Single-qubit gates · Kets & Gates

1 / 11 · Learn

Before you start

So far you have written states down and measured them. A gate changes a state without reading anything out, and every computation in the rest of this course is built from gates.

By the end of this lesson you can:

apply $X$ , $Z$ and $H$ to a state by matrix multiplication;
read a gate's two columns as what it does to $\ket0$ and $\ket1$ ;
test whether a matrix is unitary, and say why every gate must be;
turn a circuit into a matrix product in the right order;
picture $X$ , $Z$ and $H$ as half turns of the Bloch sphere.

Check yourself

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

A qubit starts in $\ket0$ . One Hadamard gate $H$ would make a measurement in the 0/1 basis come out 0 or 1 with equal odds. Instead of measuring, you apply $H$ a second time, and only then measure in that basis. What is the probability of getting 0?

A $1$
B $\tfrac12$
C $\tfrac14$
D $0$

Show the answer

Answer: A. $1$

$H\ket0 = \ket+$ and $H\ket+ = \ket0$ , so $HH\ket0 = \ket0$ . A measurement in the 0/1 basis then gives 0 with certainty, so $P(0) = 1$ .

Why not the others

B. A common picture is that $H$ randomizes a qubit, so a second $H$ leaves it random. But after one $H$ the qubit is in the definite state $\ket+$ , not a coin, and the second $H$ makes the two contributions to $\ket1$ cancel. So $HH\ket0 = \ket0$ : the 50/50 appears only if you measure in the 0/1 basis between the gates.
C. It is tempting to multiply two 50/50 steps, $\tfrac12 \times \tfrac12$ . But a gate does not add randomness step by step: it changes amplitudes, and amplitudes can cancel. Here the cancellation is complete, because $HH = I$ .
D. This treats $H$ as a "square root of NOT", as if two of them made an $X$ and sent $\ket0$ to $\ket1$ . They don't: $HH = I$ , so the qubit ends in $\ket0$ , and the outcome is 0 every time.

Check yourself

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

Is there a single-qubit gate that turns every input state into $\ket0$ , whatever the input was?

AYes: the matrix $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ keeps the $\ket0$ part, and you rescale the result.
BNo: every gate can be undone, so no gate can send two different inputs to the same output.
CYes: a suitable fixed sequence of $X$ , $Z$ and $H$ .
DNo: gates never change the probabilities of the measurement outcomes 0 and 1; only measurements do.

Show the answer

Answer: B. No: every gate can be undone, so no gate can send two different inputs to the same output.

Gates are unitary, and a unitary always has an inverse. A gate that sent both $\ket0$ and $\ket1$ to $\ket0$ would have no inverse. Resetting an unknown qubit takes a measurement in the 0/1 basis, followed by $X$ if the result was 1.

Why not the others

A. Keeping only the $\ket0$ part is a natural first idea. But that matrix sends $\ket1$ to the zero vector, which is not a state, and the rescaling factor depends on an input you may not know. A gate must keep every state normalized and be undoable, so it can never send two different inputs to the same output.
C. Each of $X$ , $Z$ and $H$ can be undone, so any sequence of them can be undone too. A sequence that sent both $\ket0$ and $\ket1$ to $\ket0$ could not be undone: from the output alone you could not tell which input you had.
D. Gates change probabilities all the time: $H\ket0 = \ket+$ turns a certain outcome 0 into a 50/50 split. The real obstacle is different: a gate is invertible, so it cannot merge two different inputs into one output.

2 / 11 · Learn

A gate is a linear map

The classical NOT gate swaps 0 and 1. Its quantum version, $X$ , does the same to the basis states: $X\ket0 = \ket1$ and $X\ket1 = \ket0$ . Quantum gates are linear, so those two facts settle what $X$ does to every state:

X(\alpha\ket0 + \beta\ket1) = \alpha\ket1 + \beta\ket0.

With states written as column vectors, a gate is a $2\times2$ matrix, and you can read it straight off: the first column is the image of $\ket0$ , the second column is the image of $\ket1$ .

X = \begin{bmatrix}0&1\\1&0\end{bmatrix},\qquad X\begin{bmatrix}\alpha\\ \beta\end{bmatrix} = \begin{bmatrix}\beta\\ \alpha\end{bmatrix}.

Refresher: Matrices acting on states Matrix times vector, and why order matters, if either is rusty.

Check yourself

A gate $U$ sends $\ket0 \mapsto \tfrac{1}{\sqrt2}(\ket0 + \ket1)$ and $\ket1 \mapsto \tfrac{1}{\sqrt2}(i\ket0 - i\ket1)$ . Which matrix is $U$ ?

A $\tfrac{1}{\sqrt2}\begin{bmatrix}1 & 1\\ i & -i\end{bmatrix}$
B $\tfrac{1}{\sqrt2}\begin{bmatrix}1 & i\\ 1 & -i\end{bmatrix}$
C $\tfrac{1}{\sqrt2}\begin{bmatrix}1 & 1\\ 1 & -1\end{bmatrix}$

Hints

Which column does $\begin{bmatrix}1\\0\end{bmatrix}$ pick out when you multiply?
Column 1 is $U\ket0$ and column 2 is $U\ket1$ , written top to bottom.

