Qubits and measurement · Kets & Gates

1 / 11 · Learn

Before you start

Quantum computing starts from a big claim: computers built on quantum mechanics could solve some problems, such as factoring large numbers or simulating molecules, far faster than any known classical method. All of it rests on one object, the qubit, and on what happens when you measure it.

By the end of this lesson you can:

write a qubit state as a ket and as a column vector;
compute measurement probabilities from the amplitudes;
normalize a state;
predict the state after a measurement, and what repeating it gives;
explain why one measurement yields only one bit.

Check yourself

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

Silver atoms from the oven pass through the three analyzers above. The first points along $z$ , and only its + exit feeds the second. The second points along $x$ , and only its + exit feeds the third, which points along $z$ again. Of the atoms that reach the third analyzer, what fraction leave by its − exit?

AAbout a quarter, because each analyzer lets only half of the atoms through.
BAbout half.
CNone of them. The first analyzer already picked out atoms that are + along $z$ , and they stay that way.
DAll of them. The $x$ analyzer turns the atoms over, from + along $z$ to − along $z$ .

Show the answer

Answer: B. About half.

About half. Each analyzer acts as a measurement along its own axis and leaves the atom in the matching state. An atom leaving the $x$ analyzer’s + exit gives + or − along $z$ with equal probability, whatever happened before.

Why not the others

A. About a quarter of the oven’s atoms reach the third analyzer, since roughly half are lost at each of the first two. The question asks about the atoms that do arrive there, and those split about evenly between its two exits.
C. A natural idea is that the atoms remember which exit of the first analyzer they took. But the $x$ analyzer is itself a measurement, and it resets the atom: one leaving its + exit carries no record of the earlier $z$ result. In the cascade the last analyzer splits the atoms about evenly. The memory picture is tempting because with no middle analyzer, a second $z$ analyzer really does send every atom out of +.
D. Nothing in the cascade turns atoms over. An atom leaving the $x$ analyzer’s + exit has no preference between + and − along $z$ , so the last analyzer’s two exits fill about equally. Lesson 5 shows why, using the qubit model you build in this lesson.

Check yourself

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

A qubit is prepared in the state $\ket{+} = \tfrac{1}{\sqrt2}\ket0 + \tfrac{1}{\sqrt2}\ket1$ . You measure it in the $\{\ket0,\ket1\}$ basis, which gives 0 or 1. Straight away you measure the same qubit again, in the same basis. What is the chance that the two results agree?

AIt can’t be predicted: measurement results are random
B $\tfrac12$ : the second measurement is a fresh 50/50 toss
C $\tfrac14$ : each result has chance $\tfrac12$ , and $\tfrac12 \times \tfrac12 = \tfrac14$
D $1$ : they always agree

Show the answer

Answer: D. $1$ : they always agree

The first measurement leaves the qubit in $\ket0$ or $\ket1$ , matching its result. Measuring that basis state again in the same basis gives the same result with probability 1.

Why not the others

A. Each first result is random, but whether the second agrees with it is not. After the first measurement the qubit is in $\ket0$ or $\ket1$ , and measuring a basis state in its own basis has a certain outcome. Randomness enters only when the state is not one of the basis states.
B. A common picture is that every measurement of $\ket+$ is a new coin toss. That holds for fresh copies of $\ket+$ , but here one qubit is measured twice. The first measurement leaves it in $\ket0$ or $\ket1$ , whichever you saw, and measuring that state again gives the same result every time. The coin picture is tempting because the first result really is 50/50.
C. Multiplying treats the two results as independent, and even then they could agree on 0 or on 1, giving $\tfrac14 + \tfrac14 = \tfrac12$ . In fact they are not independent at all: the first measurement leaves the qubit in the state it reported, so the second agrees with certainty.

2 / 11 · Learn

Kets and column vectors

A classical bit is either 0 or 1. A qubit has two basis states, written $\ket0$ and $\ket1$ (“ket zero” and “ket one”), and its state is a combination

\ket\psi = \alpha\ket0 + \beta\ket1,

where $\alpha$ and $\beta$ are complex numbers called amplitudes (with one condition on their sizes, coming shortly). When both amplitudes are nonzero the state is called a superposition of $\ket0$ and $\ket1$ . The state $\ket+$ from the prequestion is the case $\alpha = \beta = \tfrac{1}{\sqrt2}$ .

A ket is a vector in a two-dimensional complex space. $\ket0$ and $\ket1$ are the two standard unit vectors, so the amplitudes are simply the entries of a column vector, with the $\ket0$ amplitude on top:

\ket0 = \begin{bmatrix}1\\0\end{bmatrix},\qquad \ket1 = \begin{bmatrix}0\\1\end{bmatrix},

so a general state is

\alpha\ket0 + \beta\ket1 = \begin{bmatrix}\alpha\\ \beta\end{bmatrix}.

Check yourself

Which column vector is $\ket\psi = -\tfrac{i\sqrt3}{2}\ket1 + \tfrac12\ket0$ ? (The terms are written in an unusual order.)

A $\begin{bmatrix}1/2\\ -i\sqrt3/2\end{bmatrix}$
B $\begin{bmatrix}1/2\\ -\sqrt3/2\end{bmatrix}$
C $\begin{bmatrix}-i\sqrt3/2\\ 1/2\end{bmatrix}$
DThe single number $\tfrac12 - \tfrac{i\sqrt3}{2}$

Show the answer

Answer: A. $\begin{bmatrix}1/2\\ -i\sqrt3/2\end{bmatrix}$

Each amplitude goes in its own slot, whatever order the terms are written in: the $\ket0$ amplitude $\tfrac12$ on top, the $\ket1$ amplitude $-\tfrac{i\sqrt3}{2}$ below.

