Harmonic Oscillator – Relativistic Correction

In this post, I will use the stationary(time-independent) first order perturbation theory, to find out the relativistic correction to the Energy of the nth state of a Harmonic Oscillator.

In order to find out the relativistic correction to the Energy, we would need to use relativistic relations.
The relativistic Kinetic Energy is given as:
T= \sqrt{p^2c^2+m^2c^4}-mc^2
The first term in the above equation is the total energy of a relativistic particle and the second term is the rest mass energy of a particle. So we get the Kinetic energy by subtracting those.

\implies T=mc^2\sqrt{1+\frac{p^2c^2}{m^2c^4}}-mc^2

Now, using the binomial expansion on the Kinetic energy we can write it as:

Note: Binomial Expansion:(1+x)^n=1+nx+n(n-1) \frac{x^2}{2} +...

\implies T=(mc^2+\frac{p^2}{2m}-\frac{p^4}{8m^3c^4}+....)-mc^2

If we ignore the rest of the higher order terms and take  only the first three terms of the Binomial expansion, then

\implies T=\frac{p^2}{2m}-\frac{p^4}{8m^3c^2}

We can now see that the Kinetic Energy is actually modified and not just \frac{p^2}{2m} as in the classical case. Since the second term would be very small due to c^2 in the denominator, we can take it as a perturbation, and use the time-independent perturbation theory to find out the correction to the energy levels.

Let perturbation, H'=-\frac{p^4}{8m^3c^2}

Then the first order energy correction to the nth level is given as:
\implies E_n^1=-\frac{1}{8m^3c^2}\left<\psi_n\right|p^4\left|\psi_n\right>

From Schrodinger’s Equation:
\implies p^2\left|\psi_n\right>=2m(E_n-V)\left|\psi_n\right>

Using the above relation,

\implies E_n^1=-\frac{1}{8m^3c^2}\left<\psi_n\right|\left(2m(E_n-V)\right)^2\left|\psi_n\right>

\implies E_n^1=-\frac{1}{2mc^2}\left<\psi_n\right|\left(E_n^2+V^2-2E_nV\right)\left|\psi_n\right>

\implies E_n^1=-\frac{1}{2mc^2}\left(E_n^2+\left<V^2\right>-2E_n\left<V\right>\right)

From Virial Theroem for Harmonic Oscillator, we know that the expectation value of V:


\implies E_n^1=-\frac{1}{2mc^2}\left(E_n^2+\left<V^2\right>-E_n^2\right)

\implies \boxed{ E_n^1=-\frac{1}{2mc^2}\left<V^2 \right> }

So it all boils down to finding the expectation value of V^2 .

To do that we would need to use the following relations, for a Harmonic Oscillator:

V=\frac{1}{2}m\omega^2 x^2

and x=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^\dagger)

where \hat{a} and \hat{a}^\dagger are annihilation(lowering) and creation(raising) operators respectively.

Substituting the above value of x in the expression for V ,


\implies V^2=\frac{1}{4}m^2\omega^4\left(\frac{\hbar}{2m\omega}\right)^2(\hat{a}+\hat{a}^\dagger)^4

\implies V^2=\frac{\hbar^2\omega^2}{16}(\hat{a}+\hat{a}^\dagger)^4

Now you might remember the following relations for the operators \hat{a} and \hat{a}^\dagger ,


where \left|n\right> is the n-th eigenstate of the Harmonic Oscillator.

Therefore, the expectation value of V^2 can be found by evaluating the following expression:

\left< V^2 \right>= \frac{\hbar^2 \omega^2}{16} \left< n|(\hat{a}+\hat{a}^\dagger)^4|n \right>

Now we don’t need to expand (\hat{a}+\hat{a}^\dagger)^4 fully and calculate for all the terms, as only the terms with equal number of raising and lowering operators, will be finite(non-zero). Different number of raising and lowering operators will lead to a different ket and bra, and since they are orthogonal, their inner product would be zero. Therefore, expanding the above equation and leaving only the non-zero terms we get:

\implies \left< V^2 \right>= \frac{\hbar^2 \omega^2}{16}   \left< n|\hat{a}\hat{a}\hat{a}^\dagger\hat{a}^\dagger+ \hat{a}\hat{a}^\dagger \hat{a}\hat{a}^\dagger + \hat{a}^\dagger \hat{a}^\dagger \hat{a} \hat{a} + \hat{a}^\dagger \hat{a}\hat{a}^\dagger \hat{a} + \hat{a}^\dagger \hat{a}\hat{a}\hat{a}^\dagger + \hat{a}\hat{a}^\dagger\hat{a}^\dagger \hat{a}|n\right>

\implies \left< V^2 \right>= \frac{\hbar^2 \omega^2}{16}   \left< n|(n+2)(n+1)+(n+1)^2+n(n-1)+n^2+n(n+1)+n(n+1)|n\right>
\implies \left< V^2 \right>= \frac{\hbar^2 \omega^2}{16} \left<n|6n^2+6n+3|n\right>

\implies \boxed{ \left< V^2 \right>= \frac{\hbar^2 \omega^2}{16} \left(6n^2+6n+3\right) }

Plugging this value of \left<V^2\right> back in E^1_n=-\frac{1}{2mc^2}\left<V^2\right> , we get the relativistic energy correction:

\implies \boxed{E^1_n=-\frac{\hbar \omega^2}{32mc^2}\left(6n^2+6n+3\right)}

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2 thoughts on “Harmonic Oscillator – Relativistic Correction

  1. Thank you sir. It is very informative.
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