Commit c7a171eb by juandaanieel

### final text adiabatic and animations

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 ... ... @@ -219,12 +219,7 @@ \$\$ \gamma_n(t) = i \int_{\vec{R}(0)}^{\vec{R}(t)} \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle \cdot d\vec{R}. \$\$ The Aharonov-Bohm effect arises as an extra phase due to the coupling of the wavefunction with the vector potential even in a region where there's no \$\mathbf{B}\$. The magnitude of this phase is proportional to the magnetic flux going through the solenoid. That is, The Aharonov-Bohm effect arises as an extra phase due to the coupling of the wavefunction with the vector potential when traveling around a solenoid. The acquired phase is proportional to the magnetic flux going through the solenoid. That is, \$\$ \gamma = \pm \frac{e \phi}{2\hbar}. \$\$ The phase acquired moving around the solenoid can be described as a Berry phase by observing that, \$\$ \langle \psi| \nabla_{\mathbf{R}} \psi \rangle = -\frac{ie}{\hbar} \mathbf{A}(\mathbf{R}). \$\$ This relation holds when \$\langle \mathbf{p} \rangle = 0\$, i.e. for a particle in a box.
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 ... ... @@ -61,7 +61,7 @@ \$\$ \Psi(t) = \sum_n c_n(t) \psi_n(t) e^{i \theta_n(t)}. \$\$ where \$\theta_n(t) = -\frac{1}{\hbar} \int_0^t E_n(t) dt\$. What we do here is expand the wave function \$\Psi(t)\$ in a different basis for every value time \$t\$ (using the set of \$\{\psi_n(t)\}\$). This is allowed, for a given value of time \$t\$ the wave function \$Psi(t)\$ can be expanded in any arbitrary basis set. where \$\theta_n(t) = -\frac{1}{\hbar} \int_0^t E_n(t) dt\$. What we do here is expand the wave function \$\Psi(t)\$ in a different basis for every value time \$t\$ (using the set of \$\{\psi_n(t)\}\$). This is allowed, for a given value of time \$t\$ the wave function \$\Psi(t)\$ can be expanded in any arbitrary basis set. We could have absorbed the exponential factor involving \$\theta\$ into the \$c_n(t)\$, but as we will see it is convenient to include it explicitly. This term may look strange, but in fact it is a straightforward generalisation of the phase factor that an eigenstate picks up, which would be \$\exp (-i E_n t/ \hbar)\$, when using a constant Hamiltonian. ... ... @@ -187,13 +187,13 @@ \$\$ c_n(T) = \frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle}}{\overline{E_m - E_n}} \frac{i\hbar }{\overline{E_m - E_n}}\left( e^{-i \overline{E_m - E_n} T /\hbar } - 1 \right). \$\$ The coefficients for \$n \neq m\$ have been neglected using the argument that they are small. This can be explicitly written by saying that they are much smaller than 1. That is, We have found an expression for the coefficients where \$n \neq m\$. The addibatic approximation assumes that these coefficients are small. This criteria can be explicitly written as, \$\$ |c_n(T)| \leq \frac{\hbar \overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_m - E_n}^2} << 1 \$\$ Observe that from the following result we recover what we first presented at this section. We recover what we first presented at this section. ## Summary ... ...
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 ... ... @@ -214,17 +214,12 @@ \$\$ ## Summary The Berry phase can be expressed in terms of an arbitrary time-independent parameter \$\mathbf{R}\$ as, The Berry phase can be expressed in terms of an arbitrary time-dependent parameter \$\mathbf{R}(t)\$ as, \$\$ \gamma_n(t) = i \int_{\vec{R}(0)}^{\vec{R}(t)} \langle{ \psi_n | \vec{\nabla_R}\psi_n} \rangle \cdot d\vec{R}. \$\$ The Aharonov-Bohm effect arises as an extra phase due to the coupling of the wavefunction with the vector potential even in a region where there's no \$\mathbf{B}\$. The magnitude of this phase is proportional to the magnetic flux going through the solenoid. That is, The Aharonov-Bohm effect arises as an extra phase due to the coupling of the wavefunction with the vector potential when traveling around a solenoid. The acquired phase is proportional to the magnetic flux going through the solenoid. That is, \$\$ \gamma = \pm \frac{e \phi}{2\hbar}. \$\$ The phase acquired moving around the solenoid can be described as a Berry phase by observing that, \$\$ \langle \psi| \nabla_{\mathbf{R}} \psi \rangle = -\frac{ie}{\hbar} \mathbf{A}(\mathbf{R}). \$\$ This relation holds when \$\langle \mathbf{p} \rangle = 0\$, i.e. for a particle in a box. \ No newline at end of file