Pauli Operators, Quantum Gates, and Qubit Transformations

Core question: How do we deliberately transform a quantum state?

A quantum computation does not only measure a state.
Before measurement, we deliberately transform the state using quantum gates.

For a state $|\psi\rangle$, a gate $U$ produces

$$|\psi'\rangle = U|\psi\rangle.$$

The central mathematical requirement is that $U$ is unitary.

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1. Why Quantum Gates Must Be Unitary

A normalized quantum state satisfies

$$\langle \psi|\psi\rangle = 1.$$

After applying $U$,

$$|\psi'\rangle = U|\psi\rangle.$$

Then

$$\begin{aligned} \langle \psi'|\psi'\rangle &= (U|\psi\rangle)^\dagger(U|\psi\rangle)\\ &= \langle\psi|U^\dagger U|\psi\rangle. \end{aligned}$$

If

$$U^\dagger U = I,$$

then

$$\begin{aligned} \langle \psi'|\psi'\rangle &= \langle\psi|I|\psi\rangle\\ &= \langle\psi|\psi\rangle\\ &= 1. \end{aligned}$$

Therefore,

$$\boxed{\|U\psi\|=\|\psi\|}.$$

Simple English: A unitary gate changes the state without destroying its normalization.
The total probability remains $1$.

A stronger statement is that unitary operators preserve inner products:

$$\langle U\phi|U\psi\rangle = \langle\phi|U^\dagger U|\psi\rangle = \langle\phi|\psi\rangle.$$

So the geometry of the state space is preserved.

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2. Computational Basis

For a single qubit,

$$|0\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \qquad |1\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}.$$

A general qubit is

$$|\psi\rangle = \alpha|0\rangle+\beta|1\rangle,$$

with normalization

$$|\alpha|^2+|\beta|^2=1.$$

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3. Pauli $X$: Bit Flip

The Pauli $X$ matrix is

$$\boxed{ X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} }.$$

3.1 Derivation of $X|0\rangle$

$$X|0\rangle = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}.$$

Compute each entry:

$$0\cdot1+1\cdot0=0,$$ $$1\cdot1+0\cdot0=1.$$

Therefore,

$$X|0\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix} = |1\rangle.$$

Hence

$$\boxed{X|0\rangle=|1\rangle}.$$

Similarly,

$$\boxed{X|1\rangle=|0\rangle}.$$

Simple English: $X$ exchanges the two computational-basis states.
It behaves like a classical NOT gate on $|0\rangle$ and $|1\rangle$.

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4. Pauli $X$ on a Superposition

Let

$$|\psi\rangle = \alpha|0\rangle+\beta|1\rangle.$$

Because $X$ is linear,

$$\begin{aligned} X|\psi\rangle &= X(\alpha|0\rangle+\beta|1\rangle)\\ &= \alpha X|0\rangle+\beta X|1\rangle\\ &= \alpha|1\rangle+\beta|0\rangle. \end{aligned}$$

Reordering the basis terms,

$$\boxed{ X|\psi\rangle = \beta|0\rangle+\alpha|1\rangle }.$$

In matrix form,

$$\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \begin{pmatrix} \alpha\\ \beta \end{pmatrix} = \begin{pmatrix} \beta\\ \alpha \end{pmatrix}.$$

Simple English: $X$ swaps the amplitudes.

Worked Example

Take

$$|\psi\rangle = \frac{\sqrt3}{2}|0\rangle + \frac12|1\rangle.$$

Before applying $X$,

$$P(0) = \left|\frac{\sqrt3}{2}\right|^2 = \frac34,$$ $$P(1) = \left|\frac12\right|^2 = \frac14.$$

After applying $X$,

$$X|\psi\rangle = \frac12|0\rangle + \frac{\sqrt3}{2}|1\rangle.$$

Therefore,

$$P(0)=\frac14, \qquad P(1)=\frac34.$$

So the two computational-basis probabilities are exchanged.

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5. Pauli $Z$: Phase Flip

The Pauli $Z$ matrix is

$$\boxed{ Z= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} }.$$

Its action on the computational basis is

$$Z|0\rangle=|0\rangle,$$ $$Z|1\rangle=-|1\rangle.$$

For a general state,

$$|\psi\rangle = \alpha|0\rangle+\beta|1\rangle,$$

we obtain

$$\boxed{ Z|\psi\rangle = \alpha|0\rangle-\beta|1\rangle }.$$

The magnitudes of the amplitudes do not change.

Only the phase of the $|1\rangle$ component changes.

