Technopedia

The technologist’s encyclopedia

Quantum States

Quantum states are the mathematical objects that encode everything knowable about a physical system. They are relevant whenever you need to describe a system whose properties are not simultaneously definite: an electron in an atom, a spin in a magnetic field, or two particles that are entangled. The state space (Hilbert space) is the foundation on which measurement, dynamics, and many-body physics are built.

Hilbert space and superposition

Definition. A quantum state is a ray (equivalence class of vectors up to a global phase) in a complex Hilbert space H\mathcal{H}. A general state is a superposition ∣ψ⟩=∑ici∣i⟩|\psi\rangle = \sum_i c_i |i\rangle where ∑i∣ci∣2=1\sum_i |c_i|^2 = 1.

Details

From classical to quantum states

In classical mechanics, the state of a particle is a point in phase space: a definite position xx and momentum pp simultaneously. Quantum mechanics replaces this with a vector in an abstract complex vector space.

Step 1: Why a vector space?

The key experimental fact is superposition: if a system can be in state A and in state B, it can also be in any linear combination α∣A⟩+β∣B⟩\alpha|A\rangle + \beta|B\rangle. This is not a mixture (“either A or B”) but a genuinely new state with interference properties. A vector space is the minimal structure that supports this.

Step 2: Why complex?

Interference requires a phase. The probability of a transition from state ∣ψ⟩|\psi\rangle to state ∣ϕ⟩|\phi\rangle is ∣⟨ϕ∣ψ⟩∣2|\langle\phi|\psi\rangle|^2, which depends on the relative phase between components. Real amplitudes cannot produce the full range of interference patterns observed in experiments (e.g., the Aharonov–Bohm effect).

Step 3: Normalization and the Born rule.

The squared norm ⟨ψ∣ψ⟩=1\langle\psi|\psi\rangle = 1 ensures that the total probability of all possible outcomes sums to 1. The coefficient ∣ci∣2|c_i|^2 gives the probability of finding the system in basis state ∣i⟩|i\rangle upon measurement (see Measurement and uncertainty).

Example: a spin-½ particle.

The Hilbert space is C2\mathbb{C}^2. The basis states are ∣↑⟩=(1,0)T|\uparrow\rangle = (1,0)^T and ∣↓⟩=(0,1)T|\downarrow\rangle = (0,1)^T. A general state is ∣ψ⟩=α∣↑⟩+β∣↓⟩|\psi\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle with ∣α∣2+∣β∣2=1|\alpha|^2 + |\beta|^2 = 1. The state ∣ψ⟩=12(∣↑⟩+∣↓⟩)|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle) is a superposition: measuring spin along z gives up or down with equal probability, but the state is not “either up or down” — it has a definite phase relationship that produces interference if measured along a different axis.

|↓⟩ |↑⟩ |ψ⟩ = α|↑⟩ + β|↓⟩ β α Hilbert space C² for a spin-½ particle

Example problem.

Problem

A qubit is in the state ∣ψ⟩=12∣↑⟩+i32∣↓⟩|\psi\rangle = \frac{1}{2}|\uparrow\rangle + \frac{i\sqrt{3}}{2}|\downarrow\rangle. What is the probability of measuring spin-up along z? What is the probability of measuring spin-up along x, where ∣+x⟩=12(∣↑⟩+∣↓⟩)|+_x\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle)?

Solution

Spin-up along z: P(↑z)=∣⟨↑∣ψ⟩∣2=∣1/2∣2=1/4P(\uparrow_z) = |\langle\uparrow|\psi\rangle|^2 = |1/2|^2 = 1/4.

Spin-up along x:

⟨+x∣ψ⟩=12(12+i32)=1+i322\langle+_x|\psi\rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right) = \frac{1+i\sqrt{3}}{2\sqrt{2}}.

P(+x)=∣1+i322∣2=1+38=12P(+_x) = \left|\frac{1+i\sqrt{3}}{2\sqrt{2}}\right|^2 = \frac{1+3}{8} = \frac{1}{2}.

The phase matters: without the ii, the x-probability would be (1/2+3/2)2/2=(2+3)/4≈0.933(1/2 + \sqrt{3}/2)^2/2 = (2+\sqrt{3})/4 \approx 0.933, very different from 1/21/2.

Observables and operators

Definition. An observable is a self-adjoint (Hermitian) operator A^\hat{A} on H\mathcal{H}. Its eigenvalues are the possible measurement outcomes; its eigenvectors form a complete orthonormal basis. The expectation value in state ∣ψ⟩|\psi\rangle is ⟨ψ∣A^∣ψ⟩\langle\psi|\hat{A}|\psi\rangle.

Details

Why Hermitian operators?

A measurement outcome must be a real number. The condition A^=A^†\hat{A} = \hat{A}^\dagger (Hermiticity) guarantees real eigenvalues. Additionally, the spectral theorem ensures that the eigenvectors of a Hermitian operator form a complete orthonormal basis, so any state can be expanded in terms of possible outcomes.

Example: position and momentum.

