Evaluating using BRA KET Notation?

Question

Evaluate $\langle 0 \mid x^3 \mid 1\rangle$, assuming that all the wave functions you encounter are normalized eigenfunctions of the harmonic oscillator Hamiltonian, without Mathematica, Maple, or software! (I am aware that this overlaps with Physics, however it's primarily the math notation and logic I'm having problems with. The hamiltonian is just a definition and to me is more of a mathematics discipline)

Steps:

$$\langle 0 \mid x^3 \mid 1\rangle= \int^x_{-x} \Psi_0^* (x) x^3 \Psi_1 (x) dx = \sqrt{\frac{2m\omega}{\hbar}}\sqrt{\frac{m\omega}{\pi\hbar}}\int^\infty_{-\infty}x^4 e^{-m\omega x^2/\hbar}dx$$

So the hint I got was to rewrite the right hand side of the equation, which I found to be:

$$\int^\infty_{-\infty}x^4 e^{-m\omega x^2/\hbar}dx = \frac{d^2}{d\alpha^2} \sqrt{\frac{\pi}{\alpha}}, \quad \alpha = \frac{m\omega}{\hbar}.$$

The second hint was to perform the derivatives, sub back the definition for $\alpha$ and don't forget the prefactors. Might be quicker to solve with raising and lowering operators.

The step I'm not understanding is how to apply the derivatives when they're on the other side of the equation? Are they movable, algebraically?

Answer 1

Do a sub $\frac {m\omega x^2}{\hbar^2} = t^2$, then $x = \frac {\hbar t}{\sqrt {m\omega}}$ and $dx = \frac {\hbar dt}{\sqrt {m\omega}} $, so integral can be rewritten as $$ \int_{-\infty}^\infty x^4 e^{-\frac {m\omega x^2}{\hbar^2}} dx = \int_{-\infty}^\infty \frac {\hbar^4 t^4}{m^2\omega^2} \frac {\hbar}{\sqrt{m\omega}}e^{-t^2}dt = \frac {\hbar^5}{\left(m\omega\right)^{\frac 52}} \int_{-\infty}^\infty t^4e^{-t^2}dt $$ So the problem reduces to determination of latter integral $$ \int_{-\infty}^\infty t^4 e^{-t^2}dt = \left .-\frac {t^3 e^{-t^2}}2\right|_{-\infty}^{+\infty} + \frac 32 \int_{-\infty}^\infty t^2 e^{-t^2} dt = \\ = \frac 32 \left(\left .-\frac {t e^{ -t^2}}2 \right|_{-\infty}^{+\infty} + \frac 12 \int_{-\infty}^\infty e^{-t^2}dt\right) = \frac 34 \sqrt \pi $$ and final integral is $$ \left <0 \left | x^3 \right | 1\right > = \frac {m\omega}\hbar \sqrt{\frac 2\pi} \frac {\hbar^5}{\left(m\omega\right)^{\frac 52}}\frac 34 \sqrt \pi = \frac {3\sqrt 2 \hbar^4}{4 \left( m\omega\right)^{\frac 32}} $$ PS: If it's too vague – I used integration by parts twice and at the end used the fact that $$ \int_{-\infty}^\infty e^{-t^2}dt = \sqrt \pi $$ It's Euler-Poisson or Gaussian integral.

Answer 2

Using your definition of $\alpha$, we get $$\langle 0 \mid x^3 \mid 1\rangle= \sqrt{\frac{2}{\pi}} \alpha \int^\infty_{-\infty}x^4 e^{-\alpha x^2}dx \ .$$

The important observation concerning the derivative with respect to $\alpha$ is that $$ \frac{\partial^2}{\partial\alpha^2} e^{-\alpha x^2} = \frac{\partial}{\partial\alpha} \left[-x^2 e^{-\alpha x^2}\right] = x^4 e^{-\alpha x^2} $$

Substituting back into your original integral, switching integration by $x$ and derivative w.r.t. $\alpha$ $$\langle 0 \mid x^3 \mid 1\rangle= \sqrt{\frac{2}{\pi}} \alpha \int^\infty_{-\infty} \frac{\partial^2}{\partial\alpha^2} e^{-\alpha x^2}dx = \sqrt{\frac{2}{\pi}} \alpha \frac{\partial^2}{\partial\alpha^2} \underbrace{\int^\infty_{-\infty} e^{-\alpha x^2}dx}_{\sqrt{\pi / \alpha}}$$ Thus, your final result is $$ \sqrt{\frac{2}{\pi}} \sqrt{\pi} \alpha\, \frac{\partial^2}{\partial\alpha^2} \frac{1}{\sqrt{\alpha}} = \sqrt{2} \alpha \frac{1}{2} \cdot \frac{3}{2} \alpha^{-5/2} = \sqrt{2}\, \alpha^{-3/2}$$

With this "derivative trick" you could also compute $\langle 0 \mid x^{33} \mid 1\rangle$, if you are able to take the 33rd derivative of $1/\sqrt{\alpha}$ w.t.r. to $\alpha$