\section{Quantisation} For the quantization of a given system with \textbf{well defined equations of motion} it is necessary to \textbf{guess a Lagrangian} $L(q_i,\dot{q}_i)$, i.e. to construct a Lagrangian by inspection. $q_i$ are the positional coordinates and $\dot{q}_i$ the generalized velocities. With the \textbf{Euler-Lagrange equations} \begin{equation} \label{eq:euler_lagrange} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}-\frac{\partial L}{\partial q_i}=0 \end{equation} it must be possible to come back to the equations of motion with the constructed Lagrangian.\footnote{~There is no standard procedure for constructing a Lagrangian like in classical mechanics where the Lagrangian is constructed with the kinetic energy $T$ and the potential $U$: \begin{equation} \nonumber L=T-U=\frac{1}{2}m\dot{x}^2-U \end{equation} The mass $m$ is the big problem - it isn't always as well defined as for example in classical mechanics.} The next step is to \textbf{get the conjugate variable} $p_i$: \begin{equation} \label{eq:conjugate_variable} p_i=\frac{\partial L}{\partial \dot q_i} \end{equation} Then the \textbf{Hamiltonian} has to be derived with a Legendre transformation: \begin{equation} \label{eq:hamiltonian} H=\sum_i{p_i\dot q_i}-L \end{equation} Now the \textbf{conjugate variables have to be replaced by the given operators in quantum mechanics}. For example the momentum $p$ with \begin{equation} \nonumber p\rightarrow -i\hbar\nabla. \end{equation} Now the Schr\"{o}dinger equation \begin{equation} \label{eq:schroedinger_simple} H\Psi(q)=E\Psi(q) \end{equation} is ready to be evaluated. \FloatBarrier \subsection{Harmonic Oscillator} The first example how a given system can be quantized is a one dimensional harmonic oscillator. It's assumed that just the equation of motion \begin{equation} \nonumber m\ddot x=-kx \end{equation} with $m$ the mass and the spring constant $k$ is known from this system. Now the Lagrangian $L$ must be constructed by inspection: \begin{equation} \nonumber L(x,\dot x)=\frac{m\dot x^2}{2}-\frac{kx^2}{2} \end{equation} It is the right one if the given equation of motion can be derived by the Euler-Lagrange eqn. (\ref{eq:euler_lagrange}), like in this example. The next step is to get the conjugate variable, the generalized momentum $p$: \begin{equation} \nonumber p=\frac{\partial L}{\partial \dot x}=m\dot x \end{equation} Then the Hamiltonian has to be constructed with the Legendre transformation: \begin{equation} \nonumber H=\sum_i{p_i\dot q_i}-L=\frac{p^2}{2m}+\frac{kx^2}{2}. \end{equation} With replacing the conjugate variable $p$ with \begin{equation} \nonumber p\rightarrow -i\hbar\frac{\partial}{\partial x} \end{equation} because of position space\footnote{position space: Ortsraum} and inserting in eqn. (\ref{eq:schroedinger_simple}) the Schr\"{o}dinger equation for the one dimensional harmonic oscillator is derived: \begin{equation} \nonumber -\frac{\hbar^2}{2m}\frac{\partial^2\Psi(x)}{\partial x^2}+\frac{kx^2}{2}\Psi(x)=E\Psi(x) \end{equation} \FloatBarrier \subsection{Magnetic Flux} The next example is the quantization of magnetic flux in a superconducting ring with a Josephson junction and a capacitor parallel to it (see fig. \ref{fig:01_superconducting_ring}). The equivalent to the