An eigenstate of a gate \(U\) is a state that the gate doesn't change direction-wise. The gate hits it, and the same state comes back out — possibly multiplied by a scalar:
$$U|\psi\rangle = \lambda|\psi\rangle$$
\(|\psi\rangle\) is an eigenstate of \(U\). The scalar \(\lambda\) is its eigenvalue.
The gate doesn't rotate the state to somewhere else on the Bloch sphere. It stays put. The only thing that can happen is multiplication by \(\lambda\).
Think of it this way: most states get moved around by a gate. But eigenstates are the fixed points — the states that are aligned with the gate's axis of rotation. A rotation around an axis doesn't move points that sit on that axis.
The eigenvalue \(\lambda\) is "just a number" — but for quantum gates, it's a very constrained number.
Quantum gates are unitary, which means they preserve norms. If \(U|\psi\rangle = \lambda|\psi\rangle\), then:
$$\|U|\psi\rangle\| = \||\psi\rangle\| \quad\text{(unitary preserves norms)}$$
But also:
$$\|U|\psi\rangle\| = \|\lambda|\psi\rangle\| = |\lambda| \cdot \||\psi\rangle\|$$
So \(|\lambda| \cdot \||\psi\rangle\| = \||\psi\rangle\|\), which forces \(|\lambda| = 1\).
What numbers have \(|\lambda| = 1\)? Exactly the points on the unit circle in the complex plane. Every such number can be written as:
$$\lambda = e^{i\theta}$$
for some angle \(\theta\). That's it — eigenvalues of unitary gates are always pure phase factors. They rotate the phase, never scale the amplitude. This is why they're called phase factors.
A gate hitting its own eigenstate can only rotate the phase. It can never make the state "bigger" or "smaller." The state's probabilities are completely unchanged.
Examples: \(\lambda = +1 = e^{i \cdot 0}\) (no phase change), \(\lambda = -1 = e^{i\pi}\) (180° phase flip), \(\lambda = i = e^{i\pi/2}\) (90° phase rotation).
A single-qubit gate is a \(2 \times 2\) matrix. A \(2 \times 2\) matrix has 2 eigenvalues (counting multiplicity), each with a corresponding eigenstate. So there are always exactly two eigenstates.
This follows from the fundamental theorem of algebra: a degree-2 characteristic polynomial has exactly 2 roots.
Unitary gates are rotations of the Bloch sphere. The two points that don't move under a rotation are the two poles of the rotation axis. Those poles are always diametrically opposite — they form an orthonormal basis.
This also follows from unitarity: eigenstates of a unitary matrix with distinct eigenvalues are always orthogonal. And orthogonal single-qubit states are antipodal on the Bloch sphere.
Since the two eigenstates are orthogonal and normalized, they form a complete basis for the single-qubit state space. Any state can be written as a superposition of a gate's two eigenstates.
This is the key to understanding what gates do: decompose the input into eigenstates, apply the eigenvalues, recombine.
The eigenvalue \(\lambda = e^{i\theta}\) is a phase factor. But whether that phase is observable depends entirely on context:
$$U|\psi\rangle = \lambda|\psi\rangle = e^{i\theta}|\psi\rangle$$
The entire state gets multiplied by \(e^{i\theta}\). This is a global phase — it multiplies every amplitude equally. Global phase has no physical effect. No measurement can detect it. The state is physically unchanged.
This is why "a gate does nothing to its eigenstates" is literally true in the physical sense.
If the input is a superposition \(|\psi\rangle = \alpha|\psi_1\rangle + \beta|\psi_2\rangle\) where \(|\psi_1\rangle\) and \(|\psi_2\rangle\) are eigenstates with eigenvalues \(\lambda_1\) and \(\lambda_2\):
$$U|\psi\rangle = \alpha\lambda_1|\psi_1\rangle + \beta\lambda_2|\psi_2\rangle$$
If \(\lambda_1 \neq \lambda_2\), the two components pick up different phases. The difference \(\lambda_1 / \lambda_2\) becomes a relative phase between the components. Relative phase IS observable — it changes interference patterns and measurement probabilities.
