Question 7.3: Returning to Example 7.2, show that diagonalizing the matrix...

We need to calculate the matrix elements of W. First,

W_{aa}=\int{\int{\psi ^{0}_{a}} }(x,y) \acute{H} \psi ^{0}_{a}(x,y)dxdy=\epsilon m\omega ^{2}\int{\left|\psi _{0}(x)\right| ^{2}}xdx \int{\left|\psi _{0}(y)\right| ^{2}}ydy =0

(the integrands are both odd functions). Similarly,W_{bb}=0 , and we need only compute

W_{ab}=\int{\int{\psi ^{0}_{a}} }(x,y) \acute {H} \psi ^{0}_{b}(x,y)dxdy=\epsilon m \omega ^{2}\int{\psi _{0}(x) }x\psi _{1}(x)dx \int {\psi _{1}(y)}y\psi _{0}(y)dy =0

These two integrals are equal, and recalling (Equation 2.70 [x=\sqrt{\frac{\hbar }{2m\omega } }(\hat{a} _{+}+\hat{a} _{-}) ; \hat{p}=i\sqrt{\frac{\hbar m\omega }{2 } }(\hat{a} _{+}-\hat{a} _{-}) ])

x=\sqrt{\frac{\hbar }{2m\omega } }(a_{+}+a_{-})

we have

W_{ab}=\epsilon m\omega ^{2}\left [\int {\psi _{0}(x) }\sqrt{\frac{\hbar }{2m\omega } } (a_{+}+a_{-}) \psi _{1}(x)dx \right]^{2} =\epsilon \frac{\hbar \omega}{2} \left[\int{\psi _{0}(x) \psi _{0}(x) } dx \right]^{2}=\epsilon \frac{\hbar \omega}{2}

Therefore, the matrix W is

W=\epsilon \frac{\hbar \omega}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

The (normalized) eigenvectors of this matrix are

\frac{1}{\sqrt{2} } \begin{pmatrix} 1 \\ 1 \end {pmatrix} and \frac{1}{\sqrt{2} } \begin {pmatrix} -1 \\ 1 \end{pmatrix}

These eigenvectors tell us which linear combination of \psi^{0}_{a} and \psi^{0}_{b} are the good states:

\psi^{0}_{\pm}=\frac{1}{\sqrt{2} } \left( \psi^{0}_{b} \pm \psi ^{0}_{a}\right)           (7.23)

just as in Equation 7.23. The eigenvalues of the matrix W ,

E^{1}=\pm\epsilon \frac{\hbar \omega }{2}

give the first-order corrections to the energy (compare 7.33).

E_{\pm}^1=\frac{1}{2}\left[W_{aa}+W_{ab}\pm \sqrt{(W_{aa}-W_{bb})^2+4|W_{ab}|^2}\right] .                        (7.33)