Show the answer

Answer: B. $\tfrac{1}{\sqrt2}\begin{bmatrix}1 & i\\ 1 & -i\end{bmatrix}$

Column 1 is $U\ket0 = \tfrac{1}{\sqrt2}(1, 1)$ and column 2 is $U\ket1 = \tfrac{1}{\sqrt2}(i, -i)$ . Multiplying by $\begin{bmatrix}1\\0\end{bmatrix}$ picks out column 1, and by $\begin{bmatrix}0\\1\end{bmatrix}$ column 2.

Why not the others

A. This writes the images as rows. Test it: multiplying by $\begin{bmatrix}1\\0\end{bmatrix}$ picks out the first column, $\tfrac{1}{\sqrt2}(1, i)$ , which is not $U\ket0$ . Rows feel natural because we read across, but the multiplication rule puts each input's image down a column.
C. That is $H$ , whose second column is $\ket-$ . Here the image of $\ket1$ is $i\ket-$ , so the second column is $\tfrac{1}{\sqrt2}(i, -i)$ . On $\ket1$ alone the $i$ would be a global phase, but next to the first column it becomes a relative phase, so dropping it changes the gate.

Check yourself

Apply $X$ to $\tfrac35\ket0 - \tfrac{4i}{5}\ket1$ . Enter the exact amplitudes of the result.

Hints

Where does the amplitude of $\ket1$ end up after $X$ ?
Linearity: $X(\alpha\ket0 + \beta\ket1) = \alpha\ket1 + \beta\ket0$ .

Show the answer

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

$X$ swaps the entries of the column vector: $X\begin{bmatrix}3/5\\ -4i/5\end{bmatrix} = \begin{bmatrix}-4i/5\\ 3/5\end{bmatrix}$ , that is $-\tfrac{4i}{5}\ket0 + \tfrac35\ket1$ .

3 / 11 · Learn

Z, H and the Bloch sphere

Two more single-qubit gates come up all the time. $Z$ leaves $\ket0$ alone and flips the sign of $\ket1$ . The Hadamard gate $H$ sends $\ket0$ to $\ket+$ and $\ket1$ to $\ket-$ . Stack those images as columns:

Z = \begin{bmatrix}1&0\\0&-1\end{bmatrix},\qquad H = \frac{1}{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}.

On the Bloch sphere each of these gates is a half turn of the whole sphere: $X$ about the $x$ axis, $Z$ about the $z$ axis, and $H$ about the axis halfway between $x$ and $z$ . A state on the rotation axis stays put; every other state moves.

Check yourself

What is $X\ket+$ ?

A $\ket+$
B $\ket1$
C $\ket-$
D $-\ket+$

Hints

What does $X$ do to the two amplitudes of a state?
$X$ swaps the amplitudes of $\ket0$ and $\ket1$ . What are the two amplitudes of $\ket+$ ?

Show the answer

Answer: A. $\ket+$

$\ket+$ lies on the $x$ axis, the axis of $X$ 's half turn, so it does not move. In vectors: $X$ swaps two equal entries.

Why not the others

B. $X$ acts as NOT on the basis states, but it does not send every state to $\ket1$ . By linearity $X\ket+ = \tfrac{1}{\sqrt2}(X\ket0 + X\ket1) = \tfrac{1}{\sqrt2}(\ket1 + \ket0) = \ket+$ .
C. A natural guess is that NOT flips $+$ into $-$ . But $X$ swaps the two amplitudes, and $\ket+$ has equal amplitudes, so nothing changes: $X\tfrac{1}{\sqrt2}(\ket0+\ket1) = \tfrac{1}{\sqrt2}(\ket1+\ket0)$ . On the sphere $\ket+$ sits on the axis of $X$ 's half turn; turning $\ket+$ into $\ket-$ is what $Z$ does.
D. No sign appears: swapping two equal amplitudes $\tfrac{1}{\sqrt2}$ gives back exactly $\ket+$ . A sign does appear for $\ket-$ : $X\ket- = \tfrac{1}{\sqrt2}(\ket1 - \ket0) = -\ket-$ , the same physical state.

Check yourself

Using only the gate buttons, move the state from $\ket+$ to $\ket1$ , the south pole.

Set the state on the Bloch sphere with the gates X, Z, H.

Hints

Which gate takes $\ket+$ to one of the poles?
$H$ swaps $\ket+$ and $\ket0$ ; $X$ swaps the two poles.

Show the answer

Answer: $|1\rangle$

Two routes work: $H$ then $X$ ( $\ket+ \to \ket0 \to \ket1$ ), or $Z$ then $H$ ( $\ket+ \to \ket- \to \ket1$ ).

