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L2-09 · Use vectors for two outcomes

A probability vector (p0, p1) lists two chances that add to 1, so it sits on a line segment. A qubit state vector (a0, a1) lists two amplitudes whose squares add to 1, so it sits on a circle. Two different state vectors can give the same chances in one measurement and still be different states.

Level 2 icon Level 2 · Grades 7–9 · about 45 minutes
Big idea. A probability vector (p0, p1) lists two chances that add to 1, so it sits on a line segment. A qubit state vector (a0, a1) lists two amplitudes whose squares add to 1, so it sits on a circle. Two different state vectors can give the same chances in one measurement and still be different states.
Before this lesson. L1-01, L1-09, L2-01
Grown-ups. The guide for adults is at the end of this page.

What you will learn

Story: A line and a circle

Maya writes a fair coin as a vector: (0.5, 0.5). The first number is the chance of heads, outcome 0. The second is the chance of tails, outcome 1. A bent coin might be (0.3, 0.7). In a probability vector, both numbers are from 0 to 1 and they add to 1.

She draws p0 across and p1 up. Every probability vector lands on the same line segment, from (1, 0) to (0, 1). That segment is the line p0 + p1 = 1. The pair (0.6, 0.6) is not on it, because 0.6 + 0.6 = 1.2.

In science class, Leo’s teacher describes a qubit. Its state is written as a state vector (a0, a1) of amplitudes. The rule is different: a0^2 + a1^2 = 1. For example, (0.6, 0.8) works, because 0.36 + 0.64 = 1. So every state vector lands on a circle of radius 1.

When the qubit is measured in the standard way, the chance of 0 is a0^2 and the chance of 1 is a1^2. For (0.6, 0.8), that is 0.36 and 0.64. Amplitudes can be negative, and (0.6, -0.8) gives the same chances. In this lesson amplitudes are real numbers. In general they can be complex numbers, which you will meet later.

Then the teacher shows (1/√2, 1/√2) and (1/√2, -1/√2). Each amplitude is about 0.71 or about -0.71, and each square is exactly 1/2. So both states give 50/50 chances, the same chances in the standard measurement. Yet they are different states. A different measurement, made after a step called a Hadamard gate, gives 0 every time for the first and 1 every time for the second.

Probability vectors sit on a line segment; qubit state vectors sit on a circle, and their squares give the chances.

What the animation shows. First, a point on the line segment p0 + p1 = 1 starts at (0.5, 0.5), a fair coin, and slides to (0.3, 0.7). Next comes a circle of radius 1. An arrow from the center shows a state vector, starting at (0.6, 0.8). A stacked bar shows the chance of 0 and the chance of 1, the squares of the amplitudes. As the arrow turns, the two parts change but the bar always reaches 1, even when a0 is negative. Back at (0.6, 0.8), the screen shows 0.36 + 0.64 = 1. Last, two arrows point to (1/√2, 1/√2) and (1/√2, -1/√2). Their bars are the same, 1/2 and 1/2: same chances, different states.

Hands-on: Points on a line, arrows on a circle

About 20 minutes.

You need:

Steps:

  1. Draw axes for p0 across and p1 up, with 10 cm standing for 1. Draw the segment from (1, 0) to (0, 1).

  2. Plot (0.5, 0.5), (0.3, 0.7), and (0.9, 0.1). Check that each pair adds to 1. Is (0.6, 0.6) on your segment?

  3. On a new page, draw axes a0 and a1 and a circle of radius 10 cm around where they cross. Here 10 cm stands for 1.

  4. With a protractor, measure angles from the right half of the a0 axis. Mark points on the circle at 0, 30, 45, 60, and 90 degrees, and at -45 degrees.

  5. For each point, measure a0 across and a1 up in cm and divide by 10. Below the axis, a1 is negative.

  6. Square each amplitude to get the chances. Add the two chances. Your totals should be close to 1.

  7. From the points you marked, find two different points with the same chances. Then mark one new point on the right half of the circle that matches another point’s chances.

Practice together

Is (0.8, -0.6) a qubit state vector? If it is, what is the chance of measuring 0 in the standard way?

Check your answer

Square each amplitude: 0.8^2 = 0.64 and (-0.6)^2 = 0.36. They add to 1, so it is a state vector. The chance of 0 is 0.64 and the chance of 1 is 0.36. The minus sign does not change the chances here.

Try it yourself

1. Is (0.5, 0.5) a probability vector? Is it a qubit state vector?

Check your answer

It is a probability vector, because 0.5 + 0.5 = 1. It is not a qubit state vector, because 0.25 + 0.25 = 0.5, not 1.

