Question

Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric field (E), so that the potential energy is shifted by an amount${\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathbf{qEx}$.(a) Show that there is no first-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.33.

(b) The Schrödinger equation can be solved directly in this case, by a change of variables${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\equiv }\mathit{x}\mathbf{-}\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right)$. Find the exact energies, and show that they are consistent with the perturbation theory approximation.

Short answer

To calculate energy corrections, one needs to find matrix elements (or expectation values) of the perturbation Hamiltonian in an unperturbed basis.

First order correction =0

Second order correction$\mathrm{E}=\mathrm{ħ\omega }\left(\mathrm{n}+\frac{1}{2}\right)-\frac{{\mathrm{q}}^2{\mathrm{E}}^2}{2{\mathrm{m\omega }}^2}$

Step-by-step solution

Prove that the diagonal matrix element vanishes for every eigen state of the quantum harmonic oscillator.

a)

See how the interaction Hamiltonian includes the position operator:

${\mathrm{H}}_{\mathrm{e}}=-\mathrm{qEx}$, prove that the diagonal matrix element (or equivalently expectation value of the position operator) vanishes for every eigen state of the quantum harmonic oscillator. Express the position operator in the terms of raising and lowering operators:$\mathit{x}\mathbf{=}\sqrt{\frac{\mathbf{ħ}}{\mathbf{2}\mathbf{m\omega }}\left(a_{+}+a_{-}\right)}$

Next, calculate the $\mathrm{n}$-th diagonal matrix element:

The latter is equal to zero because the lowering operator turns $\mathrm{n}~\left(\mathrm{n}-1\right)$and raising operator turns$\mathrm{n}~\left(\mathrm{n}+1\right)$, so the both inner products vanish since the different eigen states are mutually orthogonal. Therefore, there are no corrections of the first order in the energy levels.

Second order corrections are calculated by the useof the following relation:

Where ${\mathrm{E}}_{\mathrm{k}}^0$denote the unperturbed energy levels - energy levels of the harmonic oscillator. Start by expressing the interaction Hamiltonian in the terms of raising and lowering operators:

${\mathrm{H}}_{\mathrm{i}}=-\mathrm{qE}\sqrt{\frac{\mathrm{ħ}}{2\mathrm{m\omega }}}\left({\mathrm{a}}_{+}+{\mathrm{a}}_{-}\right)$

Proceed by doing calculation of the following matrix elements:

Second order correction.

See that the only non-vanishing matrix elements are the ones neighbouring j-th element. Therefore, the sum in equation (1) is simplified, having only two contributing terms:

${\mathrm{E}}_{\mathrm{j}}^{\left(2\right)}=\frac{{\left|\mathrm{qE}\sqrt{{\displaystyle \frac{\mathrm{ħ}}{2\mathrm{m\omega }}}}\sqrt{\mathrm{j}}\right|}^2}{{\mathrm{E}}_{\mathrm{j}-1}^0-{\mathrm{E}}_{\mathrm{j}}^0}+\frac{{\left|\mathrm{qE}\sqrt{{\displaystyle \frac{\mathrm{ħ}}{2\mathrm{m\omega }}}}\sqrt{\mathrm{j}}+1\right|}^2}{{\mathrm{E}}_{\mathrm{j}+1}^0-{\mathrm{E}}_{\mathrm{j}}^0}$

Since problem is dealing with the energy spectrum of the harmonic oscillator, the unperturbed energies are equidistant the distance between two increasing levels is equal to$ħ\omega$. Therefore, the terms in the denominators are easily evalued leading to:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{j}}^{\left(2\right)}=\frac{{\mathrm{q}}^2{\mathrm{E}}^2}{\mathrm{ħ\omega }}\frac{\mathrm{ħ}}{2\mathrm{m\omega }}\left[-\left(\mathrm{j}+1\right)+\mathrm{j}\right] \\ \Rightarrow {\mathrm{E}}_{\mathrm{j}}^{\left(2\right)}=\frac{{\mathrm{q}}^2{\mathrm{E}}^2}{2\mathrm{m\omega }} \end{gathered}$$

What is interesting is that the energy corrections of the second order do not depend on the energy level $\mathrm{j}$that we are observing - all levels are corrected by the same amount. This is exactly the reason that would lead one to think that there might be an analytical method of obtaining the energy spectrum by manipulating the original Hamiltonian and attempting to solve it in a non-perturbative manner.