Every quantum gate works by this exact mechanism:
A gate only does something observable when it hits a state that is a superposition of its eigenstates with different eigenvalues. If both eigenvalues are the same, or if the input is purely one eigenstate, nothing observable happens.
| Gate | Eigenstate 1 | \(\lambda_1\) | Eigenstate 2 | \(\lambda_2\) | Axis |
|---|---|---|---|---|---|
| X | \(|+\rangle = \tfrac{1}{\sqrt{2}}\binom{1}{1}\) | \(+1\) | \(|-\rangle = \tfrac{1}{\sqrt{2}}\binom{1}{-1}\) | \(-1\) | X-axis |
| Y | \(|{+i}\rangle = \tfrac{1}{\sqrt{2}}\binom{1}{i}\) | \(+1\) | \(|{-i}\rangle = \tfrac{1}{\sqrt{2}}\binom{1}{-i}\) | \(-1\) | Y-axis |
| Z | \(|0\rangle = \binom{1}{0}\) | \(+1\) | \(|1\rangle = \binom{0}{1}\) | \(-1\) | Z-axis |
| H | \(\cos\tfrac{\pi}{8}|0\rangle + \sin\tfrac{\pi}{8}|1\rangle\) | \(+1\) | \(\sin\tfrac{\pi}{8}|0\rangle - \cos\tfrac{\pi}{8}|1\rangle\) | \(-1\) | XZ-diagonal |
| S | \(|0\rangle\) | \(+1\) | \(|1\rangle\) | \(+i\) | Z-axis |
| T | \(|0\rangle\) | \(+1\) | \(|1\rangle\) | \(e^{i\pi/4}\) | Z-axis |
Z, S, and T are all rotations around the Z-axis — they only differ in angle (180°, 90°, 45°). Since they share the same rotation axis, they share the same eigenstates: \(|0\rangle\) and \(|1\rangle\). What differs is the eigenvalue on \(|1\rangle\):
All three leave \(|0\rangle\) with eigenvalue \(+1\) (no phase change on the north pole). The south pole \(|1\rangle\) picks up a different phase rotation for each gate.
X, Y, and Z all have eigenvalues \(+1\) and \(-1\). This is because they're all 180° rotations. A half turn either leaves a pole alone (\(+1\)) or flips it (\(-1\)). There's no in-between for a half turn.
Mathematically: Pauli gates are both unitary AND Hermitian, which forces the eigenvalues to be both on the unit circle AND real. The only numbers satisfying both: \(+1\) and \(-1\).
H's eigenstates aren't any of the named states (\(|0\rangle\), \(|1\rangle\), \(|\pm\rangle\), \(|\pm i\rangle\)). We need to compute them. Since \(H^2 = I\), the eigenvalues must satisfy \(\lambda^2 = 1\), giving \(\lambda = \pm 1\).
Solve \(H|\psi\rangle = |\psi\rangle\), i.e., \((H - I)|\psi\rangle = 0\):
$$H - I = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} - 1 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} - 1 \end{pmatrix}$$
From the first row: \((\tfrac{1}{\sqrt{2}} - 1)\alpha + \tfrac{1}{\sqrt{2}}\beta = 0\), giving \(\beta = (1 - \tfrac{1}{\sqrt{2}}) \cdot \tfrac{\sqrt{2}}{1} \cdot \alpha = (\sqrt{2} - 1)\alpha\).
After normalizing: \(|\psi_{+1}\rangle = \cos\tfrac{\pi}{8}\,|0\rangle + \sin\tfrac{\pi}{8}\,|1\rangle\)
Where \(\cos(\pi/8) \approx 0.924\) and \(\sin(\pi/8) \approx 0.383\). This state sits 22.5° from the north pole on the Bloch sphere — on the great circle in the XZ-plane, tilted slightly from \(|0\rangle\) toward \(|+\rangle\).
