Find the (lowest order) relativistic correction to the energy levels of the one-dimensional harmonic oscillator. Hint: Use the technique in Example 2.5 .

The relativistic correction to the energy levels is:

${\mathit{E}}_{\mathbf{r}}^{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}\mathbf{m}{\mathbf{c}}^{\mathbf{2}}}\left[E_n^2-2E_n⟨\hat{V}⟩+⟨{\hat{V}}^2⟩\right]$

For the harmonic oscillator

$$\begin{gathered} \hat{\mathbf{V}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathbf{\omega }}^{\mathbf{}\mathbf{2}\mathbf{}}\widehat{{\mathbf{x}}^{\mathbf{2}}\mathbf{}} \\ {\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{\hslash }\mathit{\omega }\left(n+\frac{1}{2}\right) \end{gathered}$$

Now, we can write the relativistic correction to the energy as:

$E_r^1=\frac{-1}{2m{c}^2}\left[{\hslash }^2{\omega }^2{\left(n+\frac{1}{2}\right)}^2-2\hslash \omega \left(n+\frac{1}{2}\right)\cdot \frac{1}{2}\hslash \omega \left(n+\frac{1}{2}\right)+\frac{1}{4}{m}^2{\omega }^4⟨{\hat{x}}^4⟩\right]$

From the example (2.5), we get:

$$\begin{gathered} {\hat{x}}^2=\frac{\hslash }{2m\omega }\left[a^{+}{a}^{+}+a^{+}a+a{a}^{+}+aa\right] \\ ⟨\hat{V}⟩=\frac{1}{2}\hslash \omega \left(n+\frac{1}{2}\right) \end{gathered}$$

Then

$E_r^1=\frac{-1}{2m{c}^2}\left[{\hslash }^2{\omega }^2{\left(n+\frac{1}{2}\right)}^2-{\hslash }^2{\omega }^2{\left(n+\frac{1}{2}\right)}^2+\frac{1}{4}{m}^2{\omega }^4⟨{\hat{x}}^4⟩\right]$

We need to find $⟨{\hat{x}}^4⟩$by calculating that:

$$\begin{gathered} {\hat{x}}^4=\frac{{\hslash }^2}{4m^2{\omega }^2}\left[(a^{+}{a}^{+}+a^{+}a+a{a}^{+}+aa)\cdot (a^{+}{a}^{+}+a^{+}a+a{a}^{+}+aa)\right] \\ ⟨{\hat{x}}^4⟩=⟨n\left|x{^}^4\right|n⟩ \\ =\frac{{\hslash }^2}{4m^2{\omega }^2}[⟨n|a^{+}{a}^{+}{a}^{+}{a}^{+}|n⟩+⟨n|a^{+}{a}^{+}{a}^{+}a|n⟩+⟨n|a^{+}{a}^{+}a{a}^{+}|n⟩+⟨n|a^{+}{a}^{+}aa|n⟩ \\ +⟨n|a^{+}a{a}^{+}{a}^{+}|n⟩+⟨n|a^{+}a{a}^{+}a|n⟩ \\ +⟨n|a^{+}aa{a}^{+}|n⟩+⟨n|a^{+}aaa|n⟩+⟨n|a{a}^{+}{a}^{+}{a}^{+}|n⟩n \\ +⟨n|a{a}^{+}{a}^{+}a|n⟩+⟨n|a{a}^{+}a{a}^{+}|n⟩+⟨n|a{a}^{+}aa|n⟩+⟨n|aa{a}^{+}{a}^{+}|n⟩+⟨n|aa{a}^{+}a|n⟩ \\ +⟨n|aaa{a}^{+}|n⟩+⟨n|aaaa|n⟩ \end{gathered}$$

We know that:

$$\begin{gathered} a^{+}|n⟩=\sqrt{n+1}|n+1⟩a|n⟩ \\ =\sqrt{n}|n-1⟩ \end{gathered}$$

Where, the terms which contain equal number particles due to the raising and lowering operators can survive and the other terms equal to zero from${\delta }_{nm}=0\hspace{1em};\hspace{1em}n\ne m$

