1.1.2 State and Representation#

Prompts

What is a quantum state? How does the abstract ket \(\vert \psi\rangle\) relate to its vector representation \((\alpha, \beta)\) in a particular basis?
Why does a qubit state require only two real parameters (\(\theta, \phi\)) when we started with four real numbers from two complex amplitudes?
What is Born’s rule? Why is global phase unobservable — what stays the same if we multiply a state by \(\mathrm{e}^{\mathrm{i}\gamma}\)?
The Bloch sphere maps each qubit state to a point on a 2-sphere. How do the X, Y, and Z bases appear geometrically? What does the Bloch vector \(\boldsymbol{n}\) represent physically?

Lecture Notes#

Overview#

Quantum mechanics assigns to each physical system a state vector \(\vert \psi\rangle\) that encodes all information necessary to predict measurement outcomes. This section develops the mathematical language—kets, bras, inner products, and orthonormal bases—then introduces the qubit as the simplest concrete example, and arrives at the Bloch sphere as its geometric picture.

Ket, Bra, and Inner Product#

Ket notation represents a quantum state as an abstract vector in a complex Hilbert space. For a two-level system (qubit), the state lives in \(\mathcal{H} = \mathbb{C}^2\).

Definition: Ket and Bra

The ket \(\vert \psi\rangle\) is an abstract vector in Hilbert space.
The bra \(\langle \psi\vert \) is the dual (Hermitian conjugate) of the ket: \(\langle \psi\vert = (\vert \psi\rangle)^\dagger\).
The inner product (scalar product) \(\langle\phi\vert \psi\rangle\) is a complex number satisfying:

\[\langle\phi\vert \psi\rangle = \langle\psi\vert \phi\rangle^*\]

Normalization: A state is normalized if \(\langle\psi\vert \psi\rangle = 1\).
Orthogonality: Two states are orthogonal if \(\langle\phi\vert \psi\rangle = 0\).

In vector form, once we choose a basis:

Object	Notation	Type	Representation
Ket (state)	\(\vert\psi\rangle\)	Column vector	\(\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}\)
Bra (dual)	\(\langle\psi\vert\)	Row vector	\((\psi_1^\ \ \psi_2^)\)
Inner product	\(\langle\phi\vert\psi\rangle\)	Scalar	\(\phi_1^\psi_1 + \phi_2^\psi_2\)
Outer product	\(\vert\psi\rangle\langle\phi\vert\)	Matrix	\(\begin{pmatrix} \psi_1\phi_1^* & \psi_1\phi_2^* \\ \psi_2\phi_1^* & \psi_2\phi_2^* \end{pmatrix}\)

Orthonormal Basis

A set of states \(\{\vert e_1\rangle, \vert e_2\rangle, \ldots, \vert e_n\rangle\}\) forms an orthonormal basis if

\[\langle e_i \vert e_j \rangle = \delta_{ij}\]

and the basis is complete (any state can be expanded as \(\vert \psi\rangle = \sum_i c_i \vert e_i\rangle\) with \(c_i = \langle e_i \vert \psi\rangle\)).

The Qubit State#

A qubit is defined by the existence of two orthonormal computational basis states \(\vert 0\rangle\) and \(\vert 1\rangle\):

(2)#\[\langle 0\vert 0\rangle = 1, \quad \langle 1\vert 1\rangle = 1, \quad \langle 0\vert 1\rangle = 0\]

Any qubit state can be written as a superposition of these basis states:

(3)#\[\vert\psi\rangle = \alpha \vert 0\rangle + \beta \vert 1\rangle\]

where \(\alpha, \beta \in \mathbb{C}\) are amplitudes.

Tip: Converting Between Ket and Vector Notation

To go from ket to vector: read off the coefficients of \(\vert 0\rangle\) and \(\vert 1\rangle\) and stack them as a column vector.

\[\begin{split}\vert \psi\rangle = \alpha \vert 0\rangle + \beta \vert 1\rangle \quad \longleftrightarrow \quad \begin{pmatrix} \alpha \\ \beta \end{pmatrix}\end{split}\]

To go from vector to ket: interpret the top entry as the coefficient of \(\vert 0\rangle\) and the bottom as \(\vert 1\rangle\).

Born’s Rule and Normalization#

When we measure a qubit in the computational basis, the Born rule gives the measurement probabilities:

Born Rule

If \(\vert \psi\rangle = \alpha \vert 0\rangle + \beta \vert 1\rangle\), then measurement in the computational basis \(\{\vert 0\rangle, \vert 1\rangle\}\) yields:

\[P(\text{outcome } 0) = \vert\alpha\vert^2, \quad P(\text{outcome } 1) = \vert\beta\vert^2\]

Probabilities must sum to one, so we require the normalization condition:

(4)#\[\vert\alpha\vert^2 + \vert\beta\vert^2 = 1\]

Global phase is unobservable: Two states \(\vert \psi\rangle\) and \(\mathrm{e}^{\mathrm{i}\gamma}\vert \psi\rangle\) (differing by a global phase \(\gamma \in \mathbb{R}\)) give identical probabilities for all measurements, since \(\vert\mathrm{e}^{\mathrm{i}\gamma}\alpha\vert^2 = \vert\alpha\vert^2\). Measurement probabilities are the only source of knowledge about a quantum system, so global phase has no physical meaning and can be freely removed.