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6. Why a Minus Sign Is a Phase

Euler's formula is

$$e^{i\theta} = \cos\theta+i\sin\theta.$$

For $\theta=\pi$,

$$e^{i\pi} = \cos\pi+i\sin\pi = -1.$$

Therefore,

$$\boxed{-1=e^{i\pi}}.$$

So multiplying an amplitude by $-1$ is equivalent to rotating its complex phase by $\pi$.

Simple English: $Z$ does not change the size of the amplitude.
It changes its phase.

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7. Global Phase vs. Relative Phase

7.1 Global Phase

Consider

$$|\psi\rangle \longrightarrow e^{i\theta}|\psi\rangle.$$

For any measurement state $|k\rangle$,

$$\begin{aligned} P(k) &= \left| \langle k|e^{i\theta}\psi\rangle \right|^2\\ &= \left| e^{i\theta} \langle k|\psi\rangle \right|^2\\ &= |e^{i\theta}|^2 |\langle k|\psi\rangle|^2\\ &= |\langle k|\psi\rangle|^2. \end{aligned}$$

Since

$$|e^{i\theta}|=1,$$

the measurement probabilities are unchanged.

Thus

$$\boxed{ |\psi\rangle \sim e^{i\theta}|\psi\rangle }$$

as physical pure states.

Simple English: If every amplitude rotates by the same phase, the physical state does not change.

7.2 Relative Phase

Compare

$$\alpha|0\rangle+\beta|1\rangle$$

with

$$\alpha|0\rangle-\beta|1\rangle.$$

Only one component changes phase.

That changes the phase relation between the two amplitudes.

This relative phase can affect later interference.

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8. Relative Phase Is Observable

Define

$$|+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt2},$$ $$|-\rangle = \frac{|0\rangle-|1\rangle}{\sqrt2}.$$

Apply $Z$ to $|+\rangle$:

$$\begin{aligned} Z|+\rangle &= \frac1{\sqrt2} \left( Z|0\rangle+Z|1\rangle \right)\\ &= \frac1{\sqrt2} \left( |0\rangle-|1\rangle \right)\\ &= |-\rangle. \end{aligned}$$

Therefore,

$$\boxed{Z|+\rangle=|-\rangle}.$$

Now compute the overlap:

$$\begin{aligned} \langle+|-\rangle &= \frac12 (\langle0|+\langle1|) (|0\rangle-|1\rangle)\\ &= \frac12 \left( \langle0|0\rangle - \langle0|1\rangle + \langle1|0\rangle - \langle1|1\rangle \right). \end{aligned}$$

Using orthonormality,

$$\langle0|0\rangle=1, \qquad \langle1|1\rangle=1,$$ $$\langle0|1\rangle = \langle1|0\rangle = 0.$$

Hence

$$\langle+|-\rangle = \frac12(1-1) = 0.$$

Therefore,

$$\boxed{\langle+|-\rangle=0}.$$

Simple English: Changing only one relative phase transformed $|+\rangle$ into an orthogonal state $|-\rangle$.
Relative phase becomes observable through interference and basis changes.

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9. Pauli $Y$: Bit Flip with Phase

The Pauli $Y$ matrix is

$$\boxed{ Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix} }.$$

Its basis action is

$$Y|0\rangle=i|1\rangle,$$ $$Y|1\rangle=-i|0\rangle.$$

So $Y$ flips the basis state and adds a phase.

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10. Deriving $Y=iXZ$

Compute

$$XZ = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$$

Therefore,

$$XZ = \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}.$$

Multiply by $i$:

$$iXZ = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}.$$

Thus,

$$\boxed{Y=iXZ}.$$

Also,

$$ZX = - XZ,$$

so

$$XZ=-ZX.$$

Therefore,

$$\boxed{Y=iXZ=-iZX}.$$

Simple English: $Y$ combines a bit flip and a phase change.
This also shows that $X$ and $Z$ do not commute.

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11. Pauli Matrices Are Hermitian

A matrix $A$ is Hermitian when

$$A^\dagger=A.$$

For $X$,

$$X^\dagger=X.$$

For $Z$,

$$Z^\dagger=Z.$$

For $Y$,

$$Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}.$$

Take the complex conjugate:

$$\overline{Y} = \begin{pmatrix} 0&i\\ -i&0 \end{pmatrix}.$$

Then transpose:

$$Y^\dagger = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix} = Y.$$

Therefore,

$$\boxed{ X^\dagger=X, \qquad Y^\dagger=Y, \qquad Z^\dagger=Z }.$$

Simple English: Because the Pauli matrices are Hermitian, they can represent observables.