equations of motion is here the equation for current conservation: \begin{equation} \nonumber C\ddot\Phi+I_Csin\left(\frac{2\pi\Phi}{\Phi_0}\right)=-\frac{\Phi-\Phi_e}{L_1} \end{equation} What are the terms of this equation? The charge $Q$ in the capacitor with capacity $C$ ist given by \begin{equation} \nonumber Q=C\cdot V \end{equation} with $V$ as the voltage applied. Derivate this equation after the time $t$ gives the current $I$: \begin{equation} \nonumber \dot Q=I=C\cdot\dot V \end{equation} With Faraday's law $\dot\Phi=V$ ($\Phi$ as the magnetic flux) this gives the first term for the current through the capacitor \begin{equation} \nonumber I=C\cdot\ddot\Phi. \end{equation} The current $I$ in a Josephson junction is given by \begin{equation} \nonumber I=I_c\cdot sin\left(\frac{2\pi\Phi}{\Phi_0}\right) \end{equation} with the flux quantum $\Phi_0$, which is the second term. The third term comes from the relationship between the flux $\Phi$ in an inductor with inductivity $L_1$ and a current $I$: \begin{equation} \nonumber \Phi=L_1\cdot I \end{equation} With the applied flux $\Phi_e$ and the flux in the ring this results in \begin{equation} \nonumber I=\frac{\Phi-\Phi_e}{L_1}, \end{equation} which is the third term. In this example it isn't as easy as before to construct a Lagrangian. But luckily we are quite smart and so the right Lagrangian is \begin{equation} \nonumber L(\Phi,\dot\Phi)=\frac{C\dot\Phi^2}{2}-\frac{(\Phi-\Phi_e)^2}{2L_1}+E_jcos\left(\frac{2\pi\Phi}{\Phi_0}\right). \end{equation} The conjugate variable is \begin{equation} \nonumber \frac{\partial L}{\partial \dot\Phi}=C\dot\Phi=Q. \end{equation} Then the Hamiltonian via a Legendre transformation: \begin{equation} \nonumber H=Q\dot\Phi-L=\frac{Q^2}{2C}+\frac{(\Phi-\Phi_e)^2}{2L_1}-E_jcos\left(\frac{2\pi\Phi}{\Phi_0}\right) \end{equation} Now the replacement of the conjugate variable, i.e. the quantization: \begin{equation} \nonumber Q \rightarrow -i\hbar\frac{\partial}{\partial\Phi} \end{equation} The result is the wanted Schr\"{o}dinger equation for the given system: \begin{equation} \nonumber -\frac{\hbar^2}{2C}\frac{\partial^2\Psi(\Phi)}{\partial\Phi^2}+\frac{(\Phi-\Phi_e)^2}{2L_1}\Psi(\Phi)-E_jcos\left(\frac{2\pi\Phi}{\Phi_0}\right)\Psi(\Phi)=E\Psi(\Phi) \end{equation} \begin{figure}[h]\centering \includegraphics[width=0.3\textwidth]{./pictures/01_superconducting_ring} \caption{Superconducting ring with a capacitor parallel to a Josephson junction} \label{fig:01_superconducting_ring} \end{figure} \FloatBarrier \subsection{Charged particle in a magnetic field (with spin)} For now we will neglect that the particle has spin, because in the quantisation we will treat the case of a constant magnetic field pointing in the z-direction. In this simplified case there is no contribution to the force from the spin interaction with the magnetic field (because it only adds a constant to the Hamiltonian which does not affect the equation of motion).