Exploring the 1D Wave Function: (x,t) = Ae–a|x| e–it

[Cybercole]

Ψ(x,t)=A⋅exp(A|x|)⋅exp(−iωt)


Consider the one-dimensional, time-dependent wave function for infinite motion: (x,t) = Ae–a|x| e–it where A, a, and  are positive real constants. What are: (a) normalization constant A, (b) the quantum-mechanical expectation value of coordinate x, (c) the quantum-mechanical expectation value of x2, and (d) the quantum-mechanical expectation value of the square of momentum ^p2

 

[CompuChip]

They look like they are straightforward to integrate... what is your problem exactly?
Do you not understand what a normalization constant is, or how to calculate it, or did you get stuck in the integration, or ... ?

 

[Cybercole]

I know how to normalize a funtion but i am getting stuck in the middle of it... we have never normalize somthing like this before all we have ever done was matrices, i am not very strong in this type of math

 

[Redbelly98]

I know how to normalize a funtion but i am getting stuck in the middle of it...

Please show your work, up to the part where you are stuck.

 

[paranormal]

(a) The normalization constant A can be found by integrating the square of the wave function over all space and setting it equal to 1, as the wave function must be normalized. This results in A = (√a/π)√(1/(a^2+ω^2)).

(b) The quantum-mechanical expectation value of coordinate x is given by <x> = ∫x|Ψ(x,t)|^2 dx. Plugging in the given wave function, we get <x> = 0, as the wave function is symmetric about the origin and the integral of an odd function over a symmetric interval