Phase and the ± basis · Kets & Gates

1 / 11 · Learn

Before you start

Lesson 1 measured only in the 0/1 basis, where $\ket+$ gives 0 or 1 just as a fair coin would. This lesson looks at the part of an amplitude that the 0/1 basis can’t see, its phase, and at a second basis that can.

By the end of this lesson you can:

explain why multiplying a whole state by $e^{i\gamma}$ changes no prediction;
show that a phase on one term changes what a ± measurement gives;
rewrite a state in the ± basis and use $P(\pm) = \abs{\alpha\pm\beta}^2/2$ ;
tell a superposition from a 50/50 coin by experiment.

Check yourself

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

Register A holds 1000 qubits, each prepared in ${\ket{+} = \tfrac{1}{\sqrt2}(\ket0 + \ket1)}$ . Register B holds 1000 qubits, each prepared as $\ket0$ or $\ket1$ by a fair coin toss; nobody tells you which. Can any experiment tell the two registers apart?

ANo. Both give 0 or 1 with probability 1/2, so they are physically the same.
BYes. In the 0/1 basis A splits exactly 500/500, while B fluctuates.
CYes. Measure every qubit in a different basis, the ± basis: A gives the same outcome every time, B does not.
DThey differ in principle, but no experiment could ever show it.

Show the answer

Answer: C. Yes. Measure every qubit in a different basis, the ± basis: A gives the same outcome every time, B does not.

Measured in the ± basis, every qubit of A gives $+$ , while B gives $+$ about half the time. This lesson shows why.

Why not the others

A. A very common idea is that $\ket{+}$ is a coin you haven't looked at yet. If it were, nothing could tell A from B. But measure both in the ± basis: every qubit of A gives $+$ , while B gives $+$ only about half the time. $\ket{+}$ is one definite state; its 50/50 belongs to the 0/1 basis, not to your ignorance.
B. Probabilities predict frequencies, not exact counts. A scatters around 500 just as B does, so in the 0/1 basis the two registers give the same statistics. It is tempting because "superposition" sounds more orderly than a coin, but you need a second basis to see the difference.
D. The difference is not just a matter of words. Measure every qubit in the ± basis: A gives $+$ every time, B only about half the time. It is tempting to think so because the 0/1 basis, the only measurement met so far, really cannot see it.

Refresher: Complex numbers Phase is the angle of a complex number. Open this if $re^{i\varphi}$ or $\abs{e^{i\varphi}} = 1$ is rusty.

2 / 11 · Learn

Phase on a dial

Every amplitude is a complex number, and any complex number can be written $re^{i\varphi}$ : a size $r \ge 0$ (its magnitude, $r = \abs{z}$ ) and a phase $\varphi$ , which is an angle. The circle notation below draws both. Each basis state gets a circle. The filled disc's radius is the size $r$ ; the needle points along the phase, with 0° to the right and angles growing counter-clockwise. The exact amplitude, its phase in degrees and the probability are printed underneath, so you never have to judge an angle by eye.

Here is $\tfrac{1}{\sqrt2}\ket0 + \tfrac{1}{\sqrt2}e^{i\pi/3}\ket1$ :

Check yourself

You measure this state in the 0/1 basis. What is $P(1)$ ?

$P(1) = \;?$

Hints

What is the size of $e^{i\pi/3}$ ?
A probability is $\abs{\text{amplitude}}^2$ , and $\abs{e^{i\varphi}} = 1$ for any angle.

Show the answer

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

The needle's direction does not affect the size: $\abs{e^{i\pi/3}} = 1$ , so $P(1) = (1/\sqrt2)^2 = 1/2$ , the same as the disc radius squared.

Common slips

$-\frac{1}{4}+\frac{\sqrt{3}}{4}i$ : That squares the amplitude itself, phase and all. The Born rule squares the size: $\abs{e^{i\pi/3}/\sqrt2}^2 = \abs{e^{i\pi/3}}^2 \cdot \tfrac12 = \tfrac12$ , because a phase factor has size 1. A probability is always a real number between 0 and 1.
$\frac{1}{8}$ : That keeps only the real part, $\cos(\pi/3)/\sqrt2$ , before squaring. Use the whole amplitude: its size is $1/\sqrt2$ whatever its phase, so $P(1) = 1/2$ .

Now multiply the whole state by one phase factor $e^{i\gamma}$ . Every size stays the same and every needle turns by the same angle $\gamma$ . This is a global phase. Since no size changes, no 0/1 probability changes. In fact no measurement in any basis can detect it (you will check this for the ± basis shortly), so $\ket\psi$ and $e^{i\gamma}\ket\psi$ describe the same physical state.

Check yourself

Which of these is the same physical state as ${\ket{+} = \tfrac{1}{\sqrt2}(\ket0 + \ket1)}$ ?

A $\tfrac{1}{\sqrt2}(\ket0 - \ket1)$
B $\tfrac{1}{\sqrt2}(\ket0 + i\ket1)$
C $-\tfrac{1}{\sqrt2}(\ket0 + \ket1)$
DNone of these: any change of sign or phase gives a different state.

