit of an entangled vector, say
\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ \end{pmatrix} is what I get on the end of Deutsch's algorithm. If I get it right, I should now measure the first qubit in this 2-qubit register. How can I do it?

  • $\begingroup$ You can always measure one qubit, whether in register or not does not matter. I don't understand the question. $\endgroup$
    – kludg
    Dec 1, 2019 at 13:12
  • $\begingroup$ Are you asking about matrix representing a measurement? $\endgroup$ Dec 1, 2019 at 14:27

1 Answer 1


To measure, observe that you are simply projecting a quantum state onto some basis set of vectors. First, I will note that this state is not normalized. Let us first define the following quantum state.

$$|\psi_i\rangle = \begin{pmatrix}1\\-1\\0\\0\end{pmatrix}.$$

Then, calculating the corresponding probability yields: $$|\langle \psi_i|\psi_i\rangle|^2 = (1)(1) + (-1)(-1) = 2.$$

So to normalize this state, we will simply divide by $\sqrt{2}$. Thus, we obtain the state: $$|\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\\0\\0\end{pmatrix}.$$ We now wish to measure this state in the standard basis, and so we wish to project the state onto the set of basis vectors: $$|00\rangle = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}, |01\rangle = \begin{pmatrix}0\\1\\0\\0\end{pmatrix},|10\rangle = \begin{pmatrix}0\\0\\1\\0\end{pmatrix},|11\rangle = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$.

We will now calculate the probability amplitude of the state collapsing to each of those states. That is, we wish to calculate: $$\langle00|\psi\rangle\\=\frac{1}{\sqrt{2}}\begin{pmatrix}1&0&0&0\end{pmatrix}\begin{pmatrix}1\\-1\\0\\0\end{pmatrix}\\=\frac{1}{\sqrt{2}}.$$

$$\langle01|\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0&0\end{pmatrix}\begin{pmatrix}1\\-1\\0\\0\end{pmatrix}\\=-\frac{1}{\sqrt{2}}.$$

And although it is trivial to see that the amplitudes of the two remaining states will be zero, I will include the calculations for completeness: $$\langle10|\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}0&0&1&0\end{pmatrix}\begin{pmatrix}1\\-1\\0\\0\end{pmatrix}\\=0.$$

$$\langle11|\psi\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}0&0&0&1\end{pmatrix}\begin{pmatrix}1\\-1\\0\\0\end{pmatrix}\\=0.$$

And so we see that the probability of obtaining the $|00\rangle$ and $|01\rangle$ states are 0.5 each, and so measurement of the first qubit must yield the $|0\rangle$ state. To see what would happen if you measured the second qubit, simply sample the $|00\rangle$ and $|01\rangle$ states once according to the aforementioned probabilities.

Edit: In response to a comment left on this answer, I have added the following note. If you have the state:

$$|\psi\rangle = \alpha_0|0\rangle + ... + \alpha_N|N\rangle,$$

then the probability amplitude of obtaining a component of the state, $|\psi_i\rangle$, is given by $\langle \psi_i|\psi\rangle$. Consequently, the probability of measuring a value associated with $|\psi_i\rangle$ is given by $|\langle \psi_i|\psi\rangle|^2$.

  • 1
    $\begingroup$ I feel like I'll have more questions in future, but then I'll just ask them in another thread. Thank you, Arthur, I appreciate it. You have helped me a lot. $\endgroup$
    – Penter Pro
    Dec 1, 2019 at 21:39
  • $\begingroup$ it's also not |<phi|phi>|^2, just <phi|phi>, right? Or does phi(i) means values of the vector? It makes sense then. $\endgroup$
    – Penter Pro
    Dec 6, 2019 at 14:13
  • $\begingroup$ I have edited my answer to address your question. $\endgroup$
    – Arthur-1
    Dec 6, 2019 at 14:43
  • $\begingroup$ I meant that. Just that the result won't be the same. Thus it'd be 4 instead of 2. >>>> Then, calculating the corresponding probability yields: |⟨ψi|ψi⟩|2=(1)(1)+(−1)(−1)=2. $\endgroup$
    – Penter Pro
    Dec 6, 2019 at 17:28

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