Today was Test 3. Take a breath! On Friday we continue Module 4 (Quantum) by moving from one qubit to two qubits, which is where quantum computing begins. The problems below are a warm-up for that: nothing here needs anything beyond the single-qubit rules from lecture (a state $a|0\rangle + b|1\rangle$ with $|a|^2 + |b|^2 = 1$ gives outcome $0$ with probability $|a|^2$ and outcome $1$ with probability $|b|^2$), but by the end you will have met the main new phenomenon with your own hands.

Relevant Textbook Sections:

Chapter 6 (Quantum): review single qubit states and measurement. On Friday we move on to multiple qubits.

To Do:

1. Two qubits, measured separately. Alice prepares two qubits, one after the other, and Bob measures each one in the $|0\rangle$, $|1\rangle$ basis.

   1. The first qubit is $|0\rangle$ and the second is $\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$. List the four possible pairs of outcomes (first qubit, second qubit) and the probability of each.
   2. Now both qubits are $\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$. List the four possible pairs of outcomes and their probabilities.
   3. Now the first qubit is $\frac{1}{\sqrt{5}}|0\rangle + \frac{2}{\sqrt{5}}|1\rangle$ and the second is $\frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle$. Same question.

2. Two qubits as one state. We can write a pair of qubits as a single “two-qubit state” by writing the two states side by side. For the pair in problem 1(1), that is $|0\rangle\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right)$. Multiply out as if these were ordinary products, keeping the first qubit first, and write $|0\rangle|0\rangle$ as $|00\rangle$, $|0\rangle|1\rangle$ as $|01\rangle$, and so on:

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

   The four states $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$ are the possible outcomes when both qubits are measured, and the squared lengths of the coefficients are the probabilities, just as for one qubit.

   1. Check that this agrees with your answer to 1(1).
   2. Multiply out $\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right)\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right)$ and check that it agrees with your answer to 1(2).
   3. Multiply out $\left(\frac{1}{\sqrt{5}}|0\rangle + \frac{2}{\sqrt{5}}|1\rangle\right)\left(\frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle\right)$ and check that it agrees with your answer to 1(3). Note the minus signs: they don’t affect the probabilities here, but hang on to them.
   4. Check that in each case the squared lengths of the four coefficients add up to $1$.

3. An unusual two-qubit state. Now take the rule from problem 2 as the definition: a two-qubit state is any

   $$a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle, \qquad |a|^2 + |b|^2 + |c|^2 + |d|^2 = 1,$$

   and measuring both qubits gives outcome $00$ with probability $|a|^2$, outcome $01$ with probability $|b|^2$, and so on. Consider the state

   $$|\Phi\rangle = \tfrac{1}{\sqrt{2}}|00\rangle + \tfrac{1}{\sqrt{2}}|11\rangle.$$
   1. What are the probabilities of the four outcomes $00$, $01$, $10$, $11$ when Bob measures both qubits?
   2. What is the probability that the first qubit comes out $0$? (Add up the probabilities of all outcomes whose first bit is $0$.)
   3. Suppose Bob measures only the first qubit and gets $0$. What can he now say about the second qubit, before measuring it? Compare with the state from problem 2(2), where knowing the first outcome tells you nothing about the second.
   4. Try to write $|\Phi\rangle$ as a product of two single-qubit states, $(x|0\rangle + y|1\rangle)(z|0\rangle + w|1\rangle)$, by multiplying out and comparing coefficients. Show that it is impossible. (You will need $xw = 0$ and $yz = 0$, but also $xz \neq 0$ and $yw \neq 0$.)

   A state that cannot be written as a product of single-qubit states is called entangled. Its two qubits are not independent, even if they are far apart.

4. Normalize first. Normalize the state $|00\rangle + 2i|01\rangle - 2|11\rangle$, then give the probabilities of the four outcomes when both qubits are measured. Is it a product state or entangled?