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【单选题】

考虑一维谐振子电偶极跃迁的选择定则。设谐振子的粒子质量是m,固有频率是$\omega_0$,带电荷q,在外光源的照射下它可能发生从能量本征态$|n\rangle$ 到$|n'\rangle$ 的跃迁。在长波近似下,跃迁矩阵元是$F_{n'n}=qE_0<n'|x|n>$ ,$E_0$是外光场电场强度的峰值。问题是:当外光源的频率$\omega$和量子数n,n'满足什么条件的时候可以发生共振跃迁?$F_{n'n}=?$ .提示:产生算符为 $\widehat{a}^+=\frac{1}{\sqrt{2}}(\sqrt{\frac{m w}{\hbar}}\widehat{x}-i\sqrt{\frac{1}{m w\hbar}}\widehat{p})$

A.
$\Delta n =\pm 1,F_{n'n}=qE_0\sqrt{\frac{\hbar}{2m\omega_0}}(\sqrt{n+1}\delta_{n',n+1}+\sqrt{n}\delta_{n',n-1})$
B.
$\Delta n =\pm 1,F_{n'n}=qE_0\sqrt{\frac{\hbar}{2m\omega_0}}(\sqrt{n+1}\delta_{n',n+1})$
C.
$\Delta n =0,F_{n'n}=qE_0\sqrt{\frac{\hbar}{2m\omega_0}}(\sqrt{n+1}\delta_{n',n+1}+\sqrt{n}\delta_{n',n-1})$
D.
$\Delta n = 0,F_{n'n}=qE_0\sqrt{\frac{\hbar}{2m\omega_0}}(\sqrt{n+1}\delta_{n',n+1})$
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【单选题】已知谐振子本征态为$|n\rangle$,而产生算符为 $\widehat{a}^+=\frac{1}{\sqrt{2}}(\sqrt{\frac{m w}{\hbar}}\widehat{x}-i\sqrt{\frac{1}{m w\hbar}}\widehat{p})$求$|n\rangle$与最小本征态$|n\rangle$的关系:

A.
$|n\rangle=\frac{1}{\sqrt{n!}}(\widehat{a})^n|0\rangle$
B.
$|n\rangle=\frac{1}{\sqrt{2n!}}(\widehat{a}^+)^n|0\rangle$
C.
$|n\rangle=\frac{1}{\sqrt{n!}}(\widehat{a}^+)^n|0\rangle$
D.
$|n\rangle=\frac{1}{\sqrt{2n}}(\widehat{a}^+)^n|0\rangle$

【单选题】线性谐振子偶极跃迁的选择定则正确的是:

A.
$\bigtriangleup m=m-k=\pm1$
B.
$m=\pm2k+1$
C.
$m=\pm2k$
D.
$\bigtriangleup m=m-k=\pm2$