Why not the others

B. The $i$ is part of the amplitude, not a label: the $\ket1$ amplitude is the complex number $-\tfrac{i\sqrt3}{2}$ . Dropping the $i$ gives a different state, and Lesson 2 shows an experiment that tells the two apart.
C. That copies the terms in the order they are written. The slots are fixed by the kets, not by the order: the top entry always belongs to $\ket0$ and the bottom one to $\ket1$ , because $\ket0 = (1, 0)^T$ and $\ket1 = (0, 1)^T$ .
D. Adding the amplitudes merges two separate components. $\ket0$ and $\ket1$ are different directions, like $\hat x$ and $\hat y$ in the plane, so the state needs two entries, one per basis state.

Check yourself

Now enter the column vector of $\ket\psi = \tfrac{1}{\sqrt2}\left(i\ket1 - \ket0\right)$ , entry by entry. You can type things like -1/sqrt(2) or i/sqrt(2).

Hints

Which amplitude goes on top, whatever order the terms are written in?
Multiply the factor $\tfrac{1}{\sqrt2}$ into each term, then read off the number in front of $\ket0$ and the number in front of $\ket1$ .

Show the answer

Answer: $-\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}\,i|1\rangle$

Distributing the factor gives $-\tfrac{1}{\sqrt2}\ket0 + \tfrac{i}{\sqrt2}\ket1$ , so the vector is $\begin{bmatrix}-1/\sqrt2\\ i/\sqrt2\end{bmatrix}$ .

3 / 11 · Learn

Amplitudes and probabilities

Measuring a qubit in the $\{\ket0,\ket1\}$ basis gives the result 0 or the result 1. This basis is usually called the computational basis; these lessons call it the 0/1 basis, and the measurement lab labels it $Z$ . Which result you get is random. The amplitudes set the odds:

P(0) = \abs{\alpha}^2, \qquad P(1) = \abs{\beta}^2.

This is the Born rule. $\abs{z}^2$ is the squared magnitude (or modulus) of $z$ : for ${z = a + bi}$ it is ${a^2 + b^2 = z\,z^*}$ , where $z^* = a - bi$ is the complex conjugate. It is never negative. Squaring $z$ itself is a different operation: for $z = \tfrac{i}{\sqrt2}$ , $\abs{z}^2 = \tfrac12$ but $z^2 = -\tfrac12$ .

Check yourself

A qubit is in the state $\tfrac12\ket0 + \tfrac{i\sqrt3}{2}\ket1$ . You measure it in the 0/1 basis. What is the probability of the result 1?

A $\tfrac{\sqrt3}{2}$
B $\tfrac34$
C $-\tfrac34$
D $\tfrac12$

Show the answer

Answer: B. $\tfrac34$

$P(1) = \abs\beta^2 = \beta\beta^* = \tfrac34$ . Together with $P(0) = \abs{\tfrac12}^2 = \tfrac14$ the probabilities add up to 1, as they must.

Why not the others

A. That is the size of the amplitude, $\abs\beta$ , not its square. Check: the two sizes $\tfrac12$ and $\tfrac{\sqrt3}{2}$ add up to about 1.37, more than 1, so they can’t both be probabilities. Their squares, $\tfrac14$ and $\tfrac34$ , do add up to 1.
C. That is $\beta^2 = \left(\tfrac{i\sqrt3}{2}\right)^2$ , the square of the amplitude itself, not of its magnitude. A probability can never be negative, which is the quick check. The Born rule uses $\abs\beta^2 = \beta\beta^* = \tfrac{i\sqrt3}{2}\cdot\tfrac{-i\sqrt3}{2} = \tfrac34$ .
D. A superposition does not mean even odds. The odds come from the sizes of the amplitudes: $\abs{\tfrac12}^2 = \tfrac14$ for 0 and $\abs{\tfrac{i\sqrt3}{2}}^2 = \tfrac34$ for 1. Only states whose amplitudes have equal size, like $\ket+$ , give 50/50, and they are the first examples everyone meets, which makes the idea tempting.

Refresher: Complex numbers Rusty on magnitudes and conjugates? A short refresher.

The two probabilities must add up to 1, so every state obeys

\abs{\alpha}^2 + \abs{\beta}^2 = 1,

the normalization condition. Geometrically, the column vector has length 1. People often write a vector such as ${\ket0 + i\ket1}$ and mean the normalized state that points the same way. To normalize ${a\ket0 + b\ket1}$ , divide both amplitudes by its length $\sqrt{\abs{a}^2 + \abs{b}^2}$ . For ${\ket0 + i\ket1}$ the length is $\sqrt{1 + \abs{i}^2} = \sqrt2$ , so the state is $\tfrac{1}{\sqrt2}\ket0 + \tfrac{i}{\sqrt2}\ket1$ .

Check yourself

Normalize $3\ket0 + 4i\ket1$ . Enter the amplitudes of the normalized state.

Hints

What must $\abs\alpha^2 + \abs\beta^2$ come to after you rescale?
Divide both amplitudes by the length $\sqrt{\abs{3}^2 + \abs{4i}^2}$ .
The length is $\sqrt{9 + 16} = 5$ .

Show the answer

Answer: $\frac{3}{5}|0\rangle + \frac{4}{5}\,i|1\rangle$

The length is $\sqrt{9 + 16} = 5$ , so the state is $\tfrac35\ket0 + \tfrac{4i}{5}\ket1$ . Check: $\tfrac{9}{25} + \tfrac{16}{25} = 1$ . Its probabilities are $P(0) = \tfrac{9}{25}$ and $P(1) = \tfrac{16}{25}$ .

