LOWEST-ORDER RELATIVISTIC ENERGY CORRECTION OF 1-D HARMONIC OSCILLATOR | Quantum Science Philippines
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LOWEST-ORDER RELATIVISTIC ENERGY CORRECTION OF 1-D HARMONIC OSCILLATOR

Lotis R. Racines and Edwin B. Fabillar

In quantum mechanics, relativistic correction to the energy levels of a system is used when it is introduced by a little disturbance we often recognized as [eq] \lambda [/eq]. Fine structure is an example of this where the splitting of spectral lines of atoms is due to its first-order relativistic corrections. Here is an example of finding the first-order relativistic corrections of a given system.

Our task is to find the lowest-order relativistic corrections to the energy levels of the one-dimensional harmonic oscillator.

Note: Our reference through all these is Jackson’s book of Quantum Mechanics

Start:

We begin by eq’n 6.52 ,

[eq] E_{r}^’ = – \frac {1}{2mc^2} [E^2 – 2 E\langle V \rangle + \langle V^2 \rangle] [/eq]

where [eq] E = (n + \frac {1}{2}) \hbar \omega [/eq]

and [eq] V = \frac {1}{2} m \omega ^2 x ^2 [/eq]

So,

[eq] E_{r}^’ = – \frac {1}{2mc^2} [(n + \frac{1}{2})^2 \hbar^2 \omega ^2 – 2(n + \frac{1}{2}) \hbar \omega (\frac{1}{2}m \omega ^2) \langle x^2 \rangle + \frac{1}{4} m^2 \omega ^4 \langle x^4 \rangle] [/eq]

with [eq] \langle x^2 \rangle = (n + \frac{1}{2}) \frac{\hbar}{m \omega} [/eq]

Substituting this, we get

[eq] E_{r}^’ = – \frac{1}{2mc^2} [(n + \frac{1}{2})^2 \hbar ^2 \omega ^2 – (n + \frac{1}{2})(n + \frac{1}{2}) \frac{\hbar}{m \omega} (\hbar \omega)(m \omega ^2) [/eq]

[eq]+ \frac{1}{2} m^2 \omega^4 \langle x^4 \rangle] [/eq]

[eq] E_{r}^’ = – \frac{1}{2mc^2} [(n + \frac{1}{2})^2 \hbar^2 \omega ^2 – (n + \frac{1}{2})^2 \hbar^2 \omega ^2 + \frac{1}{4} m^2 \omega ^4 \langle x^4 \rangle] [/eq]

[eq] E_{r}^’ = – \frac{m \omega ^4}{8c^2} \langle x^4 \rangle [/eq]                    (1)

We now introduce the ladder operators. That is,

[eq] a_{+} = \sqrt {n+1}|n \rangle [/eq]

[eq] a_{-} = \sqrt {n} |n-1 \rangle [/eq]

Using these, we could then derive [eq] \langle x^4 \rangle[/eq] basing on Eq’n 2.69,

[eq] x^4 = \frac{\hbar^2}{4m^2 \omega ^2} (a_{+}^2 + a_{+}a_{-} + a_{-}a_{+} + a_{-}^2)(a_{+}^2 + a_{+}a_{-} + a_{-}a_{+} + a_{-}^2)[/eq]

[eq] \langle x^4 \rangle = \frac{\hbar^2}{4m^2 \omega ^2} \langle m | (a_{+}^2 a_{-}^2 + a_{+}a_{-}a_{+}a_{-} + a_{+}a_{-}a_{-}a_{+} + a_{-}a_{+}a_{+}a_{-} [/eq]

[eq] + a_{-}a_{+}a_{-}a_{+} + a_{-}^2 a_{+}^2) | n \rangle [/eq]

Note that only equal numbers of  raising and lowering operators will survive.

By eq’n 2.66, [eq] h(\x) = h_{even} (\x) + h_{odd} (\x) [/eq]

[eq] \langle x^4 \rangle = \frac{\hbar ^2}{4m^2 \omega ^2}{\langle n| a_{+}^2[\sqrt {n(n-1)}|n-2\rangle] + a_{+} a_{-} \langle n|n \rangle + a_{+} a_{-} \langle (n+1)|n \rangle [/eq]

[eq] + a_{-} a_{+} \langle n|n \rangle + a_{-} a_{+} \langle (n+1)|n \rangle + a_{-}^2 \langle \sqrt {(n+1)(n+2)}|n+2 \rangle} [/eq]

[eq] \langle x^4 \rangle = \frac{\hbar^2}{4m^2 \omega^2} \{\langle n|[\sqrt{n(n-1)}\sqrt{n(n-1)}|n \rangle] + n \langle n|n \rangle + (n+1) \langle n|n \rangle [/eq]

[eq] + n \langle (n+1)|n \rangle + (n+1) \langle (n+1)|n \rangle + \sqrt{(n+1)(n+2)} \langle \sqrt {(n+1)(n+2)}|n \rangle \} [/eq]

[eq] \langle x^4 \rangle = \frac{\hbar ^2}{4m^2 \omega ^2} [n(n-1) + n^2 + (n+1)n + n(n+1) + (n+1)^2 + (n+1)(n+2)] [/eq]

[eq] \langle x^4 \rangle = (\frac{\hbar}{2m \omega})^2 [n^{2} – n + n^2 + n^2 + n + n^2 + n + n^2 + 2n + 1 + n^2 + 3n +2] [/eq]

[eq] \langle x^4 \rangle = (\frac{\hbar}{2m \omega})^2 [6n^2 + 6n + 3] [/eq]

Going back to (1) to get [eq] E_{r}^’ [/eq] ,

[eq] E_{r}^’= – \frac{m \omega ^4}{8c^2} (\frac{\hbar ^2}{4 m^2 \omega ^2}) (3)(3n^2 + 2n + 1) [/eq]

Thus, the lowest-order relativistic correction of one-dimensional harmonic oscillator is

[eq] E_{r}^’ = – \frac{3}{32} (\frac {\omega ^2 \hbar ^2}{mc^2}) (3n^2 + 2n +1) [/eq]

One Response to “LOWEST-ORDER RELATIVISTIC ENERGY CORRECTION OF 1-D HARMONIC OSCILLATOR”

  1. Black Says:

    It should be 2n^2 + 2n + 1

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