← All modules 03 Joint states and entanglement

SESAME · Day 1 · Module 03

Joint states
and entanglement

We can now follow one qubit through a circuit.
What changes when we use two?

01 / Describe the pair02 / Create entanglement03 / Compare correlations

Opening question / Two results on each run

We run a circuit with two qubits.

We measure both at the end. Each run starts with the same preparation.

A circuit with two qubits and two recorded results The top wire is q zero and the bottom wire is q one. Both pass through one box labelled Circuit, whose gates are unspecified. Each qubit is then measured. Its quantum wire continues, and a separate double line carries its recorded result. The first result and the second result are each zero or one with probability one half. q₀ First result 0 or 1, each with probability ½ q₁ Second result 0 or 1, each with probability ½ Circuit

Do these probabilities tell us how often the two results agree on the same run?

No. These individual probabilities allow all three:

Always agree

Always disagree

Agree half the time

To calculate agreement, we need to describe the pair.

Joint states / Controlled gates / Correlations

Knowing what each qubit does on its own is not always enough to predict what they do together.

01 Two qubits share one state with four amplitudes.

02 Controlled gates can create entanglement from a product state.

03 Entanglement supplies useful correlations without allowing signals faster than light.

01 / Describe the pair · 1.1

Two qubits have four joint amplitudes.

We write \(|q_0q_1\rangle\): q₀ is the left entry.

\[|\psi\rangle=c_{00}|00\rangle+c_{01}|01\rangle+c_{10}|10\rangle+c_{11}|11\rangle\]
\[\begin{pmatrix}c_{00}\\c_{01}\\c_{10}\\c_{11}\end{pmatrix}\]

Entry order: 00, 01, 10, 11.

\[|c_{00}|^2+|c_{01}|^2+|c_{10}|^2+|c_{11}|^2=1\]

Use the same measurement rule:

\[P(01)=|c_{01}|^2\]

\(n\) qubits: \(2^n\) amplitudes. One readout: one string of \(n\) bits.

01 / Describe the pair · 1.2

Product states multiply the individual amplitudes.

For column vectors, the tensor product is the Kronecker product.

\[(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)\]
00\(ac\)
01\(ad\)
10\(bc\)
11\(bd\)
\[=ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle\]

The two individual vectors determine the four joint entries.

01 / Describe the pair · 1.2 / Example

A superposition can still be a product state.

Plus on q₀ · Zero on q₁

\[|+\rangle\otimes|0\rangle=\frac{|00\rangle+|10\rangle}{\sqrt2}\]
\[\begin{pmatrix}1/\sqrt2\\0\\1/\sqrt2\\0\end{pmatrix}\]

Entangled: the joint state vector cannot be factored into one state vector for each qubit.

Our plus tensor zero example still factors.

01 / Describe the pair · 1.3

Gates acting separately preserve the product form.

q₀ |0⟩ q₁ |0⟩ H
\[(H\otimes I)(|0\rangle\otimes|0\rangle)=(H|0\rangle)\otimes|0\rangle=|+\rangle\otimes|0\rangle\]

I leaves the second factor unchanged.

Apply H to q₁ as well:

\[|+\rangle\otimes|0\rangle\longrightarrow|+\rangle\otimes|+\rangle\]

It is still a product.

01 / Describe the pair · 1.3 / Prediction

Can more single-qubit gates create entanglement from a product input?

Start with one state vector for each qubit.
Apply any sequence of gates on individual wires.

No. Each step changes a factor and leaves a product.

To create entanglement from this input, we need an operation involving both qubits.

02 / Create entanglement · 2.1

CNOT flips the target label when the control label is 1.

q₀ q₁

Dot: control q₀
Circled plus: target q₁

\[|00\rangle\mapsto|00\rangle\]
\[|01\rangle\mapsto|01\rangle\]
\[|10\rangle\mapsto|11\rangle\]
\[|11\rangle\mapsto|10\rangle\]

Apply the rule to every term and keep its amplitude.

One reversible operation. CNOT does not measure the control.

02 / Create entanglement · 2.2

H followed by CNOT prepares a Bell state.

q₀ |0⟩ q₁ |0⟩ H

After H on q₀

\[\frac{|00\rangle+|10\rangle}{\sqrt2}\]

The input to CNOT.

CNOT, term by term

\[|00\rangle\longmapsto|00\rangle\]
\[|10\rangle\longmapsto|11\rangle\]
\[\frac{|00\rangle+|11\rangle}{\sqrt2}=|\Phi^+\rangle\qquad\text{Bell phi plus}\]

02 / Create entanglement · 2.2 / Prediction

Does CNOT copy a plus input?