Aside: the Pauli Y gate

There is a fourth gate, $Y = \begin{bmatrix}0&-i\\i&0\end{bmatrix}$ . Together $X$ , $Y$ and $Z$ are called the Pauli matrices. On the Bloch sphere $Y$ is a half turn about the $y$ axis. You can check that $Y = iXZ$ , so for this course it adds nothing new.

4 / 11 · Learn

Why gates must be unitary

A gate has to turn every state into a state, so it must preserve length: $\abs{\alpha}^2 + \abs{\beta}^2 = 1$ before and after. For a matrix $U$ that requirement is exactly

U^\dagger U = I,

where the adjoint $U^\dagger$ is $U$ transposed with every entry complex-conjugated. A matrix with this property is unitary. In practice you check the columns: each must have length 1, and the two must be orthogonal (conjugate the first column's entries before multiplying; see the inner-product refresher). A unitary can always be undone by applying $U^\dagger$ . In the other direction, every unitary matrix is a valid gate. So there are infinitely many single-qubit gates, against one non-trivial classical one-bit gate, NOT.

Check yourself

Select every matrix that is a valid gate.

Select all that apply.

A $\tfrac{1}{\sqrt2}\begin{bmatrix}1&i\\i&1\end{bmatrix}$
B $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$
C $\begin{bmatrix}1&1\\0&1\end{bmatrix}$
D $\begin{bmatrix}1&0\\0&0\end{bmatrix}$

Hints

What length must each column have?
A unitary has columns of length 1 that are orthogonal to each other.

Show the answer

Answers: A. $\tfrac{1}{\sqrt2}\begin{bmatrix}1&i\\i&1\end{bmatrix}$ ; B. $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$

Check the columns. In the first matrix each has squared length $\tfrac12 + \tfrac12 = 1$ , and $\tfrac12\big(1\cdot i + (-i)\cdot 1\big) = 0$ , so they are orthogonal. $Z$ has columns $(1, 0)$ and $(0, -1)$ . The shear and the projector each have a second column whose length is not 1.

Why not the others

C. The shear can be undone, yet it is not unitary: its second column $(1, 1)$ has length $\sqrt2$ , so it sends $\ket1$ to an unnormalized vector. Being invertible is not enough; a gate must also preserve length.
D. This projector keeps the $\ket0$ part and sends $\ket1$ to the zero vector, which is not a state. It cannot be undone: $\tfrac35\ket0 + \tfrac45\ket1$ and $\tfrac35\ket0 - \tfrac45\ket1$ both come out as $\tfrac35\ket0$ . It describes the outcome 0 of a measurement in the 0/1 basis, and a measurement is not a gate.

Check yourself

You know a qubit is in the state $\ket+$ . Can a gate turn it into $\ket0$ ?

ANo: only a measurement can turn a superposition into a basis state.
BYes: the gate $H$ does it, since $H\ket+ = \ket0$ .
COnly half the time: measure in the 0/1 basis and hope for 0.
DYes: apply $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and multiply by $\sqrt2$ .

Hints

Which gate turns $\ket+$ into a basis state?
A gate is one fixed, invertible map. It can carry one known state anywhere; what it cannot do is send every state to the same place.

Show the answer

Answer: B. Yes: the gate $H$ does it, since $H\ket+ = \ket0$ .

$H\ket+ = \ket0$ . This is the contrast with your guess: a gate is one fixed, invertible map, so it can carry one known state to $\ket0$ (or anywhere else), but it cannot send every state to the same place.

Why not the others

A. Gates turn superpositions into basis states all the time: $H\ket+ = \tfrac12(\ket0 + \ket1) + \tfrac12(\ket0 - \ket1) = \ket0$ , with certainty and no measurement. What a gate cannot do is reset an unknown state, because it would have to send different inputs to the same output.
C. A measurement gives 0 only half the time, and you would then need $X$ after a 1. No measurement is needed here: because the state is known, the gate $H$ turns it into $\ket0$ every time.
D. That does give $\ket0$ for this one input. But the projector is not unitary, so it is not a gate, and the factor $\sqrt2$ only works because you already knew the input. On $\ket1$ the same recipe gives the zero vector.

Aside: every gate from three rotations

Every single-qubit unitary can be written as a global phase times three rotations:

\begin{aligned} U = e^{i\alpha}&\begin{bmatrix}e^{-i\beta/2}&0\\0&e^{i\beta/2}\end{bmatrix}\\ \times&\begin{bmatrix}\cos\frac{\gamma}{2}&-\sin\frac{\gamma}{2}\\ \sin\frac{\gamma}{2}&\cos\frac{\gamma}{2}\end{bmatrix}\\ \times&\begin{bmatrix}e^{-i\delta/2}&0\\0&e^{i\delta/2}\end{bmatrix} \end{aligned}

with real $\alpha$ , $\beta$ , $\gamma$ , $\delta$ . Careful: here $\alpha$ and $\beta$ are angles, not the amplitudes of a state. For now you only need the idea that a gate can be any unitary.

5 / 11 · Learn

Order matters

In a circuit diagram time runs left to right: the gate drawn first acts first. In algebra the matrix written next to the ket acts first. So a circuit that applies $X$ and then $H$ is written

HX\ket\psi \qquad (\text{read right to left}).