Refresher: Vectors and normalization More on column vectors, lengths and normalization.

Check yourself

Is $\tfrac12\ket0 + \tfrac12\ket1$ a valid qubit state?

AYes: it is $\ket+$ .
BYes: any pair of complex numbers describes a qubit state.
CNo: $\abs{\tfrac12}^2 + \abs{\tfrac12}^2 = \tfrac12$ , not 1.
DYes: its amplitudes add up to 1.

Show the answer

Answer: C. No: $\abs{\tfrac12}^2 + \abs{\tfrac12}^2 = \tfrac12$ , not 1.

$\abs{\tfrac12}^2 + \abs{\tfrac12}^2 = \tfrac14 + \tfrac14 = \tfrac12 \ne 1$ . Multiplied by $\sqrt2$ it becomes $\ket+$ .

Why not the others

A. $\ket+$ has amplitudes $\tfrac{1}{\sqrt2} \approx 0.71$ , not $\tfrac12$ . This vector points the same way as $\ket+$ but is too short: its length is $\tfrac{1}{\sqrt2}$ , and dividing by that turns it into $\ket+$ .
B. Only pairs whose squared sizes add up to 1 do, because those numbers are the probabilities of the two results and one of them must happen. A pair that fails the test can be rescaled into a valid state, but as written it is not one.
D. The condition is on the squared sizes, not on the amplitudes themselves. Here the squares add up to $\tfrac14 + \tfrac14 = \tfrac12$ , so the Born rule would leave half the probability unaccounted for. The idea is tempting if you picture amplitudes as probabilities; the valid state $\ket+$ has amplitudes that add up to $\sqrt2$ .

4 / 11 · Learn

What a measurement leaves behind

Measuring in the 0/1 basis does more than report a result. Afterwards the qubit is in the basis state that matches it: $\ket0$ if you got 0, $\ket1$ if you got 1. The amplitudes $\alpha$ and $\beta$ are gone, replaced by the basis state you saw. This is often called the collapse of the state.

\begin{aligned} \alpha\ket0 + \beta\ket1 \;&\xrightarrow{\ \text{result } 0\ }\; \ket0 \\ \alpha\ket0 + \beta\ket1 \;&\xrightarrow{\ \text{result } 1\ }\; \ket1 \end{aligned}

So a measurement is repeatable. Measure the same qubit again in the same basis, and the Born rule now applies to the new state: $\ket0$ gives 0 with probability $\abs{1}^2 = 1$ .

Check yourself

You measure $\ket+$ in the 0/1 basis and get 1. What is the state now, and what is $P(1)$ if you measure again straight away, in the same basis?

A $\ket1$ , and $P(1) = \tfrac12$
B $\ket1$ , and $P(1) = 1$
C $\tfrac{1}{\sqrt2}\ket1$ , and $P(1) = \tfrac12$
D $\ket+$ again, and $P(1) = \tfrac12$

Show the answer

Answer: B. $\ket1$ , and $P(1) = 1$

The result 1 leaves the qubit in $\ket1$ , and $\ket1$ gives 1 with probability $\abs{1}^2 = 1$ .

Why not the others

A. You have the right state, so apply the Born rule to it: $\ket1$ has amplitudes $0$ and $1$ , so $P(1) = \abs{1}^2 = 1$ . The $\tfrac12$ belonged to the state before the first measurement.
C. The qubit is left in the basis state $\ket1$ itself, with amplitude 1. Keeping the old amplitude $\tfrac{1}{\sqrt2}$ would give a vector of length $\tfrac{1}{\sqrt2}$ , which is not a valid state: its probabilities would add up to only $\tfrac12$ .
D. It is tempting to picture a measurement as a look that leaves the state alone, or as a jolt the qubit recovers from. Neither fits: the result 1 leaves the qubit in $\ket1$ , and it stays there. Try it in the lab below, where every repeat on the same qubit agrees with the first result.

Check yourself

A qubit in $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ is measured in the 0/1 basis, and the result is 0. Straight away it is measured again, in the same basis. What is $P(0)$ for this second measurement?

A $1$
B $\tfrac34$ , because the state jumps to $\ket0$ and then relaxes back to the original
C $\tfrac34$ , because measuring only reads off a value; it doesn’t change the state
D $\tfrac12$ , because the second measurement is a fresh coin flip

Show the answer

Answer: A. $1$

The result 0 leaves the qubit in $\ket0$ , so the immediate repeat gives $P(0) = \abs{1}^2 = 1$ .

Why not the others

B. Nothing pulls the state back: after the result 0 the qubit stays in $\ket0$ until something else acts on it. So the immediate repeat gives 0 with probability 1, as the repeats in the lab show. The picture is tempting because the original state feels like the qubit’s real identity, but the measurement has replaced it.
C. If the measurement left $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ untouched, the repeat would disagree a quarter of the time, yet in the lab above repeats never disagree. The result 0 replaces the state with $\ket0$ , which gives 0 with certainty. $\tfrac34$ is the right answer for a fresh copy, which is what makes it tempting.
D. Two slips meet here. Even a fresh copy would give 0 with probability $\tfrac34$ , not $\tfrac12$ , since the amplitudes have different sizes. And this is not a fresh copy: the first result left the qubit in $\ket0$ , so the repeat gives 0 with certainty.