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12. Pauli Matrices Are Also Unitary

First,

$$X^2 = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}^2 = I.$$

Similarly,

$$Y^2=I, \qquad Z^2=I.$$

Since every Pauli matrix $P$ is Hermitian,

$$P^\dagger=P.$$

Therefore,

$$P^\dagger P = P^2 = I.$$

Hence

$$\boxed{ X,\;Y,\;Z \text{ are unitary} }.$$

Important: Hermitian does not imply unitary in general.
The Pauli matrices are unitary because they are Hermitian and satisfy $P^2=I$.

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13. Why Pauli Eigenvalues Are $\pm1$

Let $P$ be a Pauli operator.

Since

$$P^2=I,$$

suppose

$$P|\psi\rangle = \lambda|\psi\rangle.$$

Apply $P$ again:

$$P^2|\psi\rangle = P\lambda|\psi\rangle.$$

Because $\lambda$ is a scalar,

$$P^2|\psi\rangle = \lambda P|\psi\rangle.$$

Using the eigenvalue equation again,

$$P^2|\psi\rangle = \lambda^2|\psi\rangle.$$

But

$$P^2=I.$$

Hence

$$|\psi\rangle = \lambda^2|\psi\rangle.$$

Since an eigenvector is nonzero,

$$\lambda^2=1.$$

Therefore,

$$\boxed{\lambda=\pm1}.$$

Simple English: Because applying a Pauli operator twice gives the identity, its eigenvalues can only square to $1$.

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14. The $Z$-Basis

We already know

$$Z|0\rangle=+|0\rangle,$$ $$Z|1\rangle=-|1\rangle.$$

Therefore,

$$\boxed{ Z\text{-basis} = \{|0\rangle,|1\rangle\} }.$$

The corresponding eigenvalues are

$$+1,\quad -1.$$

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15. Deriving the $X$-Eigenbasis

Let

$$|v\rangle = \begin{pmatrix} a\\ b \end{pmatrix}.$$

Solve

$$X|v\rangle = \lambda|v\rangle.$$

The left-hand side is

$$X|v\rangle = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} b\\ a \end{pmatrix}.$$

The right-hand side is

$$\lambda|v\rangle = \begin{pmatrix} \lambda a\\ \lambda b \end{pmatrix}.$$

Therefore,

$$b=\lambda a, \qquad a=\lambda b.$$

15.1 Eigenvalue $+1$

If

$$\lambda=1,$$

then

$$b=a.$$

So an eigenvector is

$$\begin{pmatrix} 1\\ 1 \end{pmatrix}.$$

Its norm is

$$\sqrt{1^2+1^2} = \sqrt2.$$

Therefore the normalized eigenvector is

$$\boxed{ |+\rangle = \frac1{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix} = \frac{|0\rangle+|1\rangle}{\sqrt2} }.$$

15.2 Eigenvalue $-1$

If

$$\lambda=-1,$$

then

$$b=-a.$$

A normalized eigenvector is

$$\boxed{ |-\rangle = \frac1{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix} = \frac{|0\rangle-|1\rangle}{\sqrt2} }.$$

Thus,

$$\boxed{ X|+\rangle=+|+\rangle },$$ $$\boxed{ X|-\rangle=-|-\rangle }.$$

Simple English: The $X$-basis is the eigenbasis of the Pauli $X$ operator.

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16. Hadamard Gate

The Hadamard gate is

$$\boxed{ H = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} }.$$

Apply it to $|0\rangle$:

$$\begin{aligned} H|0\rangle &= \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}\\ &= \frac1{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}\\ &= |+\rangle. \end{aligned}$$

So

$$\boxed{H|0\rangle=|+\rangle}.$$

Similarly,

$$\boxed{H|1\rangle=|-\rangle}.$$

Therefore,

$$\boxed{ H: Z\text{-basis} \longleftrightarrow X\text{-basis} }.$$

Simple English: Hadamard is more precisely a basis-change gate than merely a “superposition maker.”

