A particle of mass m is moving in one dimension in a potential V(x,t). The wave
function for the particle is: ψ = Axe^([-sqrt(km)/2h_bar]*x^2)e^([-isqrt(k/m)]*3t/2). For -infinitity < x < infinity, where k and A are constants. Normalize this wave function.
Normalization of wave function in one dimension:
∫ψ*ψdx = 1
The Attempt at a Solution
So I said that ψ* would be the same as ψ except i would be negative, so ψ*ψ would be A^2*x^2*e^([-sqrt(km)/h_bar]*x^2). To normalize, ∫ψ*ψdx = A^2*∫x^2*e^([-sqrt(km)/h_bar]*x^2)dx = 1, with the integration being from negative infinity. I am mostly just having trouble getting through this integral. I tried integration by parts:
u = x^2, du = 2xdx
v = [-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2), dv = e^-([sqrt(km/h_bar]*x^2) dx.
So ∫ψ*ψdx = A^2 [uv-∫vdu] = A^2 [[x^2*[-h_bar/sqrt(km)] * [1/2x] * e^([-sqrt(km)/h_bar]*x^2)] (evaluated from negative infinity to positive infinity) - ∫[-h_bar/sqrt(km)] * e^([-sqrt(km)/h_bar]*x^2) dx, again from negative infinity to positive infinity.
So for the first term, I don't know how to evaluate that from negative infinity to infinity because there is a 2x in the denominator and an e^(x^2) in the numerator, and the only way I really know how to evaluate is by just plugging values in. I also don't see an obvious way to evaluate that integral. So is there a way to evaluate this or is integration by parts just not going to work for me? Once I find this integral I can just take the square root of its reciprocal to get the value of A.
This is my first post so I have read the rules and am hoping I have posted in the right forum, etc. Any help on this problem would be very much appreciated.