In the position representation, x^\hat{x} acts by multiplication: (x^ψ)(x)=x ψ(x)(\hat{x}\psi)(x) = x\,\psi(x). The momentum operator is p^=−iℏ ∂/∂x\hat{p} = -i\hbar\,\partial/\partial x. These satisfy the canonical commutation relation [x^,p^]=iℏ I^[\hat{x}, \hat{p}] = i\hbar\,\hat{I}. This non-commutativity is the algebraic origin of the uncertainty principle (see Measurement and uncertainty).

Example: spin operators.

The spin-½ operators are S^i=ℏ2σi\hat{S}_i = \frac{\hbar}{2}\sigma_i where σi\sigma_i are the Pauli matrices:

σx=(0110),σy=(0−ii0),σz=(100−1)\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

They satisfy [σi,σj]=2iϵijkσk[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k and σiσj=δijI^+iϵijkσk\sigma_i\sigma_j = \delta_{ij}\hat{I} + i\epsilon_{ijk}\sigma_k. The eigenvalues of each S^i\hat{S}_i are ±ℏ/2\pm\hbar/2, giving the two possible spin measurement outcomes.

Example problem.

Problem

Show that σx\sigma_x and σz\sigma_z do not commute, and find the eigenvectors of σx\sigma_x.

Solution

σxσz=(0110)(100−1)=(0−110)\sigma_x\sigma_z = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

σzσx=(100−1)(0110)=(01−10)\sigma_z\sigma_x = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

So [σx,σz]=σxσz−σzσx=(0−220)=−2iσy≠0[\sigma_x, \sigma_z] = \sigma_x\sigma_z - \sigma_z\sigma_x = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = -2i\sigma_y \neq 0.

Eigenvectors of σx\sigma_x: solve (0110)(ab)=λ(ab)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \lambda\begin{pmatrix} a \\ b \end{pmatrix}.

For λ=+1\lambda = +1: b=ab = a, so ∣+x⟩=12(1,1)T|+_x\rangle = \frac{1}{\sqrt{2}}(1,1)^T. For λ=−1\lambda = -1: b=−ab = -a, so ∣−x⟩=12(1,−1)T|-_x\rangle = \frac{1}{\sqrt{2}}(1,-1)^T.

Entanglement

Definition. A composite system in state ∣Ψ⟩∈HA⊗HB|\Psi\rangle \in \mathcal{H}_A \otimes \mathcal{H}_B is entangled if it cannot be written as ∣ψ⟩A⊗∣ϕ⟩B|\psi\rangle_A \otimes |\phi\rangle_B. Entangled states exhibit correlations stronger than any classical (local hidden-variable) model allows.

Details

Why entanglement is non-classical

A separable (product) state ∣ψ⟩A⊗∣ϕ⟩B|\psi\rangle_A \otimes |\phi\rangle_B means each subsystem has its own independent state. An entangled state means the whole has a definite state but the parts do not — measuring one part instantaneously constrains the other, regardless of distance.

Example: the Bell state.

∣Φ+⟩=12(∣↑↑⟩+∣↓↓⟩)|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle)

Neither particle has a definite spin. If you measure particle A along z and get up, particle B is instantly in the up state. But the key non-classical feature is that the correlations hold for any measurement axis simultaneously, violating Bell inequalities. No pre-existing local assignment of “A is up, B is up” can reproduce the quantum predictions for all measurement directions.

Connection to quantum information.

Entanglement is the resource for quantum computing, quantum teleportation, and quantum cryptography. A register of nn entangled qubits lives in a 2n2^n-dimensional Hilbert space, enabling exponential parallelism in certain algorithms. This application domain builds directly on the state-space formalism described here.

Example problem.

Problem

Show that ∣Φ+⟩=12(∣↑↑⟩+∣↓↓⟩)|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle) is also maximally entangled in the x-basis, i.e., it can be rewritten as 12(∣+x+x⟩+∣−−x⟩)\frac{1}{\sqrt{2}}(|+_x+_x\rangle + |-_-x\rangle).

Solution

Using ∣↑⟩=12(∣+x⟩+∣−x⟩)|\uparrow\rangle = \frac{1}{\sqrt{2}}(|+_x\rangle + |-_x\rangle) and ∣↓⟩=12(∣+x⟩−∣−x⟩)|\downarrow\rangle = \frac{1}{\sqrt{2}}(|+_x\rangle - |-_x\rangle):

∣↑↑⟩=12(∣+x+x⟩+∣+x−x⟩+∣−x+x⟩+∣−−x⟩)|\uparrow\uparrow\rangle = \frac{1}{2}(|+_x+_x\rangle + |+_x-_x\rangle + |-_x+_x\rangle + |-_-x\rangle)

∣↓↓⟩=12(∣+x+x⟩−∣+x−x⟩−∣−x+x⟩+∣−−x⟩)|\downarrow\downarrow\rangle = \frac{1}{2}(|+_x+_x\rangle - |+_x-_x\rangle - |-_x+_x\rangle + |-_-x\rangle)

Sum: ∣↑↑⟩+∣↓↓⟩=∣+x+x⟩+∣−−x⟩|\uparrow\uparrow\rangle + |\downarrow\downarrow\rangle = |+_x+_x\rangle + |-_-x\rangle

Therefore ∣Φ+⟩=12(∣+x+x⟩+∣−−x⟩)|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|+_x+_x\rangle + |-_-x\rangle). The state is symmetric under change of basis — a hallmark of maximal entanglement.

