Starting with a coupled Harmonic Oscillator problem

$$ H = \frac{p_1^2 + p_2^2}{2m} + \frac{K}{2}\left[x_1^2 + x_2^2 + \left(x_1 - x_2\right)^2\right] = \left(\frac{p_1^2}{2m} + \frac{2K}{2}x_1^2\right) + \left(\frac{p_2^2}{2m} + \frac{2K}{2}x_2^2\right) - Kx_1x_2\,, $$ we define the usual creation and annihilation operators $$ p = i\sqrt{\frac{m}{2}\sqrt{\frac{2K}{m}}}\left(a^\dagger - a\right)\,, \\ x = \sqrt{\frac{1}{2m\sqrt{\frac{2K}{m}}}}\left(a^\dagger + a\right) $$

and obtain $$ H = \sqrt{\frac{2K}{m}}\left(a^\dagger a + \frac{1}{2}\right) + \sqrt{\frac{2K}{m}}\left(b^\dagger b + \frac{1}{2}\right) - \frac{1}{4}\sqrt{\frac{2K}{m}} \left(a^\dagger + a\right)\left(b^\dagger + b\right). $$

We know that the solution should be $$ H = \sqrt{\frac{K}{m}}\left(\alpha^\dagger \alpha + 1/2\right) + \sqrt{\frac{3K}{m}}\left(\beta^\dagger \beta + 1/2\right)\,. $$

Here, $\alpha$ and $\beta$ are related to $a$ and $b$ via a transformation. Importantly, $\alpha$ and $\beta$ are composed of $a$ and $b$, not $a^\dagger$ and $b^\dagger$.

Let us look at the zero-point energy. The latter Hamiltonian shows that it is $\frac{\left(1 + \sqrt{3}\right)}{2}\sqrt{K/m}$.

When I try to get the energy using QFT, I get different results. The derivation is below.

What am I missing? Are $a$ and $b^\dagger$ not supposed to commute? Makes sense they they would...


The action for the original Hamiltonian is

$$ S =\beta \sqrt{\frac{2K}{m}} + \sum_n \bar\phi_n\left(-i\nu_n + \sqrt{\frac{2K}{m}}\right)\phi_n+ \sum_n \bar\psi_n\left(-i\nu_n + \sqrt{\frac{2K}{m}}\right)\psi_n \\ - \frac{1}{4}\sqrt{\frac{2K}{m}}\sum_n \left(\bar\phi_n \bar\psi_{-n}+ \phi_n \psi_{-n}+ \bar\phi_n \psi_{n}+\bar\psi_n \phi_{n}\right) $$ where $\nu_n$ is the Bosonic Matsubara frequency. The complex $\psi$ and $\phi$ fields correspond to the two oscillators. The assumption was made that $a$ and $b^\dagger$ commute to get the normal ordering! Note that the harmonic indices are the same for barred-unbarred products and opposite for barred-barred and unbarred-unbarred in the last term.

Massaging the expression a bit yields $$ S =\beta \sqrt{\frac{2K}{m}} + \bar\phi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\phi_0+ \bar\psi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\psi_0 \\ - \frac{1}{4}\sqrt{\frac{2K}{m}} \left(\bar\phi_0 \bar\psi_{0}+ \phi_0 \psi_{0}+ \bar\phi_0 \psi_{0}+\bar\psi_0 \phi_{0}\right) \\ + \sum_{n> 0} \begin{pmatrix} \bar\phi_{n}\,,\phi_{-n}\,, \bar\psi_{n}\,,\psi_{-n} \end{pmatrix} \begin{pmatrix} -i\nu_n + \sqrt{\frac{2K}{m}}&0& - \frac{1}{4}\sqrt{\frac{2K}{m}} & - \frac{1}{4}\sqrt{\frac{2K}{m}} \\ 0&-i\nu_{-n} + \sqrt{\frac{2K}{m}} & - \frac{1}{4}\sqrt{\frac{2K}{m}} & - \frac{1}{4}\sqrt{\frac{2K}{m}} \\ - \frac{1}{4}\sqrt{\frac{2K}{m}} & - \frac{1}{4}\sqrt{\frac{2K}{m}} & -i\nu_n + \sqrt{\frac{2K}{m}}&0 \\ - \frac{1}{4}\sqrt{\frac{2K}{m}} & - \frac{1}{4}\sqrt{\frac{2K}{m}} & 0&-i\nu_{-n} + \sqrt{\frac{2K}{m}} \end{pmatrix} \begin{pmatrix} \phi_{n} \\ \bar\phi_{-n} \\ \psi_{n} \\ \bar\psi_{-n} \end{pmatrix} $$

Next, we obtain the partition function by exponentiating the action and integrating over the fields. This gives

$$ \mathcal{Z} = \int \mathcal{D}\left(\dots\right) e^{-S} \\ = e^{-\beta \sqrt{\frac{2K}{m}}} \prod_{n>0}\frac{1}{\beta^4\left[\nu_n^4 + 2\nu_n^2\frac{2K}{m}+\frac{3}{4}\left(\frac{2K}{m}\right)^2\right]} \\ \times \int d\left(\bar\phi_0,\phi_0\right)\,d\left(\bar\psi_0,\psi_0\right) \nonumber \\ \times \exp\left[ -\bar\phi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\phi_0 - \bar\psi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\psi_0 + \frac{1}{4}\sqrt{\frac{2K}{m}} \left(\bar\phi_0 \bar\psi_{0}+ \phi_0 \psi_{0}+ \bar\phi_0 \psi_{0}+\bar\psi_0 \phi_{0}\right) \right] $$

