Output of Quantum Phase Estimation Algorithm

The second register state stay as the state you prepared it in, that is, it is left unchanged. Note that if $|urangle$ is a eigenstate of $U$ with eigenvalue $e^{2pi i theta}$ then when you apply $U^{2^j}$ to the state $|urangle$, you will get $U^{2^j} |urangle = e^{2pi i theta 2^j}|urangle $.

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_urangle |urangle = sum_u c_u |varphi_urangle otimes |urangle $, that is they can be written as a tensor product (not entangled).

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

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

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

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

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 $|1rangle$.

You can now try to look at:

which is now the circuit:

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

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

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