'perturbations'

ch06/note01.md

@@ -1,33 +1,37 @@
 ## Perturbation Method
 
 
-```{admonition} What you need to know
+:::{admonition} What you need to know
 :class: note
--  *Perturbation method*, attempts to solve analytically intractable problem by idintifying an exactly *sovable* part and a *"small"* pertrubation. The method is similair in spirit to taylor expansion of functions familiar from calculus and is the single most important method of solving problems in quantum mechanics. 
 
-- Application of pertrubation theory proceeds in two steps. Step one identify solvable part and perturbation. Part two expand energy and eigenfunctions as series of corrections of increasing order. In particle first and second order corrections to energy suffice. 
-```
+-  **Perturbation method**, attempts to solve analytically intractable problems by idintifying an exactly *sovable* part and a *"small"* pertrubation to it. 
+- The method is similair in spirit to taylor expansion of functions familiar from calculus and is a powerful tool for solving problems in quantum mechanics. 
+
+- Application of pertrubation theory proceeds in two steps. Step one identify solvable part and perturbation. Part two expand energy and eigenfunctions as series of corrections of increasing order. In practice first and second order corrections to energy are sufficient to get quantiatively accurate results. 
+:::
 
   ![](./images/perturb1.png)
 
 
 ## Time independent perturbations
 
-We have hamiltonian $\hat{H}_0$ for some exactly solvable problem think particle in a box, harmonic oscilator, etc:  
+- We have hamiltonian $\hat{H}_0$ for some exactly solvable problem think particle in a box, harmonic oscilator, etc:  
 
 $$
 \hat{H}^0 \mid n^0\rangle=E^0_n \mid n^0\rangle
 $$
 
-- Note the 0 superscript: It indicates exactly solvable hamitlonian, eigenfunctions and eigenvalues. The  $\mid n^0\rangle$ is the eigenfunction corepsonding to the nth eigenvalue $E^0_n$. Now we perturb the hamitlonian by adding a "small" pertrubation $\hat{H_1}$. Where  $\lambda$ turns perturbation on $\lambda=1$ and off $\lambda=0$. 
+- Note the 0 superscript: It indicates exactly solvable hamitlonian, eigenfunctions and eigenvalues. The  $\mid n^0\rangle$ is the eigenfunction corepsonding to the nth eigenvalue $E^0_n$. 
+
+- Consider a problem where we have a hamitlonian which is similiar to an exactly solvable problem differing only by a "small" pertrubation $\hat{H_1}$. Small means that eiganvalues should of two systems are close relative to spacing. 
+
+- The parameter  $\lambda$ turns perturbation on $\lambda=1$ and off $\lambda=0$. 
 
 $$
 \hat{H}=\hat{H}^0+\lambda {\hat{H}^1}
 $$
 
----
-
--  The objective of  perturbation theory is to solve the following problem with new hamiltonian expressed entirely in terms of eigenvalues and eigenfunctions of exactly solvable problems. 
+-  The objective of  perturbation theory is to solve the problem with new hamiltonian expressed entirely in terms of eigenvalues and eigenfunctions of exactly solvable problems. 
 
 $$
 \hat{H}\mid n\rangle =E_n \mid n\rangle
@@ -41,24 +45,25 @@ $$
 ### It's just like Taylor expansions! 
 
 
-
-We assume that eigenvalues and eigenfunctions can be expanded in power series in the parameter $\lambda$ to be set to 1 in the end.  
+- We assume that eigenvalues and eigenfunctions can be expanded in power series in the parameter $\lambda$ to be set to 1 in the end.  
 
 
 $$
-E_n =E^0_n+\lambda E^1_n+\lambda^2 E^2_n+...
+E_n ={\color{green}E^0_n}+{\color{red}\lambda E^1_n}+{\color{blue}\lambda^2 E^2_n}+...
 $$
 
 
 $$
-\mid n\rangle = \mid n^0\rangle+\lambda\mid n^1\rangle+\lambda^2\mid n^2\rangle ...
+\mid n\rangle = {\color{green}\mid n^0\rangle}+{\color{red} \lambda\mid n^1\rangle}+{\color{blue} \lambda^2\mid n^2\rangle} ...
 $$
 
