Katas UI review - Multiqubit systems

katas/content/deutsch_algo/index.md

@@ -19,6 +19,8 @@ This kata introduces you to Deutsch algorithm - the single-qubit variant of Deut
 - Basic knowledge of single-qubit gates
 - Basic knowledge of quantum measurements
 
+If you need a refresher on these topics, you can check out the previous katas.
+
 @[section]({
     "id": "deutsch_algo__problem",
     "title": "The Problem"

katas/content/deutsch_jozsa/index.md

@@ -19,7 +19,9 @@ This kata introduces you to Deutsch-Jozsa algorithm - one of the most famous alg
 
 - Basic knowledge of single-qubit gates
 - Basic knowledge of quantum measurements
-- Deutsch algorithm - the single-qubit variant of Deutsch-Jozsa algorithm. If you're not familiar with it, you can find it in the Deutsch algorithm kata.
+- Deutsch algorithm - the single-qubit variant of Deutsch-Jozsa algorithm. 
+
+If you need a refresher on these topics, you can check out the previous katas.
 
 @[section]({
     "id": "deutsch_jozsa__problem",

katas/content/grovers_search/index.md

@@ -20,6 +20,7 @@ Note that this tutorial does not cover implementing specific classical functions
 - Basic knowledge of quantum gates and measurements
 - Basic understanding of quantum oracles
 
+If you need a refresher on these topics, you can check out the previous katas.
 
 @[section]({
     "id": "grovers_search__search_problem",

katas/content/multi_qubit_systems/bell_state_change_1/solution.md

@@ -1,15 +1,15 @@
-We recognize that the goal is another Bell state. In fact, it is one of the four Bell states.
+You recognize that the goal is another Bell state. In fact, it is one of the four Bell states.
 
-We remember from the Single-Qubit Gates kata that the Pauli Z gate will change the state of the $\ket{1}$ basis state of a single qubit, so this gate seems like a good candidate for what we want to achieve. This gate leaves the sign of the $\ket{0}$ basis state of a superposition unchanged, but flips the sign of the $\ket{1}$ basis state of the superposition.
+You remember from the Single-Qubit Gates kata that the Pauli Z gate will change the state of the $\ket{1}$ basis state of a single qubit, so this gate seems like a good candidate for what you want to achieve. This gate leaves the sign of the $\ket{0}$ basis state of a superposition unchanged, but flips the sign of the $\ket{1}$ basis state of the superposition.
 
-Don't forget that the Z gate acts on only a single qubit, and we have two here.
+Don't forget that the Z gate acts on only a single qubit, and you have two here.
 Let's also remember how the Bell state is made up from its individual qubits.
 
-If the two qubits are A and B, where A is `qs[0]` and B is `qs[1]`, we can write that
+If the two qubits are A and B, where A is `qs[0]` and B is `qs[1]`, you can write that
 $\ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \big(\ket{0_{A}0_{B}} + \ket{1_{A}1_{B}}\big)$.
-If we apply the Z gate to the qubit A, it will flip the phase of the basis state $\ket{1_A}$. As this phase is in a sense spread across the entangled state, with $\ket{1_A}$ basis state being part of the second half of the superposition, this application has the effect of flipping the sign of the whole basis state $\ket{1_A1_B}$, as you can see by running the solution below.
+If you apply the Z gate to the qubit A, it will flip the phase of the basis state $\ket{1_A}$. As this phase is in a sense spread across the entangled state, with $\ket{1_A}$ basis state being part of the second half of the superposition, this application has the effect of flipping the sign of the whole basis state $\ket{1_A1_B}$, as you can see by running the solution below.
 
-The exact same calculations can be done if we apply Z to the qubit B, so that's another possible solution.
+The exact same calculations can be done if you apply Z to the qubit B, so that's another possible solution.
 @[solution]({
 "id": "multi_qubit_systems__bell_state_change_1_solution",
 "codePath": "Solution.qs"

katas/content/multi_qubit_systems/bell_state_change_2/solution.md

@@ -1,8 +1,8 @@
-We have seen in the Single-Qubit Gates kata that the Pauli X gate flips $\ket{0}$ to $\ket{1}$ and vice versa, and as we seem to need some flipping of states, perhaps this gate may be of use. (Bearing in mind, of course, that the X gate operates on a single qubit).
+You have seen in the Single-Qubit Gates kata that the Pauli X gate flips $\ket{0}$ to $\ket{1}$ and vice versa, and as you seem to need some flipping of states, perhaps this gate may be of use. (Bearing in mind, of course, that the X gate operates on a single qubit).
 
