Hilbert space, $|\\alpha\\rangle\\to \\alpha(p,q;t)$

In an N-dimensional space, it is simplest to represent a vector $|\alpha\rangle$ by the N-tuple of its components $\{a_n\}$, with respect to a specified orthonormal basis. The inner product $\langle\alpha|\beta\rangle$ is the complex number, Linear transformations are represented by matrix.

But for functions which live in infinite-dimensional space, an infinite sum or an integral may not converge, in which case the inner product does not exist, and any argument involving inner products is immediately suspect.

The set of all squre-integrable functions consitutes a vector space, $L_2(a,b)$, or Hilbert space.

$$ f(x)\ \text{ such that }\ \int_a^b|f(x)|^2dx< \infin. $$

We define the inner product of two functions

$$ \langle f| g\rangle\equiv\int_a^bf^*(x)g(x)dx. $$

• If $f$ and $g$ are both square-integrable, their inner product is guaranted to exist.
  • prove: This follows from the integral Schwarz inequality, or Hölder's inequality.
• The inner product of $f(x)$ with itself is real and non-negative; it’s zero only when $f(x)=0$ or what is considered equivalent{ In Hilbert space, two functions that have the same square integral are considered equivalent. equivalences classeds of functions.}.

$$ \langle g|f\rangle=\langle f|g\rangle^{*}.\\\langle f|f\rangle=\int_a^b|f(x)|^2dx. $$

• normalized: $\langle f|f\rangle=1.$
• orthogonal: $\langle f|g\rangle=0.$
• orthonormal: $\langle f_m|f_n\rangle=\delta_{mn}.$
• complete: $f(x)=\sum_{n=1}^{\infin}c_nf_n(x), \forall f(x)\notin \{f_n\}.$ A set of functions is complete if any other functions in Hilbert space can be expressed as a linear combination of them.
  • if the functions $\{f_n(x)\}$ are othonormal, the coefficients are given b Fourier’s trick: $c_n=\langle f_n|f\rangle.$

Observables, $\\hat Q\\to Q(p,q)=\\langle Q\\rangle$

$$ \langle Q\rangle=\int\Psi^*\hat{Q}\Psi dx=\langle\Psi|\hat{Q}\Psi\rangle. $$

Now, the outcome of a measurement has got to be real.

• the average of any measurements: $\langle Q\rangle=\langle Q\rangle^*.$
  • $\langle\Psi|\hat{Q}\Psi\rangle=\langle\hat{Q}\Psi|\Psi\rangle^*=\langle\hat{Q}\Psi|\Psi\rangle.$And this must hold true for any wave functions.
  • Thus operators representing observables are hermitian: $\langle f|\hat{Q}f\rangle=\langle\hat{Q}f|f\rangle\quad\text{for all }f(x).\to Q=Q^\dagger$
    • Proof: what makes operators Hermitian? A equivalent definition: $\langle f|\hat{Q}f\rangle=\langle\hat{Q}f|f\rangle \Leftrightarrow \langle f|\hat{Q}g\rangle=\langle\hat{Q}f|g\rangle$
    • Is the momentum operator hermitian?
    • hermitian conjuagte of some operators

Determinate States: $\\Psi_q \\text{ of } \\hat Q$

Ordinarily, when you measure an observables $Q$ on an ensemble of identically prepared systems, all in the same state $\Psi$, you do not get the same result each time.[indeterminacy of quantum mechanics]