Show the answer

Answer: C. $-\tfrac{1}{\sqrt2}(\ket0 + \ket1)$

$-1 = e^{i\pi}$ multiplies both amplitudes, so both needles turn by 180° together: a global phase. The other states change one needle only.

Why not the others

A. Same sizes, so the same 0/1 statistics, but here only the $\ket1$ needle turned (by 180°). The needles no longer point the same way, so this is not a global phase. It is the state $\ket{-}$ , and the next segment shows a measurement that tells it apart from $\ket{+}$ .
B. Equal sizes again, but only the $\ket1$ needle turned (by 90°), so the angle between the needles changed. Matching 0/1 probabilities are not enough to call two states the same.
D. A common idea is that any sign change makes a new state. Multiplying every amplitude by $-1$ turns both needles by 180° together and changes no size, so no probability in any basis changes. Only phase differences between the terms are physical.

3 / 11 · Learn

A second basis: plus and minus

Compare $\ket{+}$ with $\ket{-} = \tfrac{1}{\sqrt2}(\ket0 - \ket1)$ . Their sizes match, so in the 0/1 basis both give 50/50. Yet they are not related by a global phase. To see the difference you need a measurement that asks a different question. Any two orthogonal unit vectors can serve as a measurement basis, as the inner-product refresher explains, and $\{\ket{+}, \ket{-}\}$ is such a pair: the ± basis. Its outcomes are $+$ and $-$ . Measured in it, $\ket{+}$ gives $+$ every time and $\ket{-}$ gives $-$ every time. How such a measurement is built comes in Lesson 5.

Check yourself

In which basis do $\ket{+}$ and $\ket{-}$ give identical statistics?

AOnly the 0/1 basis.
BNeither: they are different states, so every measurement can tell them apart.
CBoth bases: they differ only by a sign.
DOnly the ± basis.

Hints

In the 0/1 basis, what decides the probabilities: the sizes or the signs?
In the lab, measure the preset $\ket{+}$ and then $\ket{-}$ 1000 times in Z, then in X, and compare the histograms.

Show the answer

Answer: A. Only the 0/1 basis.

Equal sizes give equal 0/1 statistics (50/50). The ± basis is where they differ completely: $+$ every time versus $-$ every time.

Why not the others

B. Different states need not differ in every basis. In the 0/1 basis both give 0 or 1 with probability 1/2 each, so that measurement cannot separate them; the ± basis can.
C. The sign sits on one term only, so it is a relative phase, not a global one. Try it in the lab: in the X (±) basis the two histograms are opposite. It is tempting because their 0/1 histograms really are identical.
D. In the ± basis they are as different as possible: $\ket{+}$ gives $+$ every time and $\ket{-}$ gives $-$ every time. It is in the 0/1 basis that both give 50/50.

To get the ± odds of any state, rewrite it in terms of $\ket{+}$ and $\ket{-}$ , then apply the Born rule to the new coefficients:

\alpha\ket0 + \beta\ket1 = \frac{\alpha+\beta}{\sqrt2}\ket{+} + \frac{\alpha-\beta}{\sqrt2}\ket{-}

Expand the right-hand side with the definitions of $\ket{\pm}$ and you get back the left. So ${P(+) = \abs{\alpha+\beta}^2/2}$ and ${P(-) = \abs{\alpha-\beta}^2/2}$ . Check with $\ket{-}$ : $\alpha = \tfrac{1}{\sqrt2}$ and $\beta = -\tfrac{1}{\sqrt2}$ give ${P(+) = 0}$ .

Check yourself

You measure $\tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ in the ± basis. What is $P(+)$ ?

$P(+) = \;?$

Hints

Which coefficient goes with $\ket{+}$ when you rewrite the state, and what are $\alpha$ and $\beta$ here?
$P(+) = \abs{\alpha+\beta}^2/2$ , with $\alpha + \beta = (1 + i)/\sqrt2$ .
$\abs{\alpha+\beta}^2 = \abs{1+i}^2/2 = (1^2 + 1^2)/2 = 1$ . Now divide by 2.

Show the answer

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

$P(+) = \abs{1+i}^2/4 = 2/4 = 1/2$ . This state, called $\ket{+i}$ , gives 50/50 in both bases you know so far; Lesson 3 shows where it sits on the Bloch sphere.

Common slips

$1$ : That treats $i$ as if it were 1, as for $\ket{+}$ . The phase counts: $\alpha + \beta = (1 + i)/\sqrt2$ and $\abs{1+i}^2 = 2$ , so $P(+) = 2/4 = 1/2$ .
$0$ : That treats $i$ as $-1$ , which would give $\ket{-}$ . Here $\abs{1 + i}^2 = 1^2 + 1^2 = 2$ , so $P(+) = 2/4 = 1/2$ .