The second eigenstate is orthogonal to the first (opposite pole on the Bloch sphere):
$$|\psi_{-1}\rangle = \sin\tfrac{\pi}{8}\,|0\rangle - \cos\tfrac{\pi}{8}\,|1\rangle$$
This sits 22.5° from the south pole, on the opposite side. Together the two eigenstates straddle the XZ-diagonal that H rotates around.
This is the cleanest demonstration of how different eigenvalues on eigenstates create an observable change. Let's trace it step by step.
Z's eigenstates are \(|0\rangle\) (with \(\lambda = +1\)) and \(|1\rangle\) (with \(\lambda = -1\)).
$$|+\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$$
\(|+\rangle\) is an equal superposition of both eigenstates.
$$Z|+\rangle = Z\left(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle\right)$$
$$= \frac{1}{\sqrt{2}} \underbrace{Z|0\rangle}_{(+1)|0\rangle} + \frac{1}{\sqrt{2}} \underbrace{Z|1\rangle}_{(-1)|1\rangle}$$
$$= \frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle$$
$$Z|+\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle = |-\rangle$$
The \(|0\rangle\) component kept its sign (\(\lambda = +1\)), but the \(|1\rangle\) component got its sign flipped (\(\lambda = -1\)). That sign difference — the relative phase of \(-1\) between the two components — turned \(|+\rangle\) into \(|-\rangle\).
The two eigenvalues \(+1\) and \(-1\) are different, so they introduced a relative phase between the \(|0\rangle\) and \(|1\rangle\) components. Before Z: the two components had the same sign (constructive superposition toward \(|+\rangle\)). After Z: opposite signs (destructive superposition — now it's \(|-\rangle\)).
On the Bloch sphere: \(|+\rangle\) sits on the equator at 0°, and \(|-\rangle\) sits at 180°. Z is a 180° rotation around the Z-axis, which moves equator points to the opposite side. The eigenstates (\(|0\rangle\) and \(|1\rangle\) at the poles) don't move — but everything on the equator gets rotated.
$$Z|0\rangle = (+1)|0\rangle = |0\rangle$$
No decomposition needed — it IS an eigenstate. The eigenvalue \(+1\) multiplies the whole thing. Global phase of \(+1\) = literally nothing. The state is completely unchanged. Z does nothing observable to \(|0\rangle\).
This is the asymmetry: a gate is invisible to its eigenstates but transforms everything else.
This connects eigenstates, eigenvalues, and the rotation formula into one picture.
A gate \(U\) rotates the Bloch sphere by angle \(\vartheta\) around some axis \(\hat{n}\). It has two eigenstates (the poles of that axis), each with its own eigenvalue:
The difference between the two eigenvalue angles is:
\((+\vartheta/2) - (-\vartheta/2) = \vartheta\)
That difference IS the Bloch sphere rotation angle. The eigenvalue angles are ±half the rotation.
Why does this matter? When a gate hits a superposition of its eigenstates, each component picks up its own eigenvalue. The relative phase between them — the thing that's actually observable — is the difference of the eigenvalue angles. And that difference equals the physical rotation angle on the Bloch sphere.
Z has eigenvalues \(+1 = e^{i \cdot 0}\) and \(-1 = e^{i\pi}\).
S has eigenvalues \(+1 = e^{i \cdot 0}\) and \(+i = e^{i\pi/2}\).
T has eigenvalues \(+1 = e^{i \cdot 0}\) and \(e^{i\pi/4}\).
The Bloch sphere rotation angle = the gap between the two eigenvalue angles. Bigger gap → bigger rotation → more observable effect on superpositions.
A quantum gate does nothing to its eigenstates and transforms everything else by creating relative phase between eigenstate components — and relative phase is the only mechanism by which a quantum gate produces an observable effect.