Spin and angular momentum

Definition. Angular momentum in quantum mechanics is quantized: J^2∣j,m⟩=ℏ2j(j+1)∣j,m⟩\hat{J}^2 |j,m\rangle = \hbar^2 j(j+1)|j,m\rangle and J^z∣j,m⟩=ℏm∣j,m⟩\hat{J}_z |j,m\rangle = \hbar m |j,m\rangle where j=0,12,1,…j = 0, \frac{1}{2}, 1, \ldots and m=−j,−j+1,…,jm = -j, -j+1, \ldots, j. Spin is the intrinsic angular momentum with no classical analogue.

Details

Quantization of angular momentum

In classical mechanics, angular momentum L=r×p\mathbf{L} = \mathbf{r} \times \mathbf{p} can take any magnitude and direction. In quantum mechanics, the commutation relations [J^i,J^j]=iℏϵijkJ^k[\hat{J}_i, \hat{J}_j] = i\hbar\epsilon_{ijk}\hat{J}_k force the eigenvalues to be discrete. The total angular momentum quantum number jj can be integer (bosons) or half-integer (fermions), and the projection mm takes 2j+12j+1 values.

Why spin has no classical analogue.

Orbital angular momentum arises from spatial motion (r×p\mathbf{r} \times \mathbf{p}), but spin is an intrinsic property: an electron has spin ℏ/2\hbar/2 even at a point, with no spatial extent to rotate. The two-valuedness of spin-½ (only up or down along any axis) and the 2π2\pi rotation giving a minus sign (requiring 4π4\pi to return to the original state) have no classical counterpart.

Connection to the Standard Model.

The spin of a particle determines its statistics (integer spin → Bose-Einstein, half-integer → Fermi-Dirac; see Quantum many-body) and its role in the Standard Model gauge structure. Fermions (spin-½) are matter; bosons (spin-1) mediate forces.

Example problem.

Problem

A particle has total angular momentum j=1j = 1. How many values can mm take? What are the possible values of J^2\hat{J}^2 in units of ℏ2\hbar^2?

Solution

mm takes 2j+1=32j+1 = 3 values: m=−1,0,+1m = -1, 0, +1.

J^2=ℏ2j(j+1)=ℏ2⋅1⋅2=2ℏ2\hat{J}^2 = \hbar^2 j(j+1) = \hbar^2 \cdot 1 \cdot 2 = 2\hbar^2.

Note that J^2\hat{J}^2 is the same for all three mm states — the magnitude is fixed, only the direction (projection) varies.

Quiz

Test your understanding

Problem. A two-level system has basis states ∣0⟩,∣1⟩|0\rangle, |1\rangle and is prepared in the state ∣ψ⟩=15∣0⟩+2i5∣1⟩|\psi\rangle = \frac{1}{\sqrt{5}}|0\rangle + \frac{2i}{\sqrt{5}}|1\rangle. A measurement of the observable A^=∣0⟩⟨1∣+∣1⟩⟨0∣\hat{A} = |0\rangle\langle 1| + |1\rangle\langle 0| (= σx\sigma_x) is performed.

  1. What are the possible outcomes?
  2. What are their probabilities?
  3. What is the expectation value ⟨A^⟩\langle\hat{A}\rangle?
Solution

  1. A^=σx\hat{A} = \sigma_x has eigenvalues ±1\pm 1, so the outcomes are +1+1 and −1-1.

  2. ⟨+x∣ψ⟩=12(15+2i5)=1+2i10\langle+_x|\psi\rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{5}} + \frac{2i}{\sqrt{5}}\right) = \frac{1+2i}{\sqrt{10}}

P(+1)=∣1+2i∣210=510=12P(+1) = \frac{|1+2i|^2}{10} = \frac{5}{10} = \frac{1}{2}

⟨−x∣ψ⟩=12(15−2i5)=1−2i10\langle-_x|\psi\rangle = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{5}} - \frac{2i}{\sqrt{5}}\right) = \frac{1-2i}{\sqrt{10}}

P(−1)=∣1−2i∣210=510=12P(-1) = \frac{|1-2i|^2}{10} = \frac{5}{10} = \frac{1}{2}

  1. ⟨A^⟩=(+1)⋅12+(−1)⋅12=0\langle\hat{A}\rangle = (+1)\cdot\frac{1}{2} + (-1)\cdot\frac{1}{2} = 0.

Check: ⟨ψ∣σx∣ψ⟩=15⋅2i5+−2i5⋅15=2i−2i5=0\langle\psi|\sigma_x|\psi\rangle = \frac{1}{\sqrt{5}}\cdot\frac{2i}{\sqrt{5}} + \frac{-2i}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}} = \frac{2i - 2i}{5} = 0. ✓

Authoritative sources