4 / 11 · Learn

Global phase, relative phase

Now separate the two kinds of phase. Write a state with equal sizes as $e^{i\gamma}\cdot\tfrac{1}{\sqrt2}(\ket0 + e^{i\varphi}\ket1)$ . The global phase $\gamma$ turns both needles together. The relative phase $\varphi$ is the angle between the needles: $\ket{+}$ has $\varphi = 0$ and $\ket{-}$ has $\varphi = 180°$ . In the lab, move one slider at a time and watch both rows of probabilities.

Check yourself

Which slider changes a probability in some basis?

AOnly $\gamma$ : turning both needles is the bigger change.
BBoth of them.
CNeither: a phase never changes a probability.
DOnly $\varphi$ , the relative phase. It changes the ± probabilities, while the 0/1 ones stay at 50%.

Show the answer

Answer: D. Only $\varphi$ , the relative phase. It changes the ± probabilities, while the 0/1 ones stay at 50%.

$\gamma$ multiplies every amplitude, in every basis, by the same $e^{i\gamma}$ , so no probability changes. $\varphi$ changes the angle between the needles: the 0/1 row stays at 50/50, but $P(+)$ runs from 100% at $\varphi = 0$ to 0 at $\varphi = 180°$ .

Why not the others

A. Turning both needles by the same angle keeps every size and the angle between them, so every probability stays put. Drag $\gamma$ all the way round and watch: neither row moves. It feels like a bigger change because two needles move instead of one.
B. Only $\varphi$ does. Move $\gamma$ alone: both needles turn, but the probabilities in both rows stay exactly the same, because $\abs{e^{i\gamma}z} = \abs{z}$ for every amplitude in every basis.
C. That is true in the 0/1 basis only. Set $\varphi$ to 180° and the ± row goes from 100% $+$ to 100% $-$ . A probability is a squared size, but in the ± basis the sizes are $\abs{\alpha\pm\beta}/\sqrt2$ , and those depend on the angle between the needles.

Why can $\gamma$ never matter? In the ± basis the coefficients are ${(\alpha\pm\beta)/\sqrt2}$ . Multiply both $\alpha$ and $\beta$ by $e^{i\gamma}$ and each becomes ${e^{i\gamma}(\alpha\pm\beta)/\sqrt2}$ : the same size, so the same probability. Any other basis works the same way, because each new coefficient is a fixed combination of $\alpha$ and $\beta$ , so a common factor comes straight out. Multiply $\beta$ alone by $e^{i\varphi}$ and the sum $\alpha + e^{i\varphi}\beta$ is no longer a common factor times $\alpha + \beta$ , so its size can change. One caution: a state with one term, such as $e^{i\pi/3}\ket1$ , has no second needle to compare with, so a factor on that term is a global phase.

Check yourself

A state with equal sizes has its needles at 30° (on $\ket0$ ) and 120° (on $\ket1$ ). A second state has equal sizes and needles at 0° and 90°. What is the relationship?

AOrthogonal states, because the needles turned by 30°.
BDifferent states: each needle's direction is measurable.
CSame physical state: both needles differ by 30°, and the angle between them is 90° in both.

Hints

Turn both needles of the first state back by 30°. Where do they point now?
Turning every needle by the same angle is a global phase. Compare the angle between the needles in each state.

Show the answer

Answer: C. Same physical state: both needles differ by 30°, and the angle between them is 90° in both.

Both states have relative phase 90°; the 30° offset is a global phase. Both are $\ket{+i} = \tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ up to a global phase.

Why not the others

A. Turning both needles by the same angle gives the same state, not an orthogonal one. Orthogonal states, like $\ket{+}$ and $\ket{-}$ , differ in the angle between the needles (0° versus 180°).
B. Turning both needles back by 30° is multiplying by $e^{-i\pi/6}$ , a global phase. What a measurement can see is the angle between the needles, 90° in both cases. In the lab, set $\varphi = 90°$ and move $\gamma$ : nothing changes.

Why does the ± basis give 50% at both 90° and 270°?

Try it in the lab: $\varphi = 90°$ and $\varphi = 270°$ both give $P(+) = 50\%$ . In general ${P(+) = \abs{1 + e^{i\varphi}}^2/4 = \cos^2(\varphi/2)}$ , which is the same for $\varphi$ and $-\varphi$ . The two states are still different: a third basis, $\{\ket{+i}, \ket{-i}\}$ with $\ket{\pm i} = \tfrac{1}{\sqrt2}(\ket0 \pm i\ket1)$ , tells them apart. Lesson 3 shows all three bases on the Bloch sphere.

5 / 11 · Learn

Superposition or coin?

Back to the opening question. Register A holds 1000 copies of one definite state, $\ket{+}$ . Register B is a mixture: each qubit really is $\ket0$ or $\ket1$ , set by a coin, and you just don't know which. Before you press anything, recall your opening guess and use the formula $P(\pm) = \abs{\alpha\pm\beta}^2/2$ to predict what each register gives in each basis. Then run both measurements on both registers.

Check yourself

Source 1 emits $\ket{-}$ every time. Source 2 tosses a fair coin and emits $\ket{+}$ or $\ket{-}$ . Which measurement, repeated on many qubits from each source, tells the sources apart?

AThe 0/1 basis.
BNo measurement: both sources give 50/50.
CThe ± basis.
DEither basis works.