+- Pluggin in the expansions into $\hat{H}\mid n\rangle =E_n \mid n\rangle$ we get
 
 $$
-\Big(\hat{H}^0+\lambda \hat{H}^1 \Big)\Big(\mid n^0\rangle+\lambda\mid n^1\rangle \Big)  = \Big(E^0_n+\lambda E^1_n+\lambda^2 E^2_n \Big) \Big(\mid n^0\rangle+\lambda\mid n^1\rangle\Big)
+\Big({\color{green}\hat{H}^0}+{\color{red}\lambda \hat{H}^1} \Big)\Big({\color{green}\mid n^0\rangle}+{\color{red}\lambda\mid n^1\rangle} \Big)  = \Big({\color{green}E^0_n}+{\color{red}\lambda E^1_n}+{\color{blue}\lambda^2 E^2_n}\Big) \Big({\color{green}\mid n^0\rangle}+{\color{red}\lambda\mid n^1\rangle}\Big)
 $$
 
+- Now we expand terms and collct products according to order of lambd:
 
 $$
 \begin{split}\color{green}{\hat{H}^0 \mid n^0\rangle} + \lambda \color{red}{ \big (\hat{H}^1 \mid n^0\rangle + \hat{H}^0 \mid n^1\rangle \big)}+\lambda^2 \color{blue}{\hat{H}^1 \mid n^1\rangle }  = \\ = \color{green}{E^0_n \mid n^0\rangle} +\lambda \color{red}{ \big( E^1_n \mid n^0\rangle+ E^0_n \mid n^1\big )} +  \lambda^2\color{blue}{\big(E^1_n \mid n^1\rangle+E^2_n \mid n^0\rangle \big)}\end{split}
@@ -68,7 +73,7 @@ $$
 
 
 
-### Petrubation equations of order 0, 1 and 2
+### Pertubation equations of order $0$, $1$ and $2$.
 
 Opening the brackets and collecting different orders of $\lambda$ we have  0, 1 and 2nd order perturbation equations:
 
@@ -94,12 +99,24 @@ $$
 - Note that hamitonian only has first order expansion while eigenfunctions and eigenvalues are expanded to infinite terms. Usually going to second order is enough for most problems. 
 
 
+### Computing pertrubation correction to energy levels 
 
+$$
+\boxed{E_n = \color{green}{E^0_n} + \color{red}{\langle n^0\mid H^1\mid n^0\rangle} + \color{blue}{\sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}}}
+$$
 
+- Where the matrix elements are $H_{nk}=\langle n^0\mid H^1\mid k^0\rangle$.
+
+- The energy in the denominator of 2nd order term is the difference between energy of a given state $E_n$ from all other states $E_k$ with k being the summation index. 
+- If the matrix elements of $\hat{H}^1$ are of comparable magnitude the neighbouring levels make larger contributions that distance levels.
 
-###  Fixing the normalization
+### Derivations of 1st and 2nd order corrections
 
-If we have normalization the zero order eigenfunctions, then unperturbed eigenfunctions will be orthogonal to all higher order eigenfunctions:
+:::{admonition} **Deriving 1st order correction to energy $E^1_n$**
+
+**Fixing the normalization**
+
+If we have normalization of the zero order eigenfunctions, then unperturbed eigenfunctions will be orthogonal to all higher order eigenfunctions:
 
 
 
@@ -117,77 +134,63 @@ $$
 \langle n^0 \mid n^{k} \rangle=0\,\,\, k=1,2,..
 $$
 
+**Making use of the orthogonality**
 
+$$
+{\hat{H}^0\mid n^1\rangle +\hat{H^1}\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}
+$$
 
-
-
-### 1st order correction to energy $E^1_n$
-
-We multiply  first order pertrubation equation by $\langle n^0 \mid$.  The first terms on the right is zero becasue of rothogonality. The first term on left is alos zero because of orhtogonality and hermitian property :  $\langle n^0 \mid \hat{H}^0\mid n^1\rangle = \langle n^1 \mid \hat{H}^0\mid n^0\rangle^* = E^0_n \langle n^1 \mid n^0\rangle^* = 0$
+- We multiply  first order pertrubation equation by $\langle n^0 \mid$ and use hermitian property of Hamiltonian.  
 
