CHAPTER 04

Gates, observables, and the Bloch sphere

Learning goals. Multiply gate matrices, distinguish active transformations from coordinates, derive Bloch coordinates, and understand rotations and noncommuting operations.

4.1 Matrices act on vectors

A matrix is determined by what it does to basis vectors. Its first column is the output for $|0\rangle$; its second is the output for $|1\rangle$. Linearity then fixes the output for every superposition:

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

For $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ the result is $(a\alpha+b\beta)|0\rangle+(c\alpha+d\beta)|1\rangle$. Every row forms one output amplitude. A valid closed-system gate preserves all inner products, so it satisfies $U^\dagger U=I$. Such a matrix is unitary, with inverse $U^\dagger$.

The Pauli gates and Hadamard are

$$X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},$$ $$H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

X swaps the amplitudes, Z changes their relative sign, and Y combines a swap with phases. H maps $|0\rangle$ to $|+\rangle$ and $|1\rangle$ to $|-\rangle$. These matrices are Hermitian as well as unitary; not every gate is Hermitian.

The phase gates are

$$S=\begin{pmatrix}1&0\\0&i\end{pmatrix},\qquad T=\begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix},\qquad S=T^2.$$

S has inverse $S^\dagger=\operatorname{diag}(1,-i)$, not S itself.

4.2 Gate order is matrix order in reverse time

If U acts first and V second, the output is $VU|\psi\rangle$. The rightmost matrix acts first. A circuit is conventionally read left to right, which is why translating circuits into matrix products requires care.

Worked example. Apply H to $\alpha|0\rangle+\beta|1\rangle$:

$$H|\psi\rangle=\frac{(\alpha+\beta)|0\rangle+(\alpha-\beta)|1\rangle}{\sqrt2}.$$

Applying H again gives

$$H^2|\psi\rangle =\frac{(\alpha+\beta+\alpha-\beta)|0\rangle +(\alpha+\beta-\alpha+\beta)|1\rangle}{2} =\alpha|0\rangle+\beta|1\rangle.$$

The cancellations show $H^2=I$. They also explain interference: the first H creates alternatives, and the second recombines them.

By contrast, $XZ=-ZX$. For instance, $XZ|0\rangle=|1\rangle$ but $ZX|0\rangle=-|1\rangle$. The minus sign is a global phase when these are complete single-qubit operations. In controlled versions, a phase confined to one control branch becomes relative and can matter. Never discard a phase before checking its role in a larger circuit.

4.3 Implementing a measurement basis

To measure X with a Z detector, apply H first, because $\langle0|H=\langle+|$ and $\langle1|H=\langle-|$. To measure Y, apply $S^\dagger$ and then H, so the total matrix is $HS^\dagger$:

$$\langle0|HS^\dagger =\frac{\langle0|-i\langle1|}{\sqrt2}=\langle+i|.$$

Using SH instead produces a different operation. Writing the bra of the actual measurement is an effective way to verify a basis-rotation circuit.

4.4 Bloch coordinates from amplitudes

Remove a global phase so the first nonzero coefficient is real. Every pure qubit can then be written

$$|\psi\rangle=\cos\frac\theta2|0\rangle+ e^{i\phi}\sin\frac\theta2|1\rangle,\quad 0\le\theta\le\pi.$$

Define $x=\langle X\rangle$, $y=\langle Y\rangle$, and $z=\langle Z\rangle$. Matrix multiplication gives

$$x=\alpha^*\beta+\beta^*\alpha=2\operatorname{Re}(\alpha^*\beta),$$ $$y=-i\alpha^*\beta+i\beta^*\alpha=2\operatorname{Im}(\alpha^*\beta),$$ $$z=|\alpha|^2-|\beta|^2.$$

Substituting the angular form yields

$$(x,y,z)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).$$

Its length is one. The north and south poles are $|0\rangle,|1\rangle$; the positive x and y directions are $|+\rangle,|+i\rangle$. The sphere represents pure qubit states modulo global phase, not the physical path or shape of a particle.

For $(2|0\rangle+3i|1\rangle)/\sqrt{13}$,

$$(x,y,z)=(0,12/13,-5/13),\quad x^2+y^2+z^2=(144+25)/169=1.$$

Thus equal X probabilities do not place the state at the sphere’s center. They only imply its x coordinate is zero.

4.5 Rotations and generators

For a unit vector $\boldsymbol n$, let $A=n_xX+n_yY+n_zZ$. Since $A^2=I$, the exponential series separates into even and odd powers:

$$e^{-i\theta A/2} =I\cos(\theta/2)-iA\sin(\theta/2).$$

This is $R_{\boldsymbol n}(\theta)$, rotating the Bloch vector by angle $\theta$ about $\boldsymbol n$. For example,

$$R_y(\theta)= \begin{pmatrix}\cos(\theta/2)&-\sin(\theta/2)\\ \sin(\theta/2)&\cos(\theta/2)\end{pmatrix},$$ $$R_z(\theta)=\operatorname{diag}(e^{-i\theta/2},e^{i\theta/2}).$$

The half-angle is necessary: a $2\pi$ rotation gives $-I$, physically the same pure state but a different vector representative. H corresponds to a half-turn about $(x+z)/\sqrt2$, up to global phase; it is not simply a quarter-turn about y.

4.6 Observables and uncertainty

An observable is Hermitian: $A^\dagger=A$. Its eigenvalues are real, and its orthonormal eigenvectors specify a projective measurement. A gate is unitary. Some matrices, including the Paulis, serve both roles, but “apply Z” and “measure Z” are different physical instructions.

The commutator is $[A,B]=AB-BA$. Since $[X,Y]=2iZ$, the uncertainty relation gives $\Delta X\Delta Y\ge|\langle Z\rangle|$. In $|0\rangle$, both X and Y outcomes are random, so $\Delta X=\Delta Y=1$, saturating the bound. This is a statement about measurement distributions across identically prepared copies.

4.7 Exercises