Entanglement, Bell states and no-cloning

1 / 12 · Learn

Before you start

Lesson 6 met two-qubit states that can’t be written as one state per qubit. This lesson names them, builds them with $H$ and a CNOT, and tests what they can do.

Part A builds entangled states; Part B asks what they can and can’t do.

By the end of this lesson you can:

measure one qubit of a pair in the 0/1 basis, and write the state that is left;
decide whether a two-qubit state is a product or entangled, by trying to factor it;
build the four Bell states with $H$ then CNOT, and write any of them from the $\beta_{xy}$ formula;
tell an entangled pair from a classical mix by measuring in a second basis;
explain why the CNOT copying circuit can’t copy an unknown state, and predict from the target whether a CNOT kicks a sign back onto its control, leaves it alone, or entangles the pair.

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

Guess before you learn the method. Is the two-qubit state $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$ entangled?

ANo, so each qubit has a definite value, 0 or 1.
BYes. The minus signs stop it from factoring.
CNo. It equals $\ket{-}\otimes\ket{+}$ .
DYes. It is a superposition of all four basis states.
ENo. It equals $\ket{+}\otimes\ket{-}$ .

Show the answer

Answer: E. No. It equals $\ket{+}\otimes\ket{-}$ .

$\tfrac12(\ket0 + \ket1)(\ket0 - \ket1)$ multiplies out to exactly $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$ , so it is the product $\ket{+}\otimes\ket{-}$ , not entangled. A superposition of all four basis states can still be a product.

Why not the others

A. It is a product state, but neither factor is $\ket0$ or $\ket1$ : each qubit gives 0 or 1 at random (50/50) when measured in the 0/1 basis. Product means each qubit has a state of its own, not that the state is a basis state.
B. Signs can sit inside the factors. Multiplying out $\tfrac12(\ket0 + \ket1)(\ket0 - \ket1)$ produces these minus signs exactly. The test $\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}$ gives 0 here, so the state factors.
C. The factors are in the wrong order. In $\ket a\otimes\ket b$ the first factor is the top qubit, and $\ket{-}\otimes\ket{+} = \tfrac12(\ket{00} + \ket{01} - \ket{10} - \ket{11})$ , a different state.
D. A common idea is that a state spread over several basis states must be entangled. Expand $\ket{+}\otimes\ket{-}$ and you get exactly this state, term by term. Entangled means the state cannot be written as one state per qubit, and this one can. The two ideas get mixed up because both involve several terms.

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

Guess: A CNOT whose target starts in $\ket0$ copies a basis state: $\ket0\ket0 \to \ket0\ket0$ and $\ket1\ket0 \to \ket1\ket1$ . What comes out when the input is $\ket{+}\ket0$ ?

A $\ket{+}\ket{+}$ : it copies $\ket0$ and $\ket1$ , so by linearity it copies their superposition too.
B $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ : an entangled state, not two copies.
CEither $\ket{00}$ or $\ket{11}$ , each with probability $\tfrac12$ .
D $\ket{+}\ket0$ , unchanged: the CNOT acts only when the control is definitely 1.

Show the answer

Answer: B. $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ : an entangled state, not two copies.

Term by term, $\ket{00} \to \ket{00}$ and $\ket{10} \to \ket{11}$ , so $\tfrac{1}{\sqrt2}(\ket{00} + \ket{10})$ becomes $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ . That state is entangled, and it is not $\ket{+}\ket{+}$ : the circuit copies only $\ket0$ and $\ket1$ .

Why not the others

A. A very common idea is that linearity carries copying over from $\ket0$ and $\ket1$ to their superpositions. Apply the CNOT term by term instead: $\ket{+}\ket0 = \tfrac{1}{\sqrt2}(\ket{00} + \ket{10})$ , and $\ket{10} \to \ket{11}$ , so the output is $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ . A true copy, $\ket{+}\ket{+}$ , also has $\ket{01}$ and $\ket{10}$ terms, which this output lacks. Linearity copies each term's label, not the state.
C. A gate doesn't pick an outcome; only a measurement does. The output is the single state $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ . You would see 00 or 11 at random only after measuring both qubits in the 0/1 basis, and a measurement in the ± basis can tell this state apart from a coin that picks $\ket{00}$ or $\ket{11}$ .
D. A superposition isn't “partly 1, so nothing happens”. The CNOT is linear and acts on each term separately: the $\ket{10}$ term becomes $\ket{11}$ . So the output, $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ , differs from the input.

2 / 12 · Learn

Part A · Measuring one qubit of two

Lesson 6 showed how to measure one qubit of a pair. In short, to measure the top qubit in the 0/1 basis:

the result 0 covers $\ket{00}$ and $\ket{01}$ , so $P(0) = \abs{\alpha_{00}}^2 + \abs{\alpha_{01}}^2$ ;
afterwards only those terms remain, divided by $\sqrt{P(0)}$ .

For the bottom qubit, use the second label instead. This lesson adds a question: what state is the other qubit in afterwards? For a product state, measuring one qubit leaves the other as it was. For the states below it doesn’t: the unmeasured qubit ends up in a state that depends on the result.

Check yourself

For $\ket\psi = \tfrac12\ket{00} + \tfrac12\ket{01} + \tfrac{1}{\sqrt2}\ket{11}$ , what is the probability that a 0/1 measurement of the top qubit gives 0?

$P(0) = \;?$

Hints

Which basis states have a 0 in the first (top) position?
Add the squared magnitudes of those terms’ amplitudes.
$|\tfrac12|^2 + |\tfrac12|^2 = ?$

Show the answer

Answer: $P(0) = \frac{1}{2}$

The terms with top qubit 0 are $\ket{00}$ and $\ket{01}$ : $P(0) = \tfrac14 + \tfrac14 = \tfrac12$ . The other half of the probability sits on $\ket{11}$ .

Common slips

$\frac{1}{4}$ : That is $|\alpha_{00}|^2$ alone. The result 0 for the top qubit covers every term whose first label is 0. One more term in $\ket\psi$ starts with 0: which?
$1$ : That adds the amplitudes, $\tfrac12 + \tfrac12$ . Probabilities come from squared magnitudes, so square each amplitude before you add.
$\frac{3}{4}$ : That is $\tfrac14 + \tfrac12$ , which brings in the $\ket{11}$ term. Keep only the terms whose first label (the top qubit) is 0: $\ket{00}$ and $\ket{01}$ .