The product over $\nu_{n>0}$ can be rewritten as $$ \prod_{n>0}\frac{1}{\beta^4\left[\nu_n^4 + 2\nu_n^2\frac{2K}{m}+\frac{3}{4}\left(\frac{2K}{m}\right)^2\right]} \\ =\prod_{n}\frac{1}{\beta^2\left[\left(-i\nu_n + \sqrt{\frac{K}{m}}\right) \left(-i\nu_n + \sqrt{\frac{3K}{m}}\right)\right]}\times \beta^2\left(-i\nu_0 + \sqrt{\frac{K}{m}}\right) \left(-i\nu_0 + \sqrt{\frac{3K}{m}}\right) $$ Now the product runs over all $\nu_n$. The additional factors are there because the original product did not include the zeroth harmonic. Note that $\nu_0 = 0$.

Going back to the zeroth harmonic integral $$ \int d\left(\bar\phi_0,\phi_0\right)\,d\left(\bar\psi_0,\psi_0\right) \\ \times \exp\left[ -\bar\phi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\phi_0 - \bar\psi_0\left(-i\nu_0 + \sqrt{\frac{2K}{m}}\right)\psi_0 + \frac{1}{4}\sqrt{\frac{2K}{m}}\left(\bar\phi_0 + \phi_0\right)\left(\bar\psi_0+\psi_0\right) \right] \\ = \int \frac{dx\,dy}{\pi}\frac{da\,db}{\pi} \exp\left[ -\left(x^2 + y^2 + a^2 + b^2\right)\left(\sqrt{\frac{2K}{m}}\right) + \frac{1}{4}\sqrt{\frac{2K}{m}}\left(2x\right)\left(2a\right) \right] \\ = \int \frac{dx\,dy}{\pi}\frac{da\,db}{\pi} \exp\left[ -\left(x^2 + y^2 + a^2 + b^2\right)\left( \sqrt{\frac{2K}{m}}\right) + \sqrt{\frac{2K}{m}}xa \right] \\ = \frac{1}{\beta\left(\sqrt{\frac{2K}{m}}\right)}\int \frac{dx\,da}{\pi} \exp\left[ -\left(x^2 + a^2\right)\left( \sqrt{\frac{2K}{m}}\right) + \sqrt{\frac{2K}{m}}xa \right] \\ = \frac{1}{\beta^2\sqrt{3}\frac{K}{m}} $$ Combining all this together gives $$ \mathcal{Z} = e^{-\beta\sqrt{\frac{2K}{m}}} \prod_{n}\frac{1}{\beta^2\left[\left(-i\nu_n + \sqrt{\frac{K}{m}}\right) \left(-i\nu_n + \sqrt{\frac{3K}{m}}\right)\right]}\,. $$

To get the energy of the system, we do $-T\ln\mathcal{Z}$:

$$ E = \sqrt{\frac{2K}{m}} + T\sum_n \ln\left[\beta\left(-i\nu_n + \sqrt{\frac{K}{m}}\right) \right] + \ln\left[\beta\left(-i\nu_n + \sqrt{\frac{3K}{m}}\right) \right]\,. $$

The last two terms can be summed the usual way to give $$ E = \sqrt{\frac{2K}{m}} + T \ln\left[1 - e^{-\beta\sqrt{\frac{K}{m}}}\right] + T \ln\left[1 - e^{-\beta\sqrt{\frac{3K}{m}}}\right]\,, $$ which vanish at $T = 0$. The energies inside these logarithms are the correct energies of the coupled oscillator modes. However, the term in the front, the vacuum energy that should be the only thing present at $T = 0$ is wrong...

  • 1
    $\begingroup$ I didn't understand as to how the original Hamiltonian (also, to confirm, which expression is supposed to mean the original Hamiltonian--the one with $a,b$ operators, right?) yields the ground-state energy to be $\sqrt{\frac{2K}{m}}$. $\endgroup$ – Dvij Mankad Mar 6 at 10:53
  • $\begingroup$ Yes, the "original" one is with $a$ and $b$. What I meant is that the scalar quantity not attached to the operator is $\sqrt{2K/m}$ $\endgroup$ – IcyOtter Mar 6 at 11:18
  • $\begingroup$ Okay, I see--that is not the way to find the ground-state energy. I will shortly write an answer to explain the point in a bit more detail. $\endgroup$ – Dvij Mankad Mar 6 at 11:20
  • 1
    $\begingroup$ Thank you! Let me clarify where the issue is for me. I know that we are not supposed to grab the scalar part and call it the zero-point energy. I have been trying to obtain the energy of the system using the path integral approach. At the end, I appear to be getting the correct energies for the excitations (the $\sqrt{K/m}\alpha^\dagger \alpha$ and $\sqrt{3K/m}\beta^\dagger \beta$), but my vacuum energy is still wrong. I was thinking that the problem was the commutation of $a$ with $b$, but that makes little sense... I'll post my derivation here to show what I mean. $\endgroup$ – IcyOtter Mar 6 at 11:33
  • $\begingroup$ Yes, that would be much better! :) $\endgroup$ – Dvij Mankad Mar 6 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.