note-math

Commutation relation of raising and lowering operators $\text{a }\left(\pm \text{ i}\right)=\frac{1}{2}\left(\text{x }\pm \frac{1}{\text{i }𝑚𝜔}\text{ p}\right)$ and energy operator

$$\begin{array}{ccc}\left[\text{H},\text{a }\left(\text{i }\right)\right]& =& -\mathrm{\hslash }𝜔\text{ a }\left(\text{ i }\right)\\ \left[\text{H},\text{a }\left(-\text{ i}\right)\right]& =& \mathrm{\hslash }𝜔\text{ a }\left(-\text{i}\right)\end{array}$$

let $\text{a }=\text{ a }\left(\text{i}\right)$. ${\text{a }}^{†}=\text{ a }\left(-\text{i}\right)$

let raising and lowering operators simplify to $\text{a }\left(\pm \text{ i}\right)=\text{ x }\pm \frac{1}{\text{ i }}\text{ p}$

Eigenfunctions of the energy operator

Starting from the ground state $\text{a }𝜓=0⟹𝜓={𝜓}_0=\text{exp}\left(-\frac{1}{2}{𝑦}^2\right)$

Obtain the eigenfunctions of the next energy level through the raising operator ${\text{a}}^{†}$

$$\begin{array}{ccc}{𝜓}_{𝑛}={\text{ a }}^{†𝑛}𝜓& =& {\left(-1\right)}^{𝑛}{𝑒}^{\frac{1}{2}{𝑦}^2}\frac{{𝑑}^{𝑛}}{𝑑{𝑦}^{𝑛}}{𝑒}^{-\frac{1}{2}{𝑦}^2}𝜓\\ & =& {𝐻}_{𝑛}\left(𝑦\right)\text{exp}\left(-\frac{1}{2}{𝑦}^2\right)\end{array}$$

where

$${𝐻}_{𝑛}\left(𝑦\right)={\left(-1\right)}^{𝑛}{𝑒}^{{𝑦}^2}\frac{{𝑑}^{𝑛}}{𝑑{𝑦}^{𝑛}}{𝑒}^{-{𝑦}^2}$$

is the Hermite polynomial

Eigenfunction normalization

$${\left(\frac{1}{2^{𝑛}𝑛!}{\left(\frac{𝑚𝜔}{\mathrm{\hslash }𝜋}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}{𝑒}^{-\frac{1}{2}\frac{𝑚𝜔}{\mathrm{\hslash }{𝑞}^2}}{𝐻}_{𝑛}\left({\left(\frac{𝑚𝜔}{\mathrm{\hslash }𝜋}𝑞\right)}^{\frac{1}{2}}\right)$$

Propagator $𝐾$ represents constructing a unitary using the path integral Lagrangian. For the harmonic oscillator, use the Fourier transform method. cf. wiki:Path_integral_formulation

let $𝑇={𝑡}_1-{𝑡}_0$

For endpoints fixed but offset from the classical path, perform Fourier expansion $𝑥-{𝑥}_{\text{ cl }}={\sum }_{𝑛=1}^{\infty }{𝑏}_{𝑛}\text{sin}\left(𝑛𝜋\frac{𝑡}{𝑇}\right)$, action $𝑆\left({𝑏}_1,\dots ,{𝑏}_{𝑛},\dots \right)={𝑆}_{\text{ cl }}+\sum \frac{1}{2}|{𝑏}_{𝑛}{|}^2\frac{𝑚}{2}𝑇\left(\frac{{𝑛}^2{𝜋}^2}{{𝑇}^2}-{𝜔}^2\right)$