\\ We start with the equation of motion of a charged particle (without spin) in an EM-field, which is given by the Lorentz force law. \begin{equation} \mathbf{F}=m\frac{\partial^2\mathbf{r}(t)}{\partial t^2}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) \label{eq:Lorentz force law} \end{equation} One can check that a suitable Lagrangian is \begin{equation} L(\mathbf{r},t)=\frac{m\mathbf{v}^2}{2}-q\phi +q\mathbf{v}\cdot\mathbf{A} \label{eq:EM lagrangian} \end{equation} where $\phi$ and $\mathbf{A}$ are the scalar and vector Potential, respectively. From the Lagrangian we get the generalized momentum (canonical conjugated variable) \begin{align} \notag p_i&=\frac{\partial L}{\partial \dot{r}_x}=mv_x+qA_x \end{align} \begin{equation} \mathbf{p}=m\mathbf{v}+q\mathbf{A} \label{eq:generalized momentum} \end{equation} \begin{align} \notag \mathbf{v}(\mathbf{p})&=\frac{1}{m}(\mathbf{p}-q\mathbf{A}) \end{align} The conjugated variable to position has now two components, the first therm in eqn.(\ref{eq:generalized momentum}) $m\mathbf{v}$ which is the normal momentum. The second term $q\mathbf{A}$ which is called the field momentum. It enters in this equation because when a charged particle is accelerated it also creates a Magnetic field. The creation of the magnetic field takes some energy, which is then conserved in the field. When one then tries to stop the moving particle one has to overcome the kinetic energy plus the energy that is conserved in the magnetic field, because the magnetic field will keep pushing the charged particle in the direction it was going ( this is when you get back the energy that went into creating the field, also called self inductance).\\ To construct the Hamiltonian we have to perform a Legendre transformation from the velocity $\mathbf{v}$ to the generalized momentum $\mathbf{p}$. \begin{align} \notag H&=\sum_{i=1}^3 p_iv_i(p_i)-L(\mathbf{r},\mathbf{v}(\mathbf{p},t)\\ \notag H&=\mathbf{p}\cdot\frac{1}{m}(\mathbf{p}-q\mathbf{A})-\frac{1}{2m}(\mathbf{p}-q\mathbf{A})^2+q\phi-\frac{q}{m}(\mathbf{p}-q\mathbf{A})\cdot\mathbf{A} \end{align} Which then reduces to \begin{equation} H=\frac{1}{2m}(\mathbf{p}-q\mathbf{A})^2+q\phi\\ \label{eq:Hamilton electron in EM field} \end{equation} Eqn.(\ref{eq:Hamilton electron in EM field}) is the Hamiltonian for a charged particle in an EM-field without Spin. The QM Hamiltonian of a single Spin in a magnetic field is given by \begin{equation} H=-\hat{\mathbf{m}}\cdot\mathbf{B} \label{eq:Spin Hamilton} \end{equation} where \begin{equation} \hat{\mathbf{m}}=\frac{2g\mu_B}{\hbar}\hat{\mathbf{S}} \label{eq: spin magnet moment} \end{equation} $\mu_B=\frac{q\hbar}{2m}$ the Bohr Magnetron and $\hat{\mathbf{S}}$ is the QM Spin operator. By Replacing the momentum and location in the Hamiltonian by their respective QM operators we obtain the QM Hamiltonian for our System \begin{equation} \hat{H}=\frac{1}{2m}(\hat{\mathbf{p}}-q\mathbf{A}(\hat{\mathbf{r}}))^2+q\phi(\hat{\mathbf{r}})-\frac{2g\mu_B}{\hbar}\hat{\mathbf{S}}\cdot \mathbf{B}(\hat{\mathbf{r}}) \label{eq:QM Hamilton} \end{equation} We consider the simple case of constant magnetic field pointing in the z-direction $\mathbf{B}=(0,0,B_z)$ and $\phi=0$. Plugging this simplifications in eqn.