5 / 11 · Learn

One shot, one bit

The amplitudes are continuous: $\abs\alpha^2$ could be $0.36$ or $0.3601$ . Does one qubit then hold an unlimited amount of information? Not in any form you can get out. A measurement in the 0/1 basis returns a single 0 or 1, which is one bit, and leaves the qubit in that basis state, so measuring it again in the same basis only repeats the answer. To estimate $\abs\alpha^2$ you need many identically prepared copies, measured one by one, and the estimate sharpens only slowly as the number of copies grows.

Check yourself

Someone hands you one qubit in a state $\alpha\ket0 + \beta\ket1$ that they don’t tell you. You measure it in the 0/1 basis and get 0. What can you conclude about the state it was in?

AThe values of $\alpha$ and $\beta$ , which the measurement has now read out.
BThat the qubit was $\ket0$ all along.
CThat $\abs\alpha^2 > \tfrac12$ : the more likely result came up.
DOnly that $\alpha \ne 0$ : a result whose amplitude is 0 never happens.

Show the answer

Answer: D. Only that $\alpha \ne 0$ : a result whose amplitude is 0 never happens.

The result 0 has probability $\abs\alpha^2$ , so seeing it rules out only $\alpha = 0$ . Any state with $\alpha \ne 0$ could have produced it.

Why not the others

A. The measurement has only two possible results, so it carries one bit, far too little to pin down two complex numbers. It also replaced the state with $\ket0$ , so measuring again in the same basis just gives 0 and tells you nothing more.
B. Many states can give 0: $\ket+$ does half the time, and $\tfrac35\ket0 + \tfrac{4i}{5}\ket1$ does 36% of the time. The result tells you the qubit is in $\ket0$ now, after the measurement, not what it was before. Reading a measurement as the reveal of a stored value is the trap from the previous segment.
C. One result doesn’t tell you which outcome was more likely. A state with $P(0) = 0.1$ gives 0 one time in ten, and that time looks exactly like this one. Telling 0.1 from 0.9 takes many copies.

Would a cleverer measurement get more out?

No. Measuring in a different basis, as in Lessons 2 and 5, still has just two outcomes. A general theorem, Holevo’s bound, shows that a single qubit on its own can deliver at most one bit of classical information, however it is prepared and measured.

6 / 11 · Explore

Many copies

One measurement gives one random bit. Many fresh copies give statistics. Predict first, then run the lab.

Predict, then see

You prepare 100 fresh copies of $\tfrac35\ket0 + \tfrac{4i}{5}\ket1$ , the state you normalized earlier, and measure each one in the 0/1 basis. Then you do a second run of 100. What do you expect?

AExactly 36 results of 0 in every run
BAbout 36 results of 0 in each run, with a slightly different count each time
CAbout 50 results of 0 in each run: a superposition gives each result half the time
DAbout 60 results of 0 in each run: the amplitude $\tfrac35$ is the chance of 0

Show what happens

What happens: B. About 36 results of 0 in each run, with a slightly different count each time

Each run gives roughly 36 zeros, and the exact count changes from run to run. The Born rule gives a probability, $\abs{3/5}^2 = \tfrac{9}{25} = 0.36$ , which predicts the long-run fraction of 0s, not the count in any one run. With 100 copies the count typically lands within about 5 of 36, and the fraction settles closer to 0.36 as the number of copies grows. A count of 50 or more turns up in fewer than 3 runs in 1000, and 60 practically never, so the data rule out both of those predictions.

Why not the others

A. A probability predicts a long-run fraction, not an exact count. Each copy is an independent random event, so a run of 100 gives a count near 36 that changes from run to run, much as 100 coin tosses rarely give exactly 50 heads.
C. Even odds need amplitudes of equal size. Here $\abs{3/5}^2 = 0.36$ and $\abs{4i/5}^2 = 0.64$ , so 0 comes up about 36 times in 100.
D. The amplitude is not the probability; its squared size is. $\abs{3/5}^2 = \tfrac{9}{25}$ , so about 36 zeros, and 60 lies far outside the range real runs land in.

Why within about 5?

The number of 0s in $N$ copies has a standard deviation of $\sqrt{N p (1-p)}$ . For ${N = 100}$ and ${p = 0.36}$ that is ${\sqrt{23.04} = 4.8}$ , and about two runs in three land within one standard deviation of $Np = 36$ . So the fraction of 0s wobbles by about $\pm 0.05$ with 100 copies, and by about $\pm 0.005$ with 10 000.

Check yourself

Type the amplitudes $1$ and $3i$ into the lab (they are not normalized). Before you press Show theory: what probability of the result 0 will the theory bar show? Give it as a number between 0 and 1.

$P(0) = \;?$

Hints

What is the length of the vector with entries $1$ and $3i$ ?
Divide both amplitudes by $\sqrt{\abs{1}^2 + \abs{3i}^2}$ , then square the size of the first.

Show the answer

Answer: $P(0) = \frac{1}{10}$

The length is $\sqrt{\abs{1}^2 + \abs{3i}^2} = \sqrt{10}$ , so $\alpha = \tfrac{1}{\sqrt{10}}$ and $P(0) = \tfrac{1}{10}$ : the theory bar shows 10%.

Common slips

$\frac{1}{4}$ : Dividing by $1 + 3 = 4$ makes the sizes add up to 1, but normalization is about their squares. The length is $\sqrt{1^2 + 3^2} = \sqrt{10}$ .
$1$ : That is $\abs{1}^2$ before normalizing. The squared sizes add up to $1 + 9 = 10$ , so divide by the length $\sqrt{10}$ first.
$0.316$ : That is the normalized amplitude, $\tfrac{1}{\sqrt{10}}$ . The probability is its squared size.
$10$ : Give the probability as a number between 0 and 1: 10% is 0.1.
$\frac{1}{2}$ : The two amplitudes have different sizes, 1 and 3, so the odds are not even. Normalize first, then square the size of the $\ket0$ amplitude.

