Title: Quantum Mechanics

Q: Observables
A: We call each property of the particle which we can measure an \emph{observable}. Each observable is represented by one of the following \emph{operators} acting on the wavefunction $\psi(x)$:
\begin{center}
  \begin{tabular}{rll}
    position & $\hat{x} = x$ & $\hat{x} \psi = x\psi(x)$\\
    momentum & $\hat{p} = -i\hbar \frac{\partial}{\partial x}$ & $\hat{p}\psi = -i\hbar \psi'(x)$\\
    energy & $H = \frac{\hat{p}^2}{2m} + V(\hat{x})$ & $H\psi = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi + V(x)\psi(x)$
  \end{tabular}
\end{center}
A \emph{definite} value $\lambda$ is obtained if and only if $\psi$ is an eigenstate, or eigenfunction, of the operator and $\lambda$ is the corresponding eigenvalue. This results in quantisation of the observable to the eigenvalues of the operator.
The \emph{expected} value of an observable with operator $Q$ is $\bra Q \ket_\psi = \bra \psi \mid Q\ket_\psi=\int_{-\infty}^\infty \psi^*Q\psi$.

Q: Position operator
A: The position operator is
\[
\hat{x} = x.
\]
If $\hat{x}\psi(x)=a\psi(x)$, then $\psi(x)=0$ for $x\neq a$. So there is no normalisable eigenfunction of the position operator.
The probability distribution of the position is $|\psi|^2$.

Q: Expected observable
A: The \emph{expected} value of an observable with operator $Q$ is
\[
\bra Q \ket_\psi = \bra \psi \mid Q\ket_\psi=\int_{-\infty}^\infty \psi^*Q\psi.
\]

Q: Momentum operator
A: The momentum operator is
\[
\hat{p} = -i\hbar \frac{\partial}{\partial x}
\]
The eigenvalue equation is
\[
-i\hbar \frac{\partial\psi}{\partial x} = p\psi(x).
\]
Hence $\psi=Ce^{ikx}$ is an eigenfunction with eigenvalue $p=\hbar k$ which it is not normalisable on $\R$ but is normalisable on a closed bounded interval.

Q: Time-independent Schr\"odinger equation
A: The Hamiltonian operator is
\[
H = \frac{\hat{p}^2}{2m} + V(\hat{x}).
\]
  The \emph{time-independent Schr\"odinger equation} is the Hamiltonian eigenvalue equation
  \[
    H\psi = E\psi,
  \]
  or
  \[
    -\frac{\hbar^2}{2m}\psi'' + V(x) \psi = E\psi.
  \]

Q: Harmonic oscillator
A: The $t-$dep SE with $V(x) = \frac{1}{2}m\omega^2 x^2$ gives
\[
  H\psi = -\frac{\hbar^2}{2m}\psi'' + \frac{1}{2}m\omega^2 x^2 \psi = E\psi.
\]
To simplify constants, we define
\[
  y = \left(\frac{m\omega}{\hbar}\right)^{\frac{1}{2}}x,\quad \mathcal{E} = \frac{2E}{\hbar \omega},
\]
both of which are dimensionless. Then we are left with
\[
  -\frac{\d^2 \psi}{\d y^2} + y^2\psi = \mathcal{E} \psi.
\]
Define $\psi = f(y) e^{-\frac{1}{2}y^2}$ to obtain \emph{Hermite's equation}
\[
  \frac{\d^2 f}{\d y^2} - 2y\frac{\d f}{\d y} + (\mathcal{E} - 1) = 0.
\]
We try a series solution $f(y) = \sum_{r \geq 0} a_r y^r$ to obtain
\[
  \sum_{r \geq 0} \big( (r + 2)(r + 1)a_{n + 2} + (\mathcal{E} - 1 - 2r)a_r\big) y^r = 0.
\]
This holds if and only if
\[
  a_{r + 2} = \frac{2 r + 1 - \mathcal{E}}{(r + 2)(r + 1)} a_r,\quad r \geq 0.
\]
This gives $2$ LI solutions of opposite parity. Unless the coefficients vanish, we get
\[
  a_{p}/a_{p - 2} \sim \frac{1}{p}
\]
so $f(y)\sim y^\alpha e^{y^2}$ and $\psi\sim y^\alpha$ is not normalisable. Hence the series must terminate and thus $\mathcal{E} = 2n + 1$ for some $n$. The solutions are thus $f(y) = h_n(y)$, where $h_n$ is the \emph{Hermite polynomial} of degree $n$ with $h_n(-y) = (-1)^n h_n(y)$. With the constant restored, the possible energy eigenvalues are
\[
  E_n = \hbar \omega \left(n + \frac{1}{2}\right),
\]
for $n \geq 0$. The wavefunctions are
\[
  \psi_n(x) = h_n \left(\left(\frac{m\omega}{\hbar}\right)^{\frac{1}{2}} x\right) \exp\left(-\frac{1}{2}\frac{m\omega}{\hbar} x^2\right),
\]
where normalization fixes $a_0$ and $a_1$.

