Before solving the equation we are going to see how the solutions behave at large |y| (and also large |x|, since these variable are proportional!). For |y| very large, whatever the value of ϵ, ϵ ≪ {y}^{2}, and thus we have to solve
|
{{d}^{2}u\over
d{y}^{2}} = {y}^{2}u(y).
| (7.11) |
This has two type of solutions, one proportional to {e}^{{y}^{2}∕2 } and one to {e}^{−{y}^{2}∕2 }. We reject the first one as being not normalisable.
Question: Check that these are the solutions. Why doesn’t it matter that they don’t exactly solve the equations?
Substitute u(y) = H(y){e}^{−{y}^{2}∕2 }. We find
|
{{d}^{2}u\over
d{y}^{2}} = \left (H''(y) − 2yH'(y) + {y}^{2}H(y)\right ){e}^{−{y}^{2}∕2
}
| (7.12) |
so we can obtain a differential equation for H(y) in the form
|
H''(y) − 2yH'(y) + (2ϵ − 1)H(y) = 0.
| (7.13) |
This equation will be solved by a substitution and infinite series (Taylor series!), and showing that it will have to terminates somewhere, i.e., H(y) is a polynomial!