Solve the Schrödinger equation and find the exact energies.

b)

To solve the Schrodinger analytically, begin by the use of the following substitution:

$$\begin{gathered} \dot{\mathrm{x}}=\mathrm{x}-\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right) \\ \Rightarrow \mathrm{x}=\dot{\mathrm{x}}+\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right) \end{gathered}$$

Next step is to insert this substitution into the original Schrodinger equation:

$$\begin{gathered} -\frac{\mathrm{ħ}}{2\mathrm{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^2}\mathrm{\psi }\left(\mathrm{x}\right)+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\mathrm{\psi }\left(\mathrm{x}\right)-\mathrm{qEx\psi }\left(\mathrm{x}\right) \\ =\mathrm{E\psi }\left(\mathrm{x}\right)-\frac{{\mathrm{ħ}}^2}{2\mathrm{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^{\text{'}2}}\mathrm{\psi }\left(\dot{\mathrm{x}}\right)+\frac{1}{2}{\mathrm{m\omega }}^2{\left(\dot{\mathrm{x}}+\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right)\right)}^2\mathrm{\psi }\left(\dot{\mathrm{x}}\right)-\mathrm{qE}\left(\dot{\mathrm{x}}+\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right)\right)\mathrm{\omega }\left(\mathrm{x}\right) \\ =\mathrm{E\psi }\left(\dot{\mathrm{x}}\right) \end{gathered}$$

It is important to notice that the change in derivative yields no extra factors, since$\frac{\partial \dot{x}}{\partial x}=1$. Squaring the term in the brackets, obtain the following:

$$\begin{gathered} -\frac{{ħ}^2}{2\mathrm{m}}\frac{{\partial }^2}{\partial x^{\text{'}2}}\psi \left(x^{\text{'}}\right)+\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{x}}^{\text{'}2}+2{\mathrm{x}}^{\text{'}}\mathrm{qE}/{\mathrm{m\omega }}^2+{\mathrm{q}}^2{\mathrm{E}}^2/{\mathrm{m}}^2{\mathrm{\omega }}^4\right)-\mathrm{qE}\left({\mathrm{x}}^{\text{'}}\mathrm{b}+\left(\mathrm{qE}/{\mathrm{m\omega }}^2\right)\right)\mathrm{\psi }\left(\mathrm{x}\right) \\ =\mathrm{E\psi }\left({\mathrm{x}}^{\text{'}}\right) \end{gathered}$$

see that many terms cancel out, leaving this:

$-\frac{{ħ}^2}{2\mathrm{m}}\frac{{\partial }^2}{\partial x^{\text{'}2}}\psi \left(x^{\text{'}}\right)+\frac{1}{2}{\mathrm{m\omega }}^2{x}^{\text{'}2}=\left(\mathrm{E}+{\mathrm{q}}^2{\mathrm{E}}^2/2{\mathrm{m\omega }}^2\right)\psi \left(\dot{x}\right)$

By defination ${\mathrm{E}}^{\text{'}}\equiv \mathrm{E}+{\mathrm{q}}^2{\mathrm{E}}^2/2{\mathrm{m\omega }}^2$the Schrodinger equation reads:

$-\frac{\mathrm{ħ}}{2\mathrm{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^2}\mathrm{\psi }\left({\mathrm{x}}^{\text{'}}\right)+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^{\text{'}2}={\mathrm{E}}^{\text{'}}\mathrm{\psi }\left(\mathrm{x}\right)$

Which is simply an equation of a quantum harmonic oscillator, with known solutions. Therefore, the energies of the original problem,$\mathrm{E}$, are given by a simple relation:

$$\begin{gathered} \mathrm{ħ\omega }\left(\mathrm{n}+\frac{1}{2}\right)=\mathrm{E}+{\mathrm{q}}^2{\mathrm{E}}^2/2{\mathrm{m}}^2 \\ \Rightarrow \mathrm{E}=\mathrm{ħ\omega }\left(\mathrm{n}+\frac{1}{2}\right)-\frac{{\mathrm{q}}^2{\mathrm{E}}^2}{2{\mathrm{m\omega }}^2} \end{gathered}$$

This is exactly the same result that is obtained by calculating the energy corrections of the second order. This implies that perturbative method actually provided the exact solution. This is usually not the case - usually one would have to calculate infinite corrections to approach the exact value.