The target starts in zero. Our calculation gives:

\[\mathrm{CNOT}(|+\rangle\otimes|0\rangle)=\frac{|00\rangle+|11\rangle}{\sqrt2}\]

No. Two copies of plus would require:

\[|+\rangle\otimes|+\rangle=\frac{|00\rangle+|01\rangle+|10\rangle+|11\rangle}{2}\]

Four nonzero amplitudes. The Bell state has two.

Can one state vector for each qubit still describe our Bell pair?

02 / Create entanglement · 2.3

The Bell state cannot factor into two individual vectors.

For a product state, both amplitude products must agree:

\[c_{00}c_{11}=(ac)(bd)=abcd=(ad)(bc)=c_{01}c_{10}\]

Bell state / 00 × 11 amplitudes

\[c_{00}c_{11}=\frac1{\sqrt2}\frac1{\sqrt2}=\frac12\]

Bell state / 01 × 10 amplitudes

\[c_{01}c_{10}=0\cdot0=0\]

\(1/2\ne0\). No product of two individual vectors gives these amplitudes.

02 / Create entanglement · 2.3 / Check the input

Does CNOT always create entanglement?

\[|+\rangle\otimes|+\rangle=\frac{|00\rangle+|01\rangle+|10\rangle+|11\rangle}{2}\]
Amplitude00011011
Before CNOT\(1/2\)\(1/2\)\(1/2\)\(1/2\)
After CNOT\(1/2\)\(1/2\)\(1/2\)\(1/2\)

CNOT exchanges the 10 and 11 terms. Their amplitudes are equal.

\[\mathrm{CNOT}|++\rangle=|++\rangle\]

The state remains a product. The input matters.

03 / Compare correlations · 3.1

The direct readout gives 00 or 11.

Measure both qubits in the computational basis.

Joint amplitudes

\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}\]
\[\begin{pmatrix}1/\sqrt2\\0\\0\\1/\sqrt2\end{pmatrix}\]

Squared magnitudes / Probabilities

The two disagreeing outcomes have zero amplitude.

03 / Compare correlations · 3.1 / Two different questions

Each result is uncertain. Their agreement is certain.

\[\bigl(P(00),P(01),P(10),P(11)\bigr)=\left(\frac12,0,0,\frac12\right)\]

One qubit at a time

\[P(q_0=0)=P(00)+P(01)=\frac12\]
\[P(q_1=0)=P(00)+P(10)=\frac12\]

Each qubit gives 0 or 1 with equal probability.

The results agree

\[P(q_0=q_1)=P(00)+P(11)\]
\[=\frac12+\frac12=1\]

Both recorded labels are the same.

For distinct recorded outcomes, add probabilities.

03 / Compare correlations · 3.1 / Measure one qubit

The next zero-or-one readout on q₁ agrees with q₀’s result.

Start with the Bell pair. Measure only q₀.

q₀ q₁ c₀

q₀ records 0 / Probability ½

\[\text{State afterwards: }|00\rangle\]

The next q₁ readout gives 0.

q₀ records 1 / Probability ½

\[\text{State afterwards: }|11\rangle\]

The next q₁ readout gives 1.

Use the same computational-basis readout, with no intervening gate.

03 / Compare correlations · 3.2 / Two preparations

Could a random preparation explain the same results?

Prepare the Bell state

\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}\]

One superposition.

Direct readout: 00 or 11, each with probability ½.

Randomly choose a product preparation

Prepare \(|00\rangle\) with probability ½.
Prepare \(|11\rangle\) with probability ½.

Choose one on each run.

Direct readout: 00 or 11, each with probability ½.

The same direct measurement probabilities. We need another readout to distinguish them.

03 / Compare correlations · 3.2 / Change the readout

Apply H to both qubits before measuring.

Input / The Bell state

\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}\]
q₀ q₁ H H c₀ c₁
\[(H\otimes H)|\Phi^+\rangle=\frac{|++\rangle+|--\rangle}{\sqrt2}\]

Follow the joint amplitudes before calculating the readout probabilities.

