I'm currently working through Griffith's Introduction to Quantum Mechanics to prepare for an exam. My question is regarding problem 3.35.
3.35, part a) Calculate $\left< x \right>$, $\left< x^2 \right>$,$\left< p \right>$, $\left< p^2 \right>$ in the state $\left| \alpha \right>$. Remember that $a_+$ is the hermitian conjugate of $a_-$. Do not assume $\alpha$ is real.
The book says to follow example 2.5, which gives the following two equations. $$ x = \sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-) \, \, , \, \, p = i\sqrt{\frac{\hbar}{2m\omega}}(a_+ - a_-) $$
Where $a_+$ and $a_-$ are ladder operators given by: $$ a_\pm = \frac{1}{\sqrt{2\hbar m \omega}} (\mp ip + m \omega x) $$ $$ a_+a_-= \frac{H}{\hbar \omega} - \frac{1}{2} \, \, , \, \, a_-a_+=\frac{H}{\hbar \omega} + \frac{1}{2} $$ $$ [a_-,a_+] = 1 $$
Where $H$ is the Hamiltonian.
For $x^2$,
$$ x^2 = \frac{\hbar}{2m\omega} (a_+^2 +a_+a_- + a_-a_+ + a_-^2) = \frac{\hbar}{2m\omega} (a_+^2 +2a_+a_- + 1 + + a_-^2)$$
This is due to $ a_-a_+ = [a_-,a_+] + a_+ a_- = 1 + a_+a_- $. The following calculation for $\left< x^2 \right>$ is where I get confused, I'll state exactly where momentarily.
$$\left< x^2 \right> = \frac{\hbar}{2m\omega}\left < \alpha | (a_+^2 + 2a_+a_- + 1 + a_-^2) \alpha \right> \\ = \frac{\hbar}{2m\omega}(\left<a_-^2 \alpha | \alpha \right> +2\left<a_- \alpha |a_- \alpha \right> + \left<\alpha |\alpha \right> + \left< \alpha |a_-^2 \alpha \right>) $$
The following step is where my confusion is. I'm not sure why $\alpha^*$ appears.
$$ = \frac{\hbar}{2m\omega} \left[(\alpha^*)^2 + 2(\alpha^*)\alpha + 1 + \alpha^2 \right] \\ = \frac{\hbar}{2m\omega} \left[ 1+ (\alpha +\alpha^*)^2 \right] $$
I'm confused about the mathematics behind the switch to $\alpha^*$, I'm not sure about the significance of that step. I know that $\left< \alpha \right| = \left| \alpha \right>^*$, but I'm not sure how that plays a roll in the calculation.
Thank you!
This is basically using the complex version of the inner product. When entries are moved from one side of an inner product to the other they undergo a $ * operation $.
The product can be shown as an integral of $ \int \phi_1 ^* \phi_2 d^3r $ $ = (\phi_1, \phi_2) $
Also, $ (\phi_1,\phi_2) = (\phi_2, \phi_1)^* $
$ < \alpha | \alpha > = (\phi, \phi) $ $ = \int \phi^*\phi {d^3r} $
$ a_- |\alpha> = \alpha | \alpha > $
When this operation takes place to the left, the $ * $ is used. It is best to think of the $ \alpha > $ or $ < \alpha $ on each side as the $ dxdydz $ in an integral, or as a summation index. Entries on the left get a $ * $.
The idea is to get the expectation of the operator $ X^2 $ which, because $ X^2 $ is hermitian, is a real number, as you can confirm by applying the $ * $ operator to it.