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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in Jan 2019
Quantum Simple Harmonic Oscillator
Ref:
$\displaystyle\large {p^2\over 2m}+{1\over 2}\kappa x^2=E~,$ where $E$ is the energy of the system, and $\kappa$ is the stiffness. The objective is to solve $p$ and $x$ in time $t$.
We know that $p(t)=mv=m{d\over dt}x$. Let the maximum displacement be $A$. We will have $x(t)=A\cos(\omega t+\phi)$, and $p(t)=m\omega A\sin(\omega t+\phi).$
The equation becomes $\displaystyle {(m\omega A)^2\over 2m}\sin^2(\omega t+\phi)+{\kappa A^2\over 2}\cos^2(\omega t+\phi)=E ,$ which is a constant.
It follows that $\displaystyle E={(m\omega A)^2\over 2m}={\kappa A^2\over 2},$ stiffness $\kappa=m\omega^2,$ and energy $E={1\over 2}m\omega^2A^2.$
The relationship still holds but instead of scaler, $p$ and $x$ become operators $\HAT p$ and $\HAT x$.
Classical SHO: $\displaystyle {p^2\over 2m}+{1\over 2}m\omega^2x^2=E.~~~~$ Quantum SHO: $\displaystyle \HAT H\psi(x) ={\HAT p^2\over 2m}\psi(x)+{1\over 2}m\omega^2\HAT x^2\psi(x) =E~\psi(x) .$
As $\displaystyle\HAT x=x,$ and $\displaystyle\HAT p=-i\hbar{d\over dx},~$ we have $\displaystyle~\boxed{ \HAT H\psi(x) =-{\hbar^2\over 2m}{d^2\over dx^2}\psi(x)+{m\omega^2\over 2}x^2\psi(x) =E~\psi(x) .}$
The objective is to solve for all solutions of $\psi_n(x)$ and its corresponding energy $E_n$.
This method starts with finding operators to give alternative solution for higher or lower energy, then reducing the energy to zero to find the first solution, before reaching a general solution for $\psi(x)$ (taking $t=0$).
Let us start with energy in the form of $\hbar\omega$ and guess that the equation would be like $\hbar\omega\left(\ldots\right)\psi(x)=E~\psi(x).$
Inspired by the classical SHO, we put $\alpha^2p^2+\beta^2x^2$ in the bracket, which can be factorised into $(-i\alpha p+\beta x)(+i\alpha p+\beta x)$.
If $\displaystyle \hbar\omega\left(\alpha^2p^2+\beta^2x^2\right)={p^2\over 2m}+{m\omega^2x^2\over 2},~~ \boxed{ \alpha=\sqrt{1\over 2m\hbar\omega},~~ \beta=\sqrt{m\omega\over 2\hbar}~~\text{and}~~ \alpha\beta={1\over 2\hbar} .}$
If $p$ and $x$ are taken as operators, the relationship still hold. So $\HAT H=\hbar\omega\left(\alpha^2\HAT p^2+\beta^2\HAT x^2\right).$
However, $\HAT p$ and $\HAT x$ are operators and
$(-i\alpha\HAT p+\beta\HAT x)(+i\alpha\HAT p+\beta\HAT x) =\alpha^2\HAT p^2-i\alpha\HAT p\beta\HAT x+\beta\HAT x i\alpha\HAT p+\beta^2\HAT x^2 =\alpha^2\HAT p^2+\beta^2\HAT x^2+i\alpha\beta\left[\HAT x,\HAT p\right] ={\HAT p^2\over 2m\hbar\omega}+{m\omega\HAT x^2\over 2\hbar}+{i\over 2\hbar}\left[\HAT x,\HAT p\right] .$
Substitute it into the bracket, $\hbar\omega\left(\ldots\right)\psi(x)=E~\psi(x)$ will becomes $\displaystyle {\HAT p^2\over 2m}\psi(x) +{m\omega^2\HAT x^2\over 2}\psi(x) +\hbar\omega{i\over 2\hbar}\left[\HAT x,\HAT p\right]\psi(x) =E\psi(x) .$
i.e. $\displaystyle \HAT H\psi(x) =E\psi(x)-\hbar\omega{i\over 2\hbar}\left[\HAT x,\HAT p\right]\psi(x) .$
To find $\left[\HAT x,\HAT p\right]$, let us apply it to an arbitrary function $f(x)$. $\displaystyle \left[\HAT x,\HAT p\right]f(x) =x\left(-i\hbar\right){d\over dx}f(x) -\left(-i\hbar{d\over dx}\left(x~f(x)\right)\right) =-i\hbar x~f'(x) +i\hbar\left(x~f'(x)+f(x)\right) =i\hbar~f(x) .~~$ So $\displaystyle\boxed{\left[\HAT x,\HAT p\right]=i\hbar.}$
This brings us $\displaystyle \HAT H\psi(x) =\left(E+{1\over 2}\hbar\omega\right)\psi(x) .$
Before we go further, let us investigate where the ${1\over 2}\hbar\omega$ is coming from? Isn't $\HAT H\psi(x)=\hbar\omega\left(\alpha^2\HAT p^2+\beta^2\HAT x^2\right)\psi(x)=E~\psi(x)$ already?
