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In section 5.2.1 of Nielsen Chuang, Performance and Requirements, there is an idea, that what happens if we can't prepare eigen state $|u\rangle$ and instead have a state $|\psi\rangle$ which is represented by $\sum_{u} c_{u}|u\rangle$. Output state is $\sum_{u} c_{u} |\phi_{u}\rangle|u\rangle$

Now we will measure the first qubit and it will turn out to be $|\phi_{u} \rangle$ with probability proportional to $c_{u}^{2}$.

But I am curious, what will be the state of the second qubit? (Is it entangled with the first qubit?)

Some excerpts from the book itself: enter image description here enter image description here

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    $\begingroup$ They seem to make the implicit assumption that $|u\rangle$ is measured as well since $|\varphi_u\rangle$ is obtained with probability $|c_u|^2$. If you measure only the first qubit, I guess that you are left with a superposition of all the $|u\rangle$ that have eigenvalue $e^{2\pi i\varphi_u}$. $\endgroup$ – lamontap Jan 15 at 21:38
  • $\begingroup$ @lamontap Where do you find that assumption? yes so if there is just one such $|u\rangle$ then it will be just single state else superposition of all two or three etc. $\endgroup$ – user27286 Jan 15 at 23:01
  • $\begingroup$ The book says "where $u$ is chosen at random with probability $|c_u|^2$". This happens when you measure the second register of $\sum_u c_u|\varphi_u\rangle|u\rangle$. If you were to measure only the first register, you would only get the same probability distribution if the $|\varphi_u\rangle$'s are orthogonal (which may not necessarily be the case). $\endgroup$ – lamontap Jan 22 at 16:40
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Quick side note before proceeding to answering your question: $|\phi_u\rangle$ and $|u\rangle$ are registers of qubits since they can represent a set of qubits (and not necessarily a single qubit).

Now on to answering your question: at the end of the phase estimation algorithm, if we make a measurement and the first register collapses to $|\phi_u\rangle$ then the second register will collapse to $|u\rangle$. This is because, as you mentioned, they are entangled. To make this more apparent, here is the state before the measurement written out explicitly: $$\sum_{u=1}^n c_u |\phi_u\rangle |u\rangle = c_1|\phi_1\rangle|1\rangle + c_2|\phi_2\rangle|2\rangle + ... + c_n|\phi_n\rangle|n\rangle.$$ Therefore whichever $|\phi_u\rangle$ you measure in the first register, the second register will collapse to the corresponding $|u\rangle$

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  • $\begingroup$ How do you know it is entangled? This is a general question, how do we know if a state is entangled or not? $\endgroup$ – user27286 Jan 27 at 15:59
  • $\begingroup$ In general two states are entangled if you can't write them as a product state. A simple example: $\frac{1}{2}\left(|00\rangle+|01\rangle+|10\rangle+|11\rangle\right)$ is not entangled because you can write it as $\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right) \otimes \frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)$. However a state like $\frac{1}{\sqrt{2}}\left(|00\rangle+|11\rangle\right)$ cannot be written as $\left(|a|0\rangle+b|1\rangle\right) \otimes \left(c|0\rangle+d|1\rangle\right)$. A quick way to check is see if measuring one state will give you info about the second $\endgroup$ – Rajiv Krishnakumar Jan 29 at 13:46
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The second register state stay as the state you prepared it in, that is, it is left unchanged. Note that if $|u\rangle$ is a eigenstate of $U$ with eigenvalue $e^{2\pi i \theta}$ then when you apply $U^{2^j}$ to the state $|u\rangle$, you will get $U^{2^j} |u\rangle = e^{2\pi i \theta 2^j}|u\rangle $.

enter image description here

And no, the state in the first register is not entangled to the state in the second register as you can see the output state is written as $\sum_u c_u |\varphi_u\rangle |u\rangle = \sum_u c_u |\varphi_u\rangle \otimes |u\rangle $, that is they can be written as a tensor product (not entangled).


For a quick example, consider the first part of the circuit:

enter image description here

In particular, let's suppose $U$ is the Pauli-Z operator and $|u\rangle$ is just the state $|1\rangle$. More specifically, we are considering the circuit below:

enter image description here

Note that the second qubit is in the state $|1\rangle$ after application of the $X$ gate. The first qubit is in the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ after the application of the Hadamard gate. Then we applied the Controlled-Z gate. Note that $|1\rangle$ is an eigenstate of the Pauli-Z operator with $Z|1\rangle = -|1\rangle$. The state of the overall system is now: $$ |\psi \rangle = \dfrac{|01\rangle - |11\rangle}{\sqrt{2}} $$ note the negative is resulted from the eigenvalue of $-1$ when we apply Pauli $Z$ to the state $|1\rangle$. It might be tempted to say that this state is entangled but it is not... because we can rewrite it as follow:

$$ |\psi \rangle = \overbrace{\bigg( \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg)}^{\textrm{first qubit}} \otimes \overbrace{ |1\rangle}^{\textrm{second qubit}} $$

So the state of the second qubit stays the same. No changes. The state of the first qubit pick up a relative phase factor coming from the eigenvalue of $-1$ when we apply $Z$ to $|1\rangle$.

You can now try to look at:

enter image description here

which is now the circuit:

enter image description here

Once you worked it out, you will see that you can write the state of the system as follow:

$$ |\psi \rangle = \overbrace{\bigg( \dfrac{|0\rangle + |1\rangle}{\sqrt{2}} \bigg)}^{\textrm{first qubit}} \otimes \overbrace{\bigg( \dfrac{|0\rangle - |1\rangle}{\sqrt{2}} \bigg)}^{\textrm{2nd qubit}} \otimes \overbrace{ |1\rangle}^{\textrm{3rd qubit}}$$

As you can see, the state of the qubit $q_2$ stays the same. And they are not entangled to one another at all.

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  • $\begingroup$ But here in the second register we pass $\sum_{u} c_{u}|u\rangle$..it can't stay the same isn't it? $\endgroup$ – user27286 Jan 15 at 9:51

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