Step 1 — Convert each state to a Bloch vector. A single-qubit state in standard form:
$$\ket{\psi} = \cos\frac{\theta}{2}\ket{0} + e^{i\varphi}\sin\frac{\theta}{2}\ket{1}$$
maps to the Bloch vector:
$$\vec{v} = (\sin\theta\cos\varphi,\;\sin\theta\sin\varphi,\;\cos\theta)$$
Step 2 — Dot product. Compute the dot product of the two Bloch vectors:
$$\vec{v}_1 \cdot \vec{v}_2 = x_1 x_2 + y_1 y_2 + z_1 z_2$$
Step 3 — Angle. The angle between the states on the Bloch sphere is:
$$\alpha = \arccos(\vec{v}_1 \cdot \vec{v}_2)$$
You can skip Bloch vectors entirely. If you can compute the inner product \(\braket{\psi_1|\psi_2}\), the Bloch sphere angle is:
$$\alpha = \arccos\!\bigl(2|\braket{\psi_1|\psi_2}|^2 - 1\bigr)$$
This works because \(|\braket{\psi_1|\psi_2}|^2 = \cos^2(\alpha/2)\), so \(2\cos^2(\alpha/2) - 1 = \cos\alpha\).
\(\ket{0} \;\to\; (0, 0, 1)\) — north pole
\(\ket{1} \;\to\; (0, 0, -1)\) — south pole
\(\ket{+} \;\to\; (1, 0, 0)\) — positive \(x\)
\(\ket{-} \;\to\; (-1, 0, 0)\) — negative \(x\)
\(\ket{+i} \;\to\; (0, 1, 0)\) — positive \(y\)
\(\ket{-i} \;\to\; (0, -1, 0)\) — negative \(y\)
Opposite states on the sphere are orthogonal quantum states. Adjacent axes are at \(90°\) on the Bloch sphere (\(= 45°\) Hilbert space angle).
These appear constantly when extracting \(\theta\) and \(\varphi\) from state coefficients:
\(e^{i\pi/6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i\)
\(e^{i\pi/4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i = \frac{1+i}{\sqrt{2}}\)
\(e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i\)
\(e^{i\pi/2} = i\)
General rule: \(e^{i\varphi} = \cos\varphi + i\sin\varphi\). To identify \(\varphi\) from a complex number \(a + bi\), compute \(\varphi = \text{atan2}(b, a)\).
Find the angle between \(\ket{0}\) and \(\ket{+}\) on the Bloch sphere.
Method A — Bloch vectors
$$\ket{0} \;\to\; \vec{v}_1 = (0, 0, 1) \qquad \ket{+} \;\to\; \vec{v}_2 = (1, 0, 0)$$
$$\vec{v}_1 \cdot \vec{v}_2 = 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0$$
$$\alpha = \arccos(0) = \frac{\pi}{2} = 90°$$
Method B — Inner product shortcut
$$\braket{0|+} = \bra{0}\!\left(\frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}\right) = \frac{1}{\sqrt{2}}$$
$$|\braket{0|+}|^2 = \frac{1}{2}$$
$$\alpha = \arccos\!\left(2 \cdot \frac{1}{2} - 1\right) = \arccos(0) = \frac{\pi}{2} = 90°$$
Both methods agree: \(90°\). The north pole and the positive \(x\)-axis are perpendicular on the Bloch sphere.
Find the Bloch vector of \(\ket{\psi} = \frac{1+i}{2}\ket{0} + \frac{1-i}{2}\ket{1}\), then find the angle between \(\ket{\psi}\) and \(\ket{0}\).
Step 1 — Remove global phase
The coefficient of \(\ket{0}\) is \(\frac{1+i}{2}\). Its magnitude is \(\left|\frac{1+i}{2}\right| = \frac{\sqrt{2}}{2}\), and its phase is \(e^{i\pi/4}\).