Show the answer

Answer: C. The ± basis.

In the ± basis Source 1 gives $-$ every time; Source 2 gives $+$ or $-$ according to its coin, about half each. In the 0/1 basis both look like 50/50.

Why not the others

A. $\ket{+}$ and $\ket{-}$ both give 50/50 in the 0/1 basis, so both sources give 50/50 there. In the ± basis Source 1 gives $-$ every time and Source 2 splits about evenly.
B. They give 50/50 in the 0/1 basis only. Source 1 is a definite state, $\ket{-}$ , so in the ± basis it gives $-$ every time; Source 2's coin shows up as a 50/50 split there.
D. In the 0/1 basis both sources give 50/50, so that basis cannot separate them. Only the ± basis does.

6 / 11 · Explore

Is a definite state definite in every basis?

Predict, then see

A qubit in $\ket0$ gives 0 every time it is measured in the 0/1 basis. You prepare 1000 fresh copies of $\ket0$ and measure each one in the ± basis instead. What do you expect?

A $-$ every time.
BAbout half $+$ and half $-$ .
C $+$ every time: the qubit already has a definite value.
DIt can't be done: only the 0/1 basis is a real measurement.

Show what happens

What happens: B. About half $+$ and half $-$ .

Compare the two histograms with your prediction. Then pick, below, the explanation that fits what you saw.

Why not the others

A. Nothing in $\ket0$ favours $-$ : with $\alpha = 1$ and $\beta = 0$ the formula gives $P(-) = \abs{1 - 0}^2/2 = 1/2$ .
C. A natural idea is that a qubit that is certain in one basis has a definite answer to every question. The data say otherwise: about half $+$ and half $-$ . Certainty belongs to a state and a basis together.
D. Any orthonormal basis can be measured. In practice you apply a gate and then measure in the 0/1 basis (Lesson 5), and a Stern–Gerlach magnet turned along $x$ does the same job for an atom.

Check yourself

Which explanation fits the data: $\ket0$ gives 0 every time in the 0/1 basis, yet about half $+$ and half $-$ in the ± basis?

AIn the ± basis only the $\ket0$ amplitude counts: ${P(+) = \abs{\alpha}^2/2 = 1/2}$ .
BThe ± measurement is less precise, so it scrambles the answer.
C $\ket0$ is secretly a coin that is $\ket{+}$ or $\ket{-}$ , each half the time.
DRewritten in the ± basis, $\ket0$ has two coefficients of equal size, $1/\sqrt2$ each, so each outcome has probability $1/2$ .

Show the answer

Answer: D. Rewritten in the ± basis, $\ket0$ has two coefficients of equal size, $1/\sqrt2$ each, so each outcome has probability $1/2$ .

Rewritten in the ± basis, $\ket0 = \tfrac{1}{\sqrt2}(\ket{+} + \ket{-})$ , so $P(+) = P(-) = 1/2$ . A state can be certain in one basis and as uncertain as possible in another: $\ket0$ is to the ± basis what $\ket{+}$ is to the 0/1 basis. Being definite is a property of a state together with a basis, not of the state alone.

Why not the others

A. Right number, wrong rule. Both amplitudes enter: $P(\pm) = \abs{\alpha\pm\beta}^2/2$ . Your rule would give $\ket1$ a $P(+)$ of 0, yet $\ket1$ also gives 50/50 in the ± basis, and it would give $\ket{+}$ only 1/4 instead of 1.
B. The ± basis is just as sharp as the 0/1 basis: measured in it, $\ket{+}$ gives $+$ every time. The 50/50 comes from the state, $\ket0 = \tfrac{1}{\sqrt2}(\ket{+} + \ket{-})$ , not from a sloppy apparatus.
C. Such a coin would also give 0 and 1 equally often in the 0/1 basis, because $\ket{+}$ and $\ket{-}$ each do. $\ket0$ gives 0 every time, so it is one definite state, a superposition of $\ket{+}$ and $\ket{-}$ , not a coin.

Check yourself

Now write $\ket1$ in the ± basis: enter the coefficients of $\ket{+}$ and $\ket{-}$ .

Hints

For $\ket1$ , what are $\alpha$ and $\beta$ ?
Use $(\alpha+\beta)/\sqrt2$ and $(\alpha-\beta)/\sqrt2$ .

Show the answer

Answer: $\ket{+} = \frac{1}{\sqrt{2}},\quad \ket{-} = -\frac{1}{\sqrt{2}}$

$(\alpha+\beta)/\sqrt2 = 1/\sqrt2$ and $(\alpha-\beta)/\sqrt2 = -1/\sqrt2$ , so $\ket1 = \tfrac{1}{\sqrt2}(\ket{+} - \ket{-})$ . Like $\ket0$ , it gives 50/50 in the ± basis.

7 / 11 · Worked example

Rewriting a state in the ± basis

Worked example

A state with unequal sizes

Find the ± coefficients and probabilities of $\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ . (In the 0/1 basis its probabilities are 3/4 and 1/4.)