7 / 11 · Worked example

Worked example

Watch the method once, finish the example yourself, then try one on your own.

Worked example

Probabilities and the state afterwards

Measure $\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac{i}{2}\ket1$ in the 0/1 basis. Find the probability of each result, and what the qubit is left in.

Read off the amplitudes
$\alpha = \tfrac{\sqrt3}{2}$ and $\beta = \tfrac{i}{2}$ . As a column vector, $\ket\psi = \begin{bmatrix}\sqrt3/2\\ i/2\end{bmatrix}$ .
Probability of 0
$\alpha$ is real, so its squared magnitude is just its square: $P(0) = \left(\tfrac{\sqrt3}{2}\right)^2 = \tfrac34$ .
A complex amplitude
$\beta$ is complex, so use $\abs\beta^2 = \beta\,\beta^*$ . The conjugate flips the sign of $i$ : $\beta^* = -\tfrac{i}{2}$ .
Your turn
Compute $P(1) = \beta\,\beta^*$ .
$P(1) = \;?$
Hints
1. Which two numbers does $\beta\,\beta^*$ multiply together?
2. They are $\beta = \tfrac{i}{2}$ and its conjugate $\beta^* = -\tfrac{i}{2}$ . Multiply the numbers and the $i$ ’s separately.
3. The $i$ ’s give $i \cdot (-i) = -i^2 = 1$ , and the numbers give $\tfrac12 \cdot \tfrac12$ .
Show the answer
Answer: $P(1) = \frac{1}{4}$
$\tfrac{i}{2}\cdot\left(-\tfrac{i}{2}\right) = -\tfrac{i^2}{4} = \tfrac14$ . Check: $\tfrac34 + \tfrac14 = 1$ .
Common slips
- $-\frac{1}{4}$ : That is $\beta^2 = \left(\tfrac{i}{2}\right)^2$ , the square without the conjugate, and a probability can never be negative. Use $\beta\beta^* = \tfrac{i}{2}\cdot\left(-\tfrac{i}{2}\right) = -\tfrac{i^2}{4} = \tfrac14$ .
- $\frac{1}{2}$ : That is $\abs\beta = \tfrac12$ , the size of the amplitude. The probability is its square, $\left(\tfrac12\right)^2 = \tfrac14$ .
- $\frac{3}{4}$ : That is $P(0)$ . This step asks for the result 1, whose amplitude is $\beta = \tfrac{i}{2}$ .
Your turn
Suppose the result is 1. You measure the same qubit again straight away, in the same basis. What is $P(1)$ now?
$P(1) = \;?$
Hint
1. What state is the qubit in after giving the result 1?
Show the answer
Answer: $P(1) = 1$
The result 1 leaves the qubit in $\ket1$ , so $P(1) = \abs{1}^2 = 1$ . Likewise, after the result 0 (3 times in 4) it is in $\ket0$ , and a repeat gives 0. Either way the amplitudes $\tfrac{\sqrt3}{2}$ and $\tfrac{i}{2}$ are gone.
Common slips
- $\frac{1}{4}$ : That is the chance for a fresh copy of $\ket\psi$ . This qubit has already given 1, so it is in $\ket1$ now, and $\ket1$ gives 1 every time.
- $0$ : A measurement does not flip the qubit. The result 1 leaves it in $\ket1$ , so the repeat gives 1 again.
- $\frac{3}{4}$ : That is $P(0)$ for a fresh copy. After the result 1 the qubit is in $\ket1$ , and the original amplitudes no longer apply.

Here is one more worked example with different numbers.

Another worked example

Second example

Measure $\ket\psi = -\tfrac{i}{\sqrt3}\ket0 + \sqrt{\tfrac23}\,\ket1$ in the 0/1 basis.

Read off the amplitudes
$\alpha = -\tfrac{i}{\sqrt3}$ and $\beta = \sqrt{\tfrac23}$ .
Probability of 0
$\alpha$ is complex, with conjugate $\alpha^* = \tfrac{i}{\sqrt3}$ . So $P(0) = \alpha\alpha^* = \left(-\tfrac{i}{\sqrt3}\right)\left(\tfrac{i}{\sqrt3}\right) = -\tfrac{i^2}{3} = \tfrac13$ . Squaring without the conjugate would give $\alpha^2 = -\tfrac13$ , which cannot be a probability.
Probability of 1
$\beta$ is real, so $P(1) = \left(\sqrt{\tfrac23}\right)^2 = \tfrac23$ . Check: $\tfrac13 + \tfrac23 = 1$ .
After the measurement
The result 0 leaves the qubit in $\ket0$ ; the result 1 leaves it in $\ket1$ . An immediate repeat in the same basis gives the same result with probability 1.

Check yourself

On your own, with new numbers: $\ket\psi = -\tfrac{i}{2}\ket0 - \tfrac{i\sqrt3}{2}\ket1$ is measured in the 0/1 basis. What is $P(1)$ ?

$P(1) = \;?$

Hints

Which amplitude belongs to the result 1?
The probability is the squared magnitude of that amplitude, $\abs{z}^2 = z\,z^*$ , not its square.
Multiply $-\tfrac{i\sqrt3}{2}$ by its conjugate $\tfrac{i\sqrt3}{2}$ .

Show the answer

Answer: $P(1) = \frac{3}{4}$

$P(1) = \abs{-\tfrac{i\sqrt3}{2}}^2 = \tfrac{3}{4}$ .