Check yourself

Suppose that measurement gave 0. Enter the state of the pair afterwards.

Hints

Which terms survive a result of 0 on the top qubit?
Keep those terms, then divide by $\sqrt{P(0)}$ so the squared magnitudes add up to 1.
Divide $\tfrac12$ by $\sqrt{1/2} = \tfrac{1}{\sqrt2}$ .

Show the answer

Answer: $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|01\rangle$

$\big(\tfrac12\ket{00} + \tfrac12\ket{01}\big) / \tfrac{1}{\sqrt2} = \tfrac{1}{\sqrt2}(\ket{00} + \ket{01})$ . That is $\ket0\otimes\ket+$ : after the measurement the bottom qubit has a state of its own, $\ket+$ . Had the result been 1, only $\ket{11}$ would remain, and the bottom qubit would be in $\ket1$ .

3 / 12 · Learn

Part A · Product or entangled?

In Lesson 6 you multiplied out tensor products (refresher). If the top qubit is $a_0\ket0 + a_1\ket1$ and the bottom one is $b_0\ket0 + b_1\ket1$ , lay the pair’s amplitudes out with one row per value of the top qubit:

\begin{bmatrix}\alpha_{00} & \alpha_{01}\\ \alpha_{10} & \alpha_{11}\end{bmatrix} = \begin{bmatrix}a_0b_0 & a_0b_1\\ a_1b_0 & a_1b_1\end{bmatrix}

Both rows are multiples of $(b_0, b_1)$ , so the two diagonal products agree: $\alpha_{00}\alpha_{11} = a_0a_1b_0b_1 = \alpha_{01}\alpha_{10}$ . The reverse also holds: when they agree, you can read off the factors. A state that fails this test can’t be written as one state per qubit. That is what entangled means.

Check yourself

Apply the test to the state from the start of the lesson, $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$ .

$\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = \;?$

Hints

Read off $\alpha_{00}, \alpha_{01}, \alpha_{10}, \alpha_{11}$ , keeping their signs.
The test multiplies the diagonal pairs of the grid: $\alpha_{00}\alpha_{11}$ , minus $\alpha_{01}\alpha_{10}$ .
$\alpha_{00}\alpha_{11} = (\tfrac12)(-\tfrac12) = -\tfrac14$ . Now compute $\alpha_{01}\alpha_{10}$ .

Show the answer

Answer: $\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = 0$

$(\tfrac12)(-\tfrac12) - (-\tfrac12)(\tfrac12) = -\tfrac14 + \tfrac14 = 0$ , so the state is a product.

Common slips

$-\frac{1}{2}$ : Watch the signs. One of the two factors in $\alpha_{01}\alpha_{10}$ is negative, so that product is negative too, and subtracting a negative number adds its size.
$-\frac{1}{4}$ : That is $\alpha_{00}\alpha_{11}$ alone. Now subtract the other diagonal product, $\alpha_{01}\alpha_{10}$ , keeping track of its sign.

Check yourself

Which of these states are entangled? Use the test, or try to factor each one.

Select all that apply.

A $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$
B $\tfrac{1}{\sqrt2}(\ket{00} + \ket{01})$
C $\tfrac{1}{\sqrt2}(\ket{01} - \ket{10})$
D $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$
E $\tfrac12(\ket{00} + \ket{01} - \ket{10} - \ket{11})$

Hints

For each state, which of the four amplitudes are zero?
A state is a product exactly when $\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}$ .
If $\alpha_{01} = \alpha_{10} = 0$ , the state is a product only if $\alpha_{00}\alpha_{11} = 0$ too.

Show the answer

Answers: A. $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ ; C. $\tfrac{1}{\sqrt2}(\ket{01} - \ket{10})$ ; D. $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$

Entangled: $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ (test value $\tfrac12$ ), $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ ( $\tfrac{\sqrt3}{4}$ ) and $\tfrac{1}{\sqrt2}(\ket{01} - \ket{10})$ ( $\tfrac12$ ). Products: $\ket0\ket+$ and $\ket-\ket+$ . Unequal weights, as in $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ , still count as entangled.

Why not the others

B. $\tfrac{1}{\sqrt2}(\ket{00} + \ket{01}) = \ket0\otimes\ket+$ : the top qubit is 0 in both terms, and the test gives $0 - 0 = 0$ . Two terms are not enough to make a state entangled.
E. $\tfrac12(\ket{00} + \ket{01} - \ket{10} - \ket{11}) = \ket-\otimes\ket+$ . Both diagonal products are $-\tfrac14$ , so the test gives 0.

4 / 12 · Learn

Part A · Making entanglement: H, then CNOT

Gates that act on one qubit at a time turn a product state into another product state: $(A\otimes B)(\ket a\otimes\ket b) = A\ket a\otimes B\ket b$ . To entangle two qubits you need a gate that acts on both. The standard recipe is $H$ on the top qubit, then a CNOT with the top qubit as control. Step through it from $\ket{00}$ and watch the per-wire readouts.

Check yourself

After the circuit, the pair is in $\ket{\beta_{00}} = \tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ . Where does the top qubit’s Bloch arrow point?

AIn the same direction as the bottom qubit’s arrow, since the two always match.
BAlong $+x$ : the $H$ made it $\ket+$ , and a CNOT never changes its control.
CNowhere: the top qubit has no single-qubit state of its own. Only the pair has a state.
DAt the north or the south pole, chosen at random.

Show the answer

Answer: C. Nowhere: the top qubit has no single-qubit state of its own. Only the pair has a state.

$\ket{\beta_{00}}$ fails the factor test, so it can’t be written as (top state) $\otimes$ (bottom state). Neither qubit has a ket, so neither has an arrow on the sphere’s surface; that is why the circuit widget swaps the sphere for a badge. (A more general tool, the density matrix, does give each qubit a “reduced state”; for $\ket{\beta_{00}}$ it sits at the centre of the sphere. That tool is beyond this course.)

Why not the others

A. This assumes each qubit still has a ket and the two kets are matched. But if the top qubit had any state $\ket a$ and the bottom any $\ket b$ , the pair would be $\ket a\otimes\ket b$ , a product. $\ket{\beta_{00}}$ fails the factor test, so there are no single-qubit kets to draw arrows for.
B. “The control is unchanged” holds for control inputs $\ket0$ and $\ket1$ , which is where the truth table comes from. Here the control is a superposition, and the CNOT acts on the $\ket{10}$ term only. If the top qubit were still $\ket+$ , the pair would factor as $\ket+\otimes(\text{something})$ , but $\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10} = \tfrac12 \neq 0$ .
D. Nothing has been measured, so no result has been chosen. The pair is in the single state $\ket{\beta_{00}}$ , not secretly in $\ket{00}$ or $\ket{11}$ ; Part B shows an experiment that tells these two situations apart.