$$\begin{array}{ccc}𝐾\left(𝑇,{𝑥}_0,𝑥\right)& =& \underset{𝑛\to \infty }{\text{ lim }}\left(\text{ unitary-factor}\right)\left(𝑛\right)\cdot {\int }_{{\mathrm{ℝ}}^{𝑛}}𝑑{𝑏}_{𝑛}\cdots 𝑑{𝑏}_1{𝑒}^{\frac{\text{i }}{\mathrm{\hslash }}𝑆\left({𝑏}_1,\dots ,{𝑏}_{𝑛}\right)}\\ & =& {\left(\frac{𝑚𝜔}{2𝜋\mathrm{\hslash }\text{ i }𝑇}\right)}^{\frac{1}{2}}{𝑒}^{\frac{\text{i }}{\mathrm{\hslash }}{𝑆}_{\text{ cl}}}\prod _{𝑛=1}^{\infty }\frac{𝑛𝜋}{2^{\frac{1}{2}}}{\int }_{\mathrm{ℝ}}𝑑{𝑏}_{𝑛}\text{exp}\left(\frac{\text{i }}{2\mathrm{\hslash }}|{𝑏}_{𝑛}{|}^2\frac{𝑚}{2}𝑇\left(\frac{{𝑛}^2{𝜋}^2}{{𝑇}^2}-{𝜔}^2\right)\right)\\ & =& {\left(\frac{𝑚𝜔}{2𝜋\mathrm{\hslash }\text{ i }\text{ sin }𝜔𝑇}\right)}^{\frac{1}{2}}\text{exp}\left(\frac{\text{i }𝑚𝜔}{2\mathrm{\hslash }}\cdot \frac{\left(|{𝑥}_1{|}^2+|{𝑥}_0{|}^2\right)\text{ cos }𝜔𝑇-2\text{ Re }⟨{𝑥}_1,{𝑥}_0⟩}{\text{sin }𝜔𝑇}\right)\end{array}$$

Used Gauss integral + infinite product ${\prod }_{𝑛=1}^{\infty }\left(1-\frac{{𝑥}^2}{{𝑛}^2}\right)=\frac{\text{sin }𝜋𝑥}{𝜋𝑥}$

Characteristic equation given by ${𝑒}^{-\frac{𝐸}{ℎ}𝑡\text{ i}}𝜓\left(𝑥\right)$

$$\text{H }𝜓=𝐸𝜓$$

Decomposition of $\text{H},{𝑒}^{-\text{ i }\frac{1}{\mathrm{\hslash }}\text{ H }𝑡},𝐾\left({𝑡}_0,𝑡,{𝑥}_0,𝑞\right)$ given by characteristic orthonormal basis $|𝑛⟩$

$$\begin{array}{ccc}\text{H }& =& \sum _{𝑛}{𝐸}_{𝑛}|𝑛⟩⟨𝑛|\\ {𝑒}^{-\text{ i }\frac{1}{\mathrm{\hslash }}\text{ H }𝑡}& =& \sum _{𝑛}{𝑒}^{-\text{ i }\frac{1}{\mathrm{\hslash }}{𝐸}_{𝑛}𝑡}|𝑛⟩⟨𝑛|\\ 𝐾& =& \sum _{𝑛}{𝑒}^{-\text{ i }\frac{1}{\mathrm{\hslash }}{𝐸}_{𝑛}𝑡}⟨𝑥|𝑛⟩⟨𝑛|{𝑥}_0⟩\end{array}$$

$𝐾={\left(\frac{𝑚𝜔}{𝜋\mathrm{\hslash }}\right)}^{\frac{1}{2}}{𝑒}^{-\text{ i }\frac{1}{2}𝜔𝑇}𝑅\left({𝑒}^{-\text{ i }𝜔𝑇}\right)$ then let $𝑅$ perform Taylor expansion, where ${𝑒}^{-\text{ i }\frac{1}{2}𝜔𝑇}{𝑒}^{-\text{ i }𝑛𝜔𝑇}={𝑒}^{-\text{ i }\left(\frac{1}{2}+𝑛\right)𝜔𝑇}$ corresponds to energy level ${𝐸}_{𝑛}=\left(\frac{1}{2}+𝑛\right)\mathrm{\hslash }𝜔$