 
 $$
-\color{red}{\hat{H}^0\mid n^1\rangle +\hat{H^1}\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}
+{{\langle n^0} \mid \hat{H}^0\mid n^1\rangle +{\langle n^0} \mid  \hat{H^1}\mid n^0\rangle = E^0_n{\langle n^0} \mid n^1 \rangle+E^1_n {\langle n^0} \mid n^0 \rangle}
 $$
 
 
-$$
-\color{red}{ \color{black}{\langle n^0} \mid \hat{H}^0\mid n^1\rangle +\color{black}{\langle n^0} \mid  \hat{H^1}\mid n^0\rangle = E^0_n\color{black}{\langle n^0} \mid n^1 \rangle+E^1_n \color{black}{\langle n^0} \mid n^0 \rangle}
-$$
+- The first terms on the right and left hand side are zero becasue of rothogonality:  $\langle n^0 \mid \hat{H}^0\mid n^1\rangle = \langle n^1 \mid \hat{H}^0\mid n^0\rangle = E^0_n \langle n^1 \mid n^0\rangle = 0$
+
+**First order correction**
 
+- We have obtained the central result of pertrubation theory: *The 1st order correction to energy* $E^1_n$. 
 
 $$
 \color{red}{E_n^1 = \langle n^0 \mid \hat{H}^1\mid n^0 \rangle}
 $$
 
----
-
-- We have obtained the central result of pertrubation theory: *The 1st order correction to energy* $E^1_n$ 
-
 
 $$
-E_n=\color{green}{E^0_n}+\color{red}{E^1_n}=\color{green}{E^0_n}+\color{red}{\langle n^0 \mid \hat{H}^1\mid n^0 \rangle}
+\boxed{E_n=\color{green}{E^0_n}+\color{red}{E^1_n}=\color{green}{E^0_n}+\color{red}{\langle n^0 \mid \hat{H}^1\mid n^0 \rangle}}
 $$
 
-
-
-
 - Note how this expression is different from expectation expressions we have seen before. Here the eigenfunctions of $\hat{H}^0$ sandwich the the pertrubation hamitlonian $\hat{H}^1$ . The two hamitlonians in general do not share eigenfunctions!
 
-   
+:::
 
-### 1st order correction to eigenfunction $\mid n^1 \rangle$
 
-We express unknown first order eigenfunctions $\mid n^1 \rangle$ in terms of known eigenfunctions $\mid k^0 \rangle$ which form complete basis set. Another conseuqence of hermitian operators. 
 
 
+:::{admonition} **Deriving 1st order correction to eigenfunction $\mid n^1 \rangle$**
+
+- We express the unknown first order eigenfunctions $\mid n^1 \rangle$ in terms of known eigenfunctions $\mid k^0 \rangle$ which becasue of Hermitian property of operators must form complete basis set. 
 
 $$
 \mid n^1 \rangle = \sum_{k \neq n} c_{k} \mid k^0 \rangle
 $$
 
 
+- The coefficients are $c_k =\langle k^0 \mid n^1 \rangle$. Becasue of orthogonality  all terms involving eigenfunctions of nth state drop out $c_n=\langle n^0 \mid n^1 \rangle=0$.    This is why we have indicated $k\neq n$ condition in summation expression.
 