The order changes the answer. Take $Z$ and $X$ on $\ket0$ :

\begin{aligned} ZX\ket0 &= Z\ket1 = -\ket1,\\ XZ\ket0 &= X\ket0 = \ket1. \end{aligned}

These two differ only by an overall sign, because $ZX = -XZ$ . For $X$ and $H$ the difference shows up in a measurement: on input $\ket+$ , " $X$ then $H$ " gives $\ket0$ , but " $H$ then $X$ " gives $\ket1$ .

Check yourself

Compute $ZX\ket1$ .

A $\ket0$
B $-\ket0$
C $\ket1$
D $-\ket1$

Hints

In $ZX\ket1$ , which gate stands next to the ket, and so acts first?
Work outward from the ket: find $X\ket1$ first, then apply $Z$ to that result.

Show the answer

Answer: A. $\ket0$

$ZX\ket1 = Z(X\ket1) = Z\ket0 = \ket0$ . In the other order $XZ\ket1 = X(-\ket1) = -\ket0$ : the two products differ by a sign because $ZX = -XZ$ .

Why not the others

B. That applies $Z$ first: $XZ\ket1 = X(-\ket1) = -\ket0$ . In $ZX\ket1$ the gate next to the ket acts first, so $X\ket1 = \ket0$ and then $Z\ket0 = \ket0$ . Here the slip costs only a sign, but for pairs such as $X$ and $H$ it changes what you measure.
C. $X$ moves $\ket1$ to $\ket0$ , and $Z$ never turns one basis state into the other, so the answer must be a multiple of $\ket0$ .
D. The minus sign is what $Z$ does to $\ket1$ , but by the time $Z$ acts, $X$ has already moved the qubit to $\ket0$ : $ZX\ket1 = Z\ket0 = \ket0$ .

Check yourself

A circuit applies $X$ , then $H$ , then $Z$ to $\ket\psi$ . Which expression gives its output?

A $XHZ\ket\psi$
B $X\ket\psi + H\ket\psi + Z\ket\psi$
C $ZHX\ket\psi$
D $HZX\ket\psi$

Hints

Which gate does the qubit meet first, and where must that gate stand in the product?
The first gate stands next to the ket; each later gate is written to the left of the one before it.

Show the answer

Answer: C. $ZHX\ket\psi$

Read the circuit left to right and write the matrices right to left: $Z\big(H(X\ket\psi)\big) = ZHX\ket\psi$ .

Why not the others

A. This writes the gates in the order they appear in the diagram. In algebra the matrix next to the ket acts first, so the first gate, $X$ , must stand on the right: $ZHX\ket\psi$ . The diagram reads like text, which is why this slip is so common.
B. Gates applied one after another compose by multiplication, not addition: each gate acts on the output of the one before it.
D. $X$ is in the right place, next to the ket, but $H$ and $Z$ are swapped. After $X$ the qubit meets $H$ , so $H$ comes next to the left, and $Z$ , the last gate, stands on the far left.

6 / 11 · Explore

Two Hadamards

Predict, then see

Two circuits start from $\ket0$ . A: $H$ , then $H$ , then measure in the 0/1 basis. B: $H$ , measure in that basis, $H$ , measure again. Over 100 runs, how often does each give 0 at the end?

AA: all 100 runs. B: about 50.
BBoth all 100: the second $H$ undoes the first whether or not you look in between.
CA: none, because two $H$ gates make an $X$ . B: about 50.
DBoth about 50: $H$ randomizes the qubit, and doing it twice leaves it random.

In a sentence or two: why does a measurement between the two $H$ gates change the result?

Show what happens

What happens: A. A: all 100 runs. B: about 50.

After one $H$ the qubit is in $\ket+$ : a definite state, not a coin. The second $H$ sends $\ket0$ to $\tfrac{1}{\sqrt2}(\ket0+\ket1)$ and $\ket1$ to $\tfrac{1}{\sqrt2}(\ket0-\ket1)$ , so the two contributions to $\ket1$ cancel and the two contributions to $\ket0$ add; this cancellation is interference, and it makes $HH\ket0 = \ket0$ . A measurement in the 0/1 basis in between replaces $\ket+$ with $\ket0$ or $\ket1$ at random, so there is nothing left to interfere. The second $H$ then gives $\ket+$ or $\ket-$ , and the final result is 50/50.

Why not the others

B. The second $H$ undoes the first only if nothing happens in between. A measurement in the 0/1 basis replaces $\ket+$ by $\ket0$ or $\ket1$ ; $H$ then gives $\ket+$ or $\ket-$ , each 50/50 in that basis. Looking changed the state.
C. $H$ is not a "square root of NOT": $HH = I$ , not $X$ , so circuit A returns the qubit to $\ket0$ .
D. That is the picture of $H$ as a coin flip. If it were right, A and B would behave the same, but A gives 0 every time. Only the measurement in B turns the qubit into a coin.

