update week 4

Author: mhjensen

doc/pub/week4/html/week4-bs.html

@@ -964,33 +964,43 @@ which can be rewritten as
 
 
 We can solve this equation iteratively keeping in mind the time-ordering of the of the operators
+!bt
 \[
 \hat{U}(t,t_0)=1-\frac{\imath}{\hbar}\int_{t_0}^t dt' \hat{H}_I(t')+\left(\frac{-\imath}{\hbar}\right)^2\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')+\dots
 \]
+!et
 The third term can be written as 
+!bt
 \[
 \int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')=
 \frac{1}{2}\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')
 +\frac{1}{2}\int_{t_0}^t dt''\int_{t''}^{t} dt' \hat{H}_I(t')\hat{H}_I(t'').
 \]
-
+!et
 
 !split
 ===== Interaction picture  =====
 
 
 We obtain this expression by changing the integration order in the second term
 via a change of the integration variables $t'$ and $t''$  in 
+!bt
 \[
 \frac{1}{2}\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'').
 \]
+!et
 We can rewrite the terms which contain the double integral as
+!bt
 \[
-\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')=\]
+\int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')=
+\]
+!et
+!bt
 \[
 \frac{1}{2}\int_{t_0}^t dt'\int_{t_0}^{t'} dt''\left[\hat{H}_I(t')\hat{H}_I(t'')\Theta(t'-t'')
 +\hat{H}_I(t')\hat{H}_I(t'')\Theta(t''-t')\right],
 \]
+!et
 with $\Theta(t''-t')$ being the standard Heavyside or step function. The step function allows us to give a specific time-ordering to the above expression.
 
 
@@ -1000,10 +1010,12 @@ with $\Theta(t''-t')$ being the standard Heavyside or step function. The step fu
 
 
 With the $\Theta$-function we can rewrite the last expression as 
+!bt
 \[
 \int_{t_0}^t dt'\int_{t_0}^{t'} dt'' \hat{H}_I(t')\hat{H}_I(t'')=
 \frac{1}{2}\int_{t_0}^t dt'\int_{t_0}^{t'} dt''\hat{T}\left[\hat{H}_I(t')\hat{H}_I(t'')\right],
 \]
+!et
 where $\Hat{T}$ is the so-called time-ordering operator. 
 
 
@@ -1013,11 +1025,13 @@ where $\Hat{T}$ is the so-called time-ordering operator.
 
 
 With this definition, we can rewrite the expression for $\hat{U}$ as 
+!bt
 \[
 \hat{U}(t,t_0)=\sum_{n=0}^{\infty}\left(\frac{-\imath}{\hbar}\right)^n\frac{1}{n!}
 \int_{t_0}^t dt_1\dots \int_{t_0}^t dt_N \hat{T}\left[\hat{H}_I(t_1)\dots\hat{H}_I(t_n)\right]=\hat{T}\exp{\left[\frac{-\imath}{\hbar}
 \int_{t_0}^t dt' \hat{H}_I(t')\right]}.
 \]
+!et
 The above time-evolution operator in the interaction picture will be used
 to derive various contributions to many-body perturbation theory. See also exercise 26 for a discussion of the various time orderings.
 
@@ -1028,59 +1042,75 @@ to derive various contributions to many-body perturbation theory. See also exerc
 
 
 We wish now to define a unitary transformation in terms of $\hat{H}$ by defining
+!bt
 \[
 |\Psi_H(t)\rangle = \exp{(\imath\hat{H}t/\hbar)}|\Psi_S(t)\rangle,
 \]
+!et
 which is again a unitary transformation carried out now at the time $t$ on the 
 wave function in the Schr\"odinger picture. If we combine this equation with 
 Schr\"odinger's equation we obtain the following equation of motion
+!bt
 \[
 \imath \hbar\frac{\partial }{\partial t}|\Psi_H(t)\rangle = 0,
 \]
+!et
 meaning that $|\Psi_H(t)\rangle$ is time independent. An operator in this picture is defined as
+!bt
 \[
 \hat{O}_H(t)=
 \exp{(\imath\hat{H}t/\hbar)}\hat{O}_S\exp{(-\imath\hat{H}t/\hbar)}.
 \]
-
+!et
 