Q: Infinite well - Particle in a box
A: \[
  V(x) =
  \begin{cases}
    0 & |x| \leq a\\
    \infty & |x| > a.
  \end{cases}
\]
To bound the energy, we must have $\psi(x)=0$ for $|x| >0$ and continuity gives $\psi(\pm a)=0$. Within $|x| \leq a$, the Schr\"odinger equation is
\[
  -\frac{\hbar^2}{2m}\psi'' = E\psi.
\]
\textbf{Case} $E>0$. Writing $ E = \frac{\hbar^2 k^2}{2m}$ we obtain
\begin{align*}
  \psi'' + k^2 \psi &= 0 \\
 \psi &= A\cos kx + B\sin kx \\
0&=A \cos ka \pm B\sin ka \\
0&=A\cos ka = B\sin ka.
\end{align*}
This leaves two possibilities:
\begin{enumerate}
  \item $B = 0$ and $ka = n\pi/2$ with $n$ odd
  \item $A = 0$ and $ka = n\pi/2$ with $n$ even.
\end{enumerate}
Hence the allowed energy levels are
\[
  E_n = \frac{\hbar^2 \pi^2}{8ma^2} n^2,
\]
where $n \in\N$, and the wavefunctions are
\[
  \psi_n(x) = \left(\frac{1}{a}\right)^{\frac{1}{2}}
  \begin{cases}
    \cos \frac{n\pi x}{2a} & n\text{ odd}\\
    \sin \frac{n\pi x}{2a} & n\text{ even}
  \end{cases}.
\]
The solutions just look like standing waves on a string with two fixed end points.
\textbf{Case} $E\leq 0$. If $E=0$, then $\psi ''=0$ so $\psi'=$ constant and BCs give $\psi=0$. If $E<0$, then letting $ E = \frac{\hbar^2 k^2}{2m}$ we obtain
\begin{align*}
  \psi'' - k^2 \psi &= 0 \\
\psi &= Ae^{kx} + Be^{-kx}
\end{align*}
and BCs again give $\psi=0$.

Q: Time-dependent Schr\"odinger Equation
A: For a time-dependent wavefunction $\Psi(x, t)$, the \emph{time-dependent Schr\"odinger equation} is
  \[
    i\hbar \frac{\partial \Psi}{\partial t} = H\Psi.\tag{$*$}
  \]
For a particle in a potential $V(x)$, this can reads
\[
  i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x) \Psi.
\]

Q: Stationary state
A: Consider a separable wavefunction $\Psi(x,t)=\psi(x)f(t)$. The t-dependent SE gives
\begin{align*}
i\hbar\psi(x)\dot{f} &= fH\psi \\
i\hbar\psi(x)\dot{f}\frac{1}{f} &= \frac{1}{\psi}H\psi \\
&=E
\end{align*}
where $E$ is a separation constant. So $H\psi=E\psi$ and $f=e^{-iEt/\hbar}$. This prompts the definition of the equation for a stationary state
  \[
    \Psi(x, t) = \psi(x) \exp\left(-i \omega t\right)
  \]
where $H\psi=E\psi$, $E=\hbar\omega$ and $|\Psi(x,t)|^2=|\psi(x)|^2$.

Q: Conservation of Probability
A: \begin{prop}
  The probability density
  \[
    P(x, t) = |\Psi(x, t)|^2
  \]
  obeys a conservation equation
  \[
    \frac{\partial P}{\partial t} = - \frac{\partial j}{\partial x},
  \]
  where
  \[
    j(x, t) = -\frac{i\hbar}{2m} \left(\Psi^*\Psi' - (\Psi^*)' \Psi\right)
  \]
  is the \emph{probability current}.
\end{prop}
\begin{proof}
The Schr\"odinger equation gives
\[
i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \Psi'' + V\Psi
\]
and taking the complex conjugate gives
\[
-i\hbar \frac{\partial \Psi^*}{\partial t} = -\frac{\hbar^2}{2m} (\Psi^*)'' + V\Psi^*.
\]
Substituting these in gives
  \begin{align*}
    \frac{\partial P}{\partial t} &= \Psi^* \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi*}{\partial t} \Psi\\
    &= \Psi^* \frac{i\hbar }{2m}\Psi'' - \frac{i\hbar}{2m}\Psi''^* \Psi\\
    &= -\frac{\partial j}{\partial x}.
  \end{align*}
\end{proof}