We may say that the "deduction" so far is not following logics. Why do we need to separate $\alpha^2+\beta^2$ into two factors? This "factorisation" is only applicable to scaler. For operators, you will necessarily encounter issues with the commutator (which is always zero for scalers).
You might say that this is a deliberate attempt to highlight the difference between classical and quantum SHO through the commutator, and observe the difference. In this case, it is ${1\over 2}\hbar\omega$.
Now let us see how this can help us solve the QSHO problem without solving the equation.
Let us define two operators: $\HAT{~~a_+}=-i\alpha\HAT p+\beta\HAT x$ and $\HAT{~~a_-}=+i\alpha\HAT p+\beta\HAT x.$
As a result, $\displaystyle \hbar\omega\hat{~a_+}\hat{~a_-} =\left({\HAT p^2\over 2m}+{m\omega^2\HAT x^2\over 2}\right)-{1\over2}\hbar\omega =\HAT H-{1\over2}\hbar\omega .~~~~ \HAT H =\hbar\omega\left(\hat{~a_+}\hat{~a_-}+{1\over2}\right) .$
For $\hat{~a_-}\hat{~a_+}$, the sign of the $\left[\HAT x,\HAT p\right]$ commutator is reversed, so we will have $\displaystyle \HAT H =\hbar\omega\left(\hat{~a_-}\hat{~a_+}-{1\over2}\right) .$
$\left[\hat{~a_-},\hat{~a_+}\right] =\hat{~a_-}\hat{~a_+}-\hat{~a_+}\hat{~a_-} =\left(\alpha^2\HAT p^2+i\alpha\beta\HAT p\HAT x -i\beta\alpha\HAT x\HAT p+\beta^2\HAT x^2\right) -\left(\alpha^2\HAT p^2-i\alpha\beta\HAT p\HAT x +i\beta\alpha\HAT x\HAT p+\beta^2\HAT x^2\right)\\ =2i\alpha\beta\left(\HAT p\HAT x-\HAT x\HAT p\right) =-2i{1\over 2\hbar}\left[\HAT x,\HAT p\right] =1 .$
$\displaystyle \therefore\boxed{\left[\hat{~a_-},\hat{~a_+}\right]=1 .}$ i.e. $\hat{~a_-}\hat{~a_+}=\hat{~a_+}\hat{~a_-}+1.$
Suppose we have a solution $\psi_n$, then $\HAT H\psi_n=E_n\psi_n$ and
$\HAT H\left(\hat{~a_+}\psi_n\right) =\hbar\omega\left(\hat{~a_+}\hat{~a_-}+{1\over2}\right)\hat{~a_+}\psi_n =\hbar\omega\left(\hat{~a_+}\hat{~a_-}\hat{~a_+}+{1\over2}\hat{~a_+}\right)\psi_n =\hat{~a_+}\left(\hbar\omega\left(\hat{~a_-}\hat{~a_+}+{1\over2}\right)\right)\psi_n =\hat{~a_+}\left(\hbar\omega\left(\hat{~a_+}\hat{~a_-}+{1\over2}+1\right)\right)\psi_n\\ =\hat{~a_+}\left(\HAT H+\hbar\omega\right)\psi_n =\hat{~a_+}\left(E_n\psi_n+\hbar\omega~\psi_n\right) =\left(E_n+\hbar\omega\right)\left(\hat{~a_+}\psi_n\right) .$
So $\hat{~a_+}\psi_n$ provides a solution with energy $E_n+\hbar\omega.$
By the same token,
$\HAT H\left(\hat{~a_-}\psi_n\right) =\hbar\omega\left(\hat{~a_-}\hat{~a_+}-{1\over2}\right)\hat{~a_-}\psi_n =\hbar\omega\left(\hat{~a_-}\hat{~a_+}\hat{~a_-}-{1\over2}\hat{~a_-}\right)\psi_n =\hat{~a_-}\left(\hbar\omega\left(\hat{~a_+}\hat{~a_-}-{1\over2}\right)\right)\psi_n =\hat{~a_-}\left(\hbar\omega\left(\hat{~a_-}\hat{~a_+}-{1\over2}-1\right)\right)\psi_n\\ =\hat{~a_-}\left(\HAT H-\hbar\omega\right)\psi_n =\hat{~a_-}\left(E_n\psi_n-\hbar\omega~\psi_n\right) =\left(E_n-\hbar\omega\right)\left(\hat{~a_-}\psi_n\right) .$
So $\hat{~a_-}\psi_n$ is also a solution with energy $E_n-\hbar\omega.$
The trick here is $\left(\hat{~a_+}\hat{~a_-}\right)\hat{~a_+}=\hat{~a_+}\left(\hat{~a_-}\hat{~a_+}\right),~~$ Like the $\hat{~a_+}$ from the right "knocks off" itself from the left, and increase the bracketed value by one.