 !split
 ===== Interaction picture  =====
 
 
 The time dependence is then in the operator itself, and this yields in turn the
 following equation of motion
+!bt
 \[
 \imath \hbar\frac{\partial }{\partial t}\hat{O}_H(t) = \exp{(\imath\hat{H}t/\hbar)}\left[\hat{O}_H\hat{H}-\hat{H}\hat{O}_H\right]\exp{(-\imath\hat{H}t/\hbar)}=\left[\hat{O}_H(t),\hat{H}\right].
 \]
+!et
 We note that an operator in the Heisenberg picture can be related to the corresponding
 operator in the interaction picture as 
+!bt
 \[
 \hat{O}_H(t)=
-\exp{(\imath\hat{H}t/\hbar)}\hat{O}_S\exp{(-\imath\hat{H}t/\hbar)}=\]
+\exp{(\imath\hat{H}t/\hbar)}\hat{O}_S\exp{(-\imath\hat{H}t/\hbar)}=
+\]
+!et
+!bt
 \[
 \exp{(\imath\hat{H}_It/\hbar)}\exp{(-\imath\hat{H}_0t/\hbar)}\hat{O}_I\exp{(\imath\hat{H}_0t/\hbar)}\exp{(-\imath\hat{H}_It/\hbar)}.
 \]
-
+!et
 
 !split
 ===== Interaction picture  =====
 
 With our definition of the time evolution operator we see that
+!bt
 \[
 \hat{O}_H(t)=\hat{U}(0,t)\hat{O}_I\hat{U}(t,0),
 \]
+!et
 which in turn implies that $\hat{O}_S=\hat{O}_I(0)=\hat{O}_H(0)$, all operators are equal at $t=0$. The wave function in the Heisenberg formalism is 
 related to the other pictures as 
+!bt
 \[
 |\Psi_H\rangle=|\Psi_S(0)\rangle=|\Psi_I(0)\rangle,
 \]
+!et
 since the wave function in the Heisenberg picture is time independent. 
 We can relate this wave function to that a given time $t$ via the time evolution operator as
+!bt
 \[
 |\Psi_H\rangle=\hat{U}(0,t)|\Psi_I(t)\rangle.
 \]
-
+!et
 
 !split
 ===== Interaction picture  =====
@@ -1090,9 +1120,11 @@ We assume that the interaction term is switched on gradually. Our wave function
 given by the solution to the unperturbed part of our Hamiltonian $\hat{H}_0$.
 We assume the ground state is given by $|\Phi_0\rangle$, which could be a Slater determinant.
 We define our Hamiltonian as
+!bt
 \[
 \hat{H}=\hat{H}_0+\exp{(-\varepsilon t/\hbar)}\hat{H}_I,
 \]
+!et
 where $\varepsilon$ is a small number. The way we write the Hamiltonian 
 and its interaction term is meant to simulate the switching of the interaction.
 
@@ -1102,46 +1134,58 @@ and its interaction term is meant to simulate the switching of the interaction.
 
 
 The time evolution of the wave function in the interaction picture is then
+!bt
 \[
 |\Psi_I(t) \rangle = \hat{U}_{\varepsilon}(t,t_0)|\Psi_I(t_0)\rangle,
 \]
+!et
 with 
+!bt
 \[
 \hat{U}_{\varepsilon}(t,t_0)=\sum_{n=0}^{\infty}\left(\frac{-\imath}{\hbar}\right)^n\frac{1}{n!}
 \int_{t_0}^t dt_1\dots \int_{t_0}^t dt_N \exp{(-\varepsilon(t_1+\dots+t_n)/\hbar)}\hat{T}\left[\hat{H}_I(t_1)\dots\hat{H}_I(t_n)\right]
 \]
-
+!et
 
 !split
 ===== Interaction picture  =====
 
 In the limit $t_0\rightarrow -\infty$, the solution ot Schr\"odinger's equation is
 $|\Phi_0\rangle$, and the eigenenergies are given by 
+!bt
 \[
 \hat{H}_0|\Phi_0\rangle=W_0|\Phi_0\rangle,
 \]
+!et
 meaning that 
+!bt
 \[
 |\Psi_S(t_0)\rangle = \exp{(-\imath W_0t_0/\hbar)}|\Phi_0\rangle,
 \]
+!et
 with the corresponding interaction picture wave function given by
+!bt
 \[
 |\Psi_I(t_0)\rangle = \exp{(\imath \hat{H}_0t_0/\hbar)}|\Psi_S(t_0)\rangle=|\Phi_0\rangle.
 \]
-
+!et
 
 
 !split
 ===== Interaction picture  =====
 
 The solution becomes time independent in the limit $t_0\rightarrow -\infty$.
 The same conclusion can be reached by looking at 
+!bt
 \[
 \imath \hbar\frac{\partial }{\partial t}|\Psi_I(t)\rangle =
 \exp{(\varepsilon |t|/\hbar)}\hat{H}_I|\Psi_I(t)\rangle 
 \]
+!et
 and taking the limit $t\rightarrow -\infty$.
 We can rewrite the equation for the wave function at a time $t=0$ as
+!bt
 \[
 |\Psi_I(0) \rangle = \hat{U}_{\varepsilon}(0,-\infty)|\Phi_0\rangle.
 \]
+!et