Question

Imagine two non interacting particles, each of mass , in the one dimensional harmonic oscillator potential (Equation 2.43). If one is in the ground state, and the other is in the first excited state, calculate $\mathbf{⟨}\mathbf{(}{\mathit{x}}_{\mathbf{1}}\mathbf{-}{\mathit{x}}_{\mathbf{2}}{\mathbf{)}}^{\mathbf{2}}\mathbf{⟩}$assuming
(a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions. Ignore spin (if this bothers you, just assume they are both in the same spin state.)

Short answer

(a) $\left(\right(x_1-x_2{)}^2)=\frac{2h}{m\omega }.$

(b)$(x_1-x_2{)}^2=\frac{h}{m\omega }$

(c)$(x_1-x_2{)}^2=\frac{3h}{m\omega }$

Step-by-step solution

(a) Determining ⟨(x1-x2)2⟩ when they are distinguishable particles.

From the Previous problem :

$$\begin{gathered} ⟨x{⟩}_0=0 \\ ⟨x{⟩}_1=0 \\ {\left(x^2\right)}_0=\frac{h}{2m\omega } \\ {x}^{2}1=\frac{3h}{2m\omega } \end{gathered}$$

We need to calculate:

$$\begin{gathered} \left(0\left|x\right|1\right)={\int }_{-\infty }^{\infty }x{\psi }_0\left(x\right){\psi }_1\left(x\right)dx \\ =\sqrt{\frac{h}{2m\omega }}\left(\sqrt{1}{\delta }_{0,0}+\sqrt{0}{\delta }_{1,-1}\right) \\ =\sqrt{\frac{h}{2m\omega }} \end{gathered}$$

From Equation 5.21:

role="math" localid="1658402226570" $$\begin{gathered} \left({\left(x_1-x_2\right)}^2\right)=\frac{h}{2m\omega }+\frac{3h}{2m\omega }-0 \\ =\frac{2h}{m\omega } \end{gathered}$$

(b) Determining ⟨(x1-x2)2⟩when they are identical bosons,

From Equation 5.21

$$\begin{gathered} \left({\left(x_1-x_2\right)}^2\right)=\frac{2ħ}{m\omega }+2\frac{h}{2m\omega } \\ =\frac{h}{m\omega } \end{gathered}$$

(c) Determining ⟨(x1-x2)2⟩ when they identical fermions,

From Equation 5.21:

$$\begin{gathered} \left({\left(x_1-x_2\right)}^2\right)=\frac{2h}{m\omega }+\frac{2h}{2m\omega } \\ =\frac{3h}{m\omega } \end{gathered}$$