The arithmetic

\begin{aligned} H\ket+ &= \tfrac{1}{\sqrt2}\big(H\ket0 + H\ket1\big)\\ &= \tfrac12(\ket0 + \ket1) + \tfrac12(\ket0 - \ket1)\\ &= \ket0. \end{aligned}

As matrices:

\begin{aligned} HH &= \tfrac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}1&1\\1&-1\end{bmatrix}\\ &= \tfrac12\begin{bmatrix}2&0\\0&2\end{bmatrix} = I. \end{aligned}

Check yourself

As a half turn about the axis between $x$ and $z$ , $H$ swaps the $x$ and $z$ axes of the Bloch sphere. Use that, or multiply the matrices: which single gate equals $HZH$ ?

A $H$
B $X$
C $Z$
D $I$

Hints

What does $HZH$ do to $\ket0$ ? Apply the gates one at a time, starting from the right.
$H\ket0 = \ket+$ and $Z\ket+ = \ket-$ . Apply the last $H$ , then do the same for $\ket1$ and compare with $X$ , $Z$ , $H$ and $I$ .

Show the answer

Answer: B. $X$

$HZH\ket0 = HZ\ket+ = H\ket- = \ket1$ and $HZH\ket1 = HZ\ket- = H\ket+ = \ket0$ , which is exactly what $X$ does. In the same way $HXH = Z$ .

Why not the others

A. Test it on $\ket0$ : $HZH\ket0 = HZ\ket+ = H\ket- = \ket1$ , while $H\ket0 = \ket+$ .
C. It is tempting to cancel the two $H$ gates, but they are not next to each other, and matrices do not commute in general. Use the sphere: the first $H$ swaps the $x$ and $z$ axes, so a half turn about $z$ between two $H$ gates acts as a half turn about $x$ .
D. Only copies of a gate standing side by side cancel. Test it on $\ket0$ : $HZH\ket0 = HZ\ket+ = H\ket- = \ket1$ , so it is not the identity.

7 / 11 · Worked example

Worked example: a gate, column by column

Worked example

H on a general state

Find $H\ket\psi$ for $\ket\psi = \tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ , and the probability of outcome 0 when you measure it in the 0/1 basis.

Use linearity
By linearity $H\ket\psi = \tfrac{\sqrt3}{2}H\ket0 + \tfrac12 H\ket1$ : the first column of $H$ times $\alpha$ , plus the second column times $\beta$ .
$\begin{aligned} H\ket\psi &= \frac{\sqrt3}{2}\cdot\frac{1}{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}\\ &\quad + \frac12\cdot\frac{1}{\sqrt2}\begin{bmatrix}1\\-1\end{bmatrix} \end{aligned}$
Add the columns
$\begin{aligned} H\ket\psi &= \frac{1}{2\sqrt2}\begin{bmatrix}\sqrt3 + 1\\ \sqrt3 - 1\end{bmatrix}\\ &\approx \begin{bmatrix}0.966\\ 0.259\end{bmatrix} \end{aligned}$
Check the length: $\frac{(\sqrt3+1)^2 + (\sqrt3-1)^2}{8} = \frac{8}{8} = 1$ , as a unitary guarantees.
Your turn
Your turn
What is the probability of outcome 0 when $H\ket\psi$ is measured in the 0/1 basis? An exact expression or a decimal is fine.
$P(0) = \;?$
Hints
1. Which entry of $H\ket\psi$ belongs to $\ket0$ ?
2. Square its magnitude: $(\sqrt3+1)^2 = 4 + 2\sqrt3$ .
Show the answer
Answer: $P(0) = 0.933$
$P(0) = \left(\frac{\sqrt3+1}{2\sqrt2}\right)^2 = \frac{4 + 2\sqrt3}{8} = \frac{2+\sqrt3}{4} \approx 0.933$ .
Common slips
- $\frac{3}{4}$ : $\tfrac34$ is $P(0)$ for the input $\ket\psi$ . The question is about the state after $H$ , whose amplitude on $\ket0$ is $\tfrac{\sqrt3+1}{2\sqrt2}$ .
- $\frac{1}{2}$ : A common idea is that a Hadamard always produces 50/50 odds. That holds for the inputs $\ket0$ and $\ket1$ , but this input already leans toward $\ket0$ , and $H$ moves it closer still. Square the top entry: $\big(\tfrac{\sqrt3+1}{2\sqrt2}\big)^2$ .
Your turn: same method, new gate
Your turn
Compute $X\big(\tfrac{\sqrt3}{2}\ket0 + \tfrac{i}{2}\ket1\big)$ and enter the exact amplitudes.
Hints
1. What are the two columns of $X$ ?
2. Column 1 is $\ket1$ and column 2 is $\ket0$ , so the output is $\alpha\ket1 + \beta\ket0$ .
Show the answer
Answer: $\frac{1}{2}\,i|0\rangle + \frac{\sqrt{3}}{2}|1\rangle$
The columns of $X$ are $\ket1$ and $\ket0$ , so the output is $\tfrac{\sqrt3}{2}\ket1 + \tfrac i2\ket0$ : the vector $\big(\tfrac i2, \tfrac{\sqrt3}{2}\big)$ .

