# Let phi_n be the properly-normalized nth energy eigenfunction of the harmonic oscillator, and let psi = hatahata^(†)phi_n. What is psi equal to ?

Mar 21, 2018

Consider the harmonic oscillator Hamiltonian...

$\hat{H} = {\hat{p}}^{2} / \left(2 \mu\right) + \frac{1}{2} \mu {\omega}^{2} {\hat{x}}^{2}$

$= \frac{1}{2 \mu} \left({\hat{p}}^{2} + {\mu}^{2} {\omega}^{2} {\hat{x}}^{2}\right)$

Now, define the substitution:

$\hat{x} \text{'} = \hat{x} \sqrt{\mu \omega}$$\text{ "" "" }$$\hat{p} \text{'} = \frac{\hat{p}}{\sqrt{\mu \omega}}$

This gives:

hatH = 1/(2mu) (hatp"'"^2 cdot muomega + mu^2omega^2 (hatx"'"^2)/(muomega))

$= \frac{\omega}{2} \left(\hat{p} {\text{'"^2 + hatx"'}}^{2}\right)$

Next, consider the substitution where:

hatx"''" = (hatx"'")/sqrt(ℏ)$\text{ "" "" }$hatp"''" = (hatp"'")/sqrt(ℏ)

so that [hatx"''", hatp"''"] = hatx"''"hatp"''" - hatp"''"hatx"''" = i. This gives:

hatH = omega/2(hatp"''"^2cdotℏ + hatx"''"^2cdotℏ)

= 1/2ℏomega(hatp"''"^2 + hatx"''"^2)

Since $\hat{p} {\text{''}}^{2}$ and $\hat{x} {\text{''}}^{2}$ can be factored into a product of complex conjugates, define the ladder operators

$\hat{a} = \frac{\hat{x} \text{''" + ihatp"''}}{\sqrt{2}}$$\text{ "" "" }$hata^(†) = (hatx"''" - ihatp"''")/sqrt2

so that:

hatahata^(†) = (hatx"''"^2 - ihatx"''"hatp"''" + ihatp"''"hatx"''" + hatp"''"^2)/2

= (hatx"''"^2 + hatp"''"^2)/2 + (i[hatp"''", hatx"''"])/2

Since $- \left[\hat{x} \text{''", hatp"''"] = [hatp"''", hatx"''}\right] = - i$, the rightmost term is $\frac{1}{2}$. By inspection,

hatH = ℏomega(hatahata^(†) - 1/2)

It can be shown that [hata, hata^(†)] = 1, so

hatahata^(†) - hata^(†)hata = 1

=> hatahata^(†) = 1 + hata^(†)hata

and so:

color(green)(hatH = ℏomega(hata^(†)hata + 1/2))

Here we recognize the form of the energy to be:

E_n = ℏomega(n + 1/2)

since it is clear from this form that with

$\hat{H} {\phi}_{n} = E {\phi}_{n}$,

we just have that

ℏomega(hata^(†)hata + 1/2)phi_n = ℏomega(n + 1/2)phi_n

Thus, the number operator can be defined as:

hatN = hata^(†)hata

whose eigenvalue is the quantum number $n$ for that eigenstate.

Hence,

color(blue)(psi_n = hatahata^(†)phi_n)

= (1 + hata^(†)hata)phi_n

$= \left(1 + \hat{N}\right) {\phi}_{n}$

$= \textcolor{b l u e}{\left(1 + n\right) {\phi}_{n}}$