Q: Bound states
A: In a bound state, the energy $E$ is quantised and in a minimal potential $V_0$ we have $V_0<E<0$.
\begin{proof}
Consider the time-independent Schr\"odinger equation for a particle of mass $m$:
\[
  H\psi = -\frac{\hbar^2}{2m}\psi'' + V(x) \psi = E\psi,
\]
with the potential $V(x) \to 0$ as $x \to \pm \infty$. This is a second order ODE for a complex function $\psi$, and hence there are two complex constants in the GS. However, since this is a linear equation, this implies that 1 complex constant corresponds to changing $\psi \mapsto \lambda \psi$, which gives no change in physical state. So we just have 1 degree of freedom to satisfy BCs.
As $|x| \to \infty$, our equation simply becomes
\[
  -\frac{\hbar^2}{2m} \psi'' = E\psi.
\]
So we get
\[
  \psi \sim
  \begin{cases}
    Ae^{ikx} + Be^{-ikx} & E = \frac{\hbar^2 k^2}{2m} > 0\\\
    Ae^{\kappa x} + Be^{-\kappa x} & E = -\frac{\hbar^2 \kappa^2}{2m} < 0.
  \end{cases}
\]
For $E<0$, if we want $\psi$ to be normalizable, then there are 2 boundary conditions for $\psi$:
\[
  \psi \sim
  \begin{cases}
    Ae^{\kappa x} & x \to -\infty\\
    B e^{-\kappa x} & x \to +\infty
  \end{cases}
\]
This is an overdetermined system, and solutions are possible only for certain values of $E$. So bound state energy levels are quantized. The lowest energy eigenstate is called the \emph{ground state}. Eigenstates with higher energies are called \emph{excited states}. Using integration by parts the energy is given by
\begin{align*}
E&=\bra H \ket_\psi \\
&= \int_{-\infty}^\infty \psi^*\left[ -\frac{\hbar^2}{2m}\psi'' + V(x) \psi \right]\d x \\
&= \int_{-\infty}^\infty \frac{\hbar^2}{2m}|\psi'|^2+ V(x) |\psi|^2 \d x \\
\geq 0 + V_0.
\end{align*}
For $E>0$, there is no normalisable solution. However there are scattering solutions.
\end{proof}

Q: Parity
A: Consider the Schr\"odinger equation for a particle of mass $m$
\[
  H\psi = -\frac{\hbar^2}{2m}\psi'' + V(x) \psi = E\psi.
\]
with potential
\[
  V(x) = V(-x).
\]
By changing variables $x \to -x$, we see that $\psi(x)$ is an eigenfunction of $H$ with energy $E$ if and only if $\psi(-x)$ is an eigenfunction of $H$ with energy $E$. There are two possibilities:
\begin{enumerate}
  \item If $\psi(x)$ and $\psi(-x)$ represent the same quantum state, then $\psi(-x) = \eta \psi(x)$ for some constant $\eta$. Since this is true for all $x$, we can do this twice and get
    \[
      \psi(x) = \eta \psi(-x) = \eta^2 \psi(x).
    \]
    So $\eta = \pm 1$ and $\psi(-x) = \pm \psi(x)$. We call $\eta$ the \emph{parity}, and say $\psi$ has even/odd parity if $\eta$ is $+1/-1$ respectively.
  \item If $\psi(x)$ and $\psi(-x)$ represent different quantum states, then we can still take linear combinations
    \[
      \psi_\pm (x) = \alpha(\psi(x) \pm \psi(-x)),
    \]
    and these are also eigenstates with energy eigenvalue $E$, where $\alpha$ is for normalization. Then by construction, $\psi_\pm (-x) = \pm \psi_\pm(x)$ and have parity $\eta = \pm 1$.
\end{enumerate}
Hence, if we are given a potential with reflective symmetry $V(-x) = V(x)$, then we can restrict our attention and just look for solutions with definite parity.

Q: Uncertainty
A: The \emph{uncertainty} in position $(\Delta x)_\psi$ and momentum $(\Delta p)_\psi$ are defined by
  \[
    (\Delta x)_\psi^2 = \bra (\hat{x} - \bra \hat{x}\ket_\psi)^2 \ket_\psi = \bra \hat{x}^2\ket_\psi - \bra \hat{x}\ket^2_\psi,
  \]
  with exactly the same expression for momentum:
  \[
    (\Delta p)_\psi^2 = \bra (\hat{p} - \bra \hat{p}\ket_\psi)^2 \ket_\psi = \bra \hat{p}^2\ket_\psi - \bra \hat{p}\ket^2_\psi,
  \]
The RHS of these equations is positive.
\begin{proof}
Let $X=\hat{x} - \alpha, P=\hat{p}-\beta$ for real constants $\alpha$ and $\beta$. Then as $X$ and $P$ are Hermitian
\begin{align*}
(\psi, X^2\psi)&=(X\psi, X\psi) \\
&= \| X\psi \|^2 \\
&\geq 0 \\
(\psi, P^2\psi)&=(P\psi, P\psi) \\
&= \| P\psi \|^2 \\
&\geq 0.
\end{align*}
Putting $\alpha= \bra \hat{x}\ket_\psi$ and $\beta=\bra \hat{p}\ket_\psi$ gives the required result.
\end{proof}