2. A qubit state vector with real amplitudes gives a chance of 0 equal to 0.25. Find two different state vectors that do this.

Check your answer

a0^2 = 0.25, so a0 is 0.5 or -0.5. Then a1^2 = 0.75, so a1 is about 0.87 or about -0.87. Two examples are (0.5, 0.87) and (0.5, -0.87). Both are rounded.

Talk about it

Check yourself

1. Which of these is a probability vector?

Show the answer

B. (0.4, 0.6)

Both entries of (0.4, 0.6) are from 0 to 1 and they add to 1. The others add to 1.2 or have a negative entry.

2. A qubit has state vector (0.6, 0.8). What is the chance of measuring 0 in the standard way?

Show the answer

B. 0.36

The chance of 0 is the square of the first amplitude: 0.6^2 = 0.36. The chance of 1 is 0.64, and 0.36 + 0.64 = 1.

3. Remember Lesson 1. Tickets cost 3 dollars each plus a 2 dollar booking fee, so y = 3x + 2. What do 4 tickets cost? (Review of L2-01)

Show the answer

B. 14 dollars

Put x = 4 into the rule: 3 times 4 is 12, plus the 2 dollar fee makes 14 dollars.

Words to know

vector
An ordered list of numbers, like (0.3, 0.7). The order matters: the first number is about outcome 0, the second about outcome 1.
amplitude
A number in a qubit state vector. It can be negative. Its square gives a chance.
qubit
A quantum object whose state is written as a state vector (a0, a1). Measuring it in the standard way gives 0 or 1.
For grown-ups: about this lesson

Why it matters. The lesson sets classical probability vectors beside quantum state vectors, the same comparison IBM’s Basics of quantum information course uses to introduce qubits Basics of quantum information, n.d.. A probability vector has nonnegative entries that sum to 1; a state vector has entries whose squared magnitudes sum to 1, and a standard-basis measurement gives outcome 0 with probability |a0|^2 Basics of quantum information, n.d.Nielsen & Chuang, 2012. OpenStax covers vectors, components and unit vectors, which supports the circle picture Jay Abramson, 2021. We keep amplitudes real so learners can use a protractor; the states (1/√2, 1/√2) and (1/√2, -1/√2) give identical standard-basis probabilities, and a Hadamard gate followed by a standard measurement distinguishes them with certainty Nielsen & Chuang, 2012.

Watch for these mix-ups.

  • The amplitudes are the chances. (The chances in the standard measurement are the squares of the amplitudes.)

  • The amplitudes must add to 1. (Their squares add to 1; for example 0.6 + 0.8 = 1.4.)

  • Two states with the same 50/50 chances must be the same state. (A different measurement, such as one after a Hadamard gate, tells (1/√2, 1/√2) and (1/√2, -1/√2) apart with certainty.)

  • A negative amplitude means a negative chance. (Squaring makes every chance zero or positive.)

To make it easier. Skip the circle drawing. Check the squares for (0.6, 0.8), (0.8, 0.6) and (1/√2, 1/√2) with a calculator, and plot the probability vectors only.

To make it harder. Show that the arrow at angle t has amplitudes (cos t, sin t), so the chances are cos^2 t and sin^2 t, which add to 1. Then multiply (1/√2, 1/√2) and (1/√2, -1/√2) by the Hadamard matrix and check the results are (1, 0) and (0, 1).

Limits. This lesson uses only real amplitudes and single-qubit state vectors. In general amplitudes are complex numbers, and the chances are the squared magnitudes. Two vectors that differ only by an overall sign, such as (0.6, 0.8) and (-0.6, -0.8), describe the same physical state, which is why the activity asks for the new point on the right half of the circle.

On the computer. Grown-ups can see where this idea leads later in Lab: Qubits and the Bloch sphere, written for older learners.

Sources for this guide. Basics of quantum information, n.d.Jay Abramson, 2021Nielsen & Chuang, 2012

Next: L2-10 · Build an evidence record

References
  1. Basics of quantum information. (n.d.). https://quantum.cloud.ibm.com/learning/en/courses/basics-of-quantum-information
  2. Nielsen, M. A., & Chuang, I. L. (2012). Quantum Computation and Quantum Information. Cambridge University Press. 10.1017/cbo9780511976667
  3. Jay Abramson. (2021). 8.8 Vectors. OpenStax, Precalculus 2e. https://openstax.org/books/precalculus-2e/pages/8-8-vectors