Check yourself

Build a circuit that turns $\ket{00}$ into $\ket{\beta_{10}} = \tfrac{1}{\sqrt2}(\ket{00} - \ket{11})$ .

Build it in the circuit builder (gates: H, X, Z, CNOT).

Hints

Which gate in the recipe can produce a minus sign?
$H\ket1 = \tfrac{1}{\sqrt2}(\ket0 - \ket1)$ has a minus sign, and $H\ket0$ doesn’t. So the top qubit should reach the $H$ as $\ket1$ .
Put $X$ on the top wire first, then $H$ on the top wire, then a CNOT from the top wire to the bottom one.

Show the answer

One circuit that works: X on q0, then H on q0, then CNOT q0→q1.

$X$ makes $\ket{10}$ ; $H$ gives $\tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ ; the CNOT gives $\tfrac{1}{\sqrt2}(\ket{00} - \ket{11})$ . Other circuits work too: $H$ then $Z$ on the top wire, then the CNOT; or build $\ket{\beta_{00}}$ and finish with $Z$ on either wire.

5 / 12 · Worked example

Part A · Worked example: the four Bell states

Worked example

$H$ then CNOT on a basis input

Input $\ket{00}$ : after $H$
$H$ acts on the top qubit only: $\ket0\ket0 \to \tfrac{1}{\sqrt2}(\ket0 + \ket1)\ket0 = \tfrac{1}{\sqrt2}(\ket{00} + \ket{10})$ . This is still a product state.
Input $\ket{00}$ : after the CNOT
The CNOT flips the bottom qubit in each term whose top qubit is 1, and leaves the other terms alone: $\ket{00} \to \ket{00}$ and $\ket{10} \to \ket{11}$ . So the output is $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11}) = \ket{\beta_{00}}$ .
Input $\ket{10}$ : after $H$
Now the top qubit starts in $\ket1$ , and $H\ket1 = \tfrac{1}{\sqrt2}(\ket0 - \ket1)$ carries a minus sign: $\ket1\ket0 \to \tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ .
Input $\ket{10}$ : after the CNOT
Your turn
Apply the CNOT to $\tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ , term by term. Enter the four amplitudes.
Hints
1. Which of the two terms has its top qubit equal to 1?
2. Only that term changes: its bottom label flips. Its sign stays with it.
Show the answer
Answer: $\frac{1}{\sqrt{2}}|00\rangle - \frac{1}{\sqrt{2}}|11\rangle$
$\ket{00} \to \ket{00}$ and $-\ket{10} \to -\ket{11}$ , so the output is $\tfrac{1}{\sqrt2}(\ket{00} - \ket{11}) = \ket{\beta_{10}}$ .

Here is one more input, worked in full, before you try one alone.

Another worked example

$H$ then CNOT on $\ket{01}$

After $H$
$H$ on the top qubit: $\ket0\ket1 \to \tfrac{1}{\sqrt2}(\ket0 + \ket1)\ket1 = \tfrac{1}{\sqrt2}(\ket{01} + \ket{11})$ . No minus sign, because the top qubit started in $\ket0$ .
After the CNOT
Term by term: $\ket{01}$ has top qubit 0, so it stays; $\ket{11}$ has top qubit 1, so its bottom qubit flips, giving $\ket{10}$ . The output is $\tfrac{1}{\sqrt2}(\ket{01} + \ket{10}) = \ket{\beta_{01}}$ .
What to notice
The bottom input decided whether the two labels in the output match ( $\ket{00}, \ket{11}$ ) or differ ( $\ket{01}, \ket{10}$ ). The top input decided the sign, through $H$ .

All four inputs give the four Bell states:

\begin{aligned}\ket{\beta_{00}} &= \tfrac{1}{\sqrt2}(\ket{00} + \ket{11}),\\ \ket{\beta_{01}} &= \tfrac{1}{\sqrt2}(\ket{01} + \ket{10}),\\ \ket{\beta_{10}} &= \tfrac{1}{\sqrt2}(\ket{00} - \ket{11}),\\ \ket{\beta_{11}} &= \tfrac{1}{\sqrt2}(\ket{01} - \ket{10}).\end{aligned}

One formula covers all four, where $\bar y$ means NOT $y$ :

\ket{\beta_{xy}} = \frac{\ket{0,y} + (-1)^x\ket{1,\bar y}}{\sqrt2}

The bottom input $y$ sets the parity (labels equal when $y = 0$ , different when $y = 1$ ). The top input $x$ sets the sign.

Check yourself

Apply $H$ to the top qubit and then a CNOT (control on top) to $\ket{10}$ . Enter the four output amplitudes.

Hints

What does $H$ do to the top qubit's starting value: which sign appears?
Apply the gates in order, term by term. The CNOT flips the bottom label only where the top label is 1.
After the $H$ the state is $\tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ . Now apply the CNOT to each term.

Show the answer

Answer: $\frac{1}{\sqrt{2}}|00\rangle - \frac{1}{\sqrt{2}}|11\rangle$

$H$ gives $\tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ ; the CNOT then flips the bottom qubit in the term whose top qubit is 1, giving $\tfrac{1}{\sqrt2}(\ket{00} - \ket{11}) = \ket{\beta_{10}}$ .

Step 1. $H$ on the top qubit: $\ket{10} \to \tfrac{1}{\sqrt2}(\ket{00} - \ket{10})$ .

Step 2. CNOT, term by term: a term whose top label is 0 is unchanged; a term whose top label is 1 has its bottom label flipped. Result: $\tfrac{1}{\sqrt2}(\ket{00} - \ket{11})$ .

Check with the formula: $\ket{\beta_{10}} = \tfrac{1}{\sqrt2}(\ket{0,0} + (-1)^{1}\ket{1,1})$ , the same state.

6 / 12 · Learn

Part B · Can a CNOT copy a qubit?

