30 May 2025
Linear Spaces, Dirac Notation and Quantum Operators
Linear vector spaces, bases, inner products, completeness, outer products, Hilbert space, coordinate representations, and operator algebra.
A complex linear vector space $V$ is closed under addition and complex scalar multiplication:
\[\lvert u\rangle,\lvert v\rangle\in V,\quad a,b\in\mathbb C \quad\Longrightarrow\quad a\lvert u\rangle+b\lvert v\rangle\in V.\]Addition is associative and commutative, a zero vector and additive inverses exist, and scalar multiplication is associative and distributive. Examples are $\mathbb C^N$, square-integrable wavefunctions, and the solution set of a homogeneous linear differential equation. The last example is a vector space because any linear combination of two solutions is again a solution.
Linear independence, basis, and dimension
The vectors $\lvert e_1\rangle,\ldots,\lvert e_N\rangle$ are linearly independent if
\[\sum_{j=1}^N c_j\lvert e_j\rangle=\lvert0\rangle \quad\Longrightarrow\quad c_j=0\quad\text{for every }j.\]A basis is an independent set that spans $V$. Hence every vector has one and only one expansion
\[\lvert\psi\rangle=\sum_{j=1}^N c_j\lvert e_j\rangle.\]If two distinct coefficient sets represented the same vector, their difference would be a nonzero linear combination equal to zero, contradicting independence. The number $N$ is the dimension. A function space can instead have a countably infinite basis, such as $e^{inx}/\sqrt{2\pi}$ on $[-\pi,\pi]$.
Bras, kets, and inner products
A state vector is a ket $\lvert\psi\rangle$. Its dual is the bra $\langle\psi\rvert$. Combining a bra and ket gives the complex number $\langle\phi\mid\psi\rangle$, subject to
\[\langle\phi\mid\psi\rangle=\langle\psi\mid\phi\rangle^*, \qquad \langle\psi\mid\psi\rangle\ge0,\] \[\langle\phi\rvert(a\lvert u\rangle+b\lvert v\rangle) =a\langle\phi\mid u\rangle+b\langle\phi\mid v\rangle.\]Positivity is strict: $\langle\psi\mid\psi\rangle=0$ only for the zero vector. Taking the adjoint conjugates scalars,
\[(a\lvert u\rangle+b\lvert v\rangle)^\dagger =a^*\langle u\rvert+b^*\langle v\rvert.\]An orthonormal basis obeys
\[\langle e_i\mid e_j\rangle=\delta_{ij}.\]Multiplying the basis expansion by $\langle e_i\rvert$ determines the coordinate
\[c_i=\langle e_i\mid\psi\rangle,\]and substitution back gives
\[\lvert\psi\rangle=\sum_i\lvert e_i\rangle\langle e_i\mid\psi\rangle.\]Because this holds for every ket,
\[\boxed{\sum_i\lvert e_i\rangle\langle e_i\rvert=I,}\]the completeness relation.
Outer products and projectors
Combining a ket and bra produces an operator:
\[(\lvert u\rangle\langle v\rvert)\lvert\psi\rangle =\lvert u\rangle\langle v\mid\psi\rangle.\]For a normalized ket, $P_u=\lvert u\rangle\langle u\rvert$ is a projector because
\[P_u^2=\lvert u\rangle\langle u\mid u\rangle\langle u\rvert =\lvert u\rangle\langle u\rvert=P_u, \qquad P_u^\dagger=P_u.\]It extracts the component of a state along $\lvert u\rangle$.
Hilbert space and coordinate representation
An inner-product space is a Hilbert space when every Cauchy sequence of its vectors converges to a vector that remains in the space. This completeness is needed for limits of state expansions.
For a discrete orthonormal basis, the ket is represented by the column
\[\lvert\psi\rangle\longleftrightarrow \begin{pmatrix}c_1\\c_2\\\vdots\end{pmatrix}, \qquad c_i=\langle e_i\mid\psi\rangle,\]and the bra by its conjugate transpose. In the continuous position basis,
\[\psi(\mathbf r)=\langle\mathbf r\mid\psi\rangle, \qquad \langle\mathbf r\mid\mathbf r'\rangle=\delta^3(\mathbf r-\mathbf r'),\] \[\int\lvert\mathbf r\rangle\langle\mathbf r\rvert\,d^3r=I, \qquad \lvert\psi\rangle=\int\lvert\mathbf r\rangle\psi(\mathbf r)\,d^3r.\]Normalization becomes $\int\lvert\psi(\mathbf r)\rvert^2d^3r=1$; therefore a three-dimensional wavefunction has SI dimension $\mathrm{m}^{-3/2}$.
Linear operators and their algebra
An operator $A$ is linear when
\[A(a\lvert u\rangle+b\lvert v\rangle)=aA\lvert u\rangle+bA\lvert v\rangle.\]Its matrix elements in an orthonormal basis are $A_{ij}=\langle e_i\rvert A\lvert e_j\rangle$. Insert completeness on both sides:
\[A=IAI =\sum_{ij}\lvert e_i\rangle A_{ij}\langle e_j\rvert.\]The sum, product, commutator, and adjoint are defined by
\[(A+B)\lvert\psi\rangle=A\lvert\psi\rangle+B\lvert\psi\rangle, \qquad (AB)\lvert\psi\rangle=A(B\lvert\psi\rangle),\] \[[A,B]=AB-BA, \qquad \langle\phi\mid A\psi\rangle =\langle A^\dagger\phi\mid\psi\rangle.\]Consequently
\[(AB)^\dagger=B^\dagger A^\dagger, \qquad [A,B]^\dagger=-[A^\dagger,B^\dagger].\]A Hermitian operator satisfies $A^\dagger=A$. Its expectation value is real:
\[\langle A\rangle^* =\langle\psi\rvert A\lvert\psi\rangle^* =\langle\psi\rvert A^\dagger\lvert\psi\rangle =\langle A\rangle.\]A unitary operator satisfies $U^\dagger U=I$ and preserves inner products,
\[\langle U\phi\mid U\psi\rangle =\langle\phi\rvert U^\dagger U\lvert\psi\rangle =\langle\phi\mid\psi\rangle.\]The outer-product reconstruction, projector identity, and operator-algebra relations are checked in the Maxima worksheet; every printed residual is zero.
Discussion