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\subsubsection*{2.5.3 Expectation values and
operators}

So far: every quantum observable is represented by
a Hermitian operator $\hat{O}$.
\begin{enumerate}[(I)]
    \item The possible outcomes of measurement of
        the observable $O$ are eigenvalues of
        $\hat{O}$.
    \item If $\hat{O}$has discrete set of
        normalised eigenfunctions $\{\psi_i\}$
        with distinct eigenvalues $\{\lambda_i\}$,
        the measurement of $O$ on a particle
        described by $\psi$ has probability
        \[ P(O = \lambda_I) = |a_i|^2 = |\langle
        \psi_i, \psi \rangle|^2 \]
        where $\psi = \sum_{i = 1}^N a_i \psi_i$.
    \item If $\{\psi_i\}$ is a set of orthonormal
        eigenfunctions of $\hat{O}$ and
        $\{\psi_i\}_{i \in I}$ complete set of
        orthonormal eigenfunctions with some
        eigenvalue $\lambda$
        \[ P(O = \lambda) = \sum_{i \in I} |a_i|^2 \]
        sanity check
        \begin{align*}
            \sum_{i = 1}^N |a_i|^2
            &= \sum_{i = 1}^N \langle a_i \psi_i,
            a_i \psi_i \rangle \\
            &= \sum_{i, j = 1}^N \langle a_i
            \psi_i, a_j \psi_j \rangle \\
            &= \langle \psi, \psi \rangle \\
            &= 1
        \end{align*}
    \item The projection postulate: If $O$
        measured on $\psi$ at time $t$ and the
        outcome of measure is $\lambda_i$ then the
        wave function of $\psi$ instantaneously
        after measurement becomes $\psi_i$
        (eigenfunction with eigenvalues) [if
        $\hat{O}$ has degenerate eigenfunction
        with some eigenvalue $\lambda$ then the
        wavefunction becomes $\psi = \sum_{i \in
        I} a_i \psi_i$]
\end{enumerate}

\begin{definition*}[Projection operator]
    Given $\psi = \sum_i a_i \psi_i = \sum_i
    \langle \psi_i, \psi \rangle \psi_i$ define
    \[ \hat{P}_i : \psi \to \langle \psi_i, \psi
    \rangle \psi_i \]
\end{definition*}

\noindent
We can now define expectation value of an
observable measured on state $\psi$
\begin{align*}
    \langle O \rangle_\psi
    &+ \sum_i \lambda_i P(O = \lambda_i) \\
    &= \sum_i \lambda_i |a_i|^2 \\
    &= \sum_i \lambda_i |\langle \psi_i, \psi
    \rangle|^2 \\
    &= \left\langle \sum_i \langle \psi_i, \psi
    \rangle \psi_i, \sum_j \lambda_j \langle
    \psi_j, \psi \rangle \psi_j \right\rangle \\
    &= \langle \psi, \hat{O} \psi \rangle \\
    &= \int \psi^*(x, t) \hat{O} \psi(x, t) \dd x
\end{align*}
Property:
\[ \langle a\hat{A} + b\hat{B} \rangle_\psi =
a\langle\hat{A}\rangle_\psi +
b\langle\hat{B}\rangle_\psi \]
$a, b \in \RR$. \myskip
Interpretation:
\begin{itemize}
    \item The physics implication of projection
        postulate is that if $O$ is measured
        twice, the outcome of second measure (of
        $\Delta t$ between measures is small) is
        the same as first with probability 1.
    \item (Born's rule) If $\phi(\bf{x}, t)$ is
        the state that gives the desired outcome
        of a measurement on a state $\psi(\bf{x},
        t)$, probability of such outcome is given
        by
        \[ |\langle \psi, \phi \rangle|^2 = \left|
        \int_{-\infty}^\infty \psi^*(x, t) \phi(x,
        t) \dd x \right|^2 \]
\end{itemize}