Multiply the entire state by \(e^{-i\pi/4}\) to make the \(\ket{0}\) coefficient real and positive:
$$e^{-i\pi/4}\ket{\psi} = \frac{\sqrt{2}}{2}\ket{0} + e^{-i\pi/4} \cdot \frac{1-i}{2}\ket{1}$$
Now compute \(e^{-i\pi/4} \cdot \frac{1-i}{2}\). Note \(\frac{1-i}{\sqrt{2}} = e^{-i\pi/4}\), so \(\frac{1-i}{2} = \frac{e^{-i\pi/4}}{\sqrt{2}}\).
$$e^{-i\pi/4} \cdot \frac{e^{-i\pi/4}}{\sqrt{2}} = \frac{e^{-i\pi/2}}{\sqrt{2}} = \frac{-i}{\sqrt{2}}$$
The state in standard form:
$$\ket{\psi} \sim \frac{1}{\sqrt{2}}\ket{0} + \frac{(-i)}{\sqrt{2}}\ket{1} = \frac{1}{\sqrt{2}}\ket{0} + e^{-i\pi/2}\frac{1}{\sqrt{2}}\ket{1}$$
Step 2 — Read off parameters
Comparing with \(\cos\frac{\theta}{2}\ket{0} + e^{i\varphi}\sin\frac{\theta}{2}\ket{1}\):
$$\cos\frac{\theta}{2} = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{2} \qquad \varphi = -\frac{\pi}{2}$$
Step 3 — Bloch vector
$$\vec{v} = (\sin\frac{\pi}{2}\cos(-\frac{\pi}{2}),\;\sin\frac{\pi}{2}\sin(-\frac{\pi}{2}),\;\cos\frac{\pi}{2}) = (0, -1, 0)$$
This is \(\ket{-i}\), on the negative \(y\)-axis.
Step 4 — Angle to |0⟩
$$\vec{v}_1 = (0, 0, 1) \qquad \vec{v}_2 = (0, -1, 0)$$
$$\vec{v}_1 \cdot \vec{v}_2 = 0 \implies \alpha = \arccos(0) = 90°$$
Find the angle between \(\ket{\psi_1} = \frac{\sqrt{3}}{2}\ket{0} + \frac{i}{2}\ket{1}\) and \(\ket{\psi_2} = \frac{1}{2}\ket{0} + \frac{\sqrt{3}}{2}e^{i\pi/4}\ket{1}\).
State 1 — Extract Bloch vector
The \(\ket{0}\) coefficient is \(\frac{\sqrt{3}}{2}\), already real and positive. Compare with \(\cos\frac{\theta_1}{2}\ket{0} + e^{i\varphi_1}\sin\frac{\theta_1}{2}\ket{1}\):
$$\cos\frac{\theta_1}{2} = \frac{\sqrt{3}}{2} \implies \frac{\theta_1}{2} = \frac{\pi}{6} \implies \theta_1 = \frac{\pi}{3}$$
The \(\ket{1}\) coefficient is \(\frac{i}{2} = \frac{1}{2}e^{i\pi/2}\), so \(\sin\frac{\theta_1}{2} = \frac{1}{2}\) (consistent) and \(\varphi_1 = \frac{\pi}{2}\).
$$\vec{v}_1 = \left(\sin\frac{\pi}{3}\cos\frac{\pi}{2},\;\sin\frac{\pi}{3}\sin\frac{\pi}{2},\;\cos\frac{\pi}{3}\right) = \left(0,\;\frac{\sqrt{3}}{2},\;\frac{1}{2}\right)$$
State 2 — Extract Bloch vector
The \(\ket{0}\) coefficient is \(\frac{1}{2}\), already real and positive.
$$\cos\frac{\theta_2}{2} = \frac{1}{2} \implies \frac{\theta_2}{2} = \frac{\pi}{3} \implies \theta_2 = \frac{2\pi}{3}$$
The \(\ket{1}\) coefficient is \(\frac{\sqrt{3}}{2}e^{i\pi/4}\), so \(\sin\frac{\theta_2}{2} = \frac{\sqrt{3}}{2}\) (consistent) and \(\varphi_2 = \frac{\pi}{4}\).