Classically, a CNOT whose target starts at 0 is a copier: bits $(a, 0)$ come out as $(a, a)$ . It does the same to the quantum basis states: $\ket0\ket0 \to \ket0\ket0$ and $\ket1\ket0 \to \ket1\ket1$ . Does it also copy an unknown state $\ket\psi = a\ket0 + b\ket1$ ? A true copy would be

\begin{aligned}\ket\psi\ket\psi &= a^2\ket{00} + ab\ket{01}\\ &\quad + ab\ket{10} + b^2\ket{11}.\end{aligned}

Step through the circuit with $\ket\psi = \ket+$ .

Check yourself

Enter the circuit’s output for the input $\ket+\ket0$ . Read it off the figure, or work it out term by term.

Hints

Write $\ket+\ket0$ as a sum of two basis states.
The CNOT is linear: apply it to each basis state separately, then add the results.
$\ket{00} \to \ket{00}$ and $\ket{10} \to \ket{11}$ .

Show the answer

Answer: $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$

$\ket+\ket0 = \tfrac{1}{\sqrt2}(\ket{00} + \ket{10}) \to \tfrac{1}{\sqrt2}(\ket{00} + \ket{11}) = \ket{\beta_{00}}$ : the same Bell state as in Part A, since $H\ket0 = \ket+$ .

Check yourself

For which states $\ket\psi = a\ket0 + b\ket1$ does the copying circuit output $\ket\psi\ket\psi$ ?

AOnly $\ket+$ and $\ket-$ , where $|a| = |b|$ .
BOnly when $ab = 0$ : the basis states $\ket0$ and $\ket1$ (up to a phase).
CEvery state: it copies $\ket0$ and $\ket1$ , and linearity extends that to every superposition.
DEvery state with real amplitudes.

Show the answer

Answer: B. Only when $ab = 0$ : the basis states $\ket0$ and $\ket1$ (up to a phase).

Matching $a\ket{00} + b\ket{11}$ with $a^2\ket{00} + ab\ket{01} + ab\ket{10} + b^2\ket{11}$ needs $ab = 0$ , so $\ket\psi$ is $\ket0$ or $\ket1$ (times a phase). The circuit copies the basis states and nothing else.

Why not the others

A. Those are the states where the missing cross terms are largest ( $|ab| = \tfrac12$ ). The output $a\ket{00} + b\ket{11}$ matches $\ket\psi\ket\psi$ only when the $\ket{01}$ and $\ket{10}$ amplitudes, $ab$ , are zero.
C. Linearity extends the gate’s action term by term, which gives $a\ket{00} + b\ket{11}$ . A copy also needs $ab$ on $\ket{01}$ and on $\ket{10}$ , so linearity is exactly what breaks copying. The idea is tempting because copying each basis state looks like copying everything built from them.
D. Try $\ket+$ , which has real amplitudes: the output is $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ , not $\ket+\ket+$ . The trouble is the missing $ab$ cross terms, and real numbers don’t remove them.

The general argument

Gates preserve inner products: if $U$ sends $\ket\psi \to \ket{\psi'}$ and $\ket\varphi \to \ket{\varphi'}$ , then $\braket{\psi'}{\varphi'} = \braket{\psi}{\varphi}$ . Suppose $U\ket\psi\ket0 = \ket\psi\ket\psi$ and $U\ket\varphi\ket0 = \ket\varphi\ket\varphi$ . The inner product of the two inputs is $\braket{\psi}{\varphi}$ ; that of the two outputs is $\braket{\psi}{\varphi}^2$ . So $\braket{\psi}{\varphi} = \braket{\psi}{\varphi}^2$ , which forces $\braket{\psi}{\varphi}$ to be 0 or 1. One gate can copy two different states only if they are orthogonal, like $\ket0$ and $\ket1$ .

7 / 12 · Explore

Part B · Correlated is not the same as entangled

The copying circuit’s output gives matching results when both qubits are measured in the 0/1 basis. So does a much plainer source. Can an experiment tell them apart?

Predict, then see

Source A sends $\ket{\beta_{00}}$ : Alice gets the top qubit, Bob the bottom one. Source B flips a hidden coin and sends either $\ket{00}$ or $\ket{11}$ . When both measure in the 0/1 basis, both sources give matching results every time. Now Alice and Bob both measure in the ± basis (the X basis). How often do their results match?

ASource A matches every time; source B matches about half the time.
BSource A drops to about half; source B still matches every time.
CBoth drop to about half: the ± basis scrambles any correlation.
DBoth sources still match every time.

In a sentence or two: why does source A still match in the ± basis, while source B doesn’t?

Show what happens

What happens: A. Source A matches every time; source B matches about half the time.

Both sources match 100% of the time in the 0/1 basis, but only $\ket{\beta_{00}}$ also matches in the ± basis, because $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11}) = \tfrac{1}{\sqrt2}(\ket{++} + \ket{--})$ . Source B always sends a product state, and the two qubits of a product state give independent results, so in the ± basis they match only by chance. Matching 0/1 results therefore don’t show that a state is entangled; a second basis can expose the difference.

Why not the others

B. This swaps the two sources. Source B’s qubits are each in a definite 0/1 basis state, and those give random ± results. It is $\ket{\beta_{00}}$ that stays matched in the ± basis.
C. That is right for source B but not for A. Rewritten in the ± basis, $\ket{\beta_{00}} = \tfrac{1}{\sqrt2}(\ket{++} + \ket{--})$ : again a matched pair, so source A still matches every time.
D. Each of source B’s pairs is a product state, $\ket{00}$ or $\ket{11}$ . A qubit in $\ket0$ or $\ket1$ gives + or − at random in the ± basis (from Lesson 2: $\ket0 = (\ket+ + \ket-)/\sqrt2$ ), and the two qubits do this independently, so they match only about half the time.

Why no source of product states can match in both bases

Give the two qubits of a product state their Bloch vectors $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$ . In the 0/1 basis their results match with probability $\tfrac12(1 + z_Az_B)$ , and in the ± basis with probability $\tfrac12(1 + x_Ax_B)$ . Since $x^2 + z^2 \le 1$ for each qubit, the Cauchy–Schwarz inequality gives $z_Az_B + x_Ax_B \le 1$ , so the two match rates add up to at most 150%. A random mix of product states averages its rates, so the same limit holds for it. Source B sits right at the limit (100% + 50%), while $\ket{\beta_{00}}$ reaches 200%. (Ruling out every classical explanation, including devices with answers agreed in advance, takes more bases and Bell’s inequality.)