\subsection{Time independent Schr\"odinger equation (TISE)}

Let's rewrite TDSE in 1D
\[ i \hbar \pfrac{\psi}{t}(x, t) =
-\frac{\hbar^2}{2m} \pfrac[2]{\psi}{x} (x,
t) + U(x) \psi(x, t) = \hat{H} \psi(x, t) \tag{1}
\]
try ansatz (try solution)
\[ \psi(x, t) = T(t) \chi(X) \tag{2} \]
Plug (2) into (1)
\[ i\hbar \pfrac{T}{t}(t) \chi(x) = T(t) \hat{H}
\chi(X) \]
divide by $T(t)\chi(x)$
\[ \frac{1}{T(t)} i\hbar \pfrac{T}{t}(t) =
\frac{\hat{H} \chi(x)}{\chi(x)} \tag{3} \]
Both LHS and RHS have to be equal to a constant
$E$, so
\[ \frac{1}{T(t)} i\hbar \pfrac{T}{t}(t) = E
\implies T(t) = e^{-iEt/\hbar} \tag{4} \]
with $E \in \RR$. So TISE is
\begin{equation*}
    \boxed{
    \begin{aligned}
        \hat{H}\chi(x)
        &= E\chi(x) \\
        -\frac{\hbar^2}{2m} \pfrac[2]{\chi}{x}(x)
        + U(x) \chi(x)
        &= E \chi (x)
    \end{aligned}
    }
    \tag{5}
\end{equation*}
\begin{hiddenflashcard}[TISE]
    \prompt{Time independent Schr\"oedinger's
    equation?}
    \[ \boxed{\cloze{\hat{H} \chi(x) \equiv
    -\frac{\hbar^2}{2m} \pfrac[2]{\chi}{x} (x) +
    U(x) \chi(x) = \cloze{E \chi(x)}}} \]
\end{hiddenflashcard}
\begin{itemize}
    \item TDSE is eigenvalue equation for
        $\hat{H}$ operator.
    \item eigenvalues of $\hat{H}$ are all
        possible outcomes of measure of energy of
        state $\psi$.
\end{itemize}

\subsection{Stationary states}

We found \emph{a} particular solution of TDSE
\begin{flashcard}[solution-of-TDSE]
    \prompt{If $\psi$ has state $\chi(x)$ at $t =
    0$, with $\chi$ an eigenstate (with energy
    $E$), what is $\psi(\bf{x}, t)$?}
    \[ \psi(x, t) = \cloze{\chi(x) e^{-iEt/\hbar}} \]
\end{flashcard}
$E$ eigenvalue associated with eigenfunction
$\chi$.

\begin{definition*}
    These solutions are called stationary states.
\end{definition*}

\noindent
Why?
\[ \rho(x, t) = |\psi(x, t)|^2 = |\chi(x)|^2 \]
If we apply theorem 2.6 to $\hat{O} = \hat{H}$

\begin{theorem}
    Every solution of TDSE can be written as a
    linear combination of stationary states.
\end{theorem}

\begin{itemize}
    \item For system that has a discrete set of
        eigenvalues of $\hat{H}$,
        \[ E_n = E_1, E_2, \dots \]
        $n \in \NN$
        \[ \psi(x, t) = \sum_n a_n \chi_n(x)
        e^{-iE_n t/\hbar} \]
    \item For system that has a continuous set of
        eigenvalues of $\hat{H}$, $E(\alpha)$
        \[ \psi(x, t) = \int A(\alpha)
        \chi_\alpha(\alpha) e^{-iE_\alpha t/\hbar}
        \dd \alpha \]
        where $A \in \CC$, $\alpha \in \RR$.
    \item $|a_n|^2$, $|A(\alpha)|^2 \dd \alpha$
        probability of measuring the particle
        energy to be $E_h = E(\alpha)$.
\end{itemize}

\noindent
Imagine a system with only 2 energy eigenvalues
$E_1 \neq E_2$ we can write the state $\psi$ at
time $t$
\[ \psi(c, t) = a_1 \chi_1(x) e^{-iE_1 t/\hbar} +
a_2 \chi_2(x) e^{-iE_2 t/\hbar} \]
\[ \implies \psi(x, 0) = a_1 \chi_1(x) + a_2
\chi_2(x) \]
if $a_1 = 0$ then $\psi(x, 0) = a_2
\chi_2(\alpha)$, $\psi(x, t) = a_2 \chi_2(x)
e^{-iE_2 t/\hbar}$ for all $t$, $|\psi(x, 0)|^2 =
|\psi(x, t)|^2$. If $a_i \neq 0$ and $a_2 \neq 0$,
\begin{align*}
    |\psi(x, t)|^2
    &= |a_1 \chi_1 e^{-i E_1 t/\hbar} + a_2 \chi_2
    e^{-iE_2 t/\hbar}|^2 \\
    &= a_1^2 |\chi_1|^2 + a_2^2 |\chi_2|^2 +
    2a_1a_2 \chi_1(x)\chi_2(x) \cos \left(
    \frac{(E_1 - E_2)t}{\hbar} \right)
\end{align*}