We have seen, how to obtain the formal solution of Harmonic oscillator using power series method. Now, we will follow somewhat a completely different kind of new approach named operator method.
We will start with defining two operators in dimensionless form with position and momentum operator as follows, $$ a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p $$ and $$ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p$$ The only physical fact we use here is the commutation relation between x and p with its corresponding representation in position basis, $$[x,p] = i\hbar $$ because, $$ [x,p]\psi = xp\psi - px\psi = -i\hbar \left[x\frac{\partial\psi}{\partial{x}} - \frac{\partial{(x\psi)}}{\partial{x}}\right] \\~\\ = \left[x\frac{\partial\psi}{\partial{x}} - x\frac {\partial\psi}{\partial{x}} - \psi\frac{\partial{x}}{\partial{x}}\right] \\~\\ = i\hbar\psi $$
Using this result, we will compute the following relations,
$$[a,a^\dagger] = \left[\sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p, \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\right] \\~\\ = \frac{1}{2\hbar}[x,ip] + \frac{1}{2\hbar} [-ip,x] $$
Using, $$ [x,x] = [p,p] = 0 $$ and $$ [x,p] = i\hbar = -[p,x] $$
$$ [a,a^\dagger] = \frac{1}{2\hbar}{-2\hbar} = -1 $$ $$\rightarrow\,\,\,\, [a,a^\dagger] = -1 $$
Similarly, the operation $aa^\dagger$ can be calculated as, $$ aa^\dagger = \frac{m\omega}{2\hbar}x^2 + \frac{i}{2\hbar}xp - \frac{i}{2\hbar}px + \frac{1}{2m\omega\hbar}p^2 = \frac{m\omega}{2\hbar}x^2 + \frac{1}{2m\omega\hbar}p^2 -\frac{1}{2} $$ Then, $$ aa^\dagger = \frac{1}{\hbar\omega} [ H - \frac{\hbar\omega}{2} ] $$ or $$ H = \hbar\omega \left[ aa^\dagger + \frac{1}{2} \right] = \hbar\omega \left[a^\dagger{a} - \frac{1}{2}\right]$$
Now, if we assume "u" is an eigen function of $aa^\dagger$ operator with eigen value $\lambda$ then we have, $$ a^\dagger{a}u = \lambda{u} $$ then we get, $$ a^\dagger{a}(au) = (1+aa^\dagger)(au) = au + aa^\dagger{a}u \\~\\= au + a\lambda{u} = (\lambda+1)u $$
Similarly, $$ a^\dagger{a}(a^\dagger{u}) = (\lambda - 1) (a^\dagger{u})$$
What these results mean is that, if "u" is an eigen function of "$a^\dagger{a}$" operation, then there exists another eigen function "au" with "$\lambda + 1 $" eigen value and "$a^\dagger{u}$" eigen function with "$\lambda -1 $" eigen value.
We will continue in the next post.
We will start with defining two operators in dimensionless form with position and momentum operator as follows, $$ a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p $$ and $$ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p$$ The only physical fact we use here is the commutation relation between x and p with its corresponding representation in position basis, $$[x,p] = i\hbar $$ because, $$ [x,p]\psi = xp\psi - px\psi = -i\hbar \left[x\frac{\partial\psi}{\partial{x}} - \frac{\partial{(x\psi)}}{\partial{x}}\right] \\~\\ = \left[x\frac{\partial\psi}{\partial{x}} - x\frac {\partial\psi}{\partial{x}} - \psi\frac{\partial{x}}{\partial{x}}\right] \\~\\ = i\hbar\psi $$
Using this result, we will compute the following relations,
$$[a,a^\dagger] = \left[\sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p, \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\right] \\~\\ = \frac{1}{2\hbar}[x,ip] + \frac{1}{2\hbar} [-ip,x] $$
Using, $$ [x,x] = [p,p] = 0 $$ and $$ [x,p] = i\hbar = -[p,x] $$
$$ [a,a^\dagger] = \frac{1}{2\hbar}{-2\hbar} = -1 $$ $$\rightarrow\,\,\,\, [a,a^\dagger] = -1 $$
Similarly, the operation $aa^\dagger$ can be calculated as, $$ aa^\dagger = \frac{m\omega}{2\hbar}x^2 + \frac{i}{2\hbar}xp - \frac{i}{2\hbar}px + \frac{1}{2m\omega\hbar}p^2 = \frac{m\omega}{2\hbar}x^2 + \frac{1}{2m\omega\hbar}p^2 -\frac{1}{2} $$ Then, $$ aa^\dagger = \frac{1}{\hbar\omega} [ H - \frac{\hbar\omega}{2} ] $$ or $$ H = \hbar\omega \left[ aa^\dagger + \frac{1}{2} \right] = \hbar\omega \left[a^\dagger{a} - \frac{1}{2}\right]$$
Now, if we assume "u" is an eigen function of $aa^\dagger$ operator with eigen value $\lambda$ then we have, $$ a^\dagger{a}u = \lambda{u} $$ then we get, $$ a^\dagger{a}(au) = (1+aa^\dagger)(au) = au + aa^\dagger{a}u \\~\\= au + a\lambda{u} = (\lambda+1)u $$
Similarly, $$ a^\dagger{a}(a^\dagger{u}) = (\lambda - 1) (a^\dagger{u})$$
What these results mean is that, if "u" is an eigen function of "$a^\dagger{a}$" operation, then there exists another eigen function "au" with "$\lambda + 1 $" eigen value and "$a^\dagger{u}$" eigen function with "$\lambda -1 $" eigen value.
We will continue in the next post.
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