Parameter Counting and the Bloch Sphere#

How many independent real parameters specify a qubit state?

Naive count: Two complex amplitudes \(\alpha, \beta\) give \(2 \times 2 = 4\) real parameters (real and imaginary parts).

Constraints:

Normalization (\(\vert\alpha\vert^2 + \vert\beta\vert^2 = 1\)) removes 1 parameter.
Global phase (\(\vert\psi\rangle \cong \mathrm{e}^{\mathrm{i}\gamma}\vert\psi\rangle\)) removes 1 parameter.

Effective count: \(4 - 1 - 1 = 2\) independent real parameters.

We can write the most general qubit state using two angles \(\theta\) and \(\phi\):

Bloch Parametrization

Every qubit state can be written (up to global phase) as

(5)#\[\vert\psi\rangle = \cos(\theta/2) \vert 0\rangle + \mathrm{e}^{\mathrm{i}\phi} \sin(\theta/2) \vert 1\rangle\]

where \(\theta \in [0, \pi]\) and \(\phi \in [0, 2\pi)\). Normalization is automatic: \(\cos^2(\theta/2) + \sin^2(\theta/2) = 1\).

At this point, \(\theta\) and \(\phi\) appear to be just mathematical parameters. But notice: two real parameters pinpoint a point on a sphere. The angles \((\theta, \phi)\) are exactly the polar and azimuthal angles of a unit sphere in 3D. This gives a one-to-one correspondence between qubit states and points on the Bloch sphere.

The X, Y, Z Bases on the Bloch Sphere

Three natural orthonormal bases correspond to antipodal pairs along the three Cartesian axes of the Bloch sphere:

Basis	Axis	Eigenstates	Bloch coordinates
Z-basis	\(\hat{z}\)	\(\vert 0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\), \(\vert 1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)	North/South pole: \((\theta, \phi) = (0, -)\) / \((\pi, -)\)
X-basis	\(\hat{x}\)	\(\vert +\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}\), \(\vert -\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\)	\((\pi/2, 0)\) / \((\pi/2, \pi)\)
Y-basis	\(\hat{y}\)	\(\vert \mathrm{i}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ \mathrm{i} \end{pmatrix}\), \(\vert \bar{\mathrm{i}}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -\mathrm{i} \end{pmatrix}\)	\((\pi/2, \pi/2)\) / \((\pi/2, 3\pi/2)\)

Each basis is associated with a Pauli operator (\(\hat{X}\), \(\hat{Y}\), \(\hat{Z}\)) whose eigenstates are that basis pair. The Pauli operators are introduced formally in §1.1.3.

Basis Change and Inverse Relations

To express a state in a different basis, invert the defining relations. For the X-basis:

(6)#\[\vert 0\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle + \vert -\rangle), \quad \vert 1\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle - \vert -\rangle)\]

For the Y-basis:

(7)#\[\vert 0\rangle = \frac{1}{\sqrt{2}}(\vert \mathrm{i}\rangle + \vert \bar{\mathrm{i}}\rangle), \quad \vert 1\rangle = -\frac{\mathrm{i}}{\sqrt{2}}(\vert \mathrm{i}\rangle - \vert \bar{\mathrm{i}}\rangle)\]

Example: To express \(\vert \psi\rangle = \alpha \vert 0\rangle + \beta \vert 1\rangle\) in the X-basis, substitute:

\[\vert \psi\rangle = \frac{\alpha + \beta}{\sqrt{2}}\vert +\rangle + \frac{\alpha - \beta}{\sqrt{2}}\vert -\rangle\]

Spin Expectation Values and the Bloch Vector#

For a qubit state \(\vert \psi\rangle\), the expectation values of the three Pauli operators form a 3D vector that points to the state’s location on the Bloch sphere:

Bloch Vector

For a state \(\vert \psi\rangle = \cos(\theta/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\vert 1\rangle\), the Bloch vector is

(8)#\[\boldsymbol{n} = (\langle\hat{X}\rangle, \langle\hat{Y}\rangle, \langle\hat{Z}\rangle) = (\sin\theta\cos\phi,\; \sin\theta\sin\phi,\; \cos\theta)\]

This is a unit vector: \(\vert\boldsymbol{n}\vert = 1\). It points from the origin to the state’s position on the Bloch sphere.