$\boxed{\HAT H\left(\hat{~a_+}\psi_n\right) =\left(E_n+\hbar\omega\right)\left(\hat{~a_+}\psi_n\right) .}~~~~ \boxed{\HAT H\left(\hat{~a_-}\psi_n\right) =\left(E_n-\hbar\omega\right)\left(\hat{~a_-}\psi_n\right) .}$
As $\hat{~a_-}$ reduces the energy of the solution, what if the energy becomes zero. Let us try this out by applying it to an unknown function $f(x)$ to obtain a zero. That function would provide a hint for the base energy state.
$\displaystyle \hat{~a_-}f(x) =\left(i\alpha\HAT p+\beta\HAT x\right)f(x) =i\sqrt{1\over 2m\hbar\omega}\left(-i\hbar{d\over dx}f(x)\right)+\sqrt{m\omega\over 2\hbar}x~f(x) =0,~~ \sqrt{1\over 2m\hbar\omega}\left(\hbar{d\over dx}f(x)\right)+\sqrt{m\omega\over 2\hbar}x~f(x)=0 ,$
$\displaystyle\large {d\over dx}f(x)=-{m\omega\over\hbar}x~f(x).~~~~\text{As}~~ {d\over dx}e^{{-m\omega x^2\over 2\hbar}+\phi}=-{m\omega\over\hbar}x~e^{{-m\omega x^2\over 2\hbar}+\phi},~~~~ f(x)=e^{{-m\omega x^2\over 2\hbar}+\phi}$ is a solution.
Let us define $\displaystyle\large\boxed{\psi_0(x)=e^{-m\omega x^2\over 2\hbar}}$, which is time-independent before normalisation.
Comparing with normal distribution $\displaystyle f(x\mid\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, one can find that $\psi_0(x)$ is a zero-centred normal distribution with $\sigma^2={\hbar\over m\omega} .$
The normalisation co-efficient $c$ can be found by $c={1\over\sqrt{2\pi\sigma^2}}=\sqrt{{m\omega}\over\hbar}.$ It is not normalised here as it should be done after superposition of all energy states $E_n$, which we have infinity number of as $n\rightarrow\infty$.
Corollary: $\displaystyle {d^2\over dx^2}e^{cx^2} ={d\over dx}\left(2cxe^{cx^2}\right) =2c{d\over dx}\left(xe^{cx^2}\right) =2c\left(e^{cx^2}+2cx^2e^{cx^2}\right) =2c\left(1+2cx^2\right)e^{cx^2} .$
$\displaystyle \HAT H\psi_0(x) =-{\hbar^2\over 2m}{d^2\over dx^2}e^{{-m\omega\over 2\hbar}x^2}+{m\omega^2\over 2}x^2\psi_0(x) =\left(-{\hbar^2\over 2m}2\left({-m\omega\over 2\hbar}\right)\left(1+2\left({-m\omega\over 2\hbar}\right)x^2\right)+{m\omega^2\over 2}x^2\right)\psi_0(x)$
$\displaystyle =\left({\hbar\omega\over 2}-{m\omega^2\over 2}x^2+{m\omega^2\over 2}x^2\right)\psi_0(x) ={1\over 2}\hbar\omega\psi_0(x)=E_0\psi_0(x).~~~~ \therefore E_0={1\over 2}\hbar\omega .$
Applying $\HAT H\left(\hat{~a_+}\psi_0\right) =\left(E_0+\hbar\omega\right)\left(\hat{~a_+}\psi_0\right) ,$ so $E_1=E_0+\hbar\omega={1\over 2}\hbar\omega+\hbar\omega={3\over 2}\hbar\omega$. By induction, $\boxed{E_n=\left(n+{1\over 2}\right)\hbar\omega.}$
If $\displaystyle [N,X]=cX,~~\text{and}~~ N\Ket n =n\Ket n,~~\text{then}~~ NX\Ket n =(XN+[N,X])\Ket n =XN\Ket n+[N,X]\Ket n =Xn\Ket n+cX\Ket n =(n+c)X\Ket n .$
If $N$ is Hermitian, its eigenvalues are real, so both $n$ and $c$ are real.\qquad $\displaystyle N=N^\dagger,~~ [N,X^\dagger] =[N^\dagger,X^\dagger] =[X,N]^\dagger =-cX^\dagger .$
By the same process, we will obtain $NX^\dagger\Ket n=(n-c)X^\dagger\Ket n.$
Here $N$ is our $\HAT H$, $X$ is our $\hat{~a_+}$ (with a positive $c$) and $X^\dagger$ our $\hat{~a_-}$.