$$\vec{v}_2 = \left(\sin\frac{2\pi}{3}\cos\frac{\pi}{4},\;\sin\frac{2\pi}{3}\sin\frac{\pi}{4},\;\cos\frac{2\pi}{3}\right)$$
$$= \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2},\;\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2},\;-\frac{1}{2}\right) = \left(\frac{\sqrt{6}}{4},\;\frac{\sqrt{6}}{4},\;-\frac{1}{2}\right)$$
Dot product
$$\vec{v}_1 \cdot \vec{v}_2 = 0 \cdot \frac{\sqrt{6}}{4} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{6}}{4} + \frac{1}{2} \cdot \left(-\frac{1}{2}\right)$$
$$= \frac{\sqrt{18}}{8} - \frac{1}{4} = \frac{3\sqrt{2}}{8} - \frac{1}{4} = \frac{3\sqrt{2} - 2}{8}$$
Angle
$$\alpha = \arccos\!\left(\frac{3\sqrt{2}-2}{8}\right)$$
Numerically: \(\frac{3\sqrt{2}-2}{8} = \frac{4.243-2}{8} \approx 0.280\), so \(\alpha \approx \arccos(0.280) \approx 73.7°\).
Verify with the shortcut: \(\braket{\psi_1|\psi_2} = \frac{\sqrt{3}}{2}\cdot\frac{1}{2} + (-\frac{i}{2})\cdot\frac{\sqrt{3}}{2}e^{i\pi/4} = \frac{\sqrt{3}}{4} - \frac{i\sqrt{3}}{4}e^{i\pi/4}\). Computing \(|\braket{\psi_1|\psi_2}|^2\) and applying \(\arccos(2|\cdot|^2 - 1)\) gives the same answer.
Find the angle between \(\ket{+}\) and \(\ket{+i}\) on the Bloch sphere.
Bloch vectors: \(\ket{+} \to (1,0,0)\), \(\ket{+i} \to (0,1,0)\).
$$\vec{v}_1 \cdot \vec{v}_2 = 1\cdot 0 + 0\cdot 1 + 0\cdot 0 = 0$$
$$\alpha = \arccos(0) = \boxed{90°}$$
Shortcut check: \(\braket{+|{+i}} = \frac{1}{2}(1+i)\), so \(|\braket{+|{+i}}|^2 = \frac{1}{2}\). Then \(\arccos(2 \cdot \frac{1}{2}-1) = \arccos(0) = 90°\). Agrees.
Find the angle between \(\ket{0}\) and \(\ket{1}\) on the Bloch sphere.
Bloch vectors: \(\ket{0} \to (0,0,1)\), \(\ket{1} \to (0,0,-1)\).
$$\vec{v}_1 \cdot \vec{v}_2 = 0+0+(-1) = -1$$
$$\alpha = \arccos(-1) = \boxed{180°}$$
Orthogonal quantum states are antipodal on the Bloch sphere — always \(180°\) apart.
Find the angle between \(\ket{\psi_1} = \ket{0}\) and \(\ket{\psi_2} = \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}\ket{1}\) on the Bloch sphere.
\(\ket{\psi_2}\): the \(\ket{0}\) coefficient is real and positive. \(\cos\frac{\theta}{2} = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}\). The \(\ket{1}\) coefficient is real positive, so \(\varphi = 0\).
$$\vec{v}_2 = \left(\sin\frac{\pi}{3}\cos 0,\;\sin\frac{\pi}{3}\sin 0,\;\cos\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2},\;0,\;\frac{1}{2}\right)$$
$$\vec{v}_1 \cdot \vec{v}_2 = 0 \cdot \frac{\sqrt{3}}{2} + 0 \cdot 0 + 1 \cdot \frac{1}{2} = \frac{1}{2}$$
$$\alpha = \arccos\!\left(\frac{1}{2}\right) = \boxed{60°}$$
Shortcut: \(\braket{0|\psi_2} = \frac{\sqrt{3}}{2}\), so \(|\braket{0|\psi_2}|^2 = \frac{3}{4}\). Then \(\arccos(2\cdot\frac{3}{4}-1) = \arccos(\frac{1}{2}) = 60°\). Agrees.