Here is the same method once more, with different numbers.

Another worked example

H on another state

Use linearity
$\begin{aligned} &H\big(\tfrac35\ket0 + \tfrac45\ket1\big)\\ &\quad= \frac35\cdot\frac{1}{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}\\ &\qquad + \frac45\cdot\frac{1}{\sqrt2}\begin{bmatrix}1\\-1\end{bmatrix} \end{aligned}$
Add the columns
$\begin{aligned} &= \frac{1}{5\sqrt2}\begin{bmatrix}3 + 4\\ 3 - 4\end{bmatrix}\\ &= \begin{bmatrix}7/(5\sqrt2)\\ -1/(5\sqrt2)\end{bmatrix}\\ &\approx \begin{bmatrix}0.990\\ -0.141\end{bmatrix} \end{aligned}$
Probabilities
In the 0/1 basis, $P(0) = \frac{49}{50} = 0.98$ and $P(1) = \frac{1}{50}$ . They add to 1, as they must.
The same idea with X
The columns of $X$ are $\ket1$ and $\ket0$ , so it just moves the amplitudes: $X\big(\tfrac35\ket0 + \tfrac45\ket1\big) = \tfrac45\ket0 + \tfrac35\ket1$ .

Check yourself

Compute $H\big(\tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1\big)$ and enter the exact amplitudes.

Hints

Which column of $H$ does $\alpha$ multiply, and which does $\beta$ ?
Linearity: $H(\alpha\ket0 + \beta\ket1) = \alpha\ket+ + \beta\ket-$ .
The top entry is $\tfrac{\alpha+\beta}{\sqrt2}$ and the bottom entry is $\tfrac{\alpha-\beta}{\sqrt2}$ .

Show the answer

Answer: $0.966|0\rangle - 0.259|1\rangle$

By linearity, $H(\alpha\ket0 + \beta\ket1) = \alpha\ket+ + \beta\ket- = \tfrac{\alpha+\beta}{\sqrt2}\ket0 + \tfrac{\alpha-\beta}{\sqrt2}\ket1$ . With $\alpha = \tfrac12$ and $\beta = \tfrac{\sqrt3}{2}$ this is $\tfrac{1+\sqrt3}{2\sqrt2}\ket0 + \tfrac{1-\sqrt3}{2\sqrt2}\ket1$ , that is, the vector $\big(\tfrac{1+\sqrt3}{2\sqrt2},\ \tfrac{1-\sqrt3}{2\sqrt2}\big)$ .

8 / 11 · Practice

Practice

Check yourself

A qubit in $\ket+$ goes through $Z$ and then $H$ . Enter the final state.

Hints

Which gate does the qubit meet first?
Apply the gates one at a time, starting with the first one in the circuit. As a product the circuit reads right to left.
After $Z$ the state is $\ket-$ .

Show the answer

Answer: $|1\rangle$

In algebra the circuit is $HZ\ket+$ . Work from the ket outward: $Z\ket+ = \ket-$ , then $H\ket- = \ket1$ . The final state is $\ket1$ . (Any overall sign is accepted: it is the same physical state.)

Check yourself

Is $M = \tfrac{1}{\sqrt2}\begin{bmatrix}1 & 1\\ i & -i\end{bmatrix}$ a valid gate?

AUnitary: a valid gate
BNot unitary: a column does not have length 1
CNot unitary: both columns have length 1, but they are not orthogonal

Hints

What two things must be true of the columns of a unitary?
Each column must have length 1, and the two must be orthogonal. For the inner product, conjugate the first column's entries before multiplying.
Each column has squared length $\tfrac12 + \tfrac12 = 1$ .

Show the answer

Answer: A. Unitary: a valid gate

It is unitary. Each column has squared length $\tfrac12 + \tfrac12 = 1$ . With the conjugate, $\tfrac12\big(1\cdot 1 + (-i)(-i)\big) = \tfrac12(1 - 1) = 0$ . Skipping the conjugate would give 1 instead.

Why not the others

B. Both tests pass here. Each column has squared length $\tfrac12 + \tfrac12 = 1$ . With the conjugate, $\tfrac12\big(1\cdot 1 + (-i)(-i)\big) = \tfrac12(1 - 1) = 0$ . Skipping the conjugate would give 1 instead.
C. Both tests pass here. Each column has squared length $\tfrac12 + \tfrac12 = 1$ . With the conjugate, $\tfrac12\big(1\cdot 1 + (-i)(-i)\big) = \tfrac12(1 - 1) = 0$ . Skipping the conjugate would give 1 instead.

Check yourself

Build a circuit that takes $\ket0$ to $\ket-$ . Any global phase is fine.

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

Hints

Which gate turns a basis state into a $\pm$ state, or back?
$H$ swaps $\ket0 \leftrightarrow \ket+$ and $\ket1 \leftrightarrow \ket-$ ; $X$ swaps $\ket0 \leftrightarrow \ket1$ ; $Z$ swaps $\ket+ \leftrightarrow \ket-$ .
One route is $X$ then $H$ .