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17. Proof That $H^2=I$

Compute

$$H^2 = \frac12 \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.$$

The entries are

$$1\cdot1+1\cdot1=2,$$ $$1\cdot1+1\cdot(-1)=0,$$ $$1\cdot1+(-1)\cdot1=0,$$ $$1\cdot1+(-1)(-1)=2.$$

Therefore,

$$H^2 = \frac12 \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix} = I.$$

Hence

$$\boxed{H^2=I},$$

and therefore

$$\boxed{H^{-1}=H}.$$

So

$$H|+\rangle=|0\rangle,$$ $$H|-\rangle=|1\rangle.$$

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18. Measuring in the $X$-Basis

Suppose

$$|\psi\rangle = \alpha|+\rangle+\beta|-\rangle.$$

Apply $H$:

$$\begin{aligned} H|\psi\rangle &= \alpha H|+\rangle+\beta H|-\rangle\\ &= \alpha|0\rangle+\beta|1\rangle. \end{aligned}$$

Now measure in the computational basis.

Therefore,

$$\boxed{ X\text{-basis measurement} = H + Z\text{-basis measurement} }.$$

Simple English: Rotate the $X$-basis into the computational basis, then perform the usual measurement.

Worked Example

Take

$$|\psi\rangle = \frac{\sqrt3}{2}|+\rangle + \frac12|-\rangle.$$

Applying $H$,

$$H|\psi\rangle = \frac{\sqrt3}{2}|0\rangle + \frac12|1\rangle.$$

Therefore,

$$P(+)=\frac34, \qquad P(-)=\frac14.$$

The $X$-basis probabilities become ordinary computational-basis probabilities after $H$.

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19. Phase Gate $S$

The $S$ gate is

$$\boxed{ S= \begin{pmatrix} 1&0\\ 0&i \end{pmatrix} }.$$

Since

$$i=e^{i\pi/2},$$

we have

$$S|0\rangle=|0\rangle,$$ $$S|1\rangle=i|1\rangle.$$

Therefore,

$$S(\alpha|0\rangle+\beta|1\rangle) = \alpha|0\rangle+i\beta|1\rangle.$$

So $S$ changes the relative phase by

$$\boxed{\pi/2}.$$

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20. $T$ Gate

The $T$ gate is

$$\boxed{ T= \begin{pmatrix} 1&0\\ 0&e^{i\pi/4} \end{pmatrix} }.$$

Thus,

$$T|0\rangle=|0\rangle,$$ $$T|1\rangle=e^{i\pi/4}|1\rangle.$$

For a general state,

$$T( \alpha|0\rangle+\beta|1\rangle ) = \alpha|0\rangle + e^{i\pi/4}\beta|1\rangle.$$

So the relative phase changes by

$$\boxed{\pi/4}.$$

The phase gates satisfy

$$\boxed{T^2=S},$$

because

$$\left(e^{i\pi/4}\right)^2 = e^{i\pi/2} = i.$$

Similarly,

$$\boxed{S^2=Z},$$

because

$$i^2=-1.$$

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21. Gate Composition

Suppose $U_1$ acts first:

$$|\psi_1\rangle = U_1|\psi\rangle.$$

Then $U_2$ acts:

$$|\psi_2\rangle = U_2|\psi_1\rangle.$$

Substitute the first equation:

$$|\psi_2\rangle = U_2(U_1|\psi\rangle).$$

By associativity,

$$|\psi_2\rangle = (U_2U_1)|\psi\rangle.$$

Therefore,

$$\boxed{ U_{\mathrm{total}} = U_2U_1 }.$$

Simple English: If $U_1$ happens first and $U_2$ happens second, the matrix product is $U_2U_1$.
The rightmost operator acts first.

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22. Gate Order Matters: $HZ\neq ZH$

Compute

$$HZ = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$$

Therefore,

$$\boxed{ HZ = \frac1{\sqrt2} \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix} }.$$

Now reverse the order:

$$ZH = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.$$

Hence

$$\boxed{ ZH = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix} }.$$

Thus,

$$\boxed{HZ\neq ZH}.$$

State Example

Start from $|0\rangle$.

First $Z$, then $H$:

$$HZ|0\rangle = H|0\rangle = |+\rangle.$$

First $H$, then $Z$:

$$ZH|0\rangle = Z|+\rangle = |-\rangle.$$

Therefore,

$$\boxed{ HZ|0\rangle=|+\rangle },$$

while

$$\boxed{ ZH|0\rangle=|-\rangle }.$$

Simple English: Quantum gate order is part of the computation.
Swapping two gates can produce a completely different state.

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23. Two-Qubit State Space

For two qubits,

$$\mathcal H = \mathbb C^2\otimes\mathbb C^2.$$

The computational basis is

$$|00\rangle,\quad |01\rangle,\quad |10\rangle,\quad |11\rangle.$$

For product states,

$$|a\rangle\otimes|b\rangle$$

describes the combined two-qubit system.

