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ψ''+ (2ε-y[itex]^{2}[/itex])ψ=0

I am letting y[itex]\rightarrow[/itex] [itex]\infty[/itex] to get...

ψ''- y[itex]^{2}[/itex]ψ=0

It says the solution to this equation in the same limit is...

ψ= Ay[itex]^{m}[/itex]e[itex]^{\pm y^{2}/2}[/itex]

The positive possibility in the exponential is ignored since it is not in the physical Hilbert space. My question is how did they solve this differential equation? I have read a couple websites and it says that you just have to "guess" it... however, is there a logical way to why you would guess this? Thank you