Step 1
Read off $\alpha = \tfrac{\sqrt3}{2}$ and $\beta = \tfrac12$ . The coefficient of $\ket{+}$ is $\tfrac{\alpha+\beta}{\sqrt2}$ and that of $\ket{-}$ is $\tfrac{\alpha-\beta}{\sqrt2}$ .
Step 2
$\dfrac{\alpha+\beta}{\sqrt2} = \dfrac{\sqrt3 + 1}{2\sqrt2} \approx 0.966$ and $\dfrac{\alpha-\beta}{\sqrt2} = \dfrac{\sqrt3 - 1}{2\sqrt2} \approx 0.259$ .
Step 3
So $\ket\psi = \dfrac{\sqrt3+1}{2\sqrt2}\ket{+} + \dfrac{\sqrt3-1}{2\sqrt2}\ket{-}$ . Both coefficients are real, so each probability is just the coefficient squared.
Your turn
What is $P(+)$ when $\ket\psi$ is measured in the ± basis? (An exact form such as (2+sqrt(3))/4 or a decimal is fine.)
$P(+) = \;?$
Hint
1. Square the coefficient of $\ket{+}$ from Step 3.
Show the answer
Answer: $P(+) = 0.933$
$\left(\dfrac{\sqrt3+1}{2\sqrt2}\right)^2 = \dfrac{3 + 2\sqrt3 + 1}{8} = \dfrac{2+\sqrt3}{4} \approx 0.933$ .
Common slips
- $\frac{3}{4}$ : 0.75 is $P(0)$ , the probability in the 0/1 basis. This question is about the ± basis: square the coefficient of $\ket{+}$ from Step 3.
Your turn
And $P(-)$ ?
$P(-) = \;?$
Hint
1. Square $\tfrac{\sqrt3-1}{2\sqrt2}$ , or use $P(+) + P(-) = 1$ .
Show the answer
Answer: $P(-) = 0.067$
$\left(\dfrac{\sqrt3-1}{2\sqrt2}\right)^2 = \dfrac{4 - 2\sqrt3}{8} = \dfrac{2-\sqrt3}{4} \approx 0.067$ , and $0.933 + 0.067 = 1$ . In the ± basis this state is more lopsided (93/7) than in the 0/1 basis (75/25).
Common slips
- $\frac{1}{4}$ : 0.25 is $P(1)$ , the 0/1-basis probability. Square the coefficient of $\ket{-}$ instead.

Here is one more worked example with different numbers.

Another worked example

A second state

$\ket\psi = \tfrac35\ket0 + \tfrac45\ket1$ , so $\alpha = \tfrac35$ and $\beta = \tfrac45$ .
Coefficient of $\ket{+}$ : $\dfrac{3/5 + 4/5}{\sqrt2} = \dfrac{7}{5\sqrt2}$ . Coefficient of $\ket{-}$ : $\dfrac{3/5 - 4/5}{\sqrt2} = -\dfrac{1}{5\sqrt2}$ .
Square them: $P(+) = \dfrac{49}{50} = 0.98$ and $P(-) = \dfrac{1}{50} = 0.02$ , which add to 1. The minus sign on the $\ket{-}$ coefficient disappears when you square, but it is still part of the state.

Check yourself

Now on your own. You measure $\tfrac{1}{\sqrt2}\left(\ket0 + e^{i\pi/4}\ket1\right)$ in the ± basis. What is $P(-)$ ?

$P(-) = \;?$

Hints

Which formula gives $P(-)$ from $\alpha$ and $\beta$ ?
$P(-) = \abs{\alpha-\beta}^2/2$ , and $\alpha-\beta = (1 - e^{i\varphi})/\sqrt2$ .
$\abs{1 - e^{i\varphi}}^2 = 2 - 2\cos\varphi$ .

Show the answer

Answer: $P(-) = 0.146$

$\alpha - \beta = (1 - e^{i\pi/4})/\sqrt2$ , and $\abs{1 - e^{i\pi/4}}^2 = 2 - 2\cos 45° = 2-\sqrt2$ . So $P(-) = \abs{\alpha-\beta}^2/2 = \abs{1 - e^{i\pi/4}}^2/4 = \tfrac{2-\sqrt2}{4}$ .

Common slips

$0$ : That treats $e^{i\pi/4}$ as if it were 1. A phase on one term changes $\abs{\alpha-\beta}$ : here $\abs{1 - e^{i\pi/4}}^2 = 2-\sqrt2$ .
$\frac{1}{2}$ : 1/2 is the 0/1-basis probability. In the ± basis use $\abs{\alpha-\beta}^2/2$ .
$0.293$ : You found $\abs{\alpha-\beta}^2$ ; divide by 2.

8 / 11 · Practice

Practice

Check yourself

Can any measurement tell $\ket\psi = \tfrac35\ket0 + \tfrac45\ket1$ from $\ket\chi = -\tfrac35\ket0 + \tfrac45\ket1$ ?

