A charged particle oscillates harmonically such that \( H = \frac{p^2}{2m} + m\omega^2 x^2 / 2 \), subjected to a constant perturbation field such that \( H' = -qE x \). Determine... A charged particle oscillates harmonically such that \( H = \frac{p^2}{2m} + m\omega^2 x^2 / 2 \), subjected to a constant perturbation field such that \( H' = -qE x \). Determine the energy shift for the n-th state to the first and second order. Read more

Steps to Solve

1. Identify the Hamiltonian components

The Hamiltonian of a harmonic oscillator is given by:

$$ H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 $$

The perturbation due to the electric field is:

$$ H' = -q E x $$

2. First Order Energy Shift Calculation

In perturbation theory, the first order energy shift for the n-th state is given by:

$$ E_n^{(1)} = \langle n | H' | n \rangle $$

To calculate this, we need to evaluate:

$$ E_n^{(1)} = \langle n | -q E x | n \rangle $$

Using the properties of the harmonic oscillator states, we know:

$$ \langle n | x | n \rangle = 0 $$

Thus,

$$ E_n^{(1)} = 0 $$

3. Second Order Energy Shift Calculation

The second order energy shift is given by:

$$ E_n^{(2)} = \sum_{m \neq n} \frac{|\langle n | H' | m \rangle|^2}{E_n^{(0)} - E_m^{(0)}} $$

The unperturbed energy levels for a harmonic oscillator are:

$$ E_n^{(0)} = \hbar \omega \left( n + \frac{1}{2} \right) $$

Now, calculate ( E_n^{(2)} ):

• First, the transition matrix element ( \langle n | H' | m \rangle ):

$$ \langle n | H' | m \rangle = -q E \langle n | x | m \rangle $$

Using the results for matrix elements of position in harmonic oscillator states:

$$ \langle n | x | m \rangle = \sqrt{\frac{\hbar}{2m\omega}} \sqrt{n} \delta_{m,n-1} + \sqrt{\frac{\hbar}{2m\omega}} \sqrt{n+1} \delta_{m,n+1} $$

This leads to:

• For ( m = n-1 ):

$$ \langle n | H' | n-1 \rangle = -q E \sqrt{ \frac{\hbar}{2m\omega}} \sqrt{n} $$

• For ( m = n+1 ):

$$ \langle n | H' | n+1 \rangle = -q E \sqrt{ \frac{\hbar}{2m\omega}} \sqrt{n+1} $$

Now substitute into the second-order expression:

• For ( m = n-1 ):

$$ E_n^{(2)} \text{ (1)} = \frac{|-q E|^2 \frac{\hbar}{2m\omega} n}{\hbar \omega \left( n - \frac{1}{2} - \left(n-1+\frac{1}{2}\right)\right)} $$

• For ( m = n+1 ):

$$ E_n^{(2)} \text{ (2)} = \frac{|-q E|^2 \frac{\hbar}{2m\omega} (n+1)}{\hbar \omega \left( n + \frac{1}{2} - \left(n+1+\frac{1}{2}\right)\right)} $$

4. Combine the Second Order Contributions

Combine both contributions to find ( E_n^{(2)} ):

$$ E_n^{(2)} = \frac{|-q E|^2 \frac{\hbar}{2m\omega}}{\hbar \omega} \left( \frac{n}{n(1)} + \frac{(n+1)}{n(2)} \right) $$

5. Final Expression for Second Order Energy Shift

This leads to:

$$ E_n^{(2)} = -\frac{q^2 E^2}{4 m \omega^2} $$