Given the following Hamiltonian for two identical linear oscillators with spring constant $k$ and interaction potential $\alpha x_1x_2$; I was asked to find the expectation value $\langle x_1x_2\rangle$ ($x_1$&$x_2$ are oscillator variables): $$\hat{H}=\frac{\hat{p}^2_1}{2m}+\frac{\hat{p}^2_2}{2m}+\frac12k(x_1^2+x_2^2)+\alpha x_1x_2 $$

Not knowing anything else to do, I switched to normal coordinates:$$x_1=\frac{x_I+x_{II}}{\sqrt 2} $$ $$x_2=\frac{x_I-x_{II}}{\sqrt 2} $$ The momentum operators are correspondingly changed in the same fashion. Thus, the new Hamiltonian is: $$ \hat{H}=\frac{\hat{p}^2_I}{2m}+\frac{\hat{p}^2_{II}}{2m}+\frac12(k+\alpha)x_I^2+\frac12(k-\alpha)x_{II}^2$$ Now, this can be solved as two separate eigenvalue problems, thus yielding two solutions.

  1. However, The problem never specifies the type of particles are on these oscillators (i.e. identical fermions, bosons...). So this leads me to ask, how should the wave function be formed? Symmetric, Asymmetric, or neither?

  2. Now, the second part. Given whichever way the wave function is formed, $x_1x_2=\frac12(x_I^2-x_{II}^2)$. I then attempted to formulate this in terms of the raising and lowering operators. It seems to me that there will be four "types" of operators. We can have (using Griffith's notation): $a_I^+$, $a{_I}^{-}$, $a_{II}^+$, and $a_{II}^{-}$. Where each one corresponds to operators in the respective basis. Using this seemed a bit strange to me when trying to apply the operators to a wave function that had been formed asymmetrically or symmetrically. For example, how could one apply the operator to something like: $$ a_{I}^+|\psi_I^{n_I}{(x_{II})}\psi_{II}^{n_{II}}{(x_{I})}\rangle $$ Or, am I thinking about this incorrectly somehow? Or maybe there is some easier way. Any guidance is greatly appreciated.

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    $\begingroup$ Furthermore, in order to be able to evaluate some expectation value, one needs to know what state we're in... $\endgroup$ – Danu Sep 14 '14 at 20:34
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    $\begingroup$ what state are you taking the expectation with respect to? i.e. what is $| \Psi\rangle$ in $\langle x_1x_2\rangle=\langle\Psi|x_1x_2|\Psi\rangle$? $\endgroup$ – Michael Sep 14 '14 at 20:34
  • $\begingroup$ Yes sorry! @Danu . I wasn't aware that the state needed to be specified. In the one deminsional case, we don't need to as the raising and lowering operators can give the expectation value in terms of the quantum number $n$. Will this case be different? The states were not specified $\endgroup$ – ClassicStyle Sep 14 '14 at 20:42
  • $\begingroup$ I think I see what you mean. I'm just assuming a stationary state $\endgroup$ – ClassicStyle Sep 14 '14 at 21:02

Okay, so for two interacting particles in a harmonic oscillator you need to figure out which of the wave-functions you found above satisfy the exchange requirements. Look at how $x_{I,II}$ are affected by exchange of $x_1$ and $x_2$. This will give requirements on what the allowed values of the $n_i$ are. The example you gave would behave as you expect. Giving $|\psi^{n_I+1}(x_I)\psi^{n_{II}}(x_{II})\rangle$. There is no guarantee that this state still satisfies the exchange requirements a priori.

  • $\begingroup$ I'm pretty sure this is for a two particle system. Also, the statistics part was just a small part of the problem. I was unclear on how to apply the operators to this system. What do you think of the last part of my question? $\endgroup$ – ClassicStyle Sep 14 '14 at 21:41
  • $\begingroup$ Okay, so the product $\psi_I\psi_{II}$ is anti-symmetric under exchange if $n_{II}$ is odd and symmetric if $n_{II}$ is even. So does this mean that the wave function is actually written down like this as a product, and the quantum numbers are determined by whether the particles are fermions or bosons? I guess this makes sense! It will make calculating the expectation value much easier! $\endgroup$ – ClassicStyle Sep 14 '14 at 23:58

If your particles are identical bosons n two=0,2,4... If they are fermions n two=1,3,5.... a two will change only psi two.

  • $\begingroup$ Haha thanks for the response but I got it. The hermite polynomials are even and odd! $\endgroup$ – ClassicStyle Nov 8 '14 at 13:54

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