The result 1 goes with the amplitude $\beta = -\tfrac{i\sqrt3}{2}$ . It is complex, so multiply it by its conjugate: $\abs{\beta}^2 = \beta\,\beta^* = \left(-\tfrac{i\sqrt3}{2}\right)\left(\tfrac{i\sqrt3}{2}\right) = \tfrac{3}{4}$ . As a check, the other probability is $\tfrac{1}{4}$ , and the two add up to 1.

Common slips

$\frac{\sqrt{3}}{2}$ : That is $\abs{\beta} = \tfrac{\sqrt3}{2}$ , the size of the amplitude. The probability is its square: $\abs{\beta}^2 = \tfrac{3}{4}$ .
$-\frac{3}{4}$ : That is $\beta^2$ , the amplitude squared without the conjugate. A probability is never negative or complex; the Born rule uses $\abs{\beta}^2 = \beta\,\beta^*$ .
$\frac{1}{4}$ : That is the probability of the other result, 0. The result 1 goes with the amplitude of $\ket{1}$ , which is $-\tfrac{i\sqrt3}{2}$ .

8 / 11 · Practice

Practice

Three problems on the lesson’s procedures. Each has hints if you need them.

Check yourself

Normalize $(1+2i)\ket0 + (1+i)\ket1$ . Enter the amplitudes of the normalized state.

Hints

What must $\abs\alpha^2 + \abs\beta^2$ equal once you are done?
Divide both amplitudes by the length of the vector, $\sqrt{\abs{1+2i}^2 + \abs{1+i}^2}$ .
Here $\abs{1+2i}^2 = 1^2 + 2^2 = 5$ and $\abs{1+i}^2 = 1^2 + 1^2 = 2$ , so the length is $\sqrt{7}$ .

Show the answer

Answer: $(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{7}}i)|0\rangle + \frac{1+i}{\sqrt{7}}|1\rangle$

The length is $\sqrt{5 + 2} = \sqrt{7}$ , so the state is $\tfrac{1+2i}{\sqrt{7}}\ket0 + \tfrac{1+i}{\sqrt{7}}\ket1$ .

The squared sizes are $\abs{1+2i}^2 = 1^2 + 2^2 = 5$ and $\abs{1+i}^2 = 1^2 + 1^2 = 2$ , so the length is $\sqrt{7}$ . Dividing both amplitudes by it gives $\tfrac{1+2i}{\sqrt{7}}\ket0 + \tfrac{1+i}{\sqrt{7}}\ket1$ . Check: $\tfrac{5}{7} + \tfrac{2}{7} = 1$ .

Check yourself

A qubit is in $\alpha\ket0 + \beta\ket1$ with $\alpha = \tfrac{1+i}{\sqrt3}$ . You are not told $\beta$ . What is $\abs\beta^2$ , the probability of the result 1 in the 0/1 basis?

$\abs{\beta}^2 = \;?$

Hints

What must $\abs\alpha^2$ and $\abs\beta^2$ add up to?
Find $\abs\alpha^2$ first. For a complex number, $\abs{a + bi}^2 = a^2 + b^2$ , with $a$ and $b$ real.
Here $\abs\alpha^2 = \tfrac{1 + 1}{3} = \tfrac{2}{3}$ .

Show the answer

Answer: $\abs{\beta}^2 = \frac{1}{3}$

$\abs\alpha^2 = \tfrac{1 + 1}{3} = \tfrac{2}{3}$ , so $\abs\beta^2 = 1 - \tfrac{2}{3} = \tfrac{1}{3}$ .

Normalization says $\abs\alpha^2 + \abs\beta^2 = 1$ , so you only need $\abs\alpha^2$ . For $\alpha = \tfrac{1+i}{\sqrt3}$ , add the squares of the real and imaginary parts of the numerator and divide by the square of the denominator: $\abs\alpha^2 = \tfrac{1 + 1}{3} = \tfrac{2}{3}$ . So $\abs\beta^2 = 1 - \tfrac{2}{3} = \tfrac{1}{3}$ , whatever the phase of $\beta$ is.

Common slips

$1$ : You may have computed $\abs{a + bi}^2$ as $a^2 + (bi)^2 = a^2 - b^2$ . The magnitude squared uses the real numbers $a$ and $b$ : $\abs{a + bi}^2 = a^2 + b^2$ .
$0.184$ : That is $1 - \abs\alpha$ . The condition is on squared sizes: $\abs\beta^2 = 1 - \abs\alpha^2$ .
$\frac{2}{3}$ : That is $\abs\alpha^2$ , the probability of the result 0. The question asks for $\abs\beta^2$ , what is left over: $1 - \abs\alpha^2$ .

Check yourself

A qubit in $\tfrac{1}{\sqrt2}\ket0 + \tfrac{i}{\sqrt2}\ket1$ is measured in the 0/1 basis, and the result is 1. It is measured again straight away, in the same basis. What is the probability that the second result is 1?

$P = \;?$

Hints

What state is the qubit in right after the first measurement?
A measurement leaves the qubit in the basis state it reported, and the Born rule then uses that new state.
After the result 1 the state is $\ket{1}$ . What is its amplitude for $\ket{1}$ ?

Show the answer

Answer: $P = 1$

The first result left the qubit in $\ket{1}$ , so $P(1) = 1$ .

The first result 1 leaves the qubit in $\ket{1}$ . In that state the amplitude of $\ket{1}$ is 1, so the second result is 1 with probability $1$ . The original amplitudes would matter only for a fresh copy.

