From our previous post harmonic-oscillator-operator-method-2,
we got the differential equation to solve for the ground state of the harmonic oscillator problem given by, $$ \frac{1}{\sqrt{2}}\left(\rho +\frac{\partial}{\partial\rho}\right)u_0 = 0 $$ with corresponding scale factors for non-dimensionalization.
Solving this we get, $$ u_0 = Ae^{\frac{-\rho^2}{2}}$$ with the normalizing constant as, $$ A = \frac{\sqrt{\alpha}}{\pi^{\frac{1}{4}}}$$ All other excited states are obtained from the ground state using creation operator as, $$ u_n = A_n a{u_0}$$ and Energy eigenvalues are obtained as $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)$$ Now, we define an operator named as Number operator and defined as $$ \hat{N} = aa^\dagger \\~\\ \hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right) \\~\\ \hat{N}u_n = nu_n \,\,\,\,\,\,where \,\,\,n=0,1,2,..$$ from the analogy $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)\,\,\,\, n = 0,1,2,..$$ Then we can work out for the normalizing constants of all eigen functions as follows,
Let us start with defining, $$ a{u_n} = c_n u_{n+1} \\~\\ a^\dagger{u_n} = d_n u_{n-1}$$ Now, we need to normalize this new eigenfunction $u_{n-1} \,\,or\,\,u_{n+1}$ from the fact that $u_n$ is normalized. So, we get, $$ \langle{u_{n-1}}\vert{u_{n-1}}\rangle = 1 = \langle \frac{a^\dagger{u_n}}{d_n}\vert\frac{a^\dagger{u_n}}{d_n}\rangle \\ \rightarrow |d_n|^2 = \langle{aa^\dagger{u_n}}\vert{u_n}\rangle \\~\\ = \langle\hat{N}u_n\vert{u_n}\rangle = ||u_n||^2 n = n $$
where we know $a,a^\dagger$ are adjoint to each other and $\hat{N}$ is hermitian $$ \left(aa^\dagger\right)^\dagger = aa^\dagger$$ and Remember $$\langle {au}\vert{au}\rangle = \langle {a^\dagger{a}u}\vert{u}\rangle = \langle{u}\vert{a^\dagger{a}u}\rangle \\~\\\neq \langle{aa^\dagger{u}}\vert{u}\rangle \neq \langle{aua^\dagger}\vert{u}\rangle $$ Similarly, $$ \langle{u_{n+1}}\vert{u_{n+1}}\rangle = 1 = \langle\frac{a{u_n}}{c_n}\vert\frac{a{u_n}}{c_n}\rangle \\ \rightarrow \,\,\,\, |c_n|^2 = \langle{u_n}\vert{a^\dagger{a}}u_n\rangle $$ where $$ a^\dagger {a}= \hat{N} + 1$$ and $$ |c_n|^2 = \langle{u_n}\vert\left(\hat{N}+1\right)u_n\rangle = (n+1)u_n\\\rightarrow \,\,\,\, c_n = \sqrt{(n+1)}$$ Then, we get$$ u_n = \left[\prod_{n=0}^{n-1} \frac{1}{\sqrt{n+1}}\right](a^n) u_0 = \frac{1}{\sqrt{n!}}(a^n)u_0$$ In general, $$ u_n = \frac{A}{\sqrt{n!2^n}} \left(\rho - \frac{\partial}{\partial\rho}\right)^nu_0 \propto H_n(\rho)e^{-\frac{\rho^2}{2}} $$ where $H_n(\rho) $ is the Hermite polynomials. And thus we get the general solution of a Simple Linear Harmonic Oscillator as, $$ u_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi}2^nn!}} H_n(\rho) e^{-\frac{\rho^2}{2}} $$
we got the differential equation to solve for the ground state of the harmonic oscillator problem given by, $$ \frac{1}{\sqrt{2}}\left(\rho +\frac{\partial}{\partial\rho}\right)u_0 = 0 $$ with corresponding scale factors for non-dimensionalization.
Solving this we get, $$ u_0 = Ae^{\frac{-\rho^2}{2}}$$ with the normalizing constant as, $$ A = \frac{\sqrt{\alpha}}{\pi^{\frac{1}{4}}}$$ All other excited states are obtained from the ground state using creation operator as, $$ u_n = A_n a{u_0}$$ and Energy eigenvalues are obtained as $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)$$ Now, we define an operator named as Number operator and defined as $$ \hat{N} = aa^\dagger \\~\\ \hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right) \\~\\ \hat{N}u_n = nu_n \,\,\,\,\,\,where \,\,\,n=0,1,2,..$$ from the analogy $$ E_n = \hbar\omega\left(n+\frac{1}{2}\right)\,\,\,\, n = 0,1,2,..$$ Then we can work out for the normalizing constants of all eigen functions as follows,
Let us start with defining, $$ a{u_n} = c_n u_{n+1} \\~\\ a^\dagger{u_n} = d_n u_{n-1}$$ Now, we need to normalize this new eigenfunction $u_{n-1} \,\,or\,\,u_{n+1}$ from the fact that $u_n$ is normalized. So, we get, $$ \langle{u_{n-1}}\vert{u_{n-1}}\rangle = 1 = \langle \frac{a^\dagger{u_n}}{d_n}\vert\frac{a^\dagger{u_n}}{d_n}\rangle \\ \rightarrow |d_n|^2 = \langle{aa^\dagger{u_n}}\vert{u_n}\rangle \\~\\ = \langle\hat{N}u_n\vert{u_n}\rangle = ||u_n||^2 n = n $$
where we know $a,a^\dagger$ are adjoint to each other and $\hat{N}$ is hermitian $$ \left(aa^\dagger\right)^\dagger = aa^\dagger$$ and Remember $$\langle {au}\vert{au}\rangle = \langle {a^\dagger{a}u}\vert{u}\rangle = \langle{u}\vert{a^\dagger{a}u}\rangle \\~\\\neq \langle{aa^\dagger{u}}\vert{u}\rangle \neq \langle{aua^\dagger}\vert{u}\rangle $$ Similarly, $$ \langle{u_{n+1}}\vert{u_{n+1}}\rangle = 1 = \langle\frac{a{u_n}}{c_n}\vert\frac{a{u_n}}{c_n}\rangle \\ \rightarrow \,\,\,\, |c_n|^2 = \langle{u_n}\vert{a^\dagger{a}}u_n\rangle $$ where $$ a^\dagger {a}= \hat{N} + 1$$ and $$ |c_n|^2 = \langle{u_n}\vert\left(\hat{N}+1\right)u_n\rangle = (n+1)u_n\\\rightarrow \,\,\,\, c_n = \sqrt{(n+1)}$$ Then, we get$$ u_n = \left[\prod_{n=0}^{n-1} \frac{1}{\sqrt{n+1}}\right](a^n) u_0 = \frac{1}{\sqrt{n!}}(a^n)u_0$$ In general, $$ u_n = \frac{A}{\sqrt{n!2^n}} \left(\rho - \frac{\partial}{\partial\rho}\right)^nu_0 \propto H_n(\rho)e^{-\frac{\rho^2}{2}} $$ where $H_n(\rho) $ is the Hermite polynomials. And thus we get the general solution of a Simple Linear Harmonic Oscillator as, $$ u_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi}2^nn!}} H_n(\rho) e^{-\frac{\rho^2}{2}} $$
No comments:
Post a Comment
Let everyone know what you think about this