Recall that $\HAT H=\hbar\omega\left(\alpha^2\HAT p^2+\beta^2\HAT x^2\right),~~ \hat{~a_+}=-i\alpha\HAT p+\beta\HAT x$ and $\hat{~a_-}=+i\alpha\HAT p+\beta\HAT x,~~\text{where}~~ \alpha=\sqrt{1\over 2m\hbar\omega},~~ \beta=\sqrt{m\omega\over 2\hbar} .$
It follows that $\alpha\beta={1\over 2\hbar}$ and $\HAT H=\hbar\omega\left(\hat{~a_+}\hat{~a_-}-i\alpha\beta\left[\HAT x,\HAT p\right]\right) .$
As $\left[\HAT x,\HAT p\right]=i\hbar~~\text{and}~~ \left[\hat{~a_-},\hat{~a_+}\right]=1,~~ \HAT H=\hbar\omega\left(\hat{~a_+}\hat{~a_-}+{1\over 2}\right)~~\text{and}~~ \HAT H=\hbar\omega\left(\hat{~a_-}\hat{~a_+}-{1\over 2}\right) .$
$\HAT H\hat{~a_+} =\hbar\omega\left(\hat{~a_+}\hat{~a_-}\hat{~a_+}+{1\over 2}\hat{~a_+}\right) =\hat{~a_+}\hbar\omega\left(\hat{~a_-}\hat{~a_+}+{1\over 2}\right),~~ \hat{~a_+}\HAT H =\hat{~a_+}\hbar\omega\left(\hat{~a_-}\hat{~a_+}-{1\over 2}\right) .$
$\left[\HAT H,\hat{~a_+}\right] =\HAT H\hat{~a_+}-\hat{~a_+}\HAT H =\hbar\omega\hat{~a_+} .$
This satisfies the $[N,X]=cX$ condition with $c=\hbar\omega$, and this results in $NX\Ket n=(n+c)X\Ket n.$
Therefore, $\HAT H\psi_n=E_n\psi_n$ will give us $\HAT H\hat{~a_+}\psi_n=(E_n+\hbar\omega)\hat{~a_+}\psi_n.~~\text{and similarly}~~ \HAT H\hat{~a_-}\psi_n=(E_n-\hbar\omega)\hat{~a_-}\psi_n.$
Given $N=N^\dagger,~~[N,X]=cX$ and $[N,X^\dagger]=-cX$, what is $[X^\dagger,X]$?
WORKING
$(X+X^\dagger) (X-X^\dagger)=X^2-XX^\dagger+X^\dagger X-X^\dagger\,^2 =X^2-X^\dagger\,^2+[X^\dagger,X] .$
$[N,X]-[N,X^\dagger]=NX-XN-NX^\dagger+X^\dagger N=2cX .$
$[X^\dagger,X]= X^\dagger X-XX^\dagger .$
$0=[N,X]+[N,X^\dagger] =NX-XN+NX^\dagger-X^\dagger N =NX-XN+N^\dagger X^\dagger-X^\dagger N^\dagger .$
$[N,X]=cX,~~ cXX^\dagger =[N,X]X^\dagger =(NX-XN)X^\dagger =NXX^\dagger-XNX^\dagger \ldots(1).$
$[N,X^\dagger]=-cX^\dagger,~~ cX^\dagger X =[N,X^\dagger]X =(NX^\dagger-X^\dagger N)X =NX^\dagger X-X^\dagger NX \ldots(2).$
$(2)-(1): c[X^\dagger,X] =N[X^\dagger,X] .$
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