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): $$\hat{H}=\frac{\hat{p_1}^2}{2m}+\frac{\hat{p_2}^2}{2m}+\frac12k(x_1^2+x_2^2)+\alpha x_1x_2 $$

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: $$ \hat{H}=\frac{\hat{p_I}^2}{2m}+\frac{\hat{p_{II}}^2}{2m}+\frac12(k+\alpha)x_I^2+\frac12(k-\alpha)x_{II}^2$$ 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.

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.

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): $$\hat{H}=\frac{\hat{p_1}^2}{2m}+\frac{\hat{p_2}^2}{2m}+\frac12k(x_1^2+x_2^2)+\alpha x_1x_2 $$ 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: $$ \hat{H}=\frac{\hat{p_I}^2}{2m}+\frac{\hat{p_{II}}^2}{2m}+\frac12(k+\alpha)x_I^2+\frac12(k-\alpha)x_{II}^2$$ Now, this can be solved as two separate eigenvalue problems, thus yielding two solutions. 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? 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.

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): $$\hat{H}=\frac{\hat{p_1}^2}{2m}+\frac{\hat{p_2}^2}{2m}+\frac12k(x_1^2+x_2^2)+\alpha x_1x_2 $$

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: $$ \hat{H}=\frac{\hat{p_I}^2}{2m}+\frac{\hat{p_{II}}^2}{2m}+\frac12(k+\alpha)x_I^2+\frac12(k-\alpha)x_{II}^2$$ 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.