(a) Construct ψ2(x). (b) Sketch ψ0, ψ1, and ψ2. (c) Check the orthogonality of ψ0, ψ1, and ψ2, by explicit integration. Hint: If you exploit the even-ness and odd-ness of the functions, there is really only one integral left to do.

Question

(a) Construct [latex] ψ_2(x) [/latex].

(b) Sketch [latex] ψ_0, ψ_1 [/latex], and [latex] ψ_2 [/latex].

(c) Check the orthogonality of [latex] ψ_0, ψ_1 [/latex], and [latex] ψ_2 [/latex], by explicit integration. Hint: If you exploit the even-ness and odd-ness of the functions, there is really only one integral left to do.

Accepted Answer

(a) Using Eqs. 2.48 and 2.60,

[latex] \hat{a}_{\pm} \equiv \frac{1}{\sqrt{2 \hbar m \omega}}(\mp i \hat{p}+m \omega x) [/latex] (2.48).

[latex] \psi_{0}(x)=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2\hbar} x^{2}} [/latex]. (2.60).

[latex] a_{+} \psi_{0}=\frac{1}{\sqrt{2 \hbar m \omega}}\left(-\hbar \frac{d}{d x}+m \omega x\right)\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-\frac{m \omega}{2 h} x^{2}} [/latex].

[latex] =\frac{1}{\sqrt{2 \hbar m \omega}}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4}\left[-\hbar\left(-\frac{m \omega}{2 \hbar}\right) 2 x+m \omega x\right] e^{-\frac{m \omega}{2 \hbar} x^{2}}=\frac{1}{\sqrt{2 \hbar m \omega}}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} 2 m \omega x e^{-\frac{m \omega}{2 \hbar} x^{2}} [/latex].

[latex] \left(a_{+}\right)^{2} \psi_{0}=\frac{1}{2 \hbar m \omega}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} 2 m \omega\left(-\hbar \frac{d}{d x}+m \omega x\right) x e^{-\frac{m \omega}{2 \hbar} x^{2}} [/latex].

[latex] =\frac{1}{\hbar}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4}\left[-\hbar\left(1-x \frac{m \omega}{2 \hbar} 2 x\right)+m \omega x^{2}\right] e^{-\frac{m \omega}{2 \hbar} x^{2}}=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4}\left(\frac{2 m \omega}{\hbar} x^{2}-1\right) e^{-\frac{m \omega}{2 \hbar} x^{2}} [/latex].

Therefore, from Eq. 2.68,

[latex] \psi_{n}=\frac{1}{\sqrt{n !}}\left(\hat{a}_{+}\right)^{n} \psi_{0} [/latex] (2.68).

[latex] \psi_{2}=\frac{1}{\sqrt{2}}\left(a_{+}\right)^{2} \psi_{0} = \frac{1}{\sqrt{2}}\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4}\left(\frac{2 m \omega}{\hbar} x^{2}-1\right) e^{-\frac{m \omega}{2 h} x^{2}} [/latex].

(b)

(c) Since [latex] ψ_0 [/latex] and [latex] ψ_2 [/latex] are even, whereas [latex] ψ_1 [/latex] is odd, [latex] \int \psi_{0}^{*} \psi_{1} d x [/latex] vanish automatically. The only one we need to check is [latex] \int \psi_{2}^{*} \psi_{0} d x [/latex];

[latex] \int \psi_{2}^{*} \psi_{0} d x=\frac{1}{\sqrt{2}} \sqrt{\frac{m \omega}{\pi \hbar}} \int_{-\infty}^{\infty}\left(\frac{2 m \omega}{\hbar} x^{2}-1\right) e^{-\frac{m \omega}{h} x^{2}} d x [/latex].

[latex] =-\sqrt{\frac{m \omega}{2 \pi \hbar}}\left(\int_{-\infty}^{\infty} e^{-\frac{m \omega}{h} x^{2}} d x-\frac{2 m \omega}{\hbar} \int_{-\infty}^{\infty} x^{2} e^{-\frac{m \omega}{h} x^{2}} d x\right) [/latex].

[latex] =-\sqrt{\frac{m \omega}{2 \pi \hbar}}\left(\sqrt{\frac{\pi \hbar}{m \omega}}-\frac{2 m \omega}{\hbar} \frac{\hbar}{2 m \omega} \sqrt{\frac{\pi \hbar}{m \omega}}\right)=0 [/latex].

Question 2.10: (a) Construct ψ2(x). (b) Sketch ψ0, ψ1, and ψ2. (c) Check th...

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