03 / Compare correlations · 3.2 / Interference

The unequal-label amplitudes cancel.

\[(H\otimes H)|\Phi^+\rangle=\frac{|++\rangle+|--\rangle}{\sqrt2}\]
Contribution after H on both00011011
From \(|00\rangle/\sqrt2\)\(\frac1{2\sqrt2}\)\(\frac1{2\sqrt2}\)\(\frac1{2\sqrt2}\)\(\frac1{2\sqrt2}\)
From \(|11\rangle/\sqrt2\)\(\frac1{2\sqrt2}\)\(-\frac1{2\sqrt2}\)\(-\frac1{2\sqrt2}\)\(\frac1{2\sqrt2}\)
Add the amplitudes\(\frac1{\sqrt2}\)\(0\)\(0\)\(\frac1{\sqrt2}\)
\[\frac{|00\rangle+|11\rangle}{\sqrt2}=|\Phi^+\rangle\]

\(P(q_0=q_1)=1\)

03 / Compare correlations · 3.2 / Random preparation

For the random preparation, average probabilities.

Follow each of the two possible preparations separately.

Choose 00 / Probability ½

\[|00\rangle\xrightarrow{H\text{ on both}}|++\rangle\]

All four amplitudes are \(1/2\).

Each outcome has probability \(1/4\).

Choose 11 / Probability ½

\[|11\rangle\xrightarrow{H\text{ on both}}|--\rangle\]

The amplitudes have different signs.

Each outcome still has probability \(1/4\).

For each recorded outcome:

\[\frac12\cdot\frac14+\frac12\cdot\frac14=\frac14\]

\(P(q_0=q_1)=P(00)+P(11)=1/4+1/4=1/2\).

03 / Compare correlations · 3.2 / Compare both readouts

The preparation and the readout both matter.

PreparationDirect readout
\(P(q_0=q_1)\)		H on both, then readout
\(P(q_0=q_1)\)
Bell phi plus\(1\)\(1\)
Randomly prepare 00 or 11, equally\(1\)\(1/2\)

Add amplitudes within the Bell superposition.
Average probabilities across the randomly selected preparations.

Measure q₀ of the Bell pair in 0/1 and ignore its result:

\[|00\rangle\text{ with probability }1/2\quad\text{or}\quad|11\rangle\text{ with probability }1/2\]

For the later H-on-both readout, this gives the random-preparation row.

Compare repeated preparations. One matching pair can come from either procedure.

03 / Compare correlations · 3.3 / Can the correlation send a message?

Alice’s choices leave Bob’s own probabilities unchanged.

Alice holds q₀. Bob holds q₁. Each case starts with a fresh Bell pair.

Alice’s actionJoint 0/1 resultsBob’s \(P(0)\)Bob’s \(P(1)\)
Do nothing00 or 11\(1/2\)\(1/2\)
Apply X01 or 10\(1/2\)\(1/2\)
Measure 0/1
Keep the record		Matching results\(1/2\)\(1/2\)

Alice’s result predicts Bob’s. She cannot choose her random result.

03 / Compare correlations · 3.3 / Prediction

Can Bob’s counts tell him whether Alice chose to measure?

Bob has his own measurement records.
Alice has not sent him hers.

No. Both choices give him the same local probabilities.

Alice’s local operations leave those probabilities unchanged when Bob includes all runs.

Alice sends her records → Bob can select runs where she got 0.

That calculation uses information Alice has sent him.

03 / Compare correlations · 3.3 / Looking ahead

Superdense coding uses a shared pair and a transmitted qubit.

Previously shared

1

Bell pair

Then transmitted

1

qubit

Message received

2

classical bits

The transmitted qubit is part of the procedure.
The shared entanglement alone does not send the message.

Module 7 develops the encoding and decoding circuit.

Joint states / Controlled gates / Correlations

Knowing what each qubit does on its own is not always enough to predict what they do together.

01 Two qubits share one state with four amplitudes.

02 Controlled gates can create entanglement from a product state.

03 Entanglement supplies useful correlations without allowing signals faster than light.

Exercise / Worked circuit

What is the probability that q₀ and q₁ agree?

Start in \(|00\rangle\). Measure both qubits in the computational basis.

q₀ |0⟩ q₁ |0⟩ H c₀ c₁
\[|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}\]
\[P(q_0=q_1)=|c_{00}|^2+|c_{11}|^2=\frac12+\frac12=\boxed{1.00}\]

Equal labels means 00 or 11. Add those two probabilities.

Exercise / Your turn

Find \(P(q_0=q_1)\) for each circuit.

Start in \(|00\rangle\). q₀ is the top wire and the left ket entry.

1 q₀ |0⟩ q₁ |0⟩ H X c₀ c₁
2 q₀ |0⟩ q₁ |0⟩ H H c₀ c₁
3 q₀ |0⟩ q₁ |0⟩ X X c₀ c₁

Submit three decimal probabilities. Explain which final states are entangled.