Expectation Value of $x^2$ in a specific state.
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!
$\endgroup$1 Answer
$\begingroup$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.
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