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Hilbert spaces and the spectral theorem
Inner products, orthonormal bases, Fourier series, and eigenvalues of operators.
An inner product gives angles and projections; an orthonormal basis expands any vector: Fourier series are exactly this in L². Self-adjoint operators have real spectrum and an eigen-decomposition, the infinite-dimensional version of diagonalising a symmetric matrix. Picture it: the symmetric matrix [[2,1],[1,2]] stretching along two perpendicular eigenvectors. Think it: quantum observables are self-adjoint operators; measurement outcomes are their spectrum.
مثال عملي · eigenvalues of [[2,1],[1,2]]
Eigenvalues of [[2,1],[1,2]]
خطوة خطوة
- \det(A - \lambda I) = 0
Eigenvalues are the roots of the characteristic polynomial.
- \det\left[\begin{matrix}2 - \lambda & 1\\1 & 2 - \lambda\end{matrix}\right] = 0
Subtract λ from the diagonal.
- \lambda^{2} - 4 \lambda + 3 = 0
Expand the determinant.
- \left(\lambda - 3\right) \left(\lambda - 1\right) = 0
Factor.
- \lambda = 3, \lambda = 1
Eigenvalues (with multiplicity).
- \lambda = 1:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}-1\\1\end{matrix}\right]
Solve (A − 1I)v = 0 for a basis eigenvector.
- \lambda = 3:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\1\end{matrix}\right]
Solve (A − 3I)v = 0 for a basis eigenvector.
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كيف: Hilbert spaces and the spectral theorem
- Eigenvalues are the roots of the characteristic polynomial.
- Subtract λ from the diagonal.
- Expand the determinant.
- Factor.
- Eigenvalues (with multiplicity).
- Solve (A − 1I)v = 0 for a basis eigenvector.
- Solve (A − 3I)v = 0 for a basis eigenvector.
أسئلة يطرحها الناس
What is a Hilbert space?
A vector space with an inner product (so lengths and angles make sense) that is complete (no missing limit points). Square-integrable functions form one; quantum states live in one.