Question

Consider the isotropic three-dimensional harmonic oscillator (Problem 4.38). Discuss the effect (in first order) of the perturbation ${\mathbf{H}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{\lambda x}}^{\mathbf{2}}\mathbf{yz}$

(for some constant $\mathit{\lambda }$) on

(a) the ground state

(b) the (triply degenerate) first excited state. Hint: Use the answers to Problems 2.12and 3.33

Short answer

a) Correction of energy in ground state is 0 .

b) Energy of first excited state splits in three energies: ,$:{\mathrm{E}}_1-\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2,{\mathrm{E}}_1\mathrm{and}{\mathrm{E}}_1+\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2.$

Step-by-step solution

Definition of hyperfine harmonic oscillator.

A harmonic oscillator is a system that experiences a restoring force F proportionate to the displacement X when it is moved from its equilibrium position, where k is a positive constant.

The ground state.

(a)

If the perturbation Hamiltonian is provided with: Find energy adjustments on the ground state and first excited state in a 3D harmonic oscillator.

${\mathrm{H}}^{\text{'}}={\mathrm{\lambda x}}^2\mathrm{yz}$

Ground state in 3D harmonic oscillator,

$|0>=|0{>}_x\left|0{>}_y\right|0{>}_z$

So first-order correction of ground state energy is:

$$\begin{gathered} {\mathrm{E}}_0^1=<0\left|{\mathrm{H}}^{\text{'}}\right|0> \\ =\mathrm{\lambda }<0\left|{\mathrm{x}}^2\right|0><0\left|\mathrm{y}\right|0><0\left|\mathrm{z}\right|0> \\ =0 \end{gathered}$$

Because$<0\left|\mathrm{y},\mathrm{z}\right|0>=0$

The (triply degenerate) first excited state.

(b)

Because the first excited state is triply degenerate, we must first acquire the perturbation matrix W and then determine the eigenvalues.

The energy shifts of the non-perturbed initial excited state are equivalent to eigenvalues.

The first aroused state is described as follows:

$$\begin{gathered} |1>=|1{>}_x\left|0{>}_y\right|0{>}_z \\ |2>=|0{>}_x\left|1{>}_y\right|0{>}_z \\ |3>=|0{>}_x\left|0{>}_y\right|1{>}_z \end{gathered}$$

There are matrix elements to compute, but the majority of them are 0 because:$⟨0|y,z|0⟩=0:$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{i}\mathrm{j}}=⟨\mathrm{i}\left|{\mathrm{H}}^{\text{'}}\right|\mathrm{j}⟩,\mathrm{i},\mathrm{j}=2,3 \\ ⟨1\left|{\mathrm{H}}^{\text{'}}\right|1⟩=\mathrm{\lambda }⟨1\left|{\mathrm{x}}^2\right|1⟩⟨0\left|\mathrm{y}\right|0⟩⟨0\left|\mathrm{z}\right|0⟩=0 \\ ⟨1\left|{\mathrm{H}}^{\text{'}}\right|2⟩=\mathrm{\lambda }⟨1\left|{\mathrm{x}}^2\right|0⟩⟨0\left|\mathrm{y}\right|1⟩⟨0\left|\mathrm{z}\right|0⟩=0 \\ ⟨2\left|{\mathrm{H}}^{\text{'}}\right|1⟩=0 \\ ⟨1\left|{\mathrm{H}}^{\text{'}}\right|3⟩=\mathrm{\lambda }⟨1\left|{\mathrm{x}}^2\right|0⟩⟨0\left|\mathrm{y}\right|0⟩⟨0\left|\mathrm{z}\right|1⟩=0 \\ ⟨3\left|{\mathrm{H}}^{\text{'}}\right|1⟩=0 \\ ⟨2\left|{\mathrm{H}}^{\text{'}}\right|2⟩=\mathrm{\lambda }⟨0\left|{\mathrm{x}}^2\right|0⟩⟨1\left|\mathrm{y}\right|1⟩⟨0\left|\mathrm{z}\right|0⟩=0 \\ ⟨3\left|{\mathrm{H}}^{\text{'}}\right|3⟩=\mathrm{\lambda }⟨0\left|{\mathrm{x}}^2\right|0⟩⟨0\left|\mathrm{y}\right|0⟩⟨1\left|\mathrm{z}\right|1⟩=0 \\ ⟨2\left|{\mathrm{H}}^{\text{'}}\right|3⟩=\mathrm{\lambda }⟨0\left|{\mathrm{x}}^2\right|0⟩⟨1\left|\mathrm{y}\right|0⟩⟨0\left|\mathrm{z}\right|1⟩=\mathrm{\lambda }\frac{\hslash }{2\mathrm{m\omega }}\sqrt{\frac{\hslash }{2\mathrm{m\omega }}}\sqrt{\frac{\hslash }{2\mathrm{m\omega }}}=\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2 \\ ⟨3\left|{\mathrm{H}}^{\text{'}}\right|2⟩=\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2 \end{gathered}$$

Perturbation matrix W is equal to:

role="math" localid="1656049433676" $\mathrm{W}=\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)$

Now calculate eigenvalues

$$\begin{gathered} \mathrm{A}=\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2 \\ \det (\mathrm{W}-\mathrm{E}.\mathrm{I})=0 \\ \left|\begin{array}{ccc}-\mathrm{E}& 0& 0\\ 0& -\mathrm{E}& \mathrm{A}\\ 0& \mathrm{A}& -\mathrm{E}\end{array}\right|=0 \\ -\mathrm{E}({\mathrm{E}}^2-{\mathrm{A}}^2)=0 \\ \mathrm{E}=0 \\ \mathrm{or} \\ {\mathrm{E}}_{12}=\pm \mathrm{A} \\ =\pm \mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2 \end{gathered}$$

Energy of first excited state splits in three energies:

$:{\mathrm{E}}_1-\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2,{\mathrm{E}}_1\mathrm{and}{\mathrm{E}}_1+\mathrm{\lambda }{\left(\frac{\hslash }{2\mathrm{m\omega }}\right)}^2.$