Q: Heisenberg's uncertainty principle
A: \begin{thm}
  If $\psi$ is any normalized state at any fixed time, then
  \[
    (\Delta x)_\psi (\Delta p)_\psi \geq \frac{\hbar}{2}.
  \]
\end{thm}
\begin{proof}
  Choose $\alpha = \bra \hat{x}\ket _\psi$ and $\beta = \bra \hat{p}\ket_\psi$, and define
  \[
    X = \hat{x} - \alpha,\quad P = \hat{p} - \beta.
  \]
  Then we have
  \begin{align*}
    (\Delta x)_\psi^2 &= (\psi, X^2 \psi) = (X\psi, X\psi) = \|X\psi\|^2\\
    (\Delta p)_\psi^2 &= (\psi, P^2 \psi) = (P\psi, P\psi) = \|P\psi\|^2
  \end{align*}
and $[\hat{x}, \hat{p}]=i\hbar \implies [X,P]=i\hbar$. Then for any $\lambda\in\R$
  \begin{align*}
    \| (X+i\lambda P)\psi \|^2 &= ((X+i\lambda P)\psi, (X+i\lambda P)\psi) \\
&= \|X\psi\|^2+ i\lambda (\psi,(XP-PX)\psi) + \lambda^2 \|P\psi\|^2 \\
&= (\Delta x)^2_\psi + i\lambda (\psi, i\hbar \psi) + \lambda^2 (\Delta p)^2_\psi \\
&= \lambda^2 (\Delta p)^2_\psi -\lambda\hbar + (\Delta x)^2_\psi.
  \end{align*}
Since the discriminant is non-positive, the result follows.
\end{proof}

Q: Ehrenfest's theorem
A: \begin{thm}[Ehrenfest]
  \begin{align*}
    \frac{\d}{\d t}\bra \hat{x}\ket_\Psi &= \frac{1}{m}\bra \hat{p}\ket_\Psi\\
    \frac{\d}{\d t}\bra \hat{p}\ket_\Psi &= -\bra V'(\hat{x})\ket_\Psi.
  \end{align*}
\end{thm}
\begin{proof}
  We have
  \begin{align*}
    \frac{\d}{\d t}\bra \hat{x}\ket_\Psi &= (\dot{\Psi}, \hat{x}\Psi) + (\Psi, \hat{x}\dot{\Psi})\\
    &= \left(\frac{1}{i\hbar}H\Psi, \hat{x}\Psi\right) + \left(\Psi, \hat{x}\left(\frac{1}{i\hbar}H\right)\Psi\right)\\
    &= -\frac{1}{i\hbar}(\Psi, H(\hat{x}\Psi)) + \frac{1}{i\hbar}(\Psi, \hat{x}(H\Psi))\\
    &= \frac{1}{i\hbar}(\Psi, (\hat{x} H - H\hat{x}) \Psi)
  \end{align*}
 where the commutator is
  \[
    (\hat{x}H - H\hat{x})\Psi = -\frac{\hbar^2}{2m}(x\Psi'' - (x\Psi)'') + (xV\Psi - Vx\Psi) = \frac{\hbar^2}{m}\Psi' = \frac{i\hbar}{m}\hat{p}\Psi.
  \]
 Similarly
  \begin{align*}
    \frac{\d}{\d t}\bra \hat{p}\ket_\Psi &= (\dot{\Psi}, \hat{p}\Psi) + (\Psi, \hat{p}\dot{\Psi})\\
    &= \left(\frac{1}{i\hbar}H\Psi, \hat{p}\Psi\right) + \left(\Psi, \hat{p}\left(\frac{1}{i\hbar}H\right)\Psi\right)\\
    &= -\frac{1}{i\hbar}(\Psi, H(\hat{p}\Psi)) + \frac{1}{i\hbar}(\Psi, \hat{p}(H\Psi))\\
    &= \frac{1}{i\hbar}(\Psi, (\hat{p} H - H\hat{p}) \Psi)
  \end{align*}
where the commutator is
  \begin{align*}
    (\hat{p}H - H\hat{p})\Psi &= -i\hbar \left(\frac{-\hbar^2}{2m}\right)((\Psi'')' - (\Psi')'') - i\hbar ((V(x)\Psi)' - V(x) \Psi') \\
    &= -i\hbar V'(x) \Psi.
  \end{align*}
\end{proof}

Q: Canonical commutation relation
A: The commutator of position and momentum is
  \[
    [\hat{x}, \hat{p}] = \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar.
  \]
\begin{proof}
From the product rule
\[
  (\hat{x}\hat{p} - \hat{p}\hat{x}) \psi = -i\hbar x\psi' - (-i\hbar(x \psi)' = i\hbar \psi.
\]
\end{proof}

Q: Scattering states
A: If $\Psi(x,t)$ is bounded but not normalisable, then it represents a \emph{scattering state}.
Then the amplitude $|\Psi(x,t)|^2$ gives the density of particles at $(x,t)$, i.e. number of particles per unit length.
The probability current density
\[
  j(x, t) = -\frac{i\hbar}{2m}(\Psi^* \Psi' - \Psi \Psi'^*)
\]
represents the \emph{flux} of particles at $(x, t)$, ie. the number of particles passing the point $x$ at unit time.

Q: Scattering state solutions
A: Consider the time-independent Schr\"odinger equation for a particle of mass $m$
\[
  H\psi = -\frac{\hbar^2}{2m}\psi'' + V(x) \psi = E\psi,
\]
with the potential $V(x) \to 0$ as $x \to \pm \infty$.
For $\psi$ not normalized but bounded and $E > 0$
\[
  \psi \sim
  \begin{cases}
    Ie^{ikx} + R e^{-ikx} & x \to -\infty\\
    Te^{ikx} & x \to +\infty
  \end{cases}
\]
where the incident wave has $p=\hbar k$ and reflected $p=-\hbar k$. So the probability flux is
\[
  j \sim
  \begin{cases}
    j_{\mathrm{inc}} + j_{\mathrm{ref}}\\
    j_{\mathrm{tr}}
  \end{cases}
  =
  \begin{cases}
    |I|^2 \frac{\hbar k}{m} - |R|^2 \frac{\hbar k}{m} & x \to -\infty\\
    |T|^2 \frac{\hbar k}{m} & x \to +\infty
  \end{cases}
\]
We also get the reflection and transmission probabilities
\begin{align*}
  P_{\mathrm{ref}} &= |A_{\mathrm{ref}}|^2 = \frac{|j_{\mathrm{ref}}|}{|j_{\mathrm{inc}}|}\\
  P_{\mathrm{tr}} &= |A_{\mathrm{tr}}|^2 = \frac{|j_{\mathrm{tr}}|}{|j_{\mathrm{inc}}|},
\end{align*}
where
\begin{align*}
  A_{\mathrm{ref}}(k) &= \frac{R}{I} \\
  A_{\mathrm{tr}}(k) &= \frac{T}{I}.
\end{align*}
are the reflection and transmission \emph{amplitudes}.

Q: Measurement axioms
A: Consider an observable $Q$ with discrete spectrum and normalized eigenstates. We can write any state $\psi$
\[
  \psi = \sum_n \alpha_n \chi_n,
\]
with $Q\chi_n = \lambda_n \chi_n$, $(\chi_m, \chi_n) = \delta_{mn}, \alpha_n = (\chi_n, \psi), \alpha_n \not= 0, \lambda_n$ distinct. Then
\begin{itemize}
  \item The outcome of a measurement is some eigenvalue of $Q$.
  \item The probability of obtaining $\lambda_n$ is
    \[
      P_n = |\alpha_n|^2,
    \]
    where $\alpha_n = (\chi_n, \psi)$ is the \emph{amplitude}.
  \item The measurement is instantaneous and forces the system into the state $\chi_n$. This is the new state immediately after the measurement is made.
\end{itemize}
We can check $\psi$ normalised is consistent with $|\alpha_n|^2$ being a probability distribution:
\begin{align*}
  (\psi, \psi) &= \left(\sum \alpha_m \chi_m, \sum \alpha_n \chi_n\right)\\
  &= \sum_{m, n} \alpha_m^* \alpha_n (\chi_m, \chi_n)\\
  &= \sum_{m, n} \alpha_m^* \alpha_n \delta_{mn}\\
  &= \sum_n |\alpha_n|^2\\
  &= \sum_n P_n\\
  &= 1.
\end{align*}

Q: Quantum systems
A: Possible states of a quantum system correspond to non-zero elements of a complex inner-product space $V$ with $\psi$ and $\alpha\psi$ are equivalent for $\alpha\neq 0$. $V$ is assumed to be complete (a Hilbert space) and infinite dimensional.
A measurable quantity or observable corresponds to a Hermitian (self-adjoint) linear operator on $V$.
The state of the system obeys the SE $i\hbar\dot{\Psi}=H\Psi$ where $H$ is a hermitian operator, the Hamiltonian, corresponding to energy.

Q: Ehrenfest's theorem (general form)
A: \begin{thm}
  If $Q$ is any operator with no explicit time dependence, then
  \[
    i\hbar \frac{\d}{\d t}\bra Q\ket_\Psi = \bra [Q, H]\ket_\Psi,
  \]
  where
  \[
    [Q, H] = QH - HQ
  \]
  is the commutator.
\end{thm}
\begin{proof}
  If $Q$ does not have time dependence, then
  \begin{align*}
    i\hbar \frac{\d}{\d t}(\Psi, Q\Psi) &= (-i\hbar \dot\Psi, Q\Psi) + (\Psi, Q i\hbar \dot\Psi)\\
    &= (-H\Psi, Q\Psi) + (\Psi, QH \Psi)\\
    &= (\Psi, (QH - HQ)\Psi)\\
    &= (\Psi, [Q, H] \Psi).
  \end{align*}
\end{proof}
If $Q$ has explicit time dependence, then we have an extra term on the right as $i\hbar \bra \dot{Q} \ket_\Psi$.

Q: Simultaneous measurability
A: To simultaneously measure two observables $A$ and $B$, we need the state to be a simultaneously an eigenstate of $A$ and $B$. In other words, simultaneous measurement is possible if and only if there is a complete basis consisting of simultaneous or joint eigenstates $\chi_n$ with
\[
  A \chi_n = \lambda_n \chi_n,\quad B \chi_n = \mu_n \chi_n.
\]
From IB Linear Algebra, a necessary and sufficient condition for $A$ and $B$ to be simultaneously measurable (ie. simultaneously diagonalizable) is for $A$ and $B$ to commute, ie.
\[
  [A, B] = AB - BA = 0.
\]

Q: Generalized uncertainty relation
A: For observables $A$ and $B$:
\[
  (\Delta A)_\psi (\Delta B)_\psi \geq \frac{1}{2} |\bra [A, B]\ket_\psi|.
\]

Q: Canonical commutation relations
A: We have
\[
  [\hat{x}_i, \hat{p}_j] = -i\hbar \delta_{ij},\quad [\hat{x}_i, \hat{x}_j] = [\hat{p}_i, \hat{p}_j] = 0.
\]
Hence the corresponding uncertainity relation is
\[
  (\Delta x_i)(\Delta p_j) \geq \frac{\hbar}{2} \delta_{ij}.
\]

Q: Structureless particle
A: A \emph{structureless particle} is one for which all observables can be written in terms of position and momentum.

Q: Conservation of probability in 3-D
A: The time-dependent Schr\"odinger equation is
\[
  i \hbar \frac{\partial \Psi}{\partial t} = H \Psi = -\frac{\hbar^2}{2m} \nabla^2\Psi + V(\mathbf{x})\Psi.
\]
The probability current is defined as
\[
  \mathbf{j} = -\frac{i\hbar}{2m} (\Psi^* \nabla \Psi - \Psi \nabla \Psi^*).
\]
This probability current obeys the conservation equation
\[
  \frac{\partial}{\partial t}|\Psi(\mathbf{x}, t)|^2 = -\nabla \cdot \mathbf{j}.
\]
This implies that for any fixed volume $V$,
\[
  \frac{\d}{\d t}\int_V |\Psi(\mathbf{x}, t)|^2 \;\d^3 x = -\int_V \nabla \cdot \mathbf{j}\;\d^3 x = -\int_{\partial V} \mathbf{j}\cdot \d \mathbf{S}.
\]
So if $|\Psi(\mathbf{x}, t)| \to 0$ sufficiently rapidly as $|\mathbf{x}| \to \infty$, then the boundary term disappears and
\[
  \frac{\d}{\d t}\int |\Psi(\mathbf{x}, t)|^2 \;\d^3 x = 0.
\]

Q: Angular momentum
A: The \emph{angular momentum} is a vector of operators
  \[
    \mathbf{L} = \hat{\mathbf{x}}\wedge \hat{\mathbf{p}} = -i\hbar \mathbf{x} \wedge \nabla.
  \]
  In components, this is given by
  \[
    L_i = \varepsilon_{ijk} \hat{x}_j\hat{p}_k = -i\hbar \varepsilon_{ijk} x_j \frac{\partial}{\partial x_k}.
  \]
  The \emph{total angular momentum} operator is
  \[
    L^2 = \mathbf{L}^2 = L_i L_i = L_1^2 + L_2^2 + L_3^3.
  \]
Both operators are Hermitian and hence observables.

Q: Commutation relations involving angular momentum
A: \begin{prop}\leavevmode
  \begin{enumerate}
    \item $[L_i, L_j] = i\hbar \varepsilon_{ijk} L_k$.
    \item $[\mathbf{L}^2, L_i] = 0$.
    \item $[L_i, \hat{x}_j] = i\hbar \varepsilon_{ijk} \hat{x}_k$ and $[L_i, \hat{p}_j] = i\hbar \varepsilon_{ijk} \hat{p}_k$
  \end{enumerate}
\end{prop}
\begin{proof}\leavevmode
  \begin{enumerate}
    \item We can compute
        \begin{align*}
        L_i L_j &= \varepsilon_{iar} \hat{x}_a \hat{p}_r \varepsilon_{jbs} \hat{x}_b \hat{p}_s\\
        &= \varepsilon_{iar} \varepsilon_{jbs} (\hat{x}_a \hat{p}_r \hat{x}_b \hat{p}_s)\\
        &= \varepsilon_{iar} \varepsilon_{jbs} (\hat{x}_a (\hat{x}_b \hat{p}_r - [\hat{p}_r, \hat{x}_b]) \hat{p}_s)\\
        &= \varepsilon_{iar} \varepsilon_{jbs} (\hat{x}_a \hat{x}_b \hat{p}_r \hat{p}_s - i\hbar \delta_{br} \hat{x}_a \hat{p}_s).
\intertext{Similarly, we have}
 L_j L_i &= \varepsilon_{iar} \varepsilon_{jbs} (\hat{x}_b \hat{x}_a \hat{p}_s \hat{p}_r - i\hbar \delta_{as} \hat{x}_b \hat{p}_r).
      \end{align*}
      Then the commutator is
      \begin{align*}
        L_i L_j - L_j L_i &= -i\hbar \varepsilon_{iar} \varepsilon_{jbs}(\delta_{br} \hat{x}_a \hat{p}_s - \delta_{as}\hat{x}_b \hat{p}_r)\\
        &= -i\hbar (\varepsilon_{iab} \varepsilon_{jbs} \hat{x}_a \hat{p}_s - \varepsilon_{iar} \varepsilon_{jba} \hat{x}_b\hat{p}_r)\\
        &= -i\hbar ((\delta_{is} \delta_{ja} - \delta_{ij} \delta_{as})\hat{x}_a \hat{p}_s - (\delta_{ib} \delta_{rj} - \delta_{ij}\delta_{rb}) \hat{x}_b \hat{p}_r)\\
        &= i\hbar (\hat{x}_i \hat{p}_j - \hat{x}_j \hat{p}_i)\\
        &= i\hbar \varepsilon_{ijk} L_k.
      \end{align*}
    \item  Using the Lebnitz property, we obtain
      \begin{align*}
        [L_i, \mathbf{L}^2] &= [L_i, L_j L_j]\\
        &= [L_i, L_j] L_j + L_j [L_i, L_j]\\
        &= i\hbar \varepsilon_{ijk} (L_k L_j + L_j L_k)\\
        &= 0
      \end{align*}
 since we are contracting an antisymmetric tensor with a symmetric tensor.
    \item Using the Leibnitz property
      \begin{align*}
        [L_i, \hat{x}_j] &= \varepsilon_{iab} [\hat{x}_a \hat{p}_b, \hat{x}_j]\\
        &= \varepsilon_{iab} ([\hat{x}_a, \hat{x}_j] \hat{p}_b + \hat{x}_a [\hat{p}_b, \hat{x}_j])\\
        &= \varepsilon_{iab} \hat{x}_a (-i\hbar \delta_{bj})\\
        &= i\hbar \varepsilon_{ija} \hat{x}_a \\
        [L_i, \hat{p}_j] &= \varepsilon_{iab} [\hat{x}_a \hat{p}_b, \hat{p}_j]\\
        &= \varepsilon_{iab} ([\hat{x}_a, \hat{p}_j] \hat{p}_b + \hat{x}_a [\hat{p}_b, \hat{p}_j])\\
        &= \varepsilon_{iab} (i\hbar \delta_{aj} \hat{p}_b)\\
        &= i\hbar \varepsilon_{ijb} \hat{p}_b.
      \end{align*}
  \end{enumerate}
\end{proof}

Q: Leibnitz Rule for commutators
A: \begin{enumerate}
\item $[A, BC] = [A, B] C + B[A, C].$
\item $[AB, C] = [A, C]B + A[B, C].$
\end{enumerate}
\begin{proof}
\begin{enumerate}
\item We can expand
\begin{align*}
 [A, B] C + B[A, C] &= (AB-BA)C+B(AC-CA) \\
&=ABC-BAC + BAC - BCA \\
&= A(BC)-(BC)A\\
&=[A, BC].
\end{align*}
\item Follows from (i) by anticommutativity \vspace{-0.5cm}
\begin{align*}
[AB,C]=-[C,AB]&=-\left( [C,A]B+A[C,B] \right) \\
&=[A,C]B +A[B,C].
\end{align*}
\end{enumerate}
\end{proof}

Q: Eigenfunctions of angular momentum in spherical polars
A: Let $L_{\pm} = L_1 \pm i L_2$. Using the chain rule and spherical polars
\begin{align*}
  L_3 &= -i\hbar \frac{\partial}{\partial{\varphi}}\\
  L_{\pm} = L_1 \pm i L_2 &= \pm \hbar e^{\pm i\varphi} \left(\frac{\partial}{\partial{\theta}} \pm i\cot \theta \frac{\partial}{\partial{\varphi}} \right)\\
  \mathbf{L}^2 &= -\hbar^2 \left(\frac{1}{\sin \theta} \frac{\partial}{\partial{\theta}} \sin \theta\frac{\partial}{\partial{\theta}} + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2}\right).
\end{align*}
This determines the angular component of the Laplacian
\[
  \nabla^2 = \frac{1}{r}\frac{\partial^2}{\partial r^2} r - \frac{1}{\hbar^2 r^2} \mathbf{L}^2.
\]
Since we know
\[
  [L_3 , \mathbf{L}^2] = 0,
\]
there are simultaneous eigenfunctions of these operators $Y_{\ell m}(\theta, \varphi)$, with $\ell = 0, 1, 2, \cdots$ and $m = 0, \pm 1, \pm 2, \cdots, \pm \ell$. These have eigenvalues $\hbar m$ for $L_3$ and $\hbar^2 \ell(\ell + 1)$ for $\mathbf{L}^2$. The numbers $\ell$ and $m$ are usually known as the angular momentum quantum numbers.

Q: Joint eigenstates for a spherically symmetric potential
A: The Hamiltonian in spherical polars is
\[
  H = -\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial^2}{\partial r^2} r + \frac{1}{2 \mu} \frac{1}{r^2} \mathbf{L}^2 + V(r).
\]
We check $[L_i, H] = [\mathbf{L}^2, H] = 0$ using Cartesians. The kinetic and potential terms are
\begin{align*}
[L_i, \hat{\mathbf{p}}^2]&= [L_i, \hat{p}_j \hat{p}_j]\\
  &= [L_i, \hat{p}_j] \hat{p}_j + \hat{p}_j [L_i, \hat{p}_j]\\
  &= i\hbar \varepsilon_{ijk} (\hat{p}_k\hat{p}_j + \hat{p}_j \hat{p}_k)\\
[L_i, V(r)] &= -i\hbar \varepsilon_{ijk}\left(x_j \frac{\partial V}{\partial x_k}\right)\\
  &= -i\hbar \varepsilon_{ijk} x_j \frac{x_k}{r} V'(r)
\end{align*}
which vanish by contracting an antisymmetric tensor with a symmetric term. So we have the joint eigenstates
\[
  \psi(\mathbf{x}) = R(r) Y_{\ell m} (\theta, \varphi).
\]
Finally, we solve the Schr\"odinger equation
\[
  -\frac{\hbar^2}{2 \mu} \frac{1}{r} \frac{\d^2}{\d r^2} (rR) + \frac{\hbar^2}{2 \mu r^2 } \ell(\ell + 1) R + VR = ER.
\]
This says radial $+$ angular $+$ potential $=$ energy. We call $R(r)$ the \emph{radial part} of the wavefunction. We define the \emph{radial wavefunction} $\chi(r) = r R(r)$. Multiplying the original SE by $r$, we obtain the \emph{radial} SE
\[
  -\frac{\hbar^2}{2 \mu} \chi'' + \frac{\hbar^2 \ell(\ell + 1)}{2 \mu r^2} \chi + V \chi = E \chi.
\]
$R$ to be finite as $r \to 0 \implies \chi = 0$ at $r = 0$. Also
\[
   \int |\psi(\mathbf{x})|^2 \;\d^3 x = \int |R(r)|^2 r^2 \;\d r \int |Y_{\ell m}(\theta, \varphi)|^2 \sin \theta \;\d \theta \;\d \varphi.
\]
Since the angular factor is fixed by a normalisation convention, $\psi$ is normalizable if and only if
\[
  \int_0^\infty |\chi(r)|^2 \;d\; r = \int_0^\infty |R(r)|^2 r^2 \;\d r < \infty.
\]