Find the angle between \(\ket{\psi_1} = \frac{1}{\sqrt{2}}\ket{0} + \frac{i}{\sqrt{2}}\ket{1}\) and \(\ket{\psi_2} = \frac{1}{\sqrt{2}}\ket{0} - \frac{i}{\sqrt{2}}\ket{1}\) on the Bloch sphere.
\(\ket{\psi_1}\): \(\theta_1 = \frac{\pi}{2}\), \(\varphi_1 = \frac{\pi}{2}\) → \(\vec{v}_1 = (0, 1, 0)\). This is \(\ket{+i}\).
\(\ket{\psi_2}\): \(\theta_2 = \frac{\pi}{2}\), \(\varphi_2 = -\frac{\pi}{2}\) → \(\vec{v}_2 = (0, -1, 0)\). This is \(\ket{-i}\).
$$\vec{v}_1 \cdot \vec{v}_2 = 0+(-1)+0 = -1$$
$$\alpha = \arccos(-1) = \boxed{180°}$$
These are orthogonal states (\(\braket{\psi_1|\psi_2} = 0\)), so they sit at opposite poles. Consistent with the rule: orthogonal \(\Leftrightarrow 180°\) on the Bloch sphere.
Find the angle between \(\ket{\psi_1} = \frac{1}{2}\ket{0} + \frac{\sqrt{3}}{2}e^{i\pi/3}\ket{1}\) and \(\ket{\psi_2} = \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}e^{i\pi/6}\ket{1}\) on the Bloch sphere.
State 1: \(\cos\frac{\theta_1}{2} = \frac{1}{2} \implies \theta_1 = \frac{2\pi}{3}\), \(\varphi_1 = \frac{\pi}{3}\).
$$\vec{v}_1 = \left(\sin\frac{2\pi}{3}\cos\frac{\pi}{3},\;\sin\frac{2\pi}{3}\sin\frac{\pi}{3},\;\cos\frac{2\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\cdot\frac{1}{2},\;\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2},\;-\frac{1}{2}\right) = \left(\frac{\sqrt{3}}{4},\;\frac{3}{4},\;-\frac{1}{2}\right)$$
State 2: \(\cos\frac{\theta_2}{2} = \frac{\sqrt{3}}{2} \implies \theta_2 = \frac{\pi}{3}\), \(\varphi_2 = \frac{\pi}{6}\).
$$\vec{v}_2 = \left(\sin\frac{\pi}{3}\cos\frac{\pi}{6},\;\sin\frac{\pi}{3}\sin\frac{\pi}{6},\;\cos\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2},\;\frac{\sqrt{3}}{2}\cdot\frac{1}{2},\;\frac{1}{2}\right) = \left(\frac{3}{4},\;\frac{\sqrt{3}}{4},\;\frac{1}{2}\right)$$
Dot product:
$$\vec{v}_1 \cdot \vec{v}_2 = \frac{\sqrt{3}}{4}\cdot\frac{3}{4} + \frac{3}{4}\cdot\frac{\sqrt{3}}{4} + \left(-\frac{1}{2}\right)\cdot\frac{1}{2}$$
$$= \frac{3\sqrt{3}}{16} + \frac{3\sqrt{3}}{16} - \frac{1}{4} = \frac{6\sqrt{3}}{16} - \frac{1}{4} = \frac{3\sqrt{3}}{8} - \frac{1}{4} = \frac{3\sqrt{3}-2}{8}$$
Angle:
$$\alpha = \arccos\!\left(\frac{3\sqrt{3}-2}{8}\right)$$
Numerically: \(\frac{3\sqrt{3}-2}{8} = \frac{5.196-2}{8} \approx 0.399\), so \(\alpha \approx \arccos(0.399) \approx \boxed{66.5°}\).