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24. Applying a Gate to Only One Qubit

The tensor-product rule is

$$\boxed{ (A\otimes B) (|a\rangle\otimes|b\rangle) = A|a\rangle\otimes B|b\rangle }.$$

To apply $X$ only to the first qubit,

$$X\otimes I.$$

For example,

$$\begin{aligned} (X\otimes I)|00\rangle &= X|0\rangle\otimes I|0\rangle\\ &= |1\rangle\otimes|0\rangle\\ &= |10\rangle. \end{aligned}$$

Therefore,

$$\boxed{ (X\otimes I)|00\rangle = |10\rangle }.$$

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25. Deriving the Matrix $X\otimes I$

We have

$$X= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}, \qquad I= \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$$

By the Kronecker product,

$$X\otimes I = \begin{pmatrix} 0I&1I\\ 1I&0I \end{pmatrix}.$$

Therefore,

$$\boxed{ X\otimes I = \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix} }.$$

Simple English: Tensor products let us extend a local one-qubit operation to the full multi-qubit state space.

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26. Controlled-NOT Gate

CNOT has

• a control qubit,
• a target qubit.

Its basis action is

$$|00\rangle\rightarrow|00\rangle,$$ $$|01\rangle\rightarrow|01\rangle,$$ $$|10\rangle\rightarrow|11\rangle,$$ $$|11\rangle\rightarrow|10\rangle.$$

If the control is $c$ and the target is $t$,

$$\boxed{ |c,t\rangle \longrightarrow |c,t\oplus c\rangle },$$

where $\oplus$ is XOR.

Simple English: If the control is $1$, flip the target.
If the control is $0$, leave the target unchanged.

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27. Building the CNOT Matrix from Basis Images

Use the ordered basis

$$|00\rangle,\; |01\rangle,\; |10\rangle,\; |11\rangle.$$

A linear operator is determined by its action on the basis.

For a matrix $M$,

$$Me_j=m_j,$$

where $m_j$ is the $j$-th column.

Therefore:

$$|00\rangle\rightarrow|00\rangle$$

gives column 1,

$$\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}.$$ $$|01\rangle\rightarrow|01\rangle$$

gives column 2,

$$\begin{pmatrix} 0\\1\\0\\0 \end{pmatrix}.$$ $$|10\rangle\rightarrow|11\rangle$$

gives column 3,

$$\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix}.$$ $$|11\rangle\rightarrow|10\rangle$$

gives column 4,

$$\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}.$$

Thus,

$$\boxed{ \operatorname{CNOT} = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix} }.$$

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28. Why CNOT Is Unitary

The columns of the CNOT matrix are

$$\begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \quad \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \quad \begin{pmatrix}0\\0\\0\\1\end{pmatrix}, \quad \begin{pmatrix}0\\0\\1\\0\end{pmatrix}.$$

These form an orthonormal basis.

Therefore,

$$\boxed{ \operatorname{CNOT}^\dagger \operatorname{CNOT} = I }.$$

Hence CNOT is unitary.

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29. Why $\operatorname{CNOT}^2=I$

Apply CNOT twice.

For example,

$$|10\rangle \rightarrow |11\rangle \rightarrow |10\rangle.$$

Similarly, every computational-basis state returns to itself after two applications.

Therefore,

$$\boxed{ \operatorname{CNOT}^2=I }.$$

Hence,

$$\boxed{ \operatorname{CNOT}^{-1} = \operatorname{CNOT} }.$$

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30. The Bell Circuit

Start from

$$|\psi_0\rangle = |00\rangle.$$

The circuit is

1. Apply $H$ to the first qubit.
2. Apply CNOT.

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30.1 Step 1: Apply $H\otimes I$

$$(H\otimes I)|00\rangle.$$

Since

$$|00\rangle = |0\rangle\otimes|0\rangle,$$

we have

$$\begin{aligned} (H\otimes I) (|0\rangle\otimes|0\rangle) &= H|0\rangle\otimes I|0\rangle\\ &= \frac{|0\rangle+|1\rangle}{\sqrt2} \otimes |0\rangle. \end{aligned}$$

Distribute the tensor product:

$$\begin{aligned} |\psi_1\rangle &= \frac1{\sqrt2} \left( |0\rangle\otimes|0\rangle + |1\rangle\otimes|0\rangle \right)\\ &= \boxed{ \frac{|00\rangle+|10\rangle}{\sqrt2} }. \end{aligned}$$

This is still a product state:

$$\boxed{ |\psi_1\rangle = |+\rangle\otimes|0\rangle }.$$

Simple English: Hadamard creates a superposition on the first qubit, but the two qubits are still separable.

