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Question 2.12: Find (x) , (p) , (x^2) , (p^2) and (T) , for the nth station...

Find \left\langle x\right\rangle , \left\langle p\right\rangle ,\left\langle x^2\right\rangle , \left\langle p^2\right\rangle   and \left\langle T\right\rangle , for the nth stationary state of the harmonic oscillator, using the method of Example 2.5. Check that the uncertainty principle is satisfied.

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From Eq. 2.70,

x=\sqrt{\frac{\hbar}{2 m \omega}}\left(\hat{a}_{+}+\hat{a}_{-}\right) ; \quad \hat{p}=i \sqrt{\frac{\hbar m \omega}{2}}\left(\hat{a}_{+}-\hat{a}_{-}\right)   (2.70).

x=\sqrt{\frac{\hbar}{2 m \omega}}\left(a_{+}+a_{-}\right), \quad p=i \sqrt{\frac{\hbar m \omega}{2}}\left(a_{+}-a_{-}\right) .

so

\langle x\rangle=\sqrt{\frac{\hbar}{2 m \omega}} \int \psi_{n}^{*}\left(a_{+}+a_{-}\right) \psi_{n} d x .

But (Eq. 2.67)

\hat{a}_{+} \psi_{n}=\sqrt{n+1} \psi_{n+1}, \quad \hat{a}_{-} \psi_{n}=\sqrt{n} \psi_{n-1}       (2.67).

a_{+} \psi_{n}=\sqrt{n+1} \psi_{n+1}, \quad a_{-} \psi_{n}=\sqrt{n} \psi_{n-1} .

So

\langle x\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left[\sqrt{n+1} \int \psi_{n}^{*} \psi_{n+1} d x+\sqrt{n} \int \psi_{n}^{*} \psi_{n-1} d x\right]= 0 (by orthogonality).

  \langle p\rangle=m \frac{d\langle x\rangle}{d t}= 0 .

x^{2}=\frac{\hbar}{2 m \omega}\left(a_{+}+a_{-}\right)^{2}=\frac{\hbar}{2 m \omega}\left(a_{+}^{2}+a_{+} a_{-}+a_{-} a_{+}+a_{-}^{2}\right) .

\left\langle x^{2}\right\rangle=\frac{\hbar}{2 m \omega} \int \psi_{n}^{*}\left(a_{+}^{2}+a_{+} a_{-}+a_{-} a_{+}+a_{-}^{2}\right) \psi_{n} .  But

\begin{cases}a_{+}^{2} \psi_{n}=a_{+}\left(\sqrt{n+1} \psi_{n+1}\right)=\sqrt{n+1} \sqrt{n+2} \psi_{n+2} & =\sqrt{(n+1)(n+2)} \psi_{n+2} . \\ a_{+} a_{-} \psi_{n}=a_{+}\left(\sqrt{n} \psi_{n-1}\right) & =\sqrt{n} \sqrt{n} \psi_{n} & =n \psi_{n} .\\ a_{-} a_{+} \psi_{n}=a_{-}\left(\sqrt{n+1} \psi_{n+1}\right)=\sqrt{n+1} \sqrt{n+1} \psi_{n} &= (n+1) ψ_n . \\ a_{-}^{2} \psi_{n}=a_{-}\left(\sqrt{n} \psi_{n-1}\right) &=\sqrt{n} \sqrt{n-1} \psi_{n-2} &=\sqrt{(n-1) n} \psi_{n-2} .\end{cases} .

\left\langle x^{2}\right\rangle=\frac{\hbar}{2 m \omega}\left[0+n \int\left|\psi_{n}\right|^{2} d x+(n+1) \int\left|\psi_{n}\right|^{2} d x+0\right]=\frac{\hbar}{2 m \omega}(2 n+1)= \left(n+\frac{1}{2}\right) \frac{\hbar}{m \omega} .

p^{2}=-\frac{\hbar m \omega}{2}\left(a_{+}-a_{-}\right)^{2}=-\frac{\hbar m \omega}{2}\left(a_{+}^{2}-a_{+} a_{-}-a_{-} a_{+}+a_{-}^{2}\right) \Rightarrow

\left\langle p^{2}\right\rangle=-\frac{\hbar m \omega}{2}[0-n-(n+1)+0]=\frac{\hbar m \omega}{2}(2 n+1)=  \left(n+\frac{1}{2}\right) m \hbar \omega .

\langle T\rangle=\left\langle p^{2} / 2 m\right\rangle= \frac{1}{2}\left(n+\frac{1}{2}\right) \hbar \omega .

\sigma_{x}=\sqrt{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}}=\sqrt{n+\frac{1}{2}} \sqrt{\frac{\hbar}{m \omega}} ;    \sigma_{p}=\sqrt{\left\langle p^{2}\right\rangle-\langle p\rangle^{2}}=\sqrt{n+\frac{1}{2}} \sqrt{m \hbar \omega} ;      \sigma_{x} \sigma_{p}=\left(n+\frac{1}{2}\right) \hbar \geq \frac{\hbar}{2} .

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