This connects the abstract state description to a concrete geometric picture. Measuring spin along direction \(\hat{\boldsymbol{n}}\) yields a definite outcome — the Bloch vector tells you where the spin “points.” The open question: what exactly are these Pauli operators, and how do we formulate physical observables in general? This is the subject of §1.1.3.

Discussion

Why does the Bloch sphere fail for systems with more than two levels?

A qubit state is specified by 2 real parameters, mapping naturally to a 2-sphere. For a qutrit (3-level system), count: \(2 \times 3 - 2 = 4\) real parameters. No simple sphere in 3D can parametrize a 4-dimensional space.

What geometric object parametrizes all pure qutrit states?
The Bloch sphere is useful because 3D rotations (\(SU(2)\)) correspond to single-qubit gates. Does a similar structure exist for qutrits?

Summary#

A quantum state \(\vert \psi\rangle\) is an abstract vector in Hilbert space; its representation as components \((\alpha, \beta)\) depends on the chosen orthonormal basis.
The Born rule gives measurement probabilities: \(P(0) = \vert\alpha\vert^2\), \(P(1) = \vert\beta\vert^2\). Global phase is unobservable.
Parameter counting (\(4 - 1 - 1 = 2\)) shows a qubit state is specified by two real angles \((\theta, \phi)\), which naturally parametrize the Bloch sphere.
Three orthonormal bases (Z, X, Y) sit at antipodal pairs on the Bloch sphere, each associated with a Pauli operator.
The Bloch vector \(\boldsymbol{n} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)\) gives the geometric direction of the spin expectation values.

Homework#

1. A qubit is in the state \(\vert \psi\rangle = \frac{1}{\sqrt{3}}\vert 0\rangle + \sqrt{\frac{2}{3}}\,\mathrm{e}^{\mathrm{i}\pi/4}\vert 1\rangle\). Find the Bloch sphere angles \((\theta, \phi)\) for this state by writing it in the standard parametrization \(\vert \psi\rangle = \cos(\theta/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\phi}\sin(\theta/2)\vert 1\rangle\).

2. Show that two states \(\vert \psi\rangle\) and \(\mathrm{e}^{\mathrm{i}\alpha}\vert \psi\rangle\) (where \(\alpha \in \mathbb{R}\)) give the same measurement probabilities for any observable. That is, show \(|\langle m \vert \psi \rangle\vert ^2 = |\langle m \vert \mathrm{e}^{\mathrm{i}\alpha}\psi \rangle\vert ^2\) for any state \(\vert m\rangle\). Why does this mean global phase is unobservable?

3. Express the state \(\vert \psi\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) in the X-basis \(\{\vert +\rangle, \vert -\rangle\}\). Then express \(\vert \phi\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \mathrm{i}\vert 1\rangle)\) in the X-basis. Compute the inner product \(\langle\phi\vert \psi\rangle\) in both the Z-basis and the X-basis representations and verify you get the same result.

4. Compute the Bloch vector \(\boldsymbol{n} = (\langle \hat{X}\rangle, \langle\hat{Y}\rangle, \langle\hat{Z}\rangle)\) for the state \(\vert \psi\rangle = \cos(\pi/8)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\pi/3}\sin(\pi/8)\vert 1\rangle\). Verify that \(\vert \boldsymbol{n}| = 1\).

5. On the Bloch sphere, orthogonal quantum states correspond to antipodal points (diametrically opposite). Verify this explicitly: show that if \(\vert \psi\rangle\) has Bloch angles \((\theta, \phi)\), then the orthogonal state \(\vert \psi^\perp\rangle\) (satisfying \(\langle\psi^\perp\vert \psi\rangle = 0\)) has Bloch angles \((\pi - \theta, \phi + \pi)\).

6. A qubit state \(\vert \psi\rangle\) has Bloch vector \(\boldsymbol{n} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)\). Show that measuring in the computational basis gives \(P(0) = \frac{1 + n_z}{2}\) and \(P(1) = \frac{1 - n_z}{2}\), where \(n_z = \cos\theta\). Interpret this geometrically: how does the measurement probability relate to the “height” of the Bloch vector?

7. The inverse relations are given in Eq. (7). Use them to express \(\vert +\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) as a linear combination of \(\vert \mathrm{i}\rangle\) and \(\vert \bar{\mathrm{i}}\rangle\). Verify normalization and \(\langle + \vert + \rangle = 1\) using your Y-basis coefficients.

8. How many real parameters are needed to specify a general \(n\)-level quantum state (a “qudit”)? Start with the naive count of \(2n\) real numbers (from \(n\) complex amplitudes), then subtract constraints from normalization and global phase. Verify your formula gives 2 for \(n = 2\) (the qubit).