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 ?

1 Answer
Mar 21, 2018

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)#