# How do you write position and momentum in terms of the raising and lowering operators?

## Something for the public to refer to. :)

Oct 5, 2016

We define the raising and lowering operators as hatb^† and $\hat{b}$ respectively. In one dimension, they are written as:

color(green)(hatb_x = sqrt((momega)/(2ℏ))(hatx + i/(momega)hatp_x))

color(green)(hatb_x^† = sqrt((momega)/(2ℏ))(hatx - i/(momega)hatp_x))

where hatb_x^† is the adjoint (complex conjugate transpose) of ${\hat{b}}_{x}$, and their commutation is [hatb_x,hatb_x^†] = hatb_xhatb_x^† - hatb_x^†hatb_x = 1.

They don't really represent anything physical, per se, but they can be convenient to use. We can add these together or subtract them to find the position and momentum operators in terms of them.

hatb_x + hatb_x^† = sqrt((momega)/(2ℏ))(hatx + i/(momega)hatp_x) + sqrt((momega)/(2ℏ))(hatx - i/(momega)hatp_x)

= sqrt((momega)/(2ℏ))[(hatx + cancel(i/(momega)hatp_x)) + (hatx - cancel(i/(momega)hatp_x))]

= sqrt((momega)/(2ℏ))(2hatx)

= sqrt((2momega)/(ℏ))hatx

Therefore:

color(blue)(hatx = sqrt(ℏ/(2momega))(hatb_x + hatb_x^†))

Similarly:

hatb_x - hatb_x^† = sqrt((momega)/(2ℏ))(hatx + i/(momega)hatp_x) - sqrt((momega)/(2ℏ))(hatx - i/(momega)hatp_x)

= sqrt((momega)/(2ℏ))[(cancel(hatx) + i/(momega)hatp_x) - (cancel(hatx) - i/(momega)hatp_x)]

= sqrt((momega)/(2ℏ))((2i)/(momega)hatp_x)

= sqrt((2)/(momegaℏ))(ihatp_x)

Therefore:

color(blue)(hatp_x) = i/i*1/isqrt((momegaℏ)/(2))(hatb_x - hatb_x^†)

= color(blue)(-isqrt((momegaℏ)/(2))(hatb_x - hatb_x^†))

These operators allow us to represent the Hamiltonian operator for the Harmonic Oscillator in one dimension as:

$\boldsymbol{{\hat{H}}_{x}} = \frac{{\hat{p}}_{x}^{2}}{2 m} + \frac{1}{2} m {\omega}^{2} {\hat{x}}^{2}$

$= \left[\ldots\right]$

= bb(ℏomega(hatb_x^†hatb_x + 1/2))

which is a convenient way to obtain each energy state as:

bb(E_(upsilon_x) = ℏomega(upsilon_x + 1/2))

where ${\upsilon}_{x}$ is the vibrational quantum number for the $x$ dimension, and goes as $0 , 1 , 2 , . . .$.