Show the answer

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

From $\ket0$ : $X\ket0 = \ket1$ , then $H\ket1 = \ket-$ . Another route is $H$ then $Z$ : $H\ket0 = \ket+$ , then $Z\ket+ = \ket-$ .

9 / 11 · Explain

Explain it

People often say that $H$ "randomizes" a qubit. If that were the whole story, why would $HH\ket0$ give $\ket0$ with certainty? Explain what actually happens.

Show a model answer

$H\ket0$ is the definite state $\ket+ = \tfrac{1}{\sqrt2}(\ket0 + \ket1)$ , not a coin that has secretly landed. When the second $H$ acts, the $\ket0$ part contributes $+\tfrac12$ to the amplitude of $\ket1$ and the $\ket1$ part contributes $-\tfrac12$ , so they cancel, while the two contributions to $\ket0$ add up to 1. Randomness appears only when you measure, and it depends on the basis: $\ket+$ gives 50/50 in the 0/1 basis but a certain result in the ± basis.

A good answer covers

Says that $H\ket0$ is a definite state with amplitudes ( $\ket+$ ), not a random mixture of 0 and 1.
Explains that the second $H$ makes the contributions to $\ket1$ cancel and those to $\ket0$ add (interference), so $HH = I$ .
Says that randomness appears only at measurement, and depends on the basis measured in.

10 / 11 · Mastery check

Mastery check

Check yourself

A qubit starts in $\ket1$ . You apply $H$ , then $H$ again, then measure in the 0/1 basis. What is the probability of outcome 0?

A $1$
B $\tfrac12$
C $\tfrac14$
D $0$

Show the answer

Answer: D. $0$

$H\ket1 = \ket-$ and $H\ket- = \ket1$ , so $HH\ket1 = \ket1$ . A measurement in the 0/1 basis then gives 1 with certainty, so $P(0) = 0$ .

Why not the others

A. This treats $H$ as a "square root of NOT", as if two of them made an $X$ and sent $\ket1$ to $\ket0$ . They don't: $HH = I$ , so the qubit ends in $\ket1$ , and the outcome is 1 every time.
B. A common picture is that $H$ randomizes a qubit, so a second $H$ leaves it random. But after one $H$ the qubit is in the definite state $\ket-$ , not a coin, and the second $H$ makes the two contributions to $\ket0$ cancel. So $HH\ket1 = \ket1$ : the 50/50 appears only if you measure in the 0/1 basis between the gates.
C. It is tempting to multiply two 50/50 steps, $\tfrac12 \times \tfrac12$ . But a gate does not add randomness step by step: it changes amplitudes, and amplitudes can cancel. Here the cancellation is complete, because $HH = I$ .

Check yourself

A gate $U$ sends $\ket0 \mapsto \tfrac{1}{\sqrt2}(\ket0 + i\ket1)$ and $\ket1 \mapsto \tfrac{1}{\sqrt2}(i\ket0 + \ket1)$ . Compute $U\ket-$ and enter its exact amplitudes, keeping any global phase factor (this asks for the vector, not just the physical state).

Show the answer

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

By linearity $U\ket- = \tfrac{1}{\sqrt2}\big(U\ket0 - U\ket1\big) = \tfrac{1-i}{2}\ket0 - \tfrac{1-i}{2}\ket1$ .

Check yourself

The expression $XH\ket1$ describes a two-gate circuit acting on $\ket1$ . Which state comes out, up to an overall sign?

A $\ket-$
B $\ket+$
C $\ket0$
D $\ket1$

Show the answer

Answer: A. $\ket-$

The gate next to the ket acts first: $H\ket1 = \ket-$ , then $X\ket- = -\ket-$ . (Got $\ket+$ ? That is the other order.)

Read $XH\ket1$ right to left. First $H$ : $H\ket1 = \ket-$ . Then $X$ : $X\ket- = -\ket-$ . Up to its sign this is $\ket-$ .

Why not the others

B. That reads the product left to right, applying $X$ first. The gate next to the ket acts first: $H\ket1 = \ket-$ , then $X\ket- = -\ket-$ . In a circuit diagram $H$ would be drawn on the left.
C. Work outward from the ket, one gate at a time: $H\ket1 = \ket-$ , then $X\ket- = -\ket-$ .
D. Work outward from the ket, one gate at a time: $H\ket1 = \ket-$ , then $X\ket- = -\ket-$ .

Check yourself

A gate $U$ sends $\ket1$ to $\ket+$ . Which of these could $U\ket0$ be, up to a global phase?

A $\ket0$
B $\ket1$
C $\ket-$
D $\ket+$

Show the answer

Answer: C. $\ket-$

$U\ket1$ and $U\ket0$ are the columns of $U$ . A unitary has orthogonal columns of length 1, and the only state on the list orthogonal to $\ket+$ is $\ket-$ .

