Question

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 ?

Answer 1

Consider the harmonic oscillator Hamiltonian...

`hatH = hatp^2/(2mu) + 1/2muomega^2hatx^2`

`= 1/(2mu) (hatp^2 + mu^2omega^2 hatx^2)`

Now, define the substitution:

`hatx"'" = hatxsqrt(muomega)` `" "" "" "` `hatp"'" = hatp/sqrt(muomega)`

This gives:

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

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

Next, consider the substitution where:

`hatx"''" = (hatx"'")/sqrt(ℏ)` `" "" "" "` `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 `hatp"''"^2` and `hatx"''"^2` can be factored into a product of complex conjugates, define the ladder operators

`hata = (hatx"''" + ihatp"''")/sqrt2` `" "" "" "` `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 `-[hatx"''", hatp"''"] = [hatp"''", hatx"''"] = -i`, the rightmost term is `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

`hatHphi_n = Ephi_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`

`= (1 + hatN)phi_n`

`= color(blue)((1 + n)phi_n)`