Then, the only terms which can be calculated are:

$$\begin{gathered} ⟨n\left|a^{+}{a}^{+}aa\right|n⟩=\sqrt{n}⟨n\left|a^{+}{a}^{+}a\right|n-1⟩ \\ =\sqrt{n}\sqrt{(n-1)}⟨n\left|a^{+}{a}^{+}\right|n-2⟩ \\ =\sqrt{n}(n-1)⟨n\left|a^{+}\right|n-1⟩ \\ =n(n-1)⟨n\mid n⟩ \\ =n(n-1){\delta }_{nn} \\ =n(n-1) \end{gathered}$$

Solve further

$$\begin{gathered} ⟨n\left|a^{+}a{a}^{+}a\right|n⟩=\surd n⟨n\left|a^{+}a{a}^{+}\right|n-1⟩ \\ =n⟨n\left|a^{+}a\right|n⟩ \\ =n\surd n⟨n\left|a^{+}\right|n-1⟩ \\ =n^2⟨n\mid n⟩ \\ =n^2{\delta }_{n}n \\ =n^2 \\ ⟨n\left|a^{+}aa{a}^{+}\right|n⟩=\surd (n+1)⟨n\left|a^{+}aa\right|n+1⟩ \\ =(n+1)⟨n\left|a^{+}a\right|n⟩ \\ =\surd n(n+1)⟨n\left|a^{+}\right|n-1⟩ \\ =n(n+1)⟨n\mid n⟩ \\ =n(n+1){\delta }_{n}n \end{gathered}$$

Solve further

$$\begin{gathered} =n(n+1)⟨n\left|a{a}^{+}{a}^{+}a\right|n⟩ \\ =\sqrt{n}⟨n\left|a{a}^{+}{a}^{+}\right|n-1⟩ \\ =n⟨n\left|a{a}^{+}\right|n⟩ \\ =n\sqrt{(n+1)}⟨n\left|a\right|n+1⟩ \\ =n(n+1)⟨n\mid n⟩ \\ =n(n+1){\delta }_{n}n \\ =n(n+1) \\ =(n+1)⟨n\left|a{a}^{+}\right|n⟩ \end{gathered}$$

Solve further

$$\begin{gathered} =(n+1)\sqrt{(n+1)}⟨n\left|a\right|n+1⟩ \\ =(n+1{)}^2⟨n\mid n⟩ \\ =(n+1{)}^2{\delta }_{nn} \\ =(n+1{)}^2⟨|aa{a}^{+}{a}^{+}|n⟩ \\ =\sqrt{(n+1)}⟨n|aa{a}^{+}|n+1⟩ \\ =\sqrt{(n+1)}\sqrt{(n+2)}⟨n|aa|n+2⟩ \\ =(n+2\left)\sqrt{(n+1}\right)⟨n\left|a\right|n+1⟩ \\ =(n+1)(n+2)⟨n\mid n⟩ \\ =(n+1)(n+2){\delta }_{nn} \\ =(n+1)(n+2) \end{gathered}$$

Then,

$$\begin{gathered} ⟨{\hat{x}}^4⟩=\frac{{\hslash }^2}{4m^2{\omega }^2}\left[n(n-1)+n^2+n(n+1)+n(n+1)+(n+1{)}^2+(n+1\left)\right(n+2)\right] \\ =\frac{{\hslash }^2}{4m^2{\omega }^2}\left[n^2-n+n^2+n^2+n+n^2+n+n^2+2n+1+n^2+2n+n+2\right] \\ =\frac{{\hslash }^2}{4m^2{\omega }^2}[6n^2+6n+3] \end{gathered}$$

Now we can calculate the correction in the energy levels:

$$\begin{gathered} E_r^1=\frac{-1}{2m{c}^2}\left[\frac{1}{4}{m}^2{\omega }^4\cdot \frac{{\hslash }^2}{4m^2{\omega }^2}(6n^2+6n+3)\right] \\ =\frac{-{\hslash }^2{\omega }^2}{32m{c}^2}(6n^2+6n+3) \end{gathered}$$

Then, the lowest-order relativistic correction to the energy levels is

$E_r^1=\frac{-3{\hslash }^2{\omega }^2}{32m{c}^2}(2n^2+2n+1)$