Why not the others

A. $U\ket1$ and $U\ket0$ are the two columns of $U$ , and a unitary has orthogonal columns. But $\abs{\braket{+}{0}} = \tfrac{1}{\sqrt2}$ , not 0.
B. $U\ket1$ and $U\ket0$ are the two columns of $U$ , and a unitary has orthogonal columns. But $\abs{\braket{+}{1}} = \tfrac{1}{\sqrt2}$ , not 0.
D. Then $U$ would send both $\ket0$ and $\ket1$ to the same state, and nothing could undo it: from the output you could not tell which input you had. A unitary keeps orthogonal inputs orthogonal.

Check yourself

Is there a single-qubit gate that turns every input state into $\ket0$ , whatever the input was?

ANo: every gate can be undone, so no gate can send two different inputs to the same output.
BYes: a suitable fixed sequence of $X$ , $Z$ and $H$ .
CYes: the matrix $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ keeps the $\ket0$ part, and you rescale the result.
DNo: gates never change the probabilities of the measurement outcomes 0 and 1; only measurements do.

Show the answer

Answer: A. No: every gate can be undone, so no gate can send two different inputs to the same output.

Gates are unitary, and a unitary always has an inverse. A gate that sent both $\ket0$ and $\ket1$ to $\ket0$ would have no inverse. Resetting an unknown qubit takes a measurement in the 0/1 basis, followed by $X$ if the result was 1.

Why not the others

B. Each of $X$ , $Z$ and $H$ can be undone, so any sequence of them can be undone too. A sequence that sent both $\ket0$ and $\ket1$ to $\ket0$ could not be undone: from the output alone you could not tell which input you had.
C. Keeping only the $\ket0$ part is a natural first idea. But that matrix sends $\ket1$ to the zero vector, which is not a state, and the rescaling factor depends on an input you may not know. A gate must keep every state normalized and be undoable, so it can never send two different inputs to the same output.
D. Gates change probabilities all the time: $H\ket0 = \ket+$ turns a certain outcome 0 into a 50/50 split. The real obstacle is different: a gate is invertible, so it cannot merge two different inputs into one output.

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.

How do you find the matrix of a gate from what it does to $\ket0$ and $\ket1$ , and how do you then apply it to $\alpha\ket0 + \beta\ket1$ ?
Show the key points
- Column 1 is the image of $\ket0$ ; column 2 is the image of $\ket1$ .
- By linearity the output is $\alpha$ times column 1 plus $\beta$ times column 2.
- Equivalently, each output entry is a row of the matrix times the input column.
What makes a matrix a valid gate, and why can no gate reset every state to $\ket0$ ?
Show the key points
- $U^\dagger U = I$ : columns of length 1 that are orthogonal (conjugate before multiplying).
- A unitary preserves normalization and can be undone by $U^\dagger$ .
- A reset would send different inputs to the same output, which cannot be undone; resetting needs a measurement.
Why does measuring $HH\ket0$ in the 0/1 basis give the outcome 0 every time, and what changes if you measure between the two $H$ gates?
Show the key points
- $H\ket0 = \ket+$ is a definite state, not a coin.
- The second $H$ makes the contributions to $\ket1$ ( $+\tfrac12$ and $-\tfrac12$ ) cancel: interference, so $HH = I$ .
- A measurement in the 0/1 basis in between replaces $\ket+$ by $\ket0$ or $\ket1$ , so the final result is 50/50.

Key ideas

A gate acts linearly, so what it does to $\ket0$ and $\ket1$ fixes everything. Its matrix has those two images as its columns.
$X$ swaps the amplitudes, $Z$ flips the sign of the $\ket1$ amplitude, and $H$ sends $\ket0 \to \ket+$ and $\ket1 \to \ket-$ . On the Bloch sphere each is a half turn: about $x$ , about $z$ , and about the axis between $x$ and $z$ .
Every gate is unitary, $U^\dagger U = I$ , and every unitary is a gate. A unitary keeps states normalized and can always be undone, so no gate resets an unknown state.
Circuits run left to right; matrix products act right to left. " $X$ then $H$ " is $HX$ . Most gates do not commute: $ZX = -XZ$ , and $HX\ket+ = \ket0$ while $XH\ket+ = \ket1$ .
$HH = I$ because the second $H$ makes amplitudes cancel (interference). Sandwiching between two $H$ gates swaps $X$ and $Z$ : $HZH = X$ , $HXH = Z$ .

Go further

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

Linear transformations and matrices3Blue1Brown · Video
Animates why a matrix’s columns are where the basis vectors land, the rule used here to build each gate’s matrix.
Quantum logic gateWikipedia · Article
Goes beyond X, Z and H to the Y, phase-shift, swap and Toffoli gates, and to which gate sets can build every circuit.
Quirk: a quantum circuit simulatorCraig Gidney · Interactive
A free in-browser circuit simulator: drag gates onto wires and watch Bloch spheres and amplitudes change as you build.

Math refreshers

The toolkit topics this lesson leans on.

Matrices acting on statesGates are matrices; circuit order vs matrix order
Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability