Lie groups and Lie algebra for physicists
Representation of groups
Stone theorem
Expression for classical quantum operators
On $\mathcal{H}$, $\ket{\psi}$ is a vector and $\hat{X}$, $\hat{V}$, $\hat{P}$ are linear operators $\mathcal{H}\to \mathcal{H}$.
We write $\ket{x}$ the eigenvectors of $\hat{X}$, which form a complete orthogonal basis of $\mathcal{H}$: $\braket{x}{x’}= \delta(x-x’)$, $\int \ket{x}\bra{x}\text{d}x=1$. Furthermore, we will assume the fundamental commutation relations $[\hat{X},\hat{P}]=i\hbar$, inherited from the Poisson braket in classical mechanics.
The Hamiltonian
$\hat{H}$ is defined to be the generator of time translations of $\ket{\psi}$, that is, for any time interval $\text{d}t$:
\[\ket{\psi(t+\text{d}t)} = e^{-\frac{i}{\hbar}\hat{H}\text{d}t}\ket{\psi(t)}\]where the factor $-\frac{i}{\hbar}$ is pure convention. Taking the limit $\text{d}t \to 0$:
\[\ket{\psi(t+\text{d}t)} \simeq_{\text{d}t\to 0} (1-\frac{i}{\hbar}\hat{H}\text{d}t)\ket{\psi(t)}\] \[\lim_{\text{d}t\to 0} \frac{\ket{\psi(t+\text{d}t)}- \ket{\psi}(t)}{\text{d}{t}} = -\frac{i}{\hbar}\hat{H} \ket{\psi}(t)\]and hence
\[\frac{\text{d}\ket{\psi}}{\text{d}t} = -\frac{i}{\hbar}\hat{H}\ket{\psi}\]The momentum
The potential
If $\hat{V}$ can only be expressed in term of the operator $\hat{X}$, which we will assume, from classical mechanics, $\hat{V}$ is self-adjoint and eigenvectors of $\hat{X}$ are also eigenvectors of $\hat{V}$:
\[\hat{V}\ket{x} = V(x)\ket{x}\]Hence:
\[\bra{x}\hat{V}\ket{\psi} = (\hat{V}\ket{x})^{\dagger}\ket{\psi} = V(x)\braket{x}{\psi}= V(x)\psi(x)\]Uncertainty relations and Fourier transform
Consider an operator $\hat{O}$ with eigenvectors $\ket{o_i}$ and eigenvalues $o_i$.
\[\hat{O} = -i\hbar\partial_y\]The eigenvalue equation thus leads:
\[o_i\ket{o_i} = -i\hbar\partial_y \ket{o_i}\]which general solutions are of the form:
\[\ket{o_i} \propto e^{i \frac{o_i y }{\hbar}}\]Furthermore, if $\ket{o_i}$ forms a basis of $\mathcal{H}$, any vector can be written as
\[\ket{\psi} = \sum \alpha_i \ket{o_i}\](or integral for continuous cases).
Thus
\[\ket{\psi} = \sum \alpha_i e^{i \frac{o_i y }{\hbar}}\ket{o_i}\]Hence the famous relation:
\[\Delta p \Delta x \geq \hbar/2\] \[\Delta E \Delta t \geq \hbar/2\]Even though there is no time observable.
Similarly we would have:
\[\Delta L \Delta \theta \geq \hbar/2\]and for gauge:
\[\Delta Q \Delta \varphi \geq \hbar/2\]where $Q$ is the electric charge and $\varphi$ the phase of the wavefunction. Superselection rules + Josephson effect