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Problem Statement

Prompt

Provide step by step solution for this question p7.12

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**P7.12)** Show by carrying out the appropriate integration that the total energy
eigenfunctions for the harmonic oscillator
\[
\psi_0(x)=\left(\frac{\alpha}{\pi}\right)^{1/4}e^{-\alpha x^2/2}
\qquad\text{and}\qquad
\psi_2(x)=\left(\frac{\alpha}{4\pi}\right)^{1/4}(2\alpha x^2-1)e^{-\alpha x^2/2}
\]
are orthogonal over the interval $$-\infty<x<\infty$$ and that $$\psi_2(x)$$ is normalized
over the same interval.

In evaluating integrals of this type:
\[
\int_{-\infty}^{\infty} f(x)\,dx=0 \quad \text{if } f(x)\text{ is odd}
\]
and
\[
\int_{-\infty}^{\infty} f(x)\,dx
=2\int_{0}^{\infty} f(x)\,dx \quad \text{if } f(x)\text{ is even}.
\]

We use the standard integrals:
\[
\int_{0}^{\infty}x^{2n}e^{-ax^2}\,dx=
\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n+1}a^n}\sqrt{\frac{\pi}{a}}
\]
and
\[
\int_{0}^{\infty}e^{-ax^2}\,dx=\left(\frac{\pi}{4a}\right)^{1/2}.
\]

**Orthogonality:**

\[
\int_{-\infty}^{\infty}\psi_2^*(x)\psi_0(x)\,dx
=\int_{-\infty}^{\infty}
\left(\frac{\alpha}{4\pi}\right)^{1/4}(2\alpha x^2-1)e^{-\alpha x^2/2}
\left(\frac{\alpha}{\pi}\right)^{1/4}e^{-\alpha x^2/2}\,dx
\]

\[
=\left(\frac{\alpha^2}{4\pi^2}\right)^{1/4}
\int_{-\infty}^{\infty}(2\alpha x^2-1)e^{-\alpha x^2}\,dx
\]

Since the integrand is even,
\[
=\left(\frac{\alpha^2}{4\pi^2}\right)^{1/4}
\cdot 2\int_{0}^{\infty}(2\alpha x^2-1)e^{-\alpha x^2}\,dx
\]

\[
=\left(\frac{\alpha^2}{4\pi^2}\right)^{1/4}
\cdot 2\left[
2\alpha\int_{0}^{\infty}x^2e^{-\alpha x^2}\,dx
-\int_{0}^{\infty}e^{-\alpha x^2}\,dx
\right]
\]

Using
\[
\int_{0}^{\infty}x^2e^{-\alpha x^2}\,dx=\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}},
\qquad
\int_{0}^{\infty}e^{-\alpha x^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}},
\]
we get
\[
=\left(\frac{\alpha^2}{4\pi^2}\right)^{1/4}
\cdot 2\left[
2\alpha\left(\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}}\right)
-\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
\right]
\]

\[
=\left(\frac{\alpha^2}{4\pi^2}\right)^{1/4}
\cdot 2\left[
\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
-\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
\right]=0.
\]

Hence, $$\psi_0$$ and $$\psi_2$$ are orthogonal.

**Normalization of $$\psi_2$$:**

\[
\int_{-\infty}^{\infty}\psi_2^*(x)\psi_2(x)\,dx
=\int_{-\infty}^{\infty}
\left(\frac{\alpha}{4\pi}\right)^{1/2}(2\alpha x^2-1)^2e^{-\alpha x^2}\,dx
\]

Since the integrand is even,
\[
=2\left(\frac{\alpha}{4\pi}\right)^{1/2}
\int_{0}^{\infty}(2\alpha x^2-1)^2e^{-\alpha x^2}\,dx
\]

Expand:
\[
(2\alpha x^2-1)^2=4\alpha^2x^4-4\alpha x^2+1
\]

So
\[
=2\left(\frac{\alpha}{4\pi}\right)^{1/2}
\left[
4\alpha^2\int_{0}^{\infty}x^4e^{-\alpha x^2}\,dx
-4\alpha\int_{0}^{\infty}x^2e^{-\alpha x^2}\,dx
+\int_{0}^{\infty}e^{-\alpha x^2}\,dx
\right]
\]

Using
\[
\int_{0}^{\infty}x^4e^{-\alpha x^2}\,dx=\frac{3}{8\alpha^2}\sqrt{\frac{\pi}{\alpha}},
\qquad
\int_{0}^{\infty}x^2e^{-\alpha x^2}\,dx=\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}},
\qquad
\int_{0}^{\infty}e^{-\alpha x^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}},
\]

we obtain
\[
=2\left(\frac{\alpha}{4\pi}\right)^{1/2}
\left[
4\alpha^2\left(\frac{3}{8\alpha^2}\sqrt{\frac{\pi}{\alpha}}\right)
-4\alpha\left(\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}}\right)
+\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
\right]
\]

\[
=2\left(\frac{\alpha}{4\pi}\right)^{1/2}
\left[
\frac{3}{2}\sqrt{\frac{\pi}{\alpha}}
-\sqrt{\frac{\pi}{\alpha}}
+\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
\right]
\]

\[
=2\left(\frac{\alpha}{4\pi}\right)^{1/2}
\left[\sqrt{\frac{\pi}{\alpha}}\right]
=2\cdot \frac{\sqrt{\alpha}}{2\sqrt{\pi}}\cdot \sqrt{\frac{\pi}{\alpha}}=1.
\]

Therefore, $$\psi_2(x)$$ is normalized.

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Problem #49