(\ref{eq:QM Hamilton}) yields \begin{equation} \hat{H}=\frac{1}{2m}(\hat{\mathbf{p}}-q\mathbf{A}(\hat{\mathbf{r}}))^2-\frac{2g\mu_B}{\hbar}B_z\hat{S}_z\equiv \hat{H}_1+\hat{S} \label{eq:QM Hamilton simple} \end{equation} We see that the Hamiltonian consists of two non interacting parts $\hat{H}_1=\frac{1}{2m}(\hat{\mathbf{p}}-q\mathbf{A}(\hat{\mathbf{r}}))^2$ and $\hat{S}=-\frac{2g\mu_B}{\hbar}B_z\hat{S}_z$. The state vector of an electron consists of a spacial part and a Spin part \begin{equation} \notag \vert \Psi\rangle = \vert \psi_{Space} \rangle \otimes \vert \chi_{Spin} \rangle \end{equation} Where the Symbol $\otimes$ denotes a so called tensor product. The Schroedinger equation reads \begin{align} \notag (\hat{H}_1\otimes\hat{\|}+\hat{S}\otimes\hat{\|})\vert \psi_{Space} \rangle \otimes \vert \chi_{Spin} \rangle &=E\vert \psi_{Space} \rangle \otimes \vert \chi_{Spin} \rangle \\ \hat{H}_1\vert \psi\rangle\otimes \vert \chi \rangle+\vert \psi\rangle\otimes\hat{S}\vert \chi \rangle &=E\vert \psi \rangle \otimes \vert \chi \rangle \label{eq:vector schroedinger total} \end{align} From eqn.(\ref{eq:QM Hamilton simple}), we see that the Spin part is essentially the $S_z$ operator. It follows that the Spin part of the state vector is $\vert \chi\rangle=\vert \pm z\rangle$. Thus, we have \begin{equation} \hat{S}\vert \chi\rangle=-\frac{2g\mu_B}{\hbar}B_z\hat{S}_z\vert \pm z\rangle=\mp g\mu_B B_z\vert \pm z\rangle \label{eq:Sz eigenvalue} \end{equation} where we used $\hat{S}_z\vert \pm z\rangle=\pm\frac{\hbar}{2}$. Using eqn.(\ref{eq:Sz eigenvalue}), we can rewrite the Schroedinger equation, eqn.(\ref{eq:vector schroedinger total}), resulting in \begin{align} \notag \hat{H}_1\vert \psi\rangle\otimes \vert \pm z\rangle\mp\vert \psi\rangle\otimes g\mu_B B_z\vert \pm z\rangle &=E\vert \psi \rangle \otimes \vert \pm z \rangle\\ \hat{H}_1\vert \psi\rangle\otimes \vert \pm z \rangle &=(E\pm g\mu_B B_z)\vert \psi\rangle\otimes\vert \pm z\rangle\equiv E^{\prime}\vert \psi\rangle\otimes\vert \pm z\rangle \label{eq:schroedinger1} \end{align} Where we define $E^{\prime}=E\pm g\mu_B B_z$. In order to obtain a differential equation in space variables, we multiply eqn.(\ref{eq:schroedinger1}) by $\langle \mathbf{r}\vert\otimes\langle\sigma\vert$ from the left. \begin{align} \notag (\langle \mathbf{r}\vert\otimes\langle\sigma\vert)(\hat{H}_1\vert \psi\rangle\otimes \vert \pm z\rangle) &=E^{\prime}(\langle \mathbf{r}\vert\otimes\langle\sigma\vert)(\vert \psi\rangle\otimes \vert \pm z\rangle) \\ \langle \mathbf{r}\vert \hat{H}_1\vert\psi\rangle\langle\sigma\vert\pm z\rangle &=E^{\prime}\langle \mathbf{r}\vert\psi\rangle\langle\sigma\vert\pm z\rangle \label{eq:schroedinger2} \end{align} Canceling the Spin wave function $\langle\sigma\vert\pm z\rangle$ on both sides of eqn.(\ref{eq:schroedinger2}) and plugging in the definition $\hat{H}_1=\frac{1}{2m}(\hat{\mathbf{p}}-q\mathbf{A}(\hat{\mathbf{r}}))^2$ from eqn.(\ref{eq:vector schroedinger total}) yields \begin{equation} \frac{1}{2m}(\hbar\nabla+q\mathbf{A}(\mathbf{r}))^2\psi(\mathbf{r},t)=E^{\prime}\psi(\mathbf{r},t) \label{eq:schroedinger3} \end{equation} For a magnetic field $\mathbf{B}=(0,0,B_z)$ we can choose the vector potential $\mathbf{A}(\mathbf{r})=B_zx\mathbf{y}$ (this is called Landau gauge). Using the Landau gauge and multiplying out the square leaves us with \begin{equation} \frac{1}{2m}(-\hbar^2(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})+2i\hbar qB_zx\frac{\partial}{\partial y}+q^2B_z^2x^2)\psi=E^{\prime}\psi \label{eq:schroedinger4} \end{equation} %%%%%%%%%%%%%%%%%%%ab hier beide versionen gleich weiter%%%%%%% In eqn.