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30.2 Step 2: Apply CNOT

Now apply CNOT:

$$\operatorname{CNOT} \frac{|00\rangle+|10\rangle}{\sqrt2}.$$

By linearity,

$$\begin{aligned} |\psi_2\rangle &= \frac1{\sqrt2} \left( \operatorname{CNOT}|00\rangle + \operatorname{CNOT}|10\rangle \right)\\ &= \frac1{\sqrt2} \left( |00\rangle+|11\rangle \right). \end{aligned}$$

Therefore,

$$\boxed{ |\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2} }.$$

This is the Bell state $|\Phi^+\rangle$.

The complete state evolution is

$$\boxed{ |00\rangle \xrightarrow{H\otimes I} \frac{|00\rangle+|10\rangle}{\sqrt2} \xrightarrow{\mathrm{CNOT}} \frac{|00\rangle+|11\rangle}{\sqrt2} }.$$

Simple English: A local Hadamard creates a superposition.
CNOT then correlates the two computational-basis branches.

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31. Numerical View of the Bell Circuit

Using basis order

$$|00\rangle,\; |01\rangle,\; |10\rangle,\; |11\rangle,$$

the initial amplitude vector is

$$|\psi_0\rangle = \begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix}.$$

After $H\otimes I$,

$$|\psi_1\rangle = \begin{pmatrix} 1/\sqrt2\\ 0\\ 1/\sqrt2\\ 0 \end{pmatrix}.$$

After CNOT,

$$|\psi_2\rangle = \begin{pmatrix} 1/\sqrt2\\ 0\\ 0\\ 1/\sqrt2 \end{pmatrix}.$$

Therefore,

$$P(00)=\frac12,$$ $$P(11)=\frac12,$$

and

$$P(01)=P(10)=0.$$

Simple English: The final state has probability only on $00$ and $11$, with equal magnitude amplitudes.

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32. What Has Changed?

Initially,

$$|00\rangle = |0\rangle\otimes|0\rangle$$

is clearly a product state.

After Hadamard,

$$\frac{|00\rangle+|10\rangle}{\sqrt2} = \left( \frac{|0\rangle+|1\rangle}{\sqrt2} \right) \otimes |0\rangle,$$

so the state is still separable.

After CNOT,

$$\boxed{ |\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2} }.$$

This state is the Bell state that motivates the next topic: entanglement.

Simple English: The important transition is
product state $\rightarrow$ local superposition $\rightarrow$ Bell state.

A formal proof that this Bell state cannot be written as a product state belongs naturally to the next discussion on entanglement.

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33. Summary of the Main Mathematical Results

$$\boxed{ X|0\rangle=|1\rangle, \qquad X|1\rangle=|0\rangle }$$ $$\boxed{ Z|+\rangle=|-\rangle }$$ $$\boxed{ X^2=Y^2=Z^2=I }$$ $$\boxed{ P^2=I \Rightarrow \lambda=\pm1 }$$ $$\boxed{ |+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt2}, \qquad |-\rangle = \frac{|0\rangle-|1\rangle}{\sqrt2} }$$ $$\boxed{ H^2=I }$$ $$\boxed{ HZ\neq ZH }$$ $$\boxed{ (A\otimes B) (|a\rangle\otimes|b\rangle) = A|a\rangle\otimes B|b\rangle }$$ $$\boxed{ \operatorname{CNOT} = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{pmatrix} }$$ $$\boxed{ |00\rangle \xrightarrow{H\otimes I} \frac{|00\rangle+|10\rangle}{\sqrt2} \xrightarrow{\operatorname{CNOT}} \frac{|00\rangle+|11\rangle}{\sqrt2} }$$

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34. Final Perspective

The mathematical progression is

$$\boxed{ \text{Pauli operators} \rightarrow \text{phase and basis control} \rightarrow \text{Hadamard} \rightarrow \text{gate composition} \rightarrow \text{tensor-product operations} \rightarrow \text{CNOT} }.$$

The most important conceptual lesson is:

Quantum gates do not directly manipulate classical probabilities.
They transform complex amplitudes and relative phases, and those amplitudes later determine measurement statistics.

The Bell circuit is the natural endpoint:

$$\boxed{ |\Phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt2} }.$$

This state leads directly to the next topic:

entanglement, separability, and quantum correlations.