Check yourself

Choose the source “Partial”: $\cos\tfrac{\pi}{8}\ket{00} + \sin\tfrac{\pi}{8}\ket{11}$ . Its test value is $\cos\tfrac{\pi}{8}\sin\tfrac{\pi}{8} \approx 0.35 \neq 0$ , so it is entangled, and in the 0/1 basis Alice and Bob always match. When both measure in the ± basis, about how often do they match?

AAbout 85% of the time.
BAbout half the time, like the classical mix.
CEvery time: entangled pairs match in every basis.

Show the answer

Answer: A. About 85% of the time.

The ± match probability is $\tfrac12(\cos\tfrac{\pi}{8} + \sin\tfrac{\pi}{8})^2 = \tfrac12(1 + \sin\tfrac{\pi}{4}) = \tfrac{2 + \sqrt2}{4} \approx 85\%$ . That equals the weight of $\ket{00}$ , $\cos^2\tfrac{\pi}{8} \approx 85\%$ , only because the angle is $\pi/8$ : for $\cos\tfrac{\pi}{6}\ket{00} + \sin\tfrac{\pi}{6}\ket{11}$ the weight is 75% but the ± match rate is about 93%. Entanglement comes in degrees, and results that match perfectly in both bases need a maximally entangled state.

Why not the others

B. Half would mean no ± correlation at all, which is what source B’s product states give. Run it: this state matches about 85% of the time, more than any source of product states can manage while also matching 100% in the 0/1 basis (see the box above).
C. Run it: the ± match rate settles near 85%, not 100%. Perfect matching in both bases needs a maximally entangled state such as $\ket{\beta_{00}}$ , whose two terms have equal weight. This state is entangled, since it fails the factor test, but less so. The idea is tempting because the Bell states are the examples everyone meets first.

Check yourself

Switch back to the source “Bell β₀₀”. Bob keeps his basis on Z and looks only at his own results. Alice switches between Z and X. What happens to Bob’s own counts?

AThey become all 0 whenever Alice measures in Z first.
BThey move away from 50/50 when Alice switches, so Alice could signal to Bob by choosing her basis.
CNothing: they stay about 50/50 whichever basis Alice uses.

Show the answer

Answer: C. Nothing: they stay about 50/50 whichever basis Alice uses.

Bob’s qubit on its own gives 0 or 1 with probability $\tfrac12$ in the 0/1 basis, whatever Alice does. Entanglement alone can’t carry a message: the correlation appears only when Alice and Bob compare notes. Lesson 8 (teleportation) builds on this.

Why not the others

A. Alice’s own 0/1 results are random, half 0 and half 1, and Bob’s match them, so his list is also half 0 and half 1. What becomes certain is Bob’s result given Alice’s result, not his result on its own.
B. Run it: Bob’s own counts stay near 50/50 for both of Alice’s choices. Bob’s results match Alice’s when their bases agree, but Alice’s results are themselves random, so Bob’s list looks the same either way. The matching only shows up when the two lists are compared, which needs an ordinary message.

8 / 12 · Learn

Part B · What the target decides: kickback or entanglement

In Lesson 6 a CNOT turned $\ket+\ket-$ into $\ket-\ket-$ : the control changed and the target didn’t. Here is the general rule. Write any control as $a\ket0 + b\ket1$ . The CNOT leaves the $\ket0$ branch alone and applies $X$ to the target in the $\ket1$ branch:

(a\ket0 + b\ket1)\ket t \;\to\; a\ket0\ket t + b\ket1\,X\ket t

Target $\ket+$ : $X\ket+ = \ket+$ , so nothing changes.
Target $\ket-$ : $X\ket- = -\ket-$ , so the output is $(a\ket0 - b\ket1)\ket-$ . The target is unchanged and the control’s relative sign flips. This is the phase kickback of Lesson 6, and it drives Deutsch’s algorithm in Lesson 9.
Target $\ket0$ : the output is $a\ket{00} + b\ket{11}$ , the copying circuit, which is entangled unless $ab = 0$ .

Check yourself

The control (top) is $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ and the target (bottom) is $\ket-$ . What does the CNOT output?

A $\left(\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1\right)\ket-$ , unchanged
B $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ , an entangled state
C $\left(\tfrac{\sqrt3}{2}\ket0 - \tfrac12\ket1\right)\ket-$
D $\left(\tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1\right)\ket-$
E $\left(\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1\right)\ket+$

Hints

Write the input as two branches, one for each value of the control.
Only the $\ket1$ branch gets $X$ on the target, and $X\ket- = -\ket-$ .
The $\ket1$ branch becomes $\tfrac12\ket1\,(-\ket-)$ ; the $\ket0$ branch is unchanged.

Show the answer

Answer: C. $\left(\tfrac{\sqrt3}{2}\ket0 - \tfrac12\ket1\right)\ket-$

Split the input by the control’s value: $\tfrac{\sqrt3}{2}\ket0\ket- + \tfrac12\ket1\ket-$ . The CNOT changes only the second term, to $\tfrac12\ket1\,X\ket- = -\tfrac12\ket1\ket-$ . So the output is $\left(\tfrac{\sqrt3}{2}\ket0 - \tfrac12\ket1\right)\ket-$ : the target is untouched, and the control’s $\ket1$ amplitude changed sign, which is a $Z$ on the control.

Why not the others

A. “The control is unchanged” is guaranteed only for control inputs $\ket0$ and $\ket1$ . Here the $\ket1$ branch gets $X$ on the target, and $X\ket- = -\ket-$ puts a minus sign on that branch: $\tfrac12\ket1$ becomes $-\tfrac12\ket1$ . Kickback works for any control, not only $\ket\pm$ .
B. That is what the copying circuit makes, with the target in $\ket0$ . With target $\ket-$ , $X$ only multiplies the target by $-1$ , so the target factors out of both branches and the pair stays a product.
D. That applies $X$ to the control, swapping its amplitudes. A CNOT never applies $X$ to its control. What reaches the control is a sign on its $\ket1$ branch: a $Z$ , not an $X$ .
E. $X$ doesn’t turn $\ket-$ into $\ket+$ : it gives $-\ket-$ , the same state with a minus sign. That sign moves onto the control’s $\ket1$ branch, and the target stays $\ket-$ .

Seen from the ± basis, the CNOT points the other way

Call $\ket+$ “0” and $\ket-$ “1”. Then the CNOT flips the top qubit exactly when the bottom qubit is $\ket-$ , and leaves the bottom alone: $\ket+\ket- \to \ket-\ket-$ and $\ket-\ket- \to \ket+\ket-$ , while $\ket\pm\ket+$ pass through unchanged. In circuit terms: $H$ on both wires, then a CNOT, then $H$ on both wires equals a CNOT with control and target swapped. Which qubit acts as the control depends on the basis you describe it in.