Common slips

$\frac{1}{2}$ : That is the chance of 1 for a fresh copy of the original state. After the first result the qubit is in $\ket{1}$ , so the original amplitudes no longer matter.
$0$ : The repeat can’t disagree with the first result: the qubit is now in $\ket{1}$ , which gives 1 every time.

9 / 11 · Explain

Explain it

Put the lesson’s central idea into your own words. Writing it out is what makes it stick.

Explain it

The amplitudes $\alpha$ and $\beta$ are continuous numbers. Why can’t you read more than one classical bit out of a single qubit?

Show a model answer

A measurement in the 0/1 basis has only two possible results, 0 or 1, with probabilities $\abs\alpha^2$ and $\abs\beta^2$ , so one shot yields one bit. The measurement also leaves the qubit in the basis state that was observed: $\alpha$ and $\beta$ are gone, and measuring again in the same basis only repeats the result. Estimating $\abs\alpha^2$ would take many identically prepared copies, and even then you get an estimate, not the exact numbers.

A good answer covers

One measurement gives one of only two results (0 or 1), so it carries one bit.
The measurement changes the state to the observed basis state, so $\alpha$ and $\beta$ cannot be read again from the same qubit.
Estimating the amplitudes would need many identically prepared copies.

10 / 11 · Mastery check

Mastery check

Check yourself

A qubit is prepared in the state $\ket{+} = \tfrac{1}{\sqrt2}\ket0 + \tfrac{1}{\sqrt2}\ket1$ . You measure it in the $\{\ket0,\ket1\}$ basis, which gives 0 or 1. Straight away you measure the same qubit again, in the same basis. What is the chance that the two results agree?

A $\tfrac12$ : the second measurement is a fresh 50/50 toss
BIt can’t be predicted: measurement results are random
C $1$ : they always agree
D $\tfrac14$ : each result has chance $\tfrac12$ , and $\tfrac12 \times \tfrac12 = \tfrac14$

Show the answer

Answer: C. $1$ : they always agree

The first measurement leaves the qubit in $\ket0$ or $\ket1$ , matching its result. Measuring that basis state again in the same basis gives the same result with probability 1.

Why not the others

A. A common picture is that every measurement of $\ket+$ is a new coin toss. That holds for fresh copies of $\ket+$ , but here one qubit is measured twice. The first measurement leaves it in $\ket0$ or $\ket1$ , whichever you saw, and measuring that state again gives the same result every time. The coin picture is tempting because the first result really is 50/50.
B. Each first result is random, but whether the second agrees with it is not. After the first measurement the qubit is in $\ket0$ or $\ket1$ , and measuring a basis state in its own basis has a certain outcome. Randomness enters only when the state is not one of the basis states.
D. Multiplying treats the two results as independent, and even then they could agree on 0 or on 1, giving $\tfrac14 + \tfrac14 = \tfrac12$ . In fact they are not independent at all: the first measurement leaves the qubit in the state it reported, so the second agrees with certainty.

Check yourself

Which column vector describes a state that gives the result 1 with probability $\tfrac14$ when measured in the 0/1 basis?

A $\begin{bmatrix}\sqrt{15}/4\\ 1/4\end{bmatrix}$
B $\begin{bmatrix}\sqrt3/2\\ -1/2\end{bmatrix}$
C $\begin{bmatrix}1/\sqrt2\\ -1/\sqrt2\end{bmatrix}$
D $\begin{bmatrix}1/2\\ i\sqrt3/2\end{bmatrix}$

Show the answer

Answer: B. $\begin{bmatrix}\sqrt3/2\\ -1/2\end{bmatrix}$

The bottom entry is the amplitude of $\ket1$ , and $\abs{-\tfrac12}^2 = \tfrac14$ .

Why not the others

A. This vector has $\tfrac14$ itself as its $\ket1$ amplitude, so $P(1) = \left(\tfrac14\right)^2 = \tfrac{1}{16}$ . The probability is the squared size of the amplitude, so the bottom entry needs size $\tfrac12$ .
C. Entries of equal size give $P(1) = \tfrac12$ . To get 1 with probability $\tfrac14$ , the bottom entry needs squared size $\tfrac14$ .
D. Here the bottom entry, the amplitude of $\ket1$ , has squared size $\tfrac34$ , so $P(1) = \tfrac34$ . The entry with squared size $\tfrac14$ sits on top, in the $\ket0$ slot.

Check yourself

The vector $2\ket0 + (2+i)\ket1$ is not normalized. Once you normalize it, what is the probability of the result 1 in the 0/1 basis?

$P(1) = \;?$

Show the answer

Answer: $P(1) = \frac{5}{9}$

$P(1) = \dfrac{\abs{2+i}^2}{\abs{2}^2 + \abs{2+i}^2} = \dfrac{5}{9}$ .

The squared sizes are $\abs{2}^2 = 4$ and $\abs{2+i}^2 = 5$ , so the length is $\sqrt{9}$ . Normalizing divides both amplitudes by it, and the probability of the result 1 is the new squared size of the $\ket1$ amplitude: $P(1) = \tfrac{5}{9}$ .

Common slips

$5$ : That is $\abs{2+i}^2$ before normalizing. The squared sizes add up to $4 + 5 = 9$ , not 1, so divide by that total first.
$\frac{\sqrt{5}}{3}$ : That is the size of the normalized amplitude, $\abs{2+i}/3$ . The probability is its square.
$\frac{4}{9}$ : That is $P(0)$ , from the amplitude of $\ket0$ . The result 1 goes with the amplitude of $\ket1$ .