(\ref{eq:schroedinger4}) only $x$ appears explicitly, which motivates the ansatz \begin{equation} \psi(\mathbf{r},t)=e^{i(k_zz+k_yy)}\phi(x) \label{eq:ansatz psi} \end{equation} After plugging this ansatz into eqn.(\ref{eq:schroedinger4}), we get \begin{equation} \frac{1}{2m}(-\hbar^2\phi^{\prime\prime}+\hbar^2(k_y^2+k_z^2)\phi-2\hbar qB_zk_yx\phi+q^2B_z^2x^2\phi)=E^{\prime}\phi \label{eq:phi1} \end{equation} By completing the square we get a term $-\hbar^2k_y^2$ which cancels with the one already there and we are left with \begin{align} \notag \frac{1}{2m}(-\hbar^2\phi^{\prime\prime}+\hbar^2 k_z^2\phi+(qB_zx-\hbar k_y)^2\phi)&=E^{\prime}\phi\\ \notag \frac{1}{2m}(-\hbar^2\phi^{\prime\prime}+q^2B_z^2 (x-\frac{\hbar k_y}{qB_z})^2\phi)&=(E^{\prime}-\frac{\hbar^2 k_z^2}{2m})\phi\equiv E^{\prime\prime}\phi\\ -\frac{\hbar^2}{2m}\phi^{\prime\prime}+\frac{1}{2}m\frac{q^2B_z^2}{m^2}(x-\frac{\hbar_y}{qB_z})^2&=E^{\prime\prime}\phi \label{eq:phi2} \end{align} We see that the $z$-part of the Energy $E_z=\frac{\hbar^2k_z^2}{2m}$ is the same as for a free electron. This is because a constant magnetic field pointing in the z-direction leaves the motion of the particle in the z-direction unchanged. Comparing eqn.(\ref{eq:phi2}) to the equation of a harmonic oscillator \begin{equation} -\frac{\hbar}{2m}\phi^{\prime\prime}+\frac{1}{2}m\omega^2(x-x_o)^2=E\phi \label{ eq:harmonic oszi} \end{equation} with energies $E_n=\hbar\omega(n+\frac{1}{2})$, we obtain \begin{align} \notag E_n^{\prime\prime}&=\hbar\frac{qB_z}{m}(n+\frac{1}{2})\\ \notag E_n^{\prime}&=\hbar\frac{qB_z}{m}(n+\frac{1}{2})+\frac{\hbar^2k_z^2}{2m}\\ E_n&=\hbar\frac{qB_z}{m}(n+\frac{1}{2})+\frac{\hbar^2k_z^2}{2m}\mp g\mu_B B_z \label{eq:Energies} \end{align} Where each energy level $E_n$ is split into two levels because of the term $\mp g\mu_B B_z$. Where the minus is for Spin up (parallel to $\mathbf{B}$), which lowers the energy because the Spin is aligned and plus for Spin down (anti parallel). It should be noted that $\omega_c=\frac{qB_z}{m}$ is also the classical angular velocity of an electron in a magnetic field. Rewriting eqn.(\ref{eq:Energies}) by using the cyclotron frequency $\omega_c$ and $\mu_B=\frac{q\hbar}{2m} $, we end up with \begin{align} \notag E_n&=\frac{\hbar^2k_z^2}{2m}+\hbar\omega_c(n+\frac{1}{2})\mp \frac{g}{2}\omega_c\\ E_n&=\frac{\hbar^2k_z^2}{2m}+\hbar\omega_c(n+\frac{1\mp g}{2}) \label{eq:energy omega c} \end{align} \subsection{Dissipation} Quantum coherence is maintained until the decoherence time. This depends on the strength of the coupling of the quantum system to other degrees of freedom. As an example: Schr\"{o}dinger's cat. The decoherence time is very short, because there is a lot of coupling with other, lots of degrees of freedom. Therefore a cat, which is dead and alive at the same time, can't be investigated. %PICTURE?? cat In solid state physics energy is exchanged between the electrons and the phonons. There are Hamiltonians $H$ for the electrons ($H_e$), the phonons ($H_{ph}$) and the coupling between both ($H_{e-ph}$). The energy is conserved, but the entropy of the entire system increases because of these interactions. \begin{equation} \nonumber H = H_e+ H_{ph}+ H_{e-ph} \end{equation} %ev. einbauen: 25.02.2009 - transmission line