Sorry, I am a newbie to quantum computing.

I am reading an overview paper (released just a week ago) titled Advances in Quantum Deep Learning: An Overview (Garg & Ramakrishnan, 2020). I am stuck on the following example from paper (shown in screenshot below).

enter image description here

I understand that one of the qubits was measured and it returned a value of $0$. Therefore, the state of entangled two-qubit quantum system has been updated as following:

$|\psi'\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle.$

But in the initial state,

$|\psi\rangle = \frac{1}{\sqrt{3}} |00\rangle + \frac{1}{\sqrt{3}} |01\rangle + \frac{1}{\sqrt{6}} |10\rangle + \frac{1}{\sqrt{6}} |11\rangle$

the superposition state $\frac{1}{\sqrt{3}} |00\rangle + \frac{1}{\sqrt{3}} |01\rangle$ has the probability $\frac{2}{3}$ which according to the statement in paper has been measured to be $0$. Then why the updated state of quantum system is

$|\psi'\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle$

which yielded $0$ instead of

$|\psi'\rangle = \frac{1}{\sqrt{2}} |10\rangle + \frac{1}{\sqrt{2}} |11\rangle$

which instead has the probability $\frac{1}{3}$ when it was not measured?

  • 1
    $\begingroup$ The first qubit is measured; and since it is measured $|0\rangle$, the post-measurement state of the second qubit is $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$. I don't quite understand what exactly is asked. $\endgroup$
    – kludg
    May 16, 2020 at 6:24

1 Answer 1


Firstly, note that the state is seperable (not entangled):

\begin{equation} |\psi\rangle = \frac{1}{\sqrt{3}} |00\rangle + \frac{1}{\sqrt{3}} |01\rangle + \frac{1}{\sqrt{6}} |10\rangle + \frac{1}{\sqrt{6}} |11\rangle= \\ = \left(\frac{\sqrt{2}}{\sqrt{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle \right) \left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \right) = |\psi_1 \rangle |\psi_2 \rangle \end{equation}

where $|\psi_1 \rangle = \left(\frac{\sqrt{2}}{\sqrt{3}}|0\rangle + \frac{1}{\sqrt{3}}|1\rangle \right)$ and $|\psi_2 \rangle= \left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \right)$ correspond to the first and the second qubit states. This $|\psi\rangle = |\psi_1 \rangle |\psi_2 \rangle$ is not possible if we have entangled states. The consequence of this is that after measuring $|0\rangle$ for the first qubit (in the example the measurement is done for the first qubit), the state of the first qubit becomes $|\psi_1 \rangle \rightarrow |0\rangle$ (the second qubit's state will not be changed) and the combined state:

$$|\psi\rangle= |0\rangle \left(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle \right) = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$$

The explanation with the projective measurements:

We should renormalize the state after the measurement (after measurement the probabilities of the state should sum to $1$). If we apply the projective measurement, we should calculate the probability of measuring $m$ outcome:

$$p(m) = \langle \psi | P_m | \psi \rangle$$

where $P_m$ is is the projector onto the eigenspace of $M$ with eigenvalue $m$, $M$ is a Hermitian operator/observable, that describes the measuremet. Then the state after the measurement outcome $m$ will be equal to:

$$\frac{P_m |\psi \rangle}{\sqrt{p_m}}$$

The division to the $\sqrt{p_m}$ is for the renormalization of the state after the action of the projector. A more rigorous definition can be found in M. Nielsen and I. Chuang's textbook page 87.

In the case of the question the observable is $M = Z\otimes I$, the projector to the $|0\rangle$ state of the first qubit is $P_{+1} = |0\rangle \langle 0| \otimes I$, the $I$ means that we are not touching the second qubit, $m$ eigenvalue is $+1$. Then the probability of measuring the first qubit $|0\rangle$:

$$p_{+1} = \langle \psi | P_{+1} | \psi \rangle = \frac{2}{3}$$

The resulting state:

$$\frac{P_{+1} |\psi \rangle}{\sqrt{p_{+1}}} = \frac{\frac{1}{\sqrt{3}}|00\rangle + \frac{1}{\sqrt{3}}|01\rangle }{\sqrt{\frac{2}{3}}} = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$$

  • $\begingroup$ Nice explanation (but projective measurement based explanation seems a bit high-level to me right now). I have one more query that if the states are separable then why paper says it is entangled? $\endgroup$
    – Rahul
    May 16, 2020 at 12:33
  • $\begingroup$ @Rahul, note that before the projective measurement explanation I added a simpler explanation. I don't know why they call this state an entangled state. It can be a matter of definition (to call every combined state entangled and regard separable states as special cases of entangled states) or perhaps they have a typo/not right example for the entangled state. From the paper: "Entanglement refers to the phenomenon by which qubits exhibit correlation with one another". I am not sure what they mean by saying "correlation" for this particular example. $\endgroup$ May 16, 2020 at 12:55
  • $\begingroup$ Okay. Sounds like even an overview paper is confusing the explanation. Thank you for your explanation. Btw what good book (except Nielsen & Chuang book, I have it and it looks difficult to read) or other resources would you recommend for studying quantum computing (such that I can pick up with using QC programming libraries like Pennylane). $\endgroup$
    – Rahul
    May 16, 2020 at 13:08
  • $\begingroup$ @Rahul, actually, I don't have experience with Pennylane, so I can't give a recommendation in that matter. I think Qiskit's textbook can give a good experience for QC programming. Also, if you haven't seen these lectures, I think it can be useful if you are interested in QML: youtube.com/… $\endgroup$ May 16, 2020 at 13:54
  • 1
    $\begingroup$ Thanks @Davit this playlist is amazing! Definitely, I am studying QC for QML. $\endgroup$
    – Rahul
    May 17, 2020 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.