9 / 12 · Practice

Part B · Practice

Three mixed items. Each one can come back later with new numbers.

Check yourself

Two qubits are in $\ket\psi = \tfrac{1}{\sqrt2}\ket{01} + \tfrac12\ket{10} + \tfrac12\ket{11}$ . Measure the top qubit in the 0/1 basis, and suppose the result is 1. Enter the state of the pair afterwards.

Hints

Which basis states have 1 in the first position (the top qubit)?
Keep only those terms, then divide them by the square root of their total probability, so the squared magnitudes add up to 1.
Here that probability is $\tfrac12$ : divide each kept amplitude by $\sqrt{\tfrac12}$ .

Show the answer

Answer: $\frac{1}{\sqrt{2}}|10\rangle + \frac{1}{\sqrt{2}}|11\rangle$

The result 1 keeps $\ket{10}$ and $\ket{11}$ , with total probability $\tfrac12$ . Dividing by $\sqrt{\tfrac12}$ gives $\tfrac{1}{\sqrt2}(\ket{10} + \ket{11})$ .

Keep: the terms whose first label is 1: $\ket{10}$ and $\ket{11}$ .

Probability: add their squared magnitudes, $P(1) = \tfrac12$ .

Renormalize: divide the kept amplitudes by $\sqrt{\tfrac12}$ , giving $\tfrac{1}{\sqrt2}(\ket{10} + \ket{11})$ .

Check yourself

Select every entangled state.

Select all that apply.

A $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$
B $\tfrac12(\ket{00} + i\ket{01} + \ket{10} + i\ket{11})$
C $\tfrac12(\ket{00} - \ket{01} - \ket{10} + \ket{11})$
D $\tfrac12(\ket{00} + i\ket{01} + i\ket{10} + \ket{11})$

Hints

For each state, do the two diagonal products of its amplitude grid agree?
Product exactly when $\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}$ ; otherwise entangled.
If two amplitudes are zero, the test is quick. With all four nonzero, watch the signs and factors of $i$ .

Show the answer

Answer: D. $\tfrac12(\ket{00} + i\ket{01} + i\ket{10} + \ket{11})$

The entangled states fail the test $\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}$ . The others factor: $\tfrac12(\ket{00} + i\ket{01} + \ket{10} + i\ket{11}) = \ket{+}\otimes\tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ ; $\tfrac12(\ket{00} - \ket{01} - \ket{10} + \ket{11}) = \ket{-}\otimes\ket{-}$ ; $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11}) = \ket{+}\otimes\ket{-}$ .

Entangled (nonzero $\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}$ ): $\tfrac12(\ket{00} + i\ket{01} + i\ket{10} + \ket{11})$ (test value $\tfrac12$ ).

Products (test value 0): $\tfrac12(\ket{00} + i\ket{01} + \ket{10} + i\ket{11}) = \ket{+}\otimes\tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ ; $\tfrac12(\ket{00} - \ket{01} - \ket{10} + \ket{11}) = \ket{-}\otimes\ket{-}$ ; $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11}) = \ket{+}\otimes\ket{-}$ .

Why not the others

A. $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$ factors: it equals $\ket{+}\otimes\ket{-}$ , and its test value is 0. A product state can spread over several basis states.
B. $\tfrac12(\ket{00} + i\ket{01} + \ket{10} + i\ket{11})$ factors: it equals $\ket{+}\otimes\tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ , and its test value is 0. A product state can spread over several basis states.
C. $\tfrac12(\ket{00} - \ket{01} - \ket{10} + \ket{11})$ factors: it equals $\ket{-}\otimes\ket{-}$ , and its test value is 0. A product state can spread over several basis states.

Check yourself

A CNOT has control (top) $\ket{-}$ and target (bottom) $\ket{1}$ . Enter the four output amplitudes.

Hints

Write the control as a sum of its $\ket0$ and $\ket1$ branches. What happens to the target in each?
The CNOT is linear: it leaves the $\ket0$ branch alone and applies $X$ to the target in the $\ket1$ branch.
In the $\ket1$ branch the target becomes $X\ket{1} = \ket0$ .

Show the answer

Answer: $\frac{1}{\sqrt{2}}|01\rangle - \frac{1}{\sqrt{2}}|10\rangle$

The input is $\tfrac{1}{\sqrt2}(\ket{01} - \ket{11})$ . Applying $X$ to the target in each term whose control label is 1 gives $\tfrac{1}{\sqrt2}(\ket{01} - \ket{10})$ , an entangled state.

Expand the control: $\ket{-}\ket{1} = \tfrac{1}{\sqrt2}(\ket0\ket{1} - \ket1\ket{1})$ .

Apply CNOT branch by branch: the $\ket0$ branch is unchanged; in the $\ket1$ branch the target becomes $X\ket{1} = \ket0$ .

Result: $\tfrac{1}{\sqrt2}(\ket{01} - \ket{10})$ , an entangled state.

10 / 12 · Explain

Explain it: the copier that entangles

Explain it

The copying circuit copies $\ket0$ and $\ket1$ perfectly. Why doesn’t it copy $\ket+$ , and what does it produce instead? Write two to four sentences, as if to a friend who expects linearity to make it work.

Show a model answer

The CNOT is linear, so it acts on each term of $\ket+\ket0 = \tfrac{1}{\sqrt2}(\ket{00} + \ket{10})$ separately: $\ket{00}$ stays and $\ket{10}$ becomes $\ket{11}$ . The output is $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11}) = \ket{\beta_{00}}$ , an entangled state. Two copies would be $\ket+\ket+ = \tfrac12(\ket{00} + \ket{01} + \ket{10} + \ket{11})$ , which has $\ket{01}$ and $\ket{10}$ terms that the output lacks. The two qubits now give matching results in the 0/1 basis, but that is a correlation, not two copies: neither qubit has a state of its own. The circuit copies $a\ket0 + b\ket1$ only when $ab = 0$ .

A good answer covers

Applies the CNOT to each term separately (linearity): $\ket{00} \to \ket{00}$ and $\ket{10} \to \ket{11}$ .
States that the result, $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ , is not $\ket+\ket+$ (which has $\ket{01}$ and $\ket{10}$ terms).
Notes that matching 0/1 results are a correlation, not two independent copies: the output is entangled and neither qubit has its own state.
Says the circuit copies $a\ket0 + b\ket1$ only when $ab = 0$ , that is, for $\ket0$ and $\ket1$ .

