We have from Harmonic oscillator operator method - 1, $$a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p \\~\\ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\\~\\ [a,a^\dagger] = -1\\~\\ aa^\dagger = \frac{1}{\hbar\omega}\left[H - \frac{\hbar\omega}{2}\right] \\~\\ a^\dagger{a}u = \lambda{u} \\~\\ a^\dagger{a} (au) = (\lambda+1) (au) \\~\\ a^\dagger{a} (a^\dagger{u}) = (\lambda - 1)(a^\dagger{u}) $$
Now, we can derive the result of Hamiltonian operator on this eigen vector, $$ H (a^\dagger{u}) = \hbar\omega \left(a^\dagger{a} - \frac{1}{2}\right)(a^\dagger{u}) = \hbar\omega \left( a^\dagger{a}a^\dagger{u} - \frac{a^\dagger{u}}{2}\right) \\~\\ = \hbar\omega\left[(\lambda -1) - \frac{1}{2} \right] (a^\dagger{u}) $$
Using the relation, $$ Hu = \hbar\omega (a^\dagger{a} - \frac{1}{2})u = \hbar\omega\left[\lambda-\frac{1}{2}\right]u = Eu $$
With its correspondence we can rewrite,
$$ H (a^\dagger{u}) = (E - \hbar\omega)(a^\dagger{u}) $$
Similarly, $$ H (au) = (E+\hbar\omega) (au)$$
or the general proof, $$ H (a^\dagger{u}) = Ha^\dagger{u} - a^\dagger{H}u + a^\dagger{H}u = [H,a^\dagger]u + a^\dagger{H}u \\~\\= [-\hbar\omega{a^\dagger} + E {a^\dagger}]u = (E - \hbar\omega)(a^\dagger{u}) $$
where, $$ [H, a^\dagger] = \hbar\omega[{aa^\dagger}, a^\dagger] +\hbar\omega[\frac{1}{2}, {a^\dagger}] \\~\\=\hbar\omega \left([ a,a^\dagger ] a^\dagger + a [ a^\dagger, a^\dagger] + [1/2, a^\dagger]\right) \\~\\ =\hbar\omega\left([a,a^\dagger]a^\dagger +0\right) = -\hbar\omega{a^\dagger} $$
As we can see, the repeated application of $"a^\dagger"$ continuously on any eigen function will give negative values for Energy. But, we have assumed the harmonic oscillator has positive energy, which restricts the negative energy states.
So, we demand a wave function such that it gets annihilated when operated by operator $"a^\dagger"$ on its lower state. Let us say, the lowest ground state is $u_0$, then we have, $$ a^\dagger{u_0} = 0 $$ or $$ \sqrt{\frac{m\omega}{2\hbar}}x{u_0} + \sqrt{\frac{1}{2m\omega\hbar}} i p{u_0} = 0 $$ Non-dimensionalization gives, $$ \frac{1}{\sqrt{2}}(\rho +\frac{\partial}{\partial{\rho}})u_0 = 0 $$ where $\rho = \alpha{x} $ and $ \alpha = \sqrt{\frac{m\omega}{\hbar}}$
By solving this, the ground state wave function can be obtained. Once we get the ground state wave function, using creation operator we can derive all other eigen states.
Now, we can derive the result of Hamiltonian operator on this eigen vector, $$ H (a^\dagger{u}) = \hbar\omega \left(a^\dagger{a} - \frac{1}{2}\right)(a^\dagger{u}) = \hbar\omega \left( a^\dagger{a}a^\dagger{u} - \frac{a^\dagger{u}}{2}\right) \\~\\ = \hbar\omega\left[(\lambda -1) - \frac{1}{2} \right] (a^\dagger{u}) $$
Using the relation, $$ Hu = \hbar\omega (a^\dagger{a} - \frac{1}{2})u = \hbar\omega\left[\lambda-\frac{1}{2}\right]u = Eu $$
With its correspondence we can rewrite,
$$ H (a^\dagger{u}) = (E - \hbar\omega)(a^\dagger{u}) $$
Similarly, $$ H (au) = (E+\hbar\omega) (au)$$
or the general proof, $$ H (a^\dagger{u}) = Ha^\dagger{u} - a^\dagger{H}u + a^\dagger{H}u = [H,a^\dagger]u + a^\dagger{H}u \\~\\= [-\hbar\omega{a^\dagger} + E {a^\dagger}]u = (E - \hbar\omega)(a^\dagger{u}) $$
where, $$ [H, a^\dagger] = \hbar\omega[{aa^\dagger}, a^\dagger] +\hbar\omega[\frac{1}{2}, {a^\dagger}] \\~\\=\hbar\omega \left([ a,a^\dagger ] a^\dagger + a [ a^\dagger, a^\dagger] + [1/2, a^\dagger]\right) \\~\\ =\hbar\omega\left([a,a^\dagger]a^\dagger +0\right) = -\hbar\omega{a^\dagger} $$
As we can see, the repeated application of $"a^\dagger"$ continuously on any eigen function will give negative values for Energy. But, we have assumed the harmonic oscillator has positive energy, which restricts the negative energy states.
So, we demand a wave function such that it gets annihilated when operated by operator $"a^\dagger"$ on its lower state. Let us say, the lowest ground state is $u_0$, then we have, $$ a^\dagger{u_0} = 0 $$ or $$ \sqrt{\frac{m\omega}{2\hbar}}x{u_0} + \sqrt{\frac{1}{2m\omega\hbar}} i p{u_0} = 0 $$ Non-dimensionalization gives, $$ \frac{1}{\sqrt{2}}(\rho +\frac{\partial}{\partial{\rho}})u_0 = 0 $$ where $\rho = \alpha{x} $ and $ \alpha = \sqrt{\frac{m\omega}{\hbar}}$
By solving this, the ground state wave function can be obtained. Once we get the ground state wave function, using creation operator we can derive all other eigen states.
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