Which entangled qubit is measured in this example?

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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?

Rahul

Posted 2020-05-16T05:07:45.257

Reputation: 13

1The 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. – kludg – 2020-05-16T06:24:37.787

Answers

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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:

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

Davit Khachatryan

Posted 2020-05-16T05:07:45.257

Reputation: 3 583

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? – Rahul – 2020-05-16T12:33:25.737

@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. – Davit Khachatryan – 2020-05-16T12:55:46.127

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). – Rahul – 2020-05-16T13:08:55.583

@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: https://www.youtube.com/watch?v=QtWCmO_KIlg&list=PLmRxgFnCIhaMgvot-Xuym_hn69lmzIokg

– Davit Khachatryan – 2020-05-16T13:54:32.597

1Thanks @Davit this playlist is amazing! Definitely, I am studying QC for QML. – Rahul – 2020-05-17T02:53:53.540