The classical energy (Hamiltonian) is
|
E = {1\over
2}m{\dot{x}}^{2} + {1\over
2}m{ω}^{2}{x}^{2} = {1\over
2m}{p}^{2} + {1\over
2}m{ω}^{2}{x}^{2}
| (7.4) |
The quantum Hamiltonian operator is thus
|
\hat{H} = {1\over
2m}\hat{{p}}^{2} + {1\over
2}m{ω}^{2}{x}^{2} = −{{ℏ}^{2}\over
2m} {{d}^{2}\over
d{x}^{2}} + {1\over
2}m{ω}^{2}{x}^{2}.
| (7.5) |
And we thus have to solve Schrödinger’s equation
|
−{{ℏ}^{2}\over
2m} {{d}^{2}\over
d{x}^{2}}ϕ(x) + {1\over
2}m{ω}^{2}{x}^{2}ϕ(x) = Eϕ(x).
| (7.6) |
In order to treat this equation it is better to remove all the physical constants, and go over to dimensionless coordinates
|
y = \sqrt{{mω\over
ℏ}} x,\qquad ϵ = {E\over
ℏω}.
| (7.7) |
Quiz What is the dimension of {mω\over ℏ} ? and of ℏω? (The dimension of ℏ is [ℏ] = \mathrm{mass} ×{\mathrm{length}}^{2}∕\mathrm{time}.)
When we substitute these new variables into the Schrödinger equation we get, using
|
{d\over
dx}f(y) = {dy\over
dx} {d\over
dy}f(y) = \sqrt{{mω\over
ℏ}} {d\over
dy}f(y),
| (7.8) |
that (ϕ(x) = u(y))
|
−{{ℏ}^{2}\over
2m} {mω\over
ℏ} {{d}^{2}\over
d{y}^{2}}u(y) + {1\over
2}m{ω}^{2} {ℏ\over
mω}{y}^{2}u(y) = ϵℏωu(y).
| (7.9) |
Cancelling the common factor ℏω we find
|
{{d}^{2}\over
d{y}^{2}}u(y) + (2ϵ − {y}^{2})u(y) = 0
| (7.10) |