11 / 12 · Mastery check

Mastery check

Check yourself

Is the two-qubit state $\tfrac12(\ket{00} + \ket{01} - \ket{10} - \ket{11})$ entangled?

AYes. It is a superposition of all four basis states.
BNo. It equals $\ket{-}\otimes\ket{+}$ .
CNo, so each qubit has a definite value, 0 or 1.
DYes. The minus signs stop it from factoring.
ENo. It equals $\ket{+}\otimes\ket{-}$ .

Show the answer

Answer: B. No. It equals $\ket{-}\otimes\ket{+}$ .

$\tfrac12(\ket0 - \ket1)(\ket0 + \ket1)$ multiplies out to exactly $\tfrac12(\ket{00} + \ket{01} - \ket{10} - \ket{11})$ , so it is the product $\ket{-}\otimes\ket{+}$ , not entangled. A superposition of all four basis states can still be a product.

Why not the others

A. A common idea is that a state spread over several basis states must be entangled. Expand $\ket{-}\otimes\ket{+}$ and you get exactly this state, term by term. Entangled means the state cannot be written as one state per qubit, and this one can. The two ideas get mixed up because both involve several terms.
C. It is a product state, but neither factor is $\ket0$ or $\ket1$ : each qubit gives 0 or 1 at random (50/50) when measured in the 0/1 basis. Product means each qubit has a state of its own, not that the state is a basis state.
D. Signs can sit inside the factors. Multiplying out $\tfrac12(\ket0 - \ket1)(\ket0 + \ket1)$ produces these minus signs exactly. The test $\alpha_{00}\alpha_{11} - \alpha_{01}\alpha_{10}$ gives 0 here, so the state factors.
E. The factors are in the wrong order. In $\ket a\otimes\ket b$ the first factor is the top qubit, and $\ket{+}\otimes\ket{-} = \tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11})$ , a different state.

Check yourself

Alice holds the top qubit and Bob the bottom qubit of $\ket{\beta_{10}} = \tfrac{1}{\sqrt2}(\ket{00} - \ket{11})$ . Alice measures in the 0/1 basis and gets 1. What does Bob get when he then measures in the 0/1 basis?

A1, but only once Alice's result has reached him.
B0, with certainty.
C1, with certainty.
D0 or 1 at random, 50/50: Alice's measurement doesn't touch Bob's qubit.

Show the answer

Answer: C. 1, with certainty.

Alice's result 1 keeps only the $\ket{11}$ term of $\ket{\beta_{10}}$ , which renormalizes to $\ket{11}$ (up to a sign). So Bob's 0/1 measurement gives 1 with certainty.

Why not the others

A. Bob's result doesn't wait for a message: after Alice's measurement the pair is in $\ket{11}$ , so he gets 1 even if he hears nothing. A message changes what Bob knows, not what he gets; without it, his own list of results still looks 50/50.
B. Look at which term survives. Alice's result 1 keeps only the term whose first label is 1, which is $\ket{11}$ ; its second label is 1. (In $\ket{\beta_{xy}}$ the two labels match when $y = 0$ and differ when $y = 1$ .)
D. Before Alice measures, Bob's result would indeed be 50/50. But the pair has one joint state, and her measurement acts on it: it keeps only the term with first label 1, leaving $\ket{11}$ . Bob's result is now fixed, wherever he is.

Check yourself

Starting from $\ket{00}$ , build a circuit whose output is $\ket{\beta_{01}} = \tfrac{1}{\sqrt2}(\ket{01} + \ket{10})$ .

Build it in the circuit builder (gates: H, X, Z, CNOT).

Show the answer

One circuit that works: X on q1, then H on q0, then CNOT q0→q1.

One way: $X$ on the bottom wire turns $\ket{00}$ into $\ket{01}$ , and then $H$ on the top qubit followed by a CNOT gives $\ket{\beta_{01}}$ . Other circuits also work, as long as the output matches up to a global phase.

Use the recipe from the worked example: H then CNOT turns $\ket{01}$ into $\ket{\beta_{01}}$ . So first prepare $\ket{01}$ with $X$ on the bottom wire, then apply $H$ to the top qubit, then a CNOT with the top qubit as control.

Check yourself

A CNOT whose target starts in $\ket0$ copies a basis state: $\ket0\ket0 \to \ket0\ket0$ and $\ket1\ket0 \to \ket1\ket1$ . Let $\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ . What comes out when the input is $\ket\psi\ket0$ ?

A $\ket\psi\ket\psi$ : it copies $\ket0$ and $\ket1$ , so by linearity it copies their superposition too.
B $\ket\psi\ket0$ , unchanged: the CNOT acts only when the control is definitely 1.
C $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ : an entangled state, not two copies.
DEither $\ket{00}$ or $\ket{11}$ , with probabilities $\tfrac34$ and $\tfrac14$ .

Show the answer

Answer: C. $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ : an entangled state, not two copies.

Term by term, $\ket{00} \to \ket{00}$ and $\ket{10} \to \ket{11}$ , so $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{10}$ becomes $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ . That state is entangled, and it is not $\ket\psi\ket\psi$ : the circuit copies only $\ket0$ and $\ket1$ .

Why not the others

A. A very common idea is that linearity carries copying over from $\ket0$ and $\ket1$ to their superpositions. Apply the CNOT term by term instead: $\ket\psi\ket0 = \tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{10}$ , and $\ket{10} \to \ket{11}$ , so the output is $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ . A true copy, $\ket\psi\ket\psi$ , also has $\ket{01}$ and $\ket{10}$ terms, which this output lacks. Linearity copies each term's label, not the state.
B. A superposition isn't “partly 1, so nothing happens”. The CNOT is linear and acts on each term separately: the $\ket{10}$ term becomes $\ket{11}$ . So the output, $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ , differs from the input.
D. A gate doesn't pick an outcome; only a measurement does. The output is the single state $\tfrac{\sqrt3}{2}\ket{00} + \tfrac12\ket{11}$ . You would see 00 or 11 at random only after measuring both qubits in the 0/1 basis, and a measurement in the ± basis can tell this state apart from a coin that picks $\ket{00}$ or $\ket{11}$ .