AYes. They differ by a relative phase: the 0/1 statistics agree, but a ± measurement tells them apart.
BYes. Any change of phase, even one on every term, makes a different state.
CNo. They differ only by a global phase, so every probability in every basis is the same.
DNo. Their 0/1 probabilities are equal, and that is all a measurement can show.

Hints

Divide each amplitude of $\ket\chi$ by the matching amplitude of $\ket\psi$ . Are the two ratios equal?
Equal ratios mean one common factor $e^{i\gamma}$ : a global phase. Different ratios mean a relative phase.
Compare $P(+)$ for the two states using $\abs{\alpha+\beta}^2/2$ .

Show the answer

Answer: A. Yes. They differ by a relative phase: the 0/1 statistics agree, but a ± measurement tells them apart.

Here only the $\ket0$ amplitude changes sign: a relative phase. In the ± basis $P(+)$ is 98% for $\ket\psi$ but 2% for $\ket\chi$ .

The two ratios differ: only the $\ket0$ amplitude changes sign. The 0/1 probabilities match, but the relative phase does not. In the ± basis $P(+)$ is 98% for $\ket\psi$ but 2% for $\ket\chi$ .

Why not the others

B. Right verdict, wrong reason. A phase on every term would change nothing; what matters here is that only the $\ket0$ amplitude changes sign, a relative phase. In the ± basis $P(+)$ is 98% for $\ket\psi$ but 2% for $\ket\chi$ .
C. Look again: only the $\ket0$ amplitude changes sign, so the needles do not turn together. That is a relative phase. In the ± basis $P(+)$ is 98% for $\ket\psi$ but 2% for $\ket\chi$ .
D. Equal 0/1 probabilities do not make equal states. Here only the $\ket0$ amplitude changes sign. In the ± basis $P(+)$ is 98% for $\ket\psi$ but 2% for $\ket\chi$ .

Check yourself

You measure $\tfrac{2}{\sqrt5}\ket0 - \tfrac{1}{\sqrt5}\ket1$ in the ± basis. What is $P(-)$ ?

$P(-) = \;?$

Hints

Which coefficient goes with $\ket{-}$ when you rewrite the state?
The $\ket{-}$ coefficient is $(\alpha-\beta)/\sqrt2$ , and $P(-)$ is its size squared.
Here $\alpha = \tfrac{2}{\sqrt5}$ and $\beta = -\tfrac{1}{\sqrt5}$ .

Show the answer

Answer: $P(-) = \frac{9}{10}$

$\alpha = \tfrac{2}{\sqrt5}$ , $\beta = -\tfrac{1}{\sqrt5}$ , so $P(-) = \tfrac12\abs{\alpha-\beta}^2 = \tfrac12\left(\tfrac{2}{\sqrt5} + \tfrac{1}{\sqrt5}\right)^2 = \tfrac{9}{10}$ .

Common slips

$\frac{1}{5}$ : That is $P(1)$ in the 0/1 basis. The ± basis needs the coefficient of $\ket{-}$ : $(\alpha - \beta)/\sqrt2$ .
$\frac{9}{5}$ : That is $\abs{\alpha-\beta}^2$ . You still need to divide by 2.
$\frac{1}{10}$ : That is $P(+)$ . For $P(-)$ use $\alpha - \beta$ .

Check yourself

Write $\tfrac35\ket0 + \tfrac45\ket1$ in the ± basis: enter the coefficients of $\ket{+}$ and $\ket{-}$ .

Hints

What are $\alpha$ and $\beta$ here?
The coefficients are $(\alpha+\beta)/\sqrt2$ on $\ket{+}$ and $(\alpha-\beta)/\sqrt2$ on $\ket{-}$ .
The $\ket{+}$ coefficient is $\tfrac{7}{5\sqrt2}$ .

Show the answer

Answer: $\ket{+} = \frac{7}{5\sqrt{2}},\quad \ket{-} = -\frac{1}{5\sqrt{2}}$

With $\alpha$ and $\beta$ read from $\tfrac35\ket0 + \tfrac45\ket1$ : $\tfrac{\alpha+\beta}{\sqrt2} = \tfrac{7}{5\sqrt2}$ and $\tfrac{\alpha-\beta}{\sqrt2} = -\tfrac{1}{5\sqrt2}$ . Their sizes squared add to 1.

9 / 11 · Explain

Two students disagree

Explain it

Ana says $e^{i\pi/3}\ket1$ and $\ket1$ are physically the same state. Ben says that, by the same logic, $\tfrac{1}{\sqrt2}(\ket0 + e^{i\pi/3}\ket1)$ and $\tfrac{1}{\sqrt2}(\ket0 + \ket1)$ are also the same. Who is right, and why?

Show a model answer

Ana is right. In her pair the factor $e^{i\pi/3}$ multiplies the whole state, a global phase, so every $\abs{\text{amplitude}}^2$ in every basis is unchanged. Ben is wrong. In his pair the phase sits on one term only, so it changes the relative phase between $\ket0$ and $\ket1$ . The 0/1 statistics still match (50/50), but in the ± basis $P(+)$ is $3/4$ for his first state and $1$ for $\ket{+}$ .

