|q⟩ Bad Qubits

beginner · Physics · Observables, Operators & Measurement

Spin Observables (Preview)

Among all physical observables in quantum mechanics, spin is special: it is intrinsically quantum, has no classical analogue, and yet its mathematics is among the simplest — a two-dimensional complex vector space. This makes it the perfect testing ground for the operator formalism developed in this module.

Spin-1/2 as a two-level system

An electron (or any spin-1/2 particle) can register only two outcomes when its spin component along any axis is measured: +/2+\hbar/2 ("spin up") or /2-\hbar/2 ("spin down"). Dividing out the overall factor /2\hbar/2, the scaled measurement outcomes are simply +1+1 and 1-1.

A spin state lives in a two-dimensional complex Hilbert space spanned by 0|0\rangle (spin up along zz) and 1|1\rangle (spin down along zz). This space is isomorphic to the qubit state space, so every spin-1/2 problem can be mapped directly onto a qubit — and conversely, every qubit can be thought of as a spin-1/2 particle.

The three Pauli operators

The spin components along the three Cartesian axes are proportional to the Pauli operators σx\sigma_x, σy\sigma_y, σz\sigma_z (also written XX, YY, ZZ):

X=(0110),Y=(0ii0),Z=(1001).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}.

Each matrix is written in the {0,1}\{|0\rangle, |1\rangle\} basis. Each is Hermitian (σ=σ\sigma^\dagger = \sigma) so, as required for observables, all eigenvalues are real. You can verify this by inspection: ZZ is already real and symmetric; XX is real and symmetric; and YY has Y12=iY_{12} = -i with Y21=i=(i)Y_{21} = i = (-i)^*, so (Y)12=Y21=i=Y12(Y^\dagger)_{12} = Y_{21}^* = -i = Y_{12}.

The physical spin-component operators are related to the Paulis by

S^x=2X,S^y=2Y,S^z=2Z.\hat{S}_x = \tfrac{\hbar}{2}\,X, \quad \hat{S}_y = \tfrac{\hbar}{2}\,Y, \quad \hat{S}_z = \tfrac{\hbar}{2}\,Z.

The scaled factor /2\hbar/2 ensures the eigenvalues come out as ±/2\pm\hbar/2, matching experiment.

Eigenvalues and eigenstates

Each Pauli matrix has eigenvalues +1+1 and 1-1. The corresponding eigenstates are:

| Operator | Eigenvalue | Eigenstate | |----------|-----------|------------| | ZZ | +1+1 | 0|0\rangle | | ZZ | 1-1 | 1|1\rangle | | XX | +1+1 | +=12(0+1)|{+}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle) | | XX | 1-1 | =12(01)|{-}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) | | YY | +1+1 | +i=12(0+i1)|{+i}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) | | YY | 1-1 | i=12(0i1)|{-i}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) |

These can each be verified by matrix-vector multiplication. For example, check that X+=+X|{+}\rangle = |{+}\rangle:

(0110)12(11)=12(11)=+.\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = |{+}\rangle. \checkmark

The Pauli algebra

The Pauli operators obey a compact set of commutation and anti-commutation relations. Their squares are all the identity:

X2=Y2=Z2=I.X^2 = Y^2 = Z^2 = I.

Their pairwise products cycle according to

XY=iZ,YZ=iX,ZX=iY,XY = iZ, \quad YZ = iX, \quad ZX = iY,

and reversing the order introduces a minus sign (e.g. YX=iZYX = -iZ). This means the Paulis do not commute: [X,Y]=XYYX=2iZ0[X, Y] = XY - YX = 2iZ \neq 0.

Non-commutativity is physically significant: it implies that spin components along different axes are incompatible observables — measuring SxS_x then SzS_z gives a different distribution than measuring SzS_z then SxS_x. No classical picture of a spinning ball can reproduce this.

Try it

A spin-1/2 particle modelled as a qubit starts in 0|0\rangle (the +1+1 eigenstate of ZZ). Prepare the +1+1 eigenstate of XX instead, which is +=12(0+1)|{+}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle). Apply the single gate that rotates the ZZ-eigenbasis into the XX-eigenbasis.

Run your code to see the quantum state.

Sign in on the full site to ask questions and join the discussion.