-- The coefficients are $c_k =\langle k^0 \mid n^1 \rangle$ . 
-
-- Becasue of orthogonality  $c_n=\langle n^0 \mid n^1 \rangle=0$   therefore we have indicates  $k\neq n$ condition in the sum.
-
-
-
----
-
-
-
-Inserting the expanstion of $\mid n^1\rangle$ and this taking dot product with bra $\langle k^0 \mid$ we find the coefficients of expansion:
+- Inserting the expanstion of $\mid n^1\rangle$ and taking dot product with bra $\langle k^0 \mid$ we find the coefficients of expansion:
 
 
 $$
-\color{red}{\hat{H}^0\mid n^1\rangle +\hat{H^1}\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}
+{\hat{H}^0\mid n^1\rangle +\hat{H^1}\mid n^0\rangle = E^0_n\mid n^1 \rangle+E^1_n\mid n^0 \rangle}
 $$
 
 
 $$
-\color{red}{  \hat{H}^0 \color{black}{\sum_{k \neq n} c_k\mid k^0\rangle} +  \hat{H^1}\mid n^0\rangle = E^0_n  \color{black} {\sum_{k \neq n} c_k \mid k^0 \rangle} +E^1_n  \mid n^0 \rangle}
+{  \hat{H}^0 \color{black}{\sum_{k \neq n} c_k\mid k^0\rangle} +  \hat{H^1}\mid n^0\rangle = E^0_n  \color{black} {\sum_{k \neq n} c_k \mid k^0 \rangle} +E^1_n  \mid n^0 \rangle}
 $$
 
 
@@ -200,28 +203,27 @@ $$
 c_k = \frac{ \langle k^0 \mid \hat{H}^1 \mid n^0\rangle}{E^0_n-E^0_k}=\frac{H_{nk}}{E^0_n-E^0_k}
 $$
 
----
-
 
 $$
-\mid n^1 \rangle = \sum_{k \neq n} c_{k} \mid k^0 \rangle = \sum_{k \neq n} \frac{H_{nk}}{E^0_n-E^0_k} \mid k^0 \rangle
+\boxed{\mid n^1 \rangle = \sum_{k \neq n} c_{k} \mid k^0 \rangle = \sum_{k \neq n} \frac{H_{nk}}{E^0_n-E^0_k} \mid k^0 \rangle}
 $$
 
-- We we have intduced convenient notaation for matrix elements of pertrubed hamitlonian:
-
-
+**Matrix element notation**
+- We we have introduced convenient notaation for matrix elements of pertrubed hamitlonian $H_{nk} = \langle k^0 \mid \hat{H}^1 \mid n^0\rangle$. Note that hamiltonian in matrix elements is always the perturbation part.
+:::
 
 
 
-### 2nd order correction. Obtaining expression for $E^2_n$
+:::{admonition} **Deriving expression for $E^2_n$**
+:class: dropdown
 
 $$
-\color{blue}{\hat{H^0}\mid n^2\rangle+\hat{H^1}\mid n^1\rangle = E_n^0\mid n^2\rangle+ E^1_n\mid n^1 \rangle+E^2_n\mid n^0 \rangle}
+{\hat{H^0}\mid n^2\rangle+\hat{H^1}\mid n^1\rangle = E_n^0\mid n^2\rangle+ E^1_n\mid n^1 \rangle+E^2_n\mid n^0 \rangle}
 $$
 
 
 $$
-\color{blue}{\color{black}{\langle n^0 \mid }\hat{H^0} \mid n^2\rangle+\color{black}{\langle n^0 \mid }\hat{H^1}\mid n^1\rangle = E^0\color{black}{\langle n^0}\mid n^2\rangle+ E^1_n\color{black}{\langle n^0 \mid } n^1 \rangle+E^2_n \color{black}{\langle n^0}\mid n^0 \rangle}
+{{\langle n^0 \mid }\hat{H^0} \mid n^2\rangle+{\langle n^0 \mid }\hat{H^1}\mid n^1\rangle = E^0{\langle n^0}\mid n^2\rangle+ E^1_n{\langle n^0 \mid } n^1 \rangle+E^2_n {\langle n^0}\mid n^0 \rangle}
 $$
 
 
@@ -234,10 +236,6 @@ $$
 - We are not done yet, the expression contains eigenfunction $\mid n^1 \rangle$ which we need to expres in terms of known solutions $\mid n^0 \rangle$  
 