Check yourself

A friend picks a number, say 0.137, and prepares a single qubit with $\abs\alpha^2$ equal to it. They send you that one qubit. How much of the number can you recover?

AAll of it: measure the qubit many times and count the 0s.
BAll of it, if the measurement is precise enough.
CAt most one bit’s worth: measured in the 0/1 basis, it gives 0 or 1 and is left in $\ket0$ or $\ket1$ .

Show the answer

Answer: C. At most one bit’s worth: measured in the 0/1 basis, it gives 0 or 1 and is left in $\ket0$ or $\ket1$ .

One shot yields one bit and replaces the state with $\ket0$ or $\ket1$ , so nothing more can be read from that qubit. Even with many copies you would only estimate the number statistically.

Why not the others

A. Only the first measurement of this qubit is informative. It leaves the qubit in $\ket0$ or $\ket1$ , so every later measurement in that basis repeats that result and counting tells you nothing new. Counting 0s works only with many separately prepared copies.
B. Precision isn’t the limit. A measurement in the 0/1 basis has only two possible results, 0 and 1, so no setting of the apparatus can return 0.137.

Check yourself

A qubit in $\ket\psi = \tfrac{1-i}{2}\ket0 - \tfrac{1}{\sqrt2}\ket1$ is measured in the 0/1 basis and gives 1. Then two more measurements are made in the same basis: one on the same qubit, and one on a fresh copy of $\ket\psi$ . What is the probability that both of these give 1?

$P = \;?$

Show the answer

Answer: $P = \frac{1}{2}$

$1 \times \tfrac{1}{2} = \tfrac{1}{2}$ : the repeat is certain, the fresh copy is not.

The repeat on the same qubit gives 1 with certainty, because the first result left it in $\ket{1}$ . The fresh copy is in $\ket\psi$ , so it gives 1 with probability $\abs{-\tfrac{1}{\sqrt2}}^2 = \tfrac{1}{2}$ . Both happen with probability $1 \times \tfrac{1}{2} = \tfrac{1}{2}$ .

Common slips

$\frac{1}{4}$ : That multiplies $\tfrac{1}{2}$ by itself, as if the repeat on the measured qubit were a fresh copy too. The first result left that qubit in $\ket{1}$ , so its repeat gives 1 with certainty; only the fresh copy contributes $\tfrac{1}{2}$ .
$1$ : The repeat on the measured qubit is certain to give 1, but the fresh copy is a new qubit in the original state. It gives 1 only with probability $\tfrac{1}{2}$ .

11 / 11 · Recap

Recap

From memory

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

You measure $\alpha\ket0 + \beta\ket1$ in the 0/1 basis. How do you get the probability of each measurement result, 0 or 1, from the amplitudes $\alpha$ and $\beta$ ?
Show the key points
- The result 0 has probability $\abs\alpha^2$ and the result 1 has probability $\abs\beta^2$ (the Born rule)
- For a complex amplitude, $\abs{z}^2 = z\,z^*$ , not $z^2$ , so a probability is never negative
- Only the size of an amplitude counts: a minus sign or a factor $i$ does not change the probability
What makes a pair of amplitudes $\alpha$ , $\beta$ a valid qubit state, and how do you fix a vector that is not one?
Show the key points
- Normalization: $\abs\alpha^2 + \abs\beta^2 = 1$ , because the two probabilities must add up to 1
- The condition is on the squared sizes, not on the amplitudes themselves: $\tfrac12\ket0 + \tfrac12\ket1$ is not a valid state
- To fix one, divide both entries by the vector’s length, $\sqrt{\abs\alpha^2 + \abs\beta^2}$
What does a measurement do to the qubit, and what happens if you repeat it?
Show the key points
- The qubit is left in the basis state that was observed
- An immediate repeat in the same basis gives the same result with probability 1
- Only fresh copies show the $\abs\alpha^2 : \abs\beta^2$ statistics

Key ideas

A qubit state is $\alpha\ket0 + \beta\ket1$ , or the column vector $\begin{bmatrix}\alpha\\ \beta\end{bmatrix}$ , with complex amplitudes $\alpha$ and $\beta$ .
Measured in the 0/1 basis, it gives 0 with probability $\abs\alpha^2$ and 1 with probability $\abs\beta^2$ . Use the magnitude: $\abs{z}^2 = z\,z^*$ .
Normalization: $\abs\alpha^2 + \abs\beta^2 = 1$ . To normalize a vector, divide it by its length.
A measurement leaves the qubit in $\ket0$ or $\ket1$ , matching the result, so an immediate repeat agrees with certainty. Only fresh copies show the $\abs\alpha^2 : \abs\beta^2$ statistics.
One measurement gives one bit. The amplitudes can only be estimated, from many identically prepared copies.

Go further

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

Quantum computing for the very curiousQuantum Country · Article
A long essay from qubits to universal gate sets with review questions built in; a free account schedules them for later.
Quantum Computing Since Democritus, lecture 9: QuantumScott Aaronson · Lecture notes
Argues that quantum mechanics is probability theory with minus signs allowed, and asks why the rule squares amplitudes.
Some light quantum mechanics (with minutephysics)3Blue1Brown · Video
Polarized light as a real two-state system: how a wave’s components become amplitudes whose squares give a photon’s odds.
Seeing Theory: a visual introduction to probability and statisticsBrown University · Interactive
Animated chapters on chance, expectation and estimating a probability from data, the statistics behind measuring many copies.

Math refreshers

The toolkit topics this lesson leans on.

Complex numbersMagnitude, phase, and why a phase factor has size 1
Vectors and normalizationKets are column vectors; probabilities must add to 1