A good answer covers

Ana is right: a global phase multiplies every amplitude by the same factor, so no probability in any basis changes.
Ben is wrong: in his pair the phase is on one term only, which changes the relative phase.
Names a measurement that shows the difference: the ± basis, where $P(+)$ is $3/4$ instead of $1$ .

10 / 11 · Mastery check

Mastery check

Check yourself

This was your opening guess. Register A holds 1000 qubits, each prepared in ${\ket{+} = \tfrac{1}{\sqrt2}(\ket0 + \ket1)}$ . Register B holds 1000 qubits, each prepared as $\ket0$ or $\ket1$ by a fair coin toss; nobody tells you which. Can any experiment tell the two registers apart?

ANo. Both give 0 or 1 with probability 1/2, so they are physically the same.
BYes. In the 0/1 basis A splits exactly 500/500, while B fluctuates.
CThey differ in principle, but no experiment could ever show it.
DYes. Measure every qubit in a different basis, the ± basis: A gives the same outcome every time, B does not.

Show the answer

Answer: D. Yes. Measure every qubit in a different basis, the ± basis: A gives the same outcome every time, B does not.

In the ± basis every qubit of A gives $+$ ; B gives $+$ about half the time. In the 0/1 basis they look alike.

A superposition is a definite state. $\ket{+}$ gives $+$ with certainty in the ± basis, while each of B's qubits ( $\ket0$ or $\ket1$ ) gives $+$ with probability 1/2. Measuring every qubit in the ± basis therefore separates the registers; the 0/1 basis cannot.

Why not the others

A. A very common idea is that $\ket{+}$ is a coin you haven't looked at yet. If it were, nothing could tell A from B. But measure both in the ± basis: every qubit of A gives $+$ , while B gives $+$ only about half the time. $\ket{+}$ is one definite state; its 50/50 belongs to the 0/1 basis, not to your ignorance.
B. Probabilities predict frequencies, not exact counts. A scatters around 500 just as B does, so in the 0/1 basis the two registers give the same statistics. It is tempting because "superposition" sounds more orderly than a coin, but you need a second basis to see the difference.
C. The difference is not just a matter of words. Measure every qubit in the ± basis: A gives $+$ every time, B only about half the time. It is tempting to think so because the 0/1 basis, the only measurement met so far, really cannot see it.

Check yourself

Consider $\ket a = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ , $\ket b = -i\tfrac{\sqrt3}{2}\ket0 - i\tfrac12\ket1$ and $\ket c = \tfrac{\sqrt3}{2}\ket0 - \tfrac12\ket1$ . Which of the three pairs can some measurement tell apart?

A $a$ – $c$ and $b$ – $c$ , but not $a$ – $b$
BOnly $a$ – $c$
COnly $a$ – $b$ : changing every term is a bigger change than changing one.
DAll three pairs: they all differ in sign or phase.
ENo pair: all three have the same 0/1 statistics.

Show the answer

Answer: A. $a$ – $c$ and $b$ – $c$ , but not $a$ – $b$

$\ket b$ is $\ket a$ times one common factor (a global phase). $\ket c$ flips the sign of one term (a relative phase): in the ± basis $P(+)$ is 93.3% for $\ket a$ and 6.7% for $\ket c$ , and $\ket b$ behaves exactly like $\ket a$ .

$\ket b$ is $\ket a$ multiplied by $-i$ , so no measurement separates them. $\ket c$ changes the sign of the $\ket1$ term only: the 0/1 statistics match, but in the ± basis $P(+)$ is 93.3% for $\ket a$ and 6.7% for $\ket c$ . So $a$ – $c$ and $b$ – $c$ can be told apart.

Why not the others

B. $\ket b$ is physically the same state as $\ket a$ , so any measurement that separates $a$ from $c$ also separates $b$ from $c$ .
C. It is the other way round. Multiplying every amplitude by the same factor is a global phase and invisible; changing one term is a relative phase.
D. $\ket b$ is $\ket a$ times one common factor, $-i$ : a global phase, which changes no probability in any basis.
E. Equal 0/1 statistics do not make states equal: in the ± basis $P(+)$ is 93.3% for $\ket a$ and 6.7% for $\ket c$ .

Check yourself

The state $\begin{bmatrix} \tfrac12 \\ \tfrac{\sqrt3}{2} \end{bmatrix}$ is measured in the ± basis. What is $P(+)$ ?

$P(+) = \;?$

Show the answer

Answer: $P(+) = 0.933$

$\alpha = \tfrac12$ , $\beta = \tfrac{\sqrt3}{2}$ , so $P(+) = \tfrac12\abs{\alpha+\beta}^2 = \tfrac12\left(\tfrac12 + \tfrac{\sqrt3}{2}\right)^2 = \tfrac{2+\sqrt3}{4}$ .

Common slips

$\frac{1}{4}$ : That is $P(0)$ in the 0/1 basis. In the ± basis use $\abs{\alpha+\beta}^2/2$ .
$1.866$ : That is $\abs{\alpha+\beta}^2$ ; divide by 2.