 
-
-### 2nd order correction. Obtaining expression for $\mid n^1 \rangle$
-
-
 $$
 \color{black}{E^2_n = \langle n^0 \mid \hat{H}^1 \mid n^1 \rangle}= \sum_{k \neq n} c_k \langle n^0 \mid \hat{H}^1 \mid k^0 \rangle = \sum_{k \neq n} c_k H_{nk}
 $$
@@ -249,18 +247,23 @@ $$
 
 
 $$
-E = \color{green}{E^0_n} + \color{red}{\langle n^0\mid H^1\mid n^0\rangle} + \color{blue}{\sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}}
+\boxed{E_n = \color{green}{E^0_n} + \color{red}{\langle n^0\mid H^1\mid n^0\rangle} + \color{blue}{\sum_{k \neq n} \frac{\mid H_{nk}\mid^2}{E^0_n-E^0_k}}}
 $$
+:::
+
+
+
 
 
-- The energy in the denominator is the difference between energy of a given state $E_n$ from all other states $E_k$ with k being the summation index. 
-- If the matrix elements of $\hat{H}^1$ are of comparable magnitude the neighbouring levels make larger contributions that distance levels.
 
 
 
-### Ground state energy perturbations
+### Applications 
 
-Let us write second order correction explicitely for the ground state for some exactly solvable hamiltonian $\hat{H^0}$ pertrubed by $\hat{H^1}$
+:::{admonition} **Example-1: Estimate ground state with second order pertrubation**
+:class: note 
+
+Write second order correction explicitely for the ground state for some exactly solvable hamiltonian $\hat{H^0}$ pertrubed by $\hat{H^1}$
 
 
 $$
@@ -274,15 +277,10 @@ $$
 
 
 - Notice that for the ground state the second order correction thereofre will always be negative because $\Delta E_{0k}=E_0-E_k<0$
+:::
 
-
-
-
-
-## Applications
-
-
-### Example-1: Magnetic field
+:::{admonition} **Example-2: Magnetic field**
+:class: note 
 
 Hydrogen atom in magnetic field problem can be seen as as a hamitonian of H atom to which we have added a small pertrubation in the form of interation with magnetic field. 
 
@@ -312,13 +310,12 @@ $$
 E=E_0+ A_{SO} \langle 0 \mid \hat{L} \hat{S}\mid 0 \rangle
 $$
 
+:::
 
 
 
-
-### Example-2: Particle in a box
-
-
+:::{admonition} **Example-3: Perturbing particle in a box**
+:class: note
 
 Estimate the energy of the ground-state and first excited-state wavefunction within first-order perturbation theory of a system with the following potential energy:
 
@@ -348,12 +345,11 @@ $$
 E_n = E^0_n+E^1_n \approx \frac{n^2 h^2}{8mL^2}+V_0
 $$
 
+:::
 
 
-
-### Exaple-3 Unharmonic oscillator
-
-
+:::{admonition} **Example-4 Unharmonic oscillator**
+:class: note
 
 Unharmonic oscillator problem can be seen as a problem fo harmonic oscillator + pertrubation in the form of unharmonic term:
 
@@ -362,21 +358,14 @@ $$
 $$
 
 
-
 - Using first order pertrubation  we find an interesting result after evaluating the integral to by using simple symmetry arguments. 
-
 
 
 $$
 E_n^1 = \langle 0\mid \gamma x^3 \mid n\rangle = \langle even/odd \mid odd \mid even /odd\rangle = 0
 $$
 
 
-
----
-
-
-
 - But energy levels surely must experience change since we added a new term to hamitlonian. To see the change we must therefore turn to second order and use ground state as an example
 
 
@@ -386,16 +375,12 @@ E^2_0 =  \sum_{k \neq 0} \frac{\mid  \langle 0\mid \gamma x^3 \mid k\rangle \mid
 $$
 
 
-
-
 $$
 E_0=E_0+E_1+E_2 = \frac{\hbar \omega}{2} + 0 + \frac{H^2_{01}}{\hbar \omega}+\frac{H^2_{03}}{2\hbar \omega}+ ...
 $$
 
-
-
 Thus we see that only terms odd terms of the sum contribute. The matrix elements need to be evaluated explicitly  using Hermite polynomials.  
-
+:::