-Let's compare the starting state $\frac{1}{\sqrt{2}} \big(\ket{0_A0_B} + \ket{1_A1_B}\big)$ with the goal state $\frac{1}{\sqrt{2}} \big(\ket{1_A0_B} + \ket{0_A1_B}\big)$ term by term and see how we need to transform it to reach the goal.
+Let's compare the starting state $\frac{1}{\sqrt{2}} \big(\ket{0_A0_B} + \ket{1_A1_B}\big)$ with the goal state $\frac{1}{\sqrt{2}} \big(\ket{1_A0_B} + \ket{0_A1_B}\big)$ term by term and see how you need to transform it to reach the goal.
 
-Using our nomenclature from "Bell state change  1", we can now see by comparing terms that $\ket{0_{A}}$ has flipped to $\ket{1_A}$ to get the first term, and $\ket{1_{A}}$ has flipped to $\ket{0_A}$ to get the second term. This allows us to say that the correct gate to use is Pauli X, applied to `qs[0]`.
+Using the nomenclature from "Bell state change  1", you can now see by comparing terms that $\ket{0_{A}}$ has flipped to $\ket{1_A}$ to get the first term, and $\ket{1_{A}}$ has flipped to $\ket{0_A}$ to get the second term. This allows you to say that the correct gate to use is Pauli X, applied to `qs[0]`.
 
 @[solution]({
 "id": "multi_qubit_systems__bell_state_change_2_solution",

katas/content/multi_qubit_systems/bell_state_change_3/solution.md

@@ -1,9 +1,9 @@
-We remember from the Single-Qubit Gates kata that the Pauli Z gate leaves the sign of the $\ket{0}$ component of the single qubit superposition unchanged but flips the sign of the $\ket{1}$ component of the superposition. We have also just seen in "Bell State Change 2" how to change our input state to the state $\frac{1}{\sqrt{2}} \big(\ket{01} + \ket{10}\big)$, which is almost our goal state (disregarding the phase change for the moment). So it would seem that a combination of these two gates will be what we need here. The remaining question is in what order to apply them, and to which qubit.
+You remember from the Single-Qubit Gates kata that the Pauli Z gate leaves the sign of the $\ket{0}$ component of the single qubit superposition unchanged but flips the sign of the $\ket{1}$ component of the superposition. You have also just seen in "Bell State Change 2" how to change our input state to the state $\frac{1}{\sqrt{2}} \big(\ket{01} + \ket{10}\big)$, which is almost your goal state (disregarding the phase change for the moment). So it would seem that a combination of these two gates will be what you need here. The remaining question is in what order to apply them, and to which qubit.
 
-First of all, which qubit? Looking back at the task "Bell state change 2", it seems clear that we need to use qubit `qs[0]`, like we did there.
+First of all, which qubit? Looking back at the task "Bell state change 2", it seems clear that you need to use qubit `qs[0]`, like you did there.
 
-Second, in what order should we apply the gates? Remember that the Pauli Z gate flips the phase of the $\ket{1}$ component of the superposition and leaves the $\ket{0}$ component alone.
-Let's experiment with applying X to `qs[0]` first. Looking at our "halfway answer" state $\frac{1}{\sqrt{2}} \big(\ket{01} + \ket{10}\big)$, we can see that if we apply the Z gate to `qs[0]`, it will leave the $\ket{0_{A}}$ alone but flip the phase of $\ket{1_{A}}$ to $-\ket{1_{A}}$, thus flipping the phase of the $\ket{11}$ component of our Bell state.
+Second, in what order should you apply the gates? Remember that the Pauli Z gate flips the phase of the $\ket{1}$ component of the superposition and leaves the $\ket{0}$ component alone.
+Let's experiment with applying X to `qs[0]` first. Looking at our "halfway answer" state $\frac{1}{\sqrt{2}} \big(\ket{01} + \ket{10}\big)$, you can see that if you apply the Z gate to `qs[0]`, it will leave the $\ket{0_{A}}$ alone but flip the phase of $\ket{1_{A}}$ to $-\ket{1_{A}}$, thus flipping the phase of the $\ket{11}$ component of our Bell state.
 
 @[solution]({
 "id": "multi_qubit_systems__bell_state_change_3_solution",