Check yourself

A CNOT has control (top) $\ket{-}$ and target (bottom) $\ket{-}$ . What comes out?

A $\ket{+}\ket{+}$
BAn entangled state: a CNOT with a superposed control always entangles.
C $\ket{+}\ket{-}$
D $\ket{-}\ket{-}$
E $\ket{-}\ket{+}$

Show the answer

Answer: C. $\ket{+}\ket{-}$

Write the control as a sum of its $\ket0$ and $\ket1$ branches. The CNOT applies $X$ to the target in the $\ket1$ branch, and $X\ket{-} = -\ket-$ . That minus sign moves onto the control's $\ket1$ branch, so $\ket{-}$ becomes $\ket{+}$ : the output is $\ket{+}\ket{-}$ .

Why not the others

A. $X$ doesn't turn $\ket+$ or $\ket-$ into its partner: $X\ket+ = \ket+$ and $X\ket- = -\ket-$ . The target always comes out as it went in; at most a sign moves onto the control.
B. A CNOT entangles a superposed control with a target in $\ket0$ or $\ket1$ . With a target in $\ket\pm$ , $X$ only multiplies the target by $\pm1$ , so the target factors out and the pair stays a product.
D. “The control is unchanged” is guaranteed only for control inputs $\ket0$ and $\ket1$ . Here the target is $\ket-$ , and $X\ket- = -\ket-$ puts a minus sign on the control's $\ket1$ branch, which turns $\ket{-}$ into $\ket{+}$ .
E. $X$ doesn't turn $\ket+$ or $\ket-$ into its partner: $X\ket+ = \ket+$ and $X\ket- = -\ket-$ . The target always comes out as it went in; at most a sign moves onto the control.

12 / 12 · Recap

Recap

From memory

Before reading the summary, write down what you remember. Pulling it back from memory is what makes it last.

How do you decide whether a two-qubit state is a product or entangled? Give one example of each.
Show the key points
- Product means it can be written as (top state) $\otimes$ (bottom state); entangled means it can’t.
- Test: product exactly when $\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}$ (the grid’s rows are proportional).
- Example: $\tfrac12(\ket{00} - \ket{01} + \ket{10} - \ket{11}) = \ket+\ket-$ is a product; $\ket{\beta_{00}}$ is entangled.
- Being a superposition of several basis states is not the test.
Give the circuit and the formula that make the four Bell states.
Show the key points
- $H$ on the top qubit, then a CNOT with the top qubit as control.
- $\ket{\beta_{xy}} = (\ket{0,y} + (-1)^x\ket{1,\bar y})/\sqrt2$ .
- The bottom input $y$ sets the parity (labels equal or different); the top input $x$ sets the sign.
Why can’t the CNOT copying circuit copy $\ket+$ , and what does it make instead?
Show the key points
- By linearity the CNOT acts on each term: $\ket{00} \to \ket{00}$ , $\ket{10} \to \ket{11}$ .
- The output $\tfrac{1}{\sqrt2}(\ket{00} + \ket{11})$ is entangled and is not $\ket+\ket+$ .
- It copies $a\ket0 + b\ket1$ only when $ab = 0$ ; no gate copies every state (no-cloning).

Key ideas

Measuring one qubit of a pair in the 0/1 basis: add $\abs{\alpha}^2$ over the terms that match the result, keep those terms, and divide by $\sqrt{P}$ . Afterwards the other qubit has a state of its own; for an entangled pair, which state depends on the result.
Entangled means the state can’t be written as one state per qubit. A two-qubit state is a product exactly when $\alpha_{00}\alpha_{11} = \alpha_{01}\alpha_{10}$ . Superposition is not the test: $\ket+\ket-$ spreads over all four basis states and is a product, yet neither qubit has a definite 0/1 value.
$H$ on the top qubit, then a CNOT, turns $\ket{xy}$ into $\ket{\beta_{xy}} = (\ket{0,y} + (-1)^x\ket{1,\bar y})/\sqrt2$ . Neither qubit of a Bell pair has a ket, or a Bloch arrow, of its own.
Matching 0/1 results can come from a coin. $\ket{\beta_{00}}$ also matches in the ± basis, which no mix of product states can do. Not every entangled state matches perfectly in every basis, and Bob’s own results never reveal Alice’s choice of basis.
The CNOT copier sends $(a\ket0 + b\ket1)\ket0$ to $a\ket{00} + b\ket{11}$ , which equals $\ket\psi\ket\psi$ only when $ab = 0$ . By linearity, no gate copies every state: the no-cloning theorem.
What a CNOT does to a superposed control depends on the target: with $\ket-$ it kicks a sign back, $(a\ket0 + b\ket1)\ket- \to (a\ket0 - b\ket1)\ket-$ , so $\ket+\ket- \to \ket-\ket-$ (phase kickback); with $\ket+$ nothing changes; with $\ket0$ the pair becomes entangled. “The control is unchanged” is guaranteed only for 0/1 control inputs.

Go further

Free resources for when you want more than this lesson covers.

Bell’s theorem: the quantum Venn diagram paradoxminutephysics · Video
Three polarizing filters show why no answers fixed in advance can match what entangled photons do: Bell’s theorem.
Entangled spin-½ particle pairs and hidden variablesQuVis, University of St Andrews · Interactive
Test whether particle pairs carrying hidden, pre-set answers could fake the statistics of an entangled pair.
No-cloning theoremWikipedia · Article
The general proof and its consequences, and how good an imperfect copy can be: a universal cloner reaches fidelity 5/6.
Entanglement made simpleFrank Wilczek, Quanta Magazine · Article
An equation-free essay that builds entanglement from everyday correlations and ends with the EPR and GHZ puzzles.

Math refreshers

The toolkit topics this lesson leans on.

Tensor productsCombining qubits: |a⟩⊗|b⟩ as a 4-entry vector
Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability

Before you start

Part A · Measuring one qubit of two

Part A · Product or entangled?

Part A · Making entanglement: H, then CNOT

Part A · Worked example: the four Bell states

HHH then CNOT on a basis input

HHH then CNOT on ∣01⟩\ket{01}∣01⟩

Part B · Can a CNOT copy a qubit?

Part B · Correlated is not the same as entangled

Part B · What the target decides: kickback or entanglement

Part B · Practice

Explain it: the copier that entangles

Mastery check

Recap

Key ideas

Go further

Math refreshers

$H$ then CNOT on a basis input

$H$ then CNOT on $\ket{01}$