91影视

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.

(a)

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

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

Ground state in 3D harmonic oscillator,

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

So first-order correction of ground state energy is:

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

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

(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:$鉄�0|y,z|0鉄�=0:$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{i}\mathrm{j}}=鉄�\mathrm{i}\left|{\mathrm{H}}^{\text{'}}\right|\mathrm{j}鉄�,\mathrm{i},\mathrm{j}=2,3 \\ 鉄�1\left|{\mathrm{H}}^{\text{'}}\right|1鉄�=\mathrm{位}鉄�1\left|{\mathrm{x}}^2\right|1鉄�鉄�0\left|\mathrm{y}\right|0鉄�鉄�0\left|\mathrm{z}\right|0鉄�=0 \\ 鉄�1\left|{\mathrm{H}}^{\text{'}}\right|2鉄�=\mathrm{位}鉄�1\left|{\mathrm{x}}^2\right|0鉄�鉄�0\left|\mathrm{y}\right|1鉄�鉄�0\left|\mathrm{z}\right|0鉄�=0 \\ 鉄�2\left|{\mathrm{H}}^{\text{'}}\right|1鉄�=0 \\ 鉄�1\left|{\mathrm{H}}^{\text{'}}\right|3鉄�=\mathrm{位}鉄�1\left|{\mathrm{x}}^2\right|0鉄�鉄�0\left|\mathrm{y}\right|0鉄�鉄�0\left|\mathrm{z}\right|1鉄�=0 \\ 鉄�3\left|{\mathrm{H}}^{\text{'}}\right|1鉄�=0 \\ 鉄�2\left|{\mathrm{H}}^{\text{'}}\right|2鉄�=\mathrm{位}鉄�0\left|{\mathrm{x}}^2\right|0鉄�鉄�1\left|\mathrm{y}\right|1鉄�鉄�0\left|\mathrm{z}\right|0鉄�=0 \\ 鉄�3\left|{\mathrm{H}}^{\text{'}}\right|3鉄�=\mathrm{位}鉄�0\left|{\mathrm{x}}^2\right|0鉄�鉄�0\left|\mathrm{y}\right|0鉄�鉄�1\left|\mathrm{z}\right|1鉄�=0 \\ 鉄�2\left|{\mathrm{H}}^{\text{'}}\right|3鉄�=\mathrm{位}鉄�0\left|{\mathrm{x}}^2\right|0鉄�鉄�1\left|\mathrm{y}\right|0鉄�鉄�0\left|\mathrm{z}\right|1鉄�=\mathrm{位}\frac{鈩�}{2\mathrm{尘蝇}}\sqrt{\frac{鈩�}{2\mathrm{尘蝇}}}\sqrt{\frac{鈩�}{2\mathrm{尘蝇}}}=\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2 \\ 鉄�3\left|{\mathrm{H}}^{\text{'}}\right|2鉄�=\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2 \end{gathered}$$

Perturbation matrix W is equal to:

role="math" localid="1656049433676" $\mathrm{W}=\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)$

Now calculate eigenvalues

$$\begin{gathered} \mathrm{A}=\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2 \\ \det (\mathrm{W}-\mathrm{E}.\mathrm{I})=0 \\ \left|\begin{array}{ccc}-\mathrm{E}& 0& 0\\ 0& -\mathrm{E}& \mathrm{A}\\ 0& \mathrm{A}& -\mathrm{E}\end{array}\right|=0 \\ -\mathrm{E}({\mathrm{E}}^2-{\mathrm{A}}^2)=0 \\ \mathrm{E}=0 \\ \mathrm{or} \\ {\mathrm{E}}_{12}=卤\mathrm{A} \\ =卤\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2 \end{gathered}$$

Energy of first excited state splits in three energies:

$:{\mathrm{E}}_1-\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2,{\mathrm{E}}_1\mathrm{and}{\mathrm{E}}_1+\mathrm{位}{\left(\frac{鈩�}{2\mathrm{尘蝇}}\right)}^2.$