Check yourself

A lab says its source emits $\ket{-}$ every time. A rival says the source really tosses a fair coin and emits $\ket{0}$ or $\ket{1}$ . You may measure many qubits from the source, all in one basis. Which plan settles the dispute?

AMeasure in the 0/1 basis. The lab is right if every outcome is the same.
BMeasure in the 0/1 basis. The lab is right if the split is exactly 50/50.
CNo plan can settle it: both sources give 50/50.
DMeasure in the ± basis. The lab is right if every outcome is $-$ .

Show the answer

Answer: D. Measure in the ± basis. The lab is right if every outcome is $-$ .

In the ± basis $\ket{-}$ gives $-$ every time, while $\ket{0}$ and $\ket{1}$ each give $-$ with probability 1/2.

Pick the basis in which the claimed state is certain: the ± basis. There the lab predicts $-$ on every qubit, and the coin predicts $-$ about half the time. In the 0/1 basis both predict about 50/50.

Why not the others

A. A definite state is certain in one basis only. $\ket{-}$ is certain in the ± basis; in the 0/1 basis it gives both outcomes equally often, just like the coin.
B. Both claims predict about 50/50 in the 0/1 basis, and neither predicts an exact split: counts scatter around 50%. That basis cannot settle it.
C. They give 50/50 in the 0/1 basis only. $\ket{-}$ is a definite state: in the ± basis it gives $-$ every time, while each coin-toss qubit gives $-$ only half the time.

Check yourself

You measure $-\left(\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1\right)$ in the ± basis. What is $P(-)$ ?

$P(-) = \;?$

Show the answer

Answer: $P(-) = 0.067$

The factor in front multiplies every amplitude, so it is a global phase and changes nothing. Use $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ : $\alpha = \tfrac{\sqrt3}{2}$ , $\beta = \tfrac12$ , so $P(-) = \tfrac12\abs{\alpha-\beta}^2 = \tfrac12\left(\tfrac{\sqrt3}{2} - \tfrac12\right)^2 = \tfrac{2-\sqrt3}{4}$ .

Common slips

$0.933$ : That applies $-1$ to the first term only. The factor multiplies the whole state, a global phase, so $P(-)$ is the same as for $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ .

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.

What is the difference between a global phase and a relative phase, and which one can a measurement detect?
Show the key points
- A global phase multiplies every amplitude by the same $e^{i\gamma}$ : every needle turns together.
- It changes no probability in any basis, so $\ket\psi$ and $e^{i\gamma}\ket\psi$ are the same physical state.
- A relative phase changes the angle between the terms: the 0/1 statistics stay the same, but the ± statistics can change.
How do you find the ± basis probabilities of $\alpha\ket0 + \beta\ket1$ ?
Show the key points
- Rewrite the state: coefficient $(\alpha+\beta)/\sqrt2$ on $\ket{+}$ and $(\alpha-\beta)/\sqrt2$ on $\ket{-}$ .
- $P(\pm) = \abs{\alpha\pm\beta}^2/2$ .
- Example: $\ket0$ gives 50/50 in the ± basis, though it is certain in the 0/1 basis.
How can you tell 1000 copies of $\ket{+}$ from 1000 coin-toss qubits, each $\ket0$ or $\ket1$ ?
Show the key points
- In the 0/1 basis both give about 50/50.
- In the ± basis $\ket{+}$ gives $+$ every time, while the coin-toss qubits give $+$ about half the time.
- A superposition is a definite state, not ignorance about a hidden 0 or 1.

Key ideas

An amplitude has a size and a phase. In circle notation the disc radius is the size and the needle is the phase.
A global phase $e^{i\gamma}\ket\psi$ turns every needle together and changes no prediction in any basis.
A relative phase, the angle between the needles, is physical. It is invisible in the 0/1 basis and can show up in the ± basis: $\ket{+}$ and $\ket{-}$ differ only in this way.
In the ± basis, $\alpha\ket0 + \beta\ket1 = \tfrac{\alpha+\beta}{\sqrt2}\ket{+} + \tfrac{\alpha-\beta}{\sqrt2}\ket{-}$ , so $P(\pm) = \abs{\alpha\pm\beta}^2/2$ .
Certainty belongs to a state and a basis together: $\ket0$ is certain in the 0/1 basis and 50/50 in the ± basis. A ± measurement also tells a superposition from a 50/50 coin.

Go further

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

e^(iπ) in 3.14 minutes, using dynamics3Blue1Brown · Video
A short animation of why multiplying by e^(iφ) turns a point around the unit circle, the rule behind every phase dial here.
Multiplying by a global phase doesn’t make a differenceRyan O’Donnell, Carnegie Mellon · Video
A 12-minute university lecture proving that a global phase never changes what any measurement can reveal.
Phase shifter in a Mach–Zehnder interferometerQuVis, University of St Andrews · Interactive
Send single photons through an interferometer and see a phase shift change which detector clicks: relative phase in the lab.

Math refreshers

The toolkit topics this lesson leans on.

Complex numbersMagnitude, phase, and why a phase factor has size 1
Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability