Title: ETH_Sem5 / AdvQC / 05_Elements_Quantum_Theory

Q: Write down the symmetrized Darwin interaction for velocities being small compared to the speed of light! Explain in words, what the basic procedure is to arrive at this solution.
A: Define two reference systems. IS: both particles are moving, IS\(\strich\): particle 2 is at rest
1. The first charge \(q_1\) thus experiences only a stationary potential \(\phi_2\strich(\pos\strich,t\strich) =\phi_2\strich(\pos\strich)\) of the resting charge \(q_2\) \(\rightarrow\) neglect retardation effects
2. Lorentz-transform the coordinates and the 4-potential to express the interaction term in terms of \(\phi_2\strich\)
3. Symmetrize the interaction term, i.e. get rid of the retardation time \(t_{12}\), by expressing everything in a power series of \(1/c\).
4. Use gauge freedom of the Euler-Lagrange eqs to arrive at:
\[{V_{12}=\frac{q_1q_2}{r_{12}}-\frac{q_1q_2}{2c^2}\left[\frac{\dpos_1\cdot\dpos_2}{r_{12}}+\frac{(\pos_{12}\cdot\dpos_1)(\pos_{12}\cdot\dpos_2)}{r_{12}^3}\right]+O(c^{-3})}\]
In this potential energy expression the magnetic and retardation effects are now expressed in terms of the velocities of the moving charges! The Coulomb interaction is recovered to be the first term of the series.

Q: How is the particle density distribution for a quantum mechanical state \(\Psi_n\) defined? What is the time evolution of the state \(\Psi_n\) and how are expectation values calculated?
A: - Particle density distribution: \[\rho_n(\pos) = N\int_{-\infty}^\infty\mathrm{d}^3r_2 \dots \int_{-\infty}^\infty\mathrm{d}^3r_N \Psi^\star_n(\pos,\pos_2,\dots,\pos_N)\Psi_n(\pos,\pos_2,\dots,\pos_N)\]
- Equation of motion: \[\imag\hbar\pdv{}{t}\Psi_n(\pos,t) = \Hop\Psi_n(\pos,t)\]
- Expectation value: \[o = \braket{\Oop}_{\Psi}=\int_{-\infty}^\infty\mathrm{d}^3r_1 \dots \int_{-\infty}^\infty\mathrm{d}^3r_N \Psi^\dagger(\{\pos_i\})\,\Oop(\{\pos_i\})\,\Psi(\{\pos_i\})\]

Q: What is a unitary transformation? What are its properties regarding eigenvalues and expectation values?
A: An operator \(U\) on Hilbert space \(\mathcal{H}\) is unitary if
\[UU^\dagger=U^\dagger U = U^{\star,T} U = \bm{1}\means U^{-1} = U^\dagger\]
\(U\) is invertible, preserves all scalar products and if \(\Oop\) is hermitian, \(U\Oop U^\dagger\) is also hermitian.
Moreover, \(\Oop\) and \(U\Oop U^\dagger\) possess the same set of eigenvalues:
\[{\Oop\Psi=a\Psi\quad\text{and}\quad(U\Oop U^\dagger)(U\Psi)=a(U\Psi)}\]
and the expectation values are invariant under unitary transformations:
\[\braket{\Oop} = \braket{\Oop}{\Psi} = \braket{U^\dagger U\Oop U^\dagger U}{\Psi} = \braket{U\Oop U^\dagger}{U\Psi}\]

Q: What's the difference between the Schrödinger and Heisenberg picture? Write down Heisenberg's equation of motion!
A: - Schrödinger picture: The state \(\Psi\) is propagated in time while the operators (observables) are stationary. The e.o.m. is the time-dependent Schrödinger equation.
- Heisenberg picture: This picture relies on stationary states but time-propagated operators.
The Heisenberg e.o.m. reads:
\[{\dv{\Oop^{(\mathrm{H})}}{t}=\frac{\imag}{\hbar}\comm{\Hop}{\Oop^{(\mathrm{H})}}}\]
This is in complete analogy to the classical e.o.m. of Hamiltonian mechanics, if the \(\comm{\Oop^{(H)}}{\Hop}/(\imag\hbar)\) is considered to be the quantum analog of the Poisson bracket \(\pb{O}{H}\). If \(\Oop^{(H)}\) commutes with \(\Hop\), it is a constant of motion. The Heisenberg e.o.m. can be derived by choosing
\[U(t)=\exp\left(-\frac{\imag}{\hbar}\Hop^{(S)}t\right) \means U^\dagger(t)=\exp\left(\frac{\imag}{\hbar}\Hop^{(S)}t\right)\]

Q: How are the momentum and position operators postulated to look like? What is the Schrödinger velocity operator?
A: One possibility for the momentum and position operators, that fulfill the commutation relations correctly are
\[\begin{aligned} \pop &= -\imag\hbar\nabla=\frac{\hbar}{\imag}\nabla \rop \\ \hat x &= \pos \end{aligned}\]
They lead to energy eigenvalues which are in remarkable agreement with experimental results.
The Schrödinger velocity operator is obtained by using the Heisenberg e.o.m.:
\[{\hat{\vel} = \dot{\hat{\pos}}=\dv{\hat{\pos}}{t}=\frac{\imag}{\hbar}\comm{\Hop}{\hat{\pos}}=\frac{\hat{\mom}}{m}}\]

Q: Write down the general formulation of the Ehrenfest theorem!
A: \[\dv{}{t}\braket{\Oop}_\Psi=\frac{\imag}{\hbar}\braket{\comm{\Hop}{\Oop}}_\Psi + \braket{\pdv{\Oop}{t}}_\Psi\]
For the case, that \(\Psi\) is an eigenfunction of \(\Hop\) or if \(\Oop\) commutes with \(\Hop\), we get:
\[\dv{}{t}\braket{\Oop}_\Psi=\braket{\pdv{\Oop}{t}}_\Psi\]

Q: Write down the Hellmann-Feynman theorem!
A: Ehrenfest theorem is actually a special case of the Hellmann-Feynman theorem, which can be generalized for first derivatives of any parameter \(\lambda\), on which the wave function may depend. If we impose that \(\Psi\) must be an eigenfunction of \(\Hop\), the theorem states:
\[{\dv{}{\lambda}\braket{\Oop}_\Psi=\braket{\pdv{\Oop}{\lambda}}_\Psi}\]

Q: How is the density operator defined? How can we use this to derive the Schrödinger current density and the continuity equation?
A: Upon relating the expectation value of the density operator \(\braket{\hat{\rho}_{\pos}}\) to the Born interpretation \(\rho(\pos,t)=\braket{\hat{\rho}_{\pos}}\) one can deduce the explicit form of the density operator:
\[{\hat{\rho}_{\pos} = \sumitoN \delta^{(3)}(\pos_i - \pos)}\]
If we now take the time derivative and use the Ehrenfest-theorem, we get the continuity equation and Schrödinger current density for a single particle:
\[\dv{\rho(\pos,t)}{t}=\dv{}{t}\braket{\hat{\rho}_{\pos}} = \frac{\imag}{\hbar}\braket{\comm{\Hop}{\hat{\rho}_{\pos}}} + \braket{\pdv{\hat{\rho}_{\pos}}{t}}\]
\[{\Rightarrow \pdv{}{t}(\Psi^\star\Psi) = -\text{div} \underbrace{\frac{\hbar}{2m\,\imag}\left\{ \Psi^\star\nabla\Psi - \left(\nabla\Psi\right)^\star\Psi \right\}}_{=\bm{j}}}\]
The current density can also be expressed using the Schrödinger velocity operator:
\[\bm{j} = \frac{1}{2}\left(\Psi^\star\dot{\rop}\Psi+(\dot{\rop}\Psi)^\star\Psi\right) = \Re\{\Psi^\star\dot{\rop}\Psi\}\]

Q: Give the motivation for deriving the Klein-Gordon equation! Write down the equation both in normal and covariant form! What are the eigenfunctions and eigenvalues?
A: The problem with the Schrödinger quantum mechanics is, that it is not Lorentz covariant, since the time (first) and the spatial (second) derivates are of different order!
By starting from the energy-dispersion relation and using the TDSE, one can derive the Klein-Gordon equation for a free particle
\[\begin{aligned}E^2 = c^2\mom^2+m_e^2c^4 \quad \xrightarrow[\mom\ \rightarrow\ -\imag\hbar\nabla]{E\ \rightarrow\ \imag\hbar\,\partial/\partial t} \\ {-\hbar^2\pdv{^2}{t^2}\Psi = \left(-\hbar^2c^2 \nabla^2 + m_e^2c^4\right)\Psi} \end{aligned}\]
or in covariant form
\[{\left[\partial_\mu\partial^\mu + \left(\frac{m_ec}{\hbar}\right)^2\right]\Psi = \left[\dAlembert + \left(\frac{m_ec}{\hbar}\right)^2\right] = 0}\]
Possible eigenfunctions are give by plane waves of the form
\[\Psi(\pos,t)=\exp{\frac{\imag}{\hbar}\left(Et-\mom\cdot\pos\right)}\]
This yields the eigenvalues
\[{E^2 = c^2\mom^2+m_e^2c^4\quad\rightarrow\quad E=\pm\sqrt{c^2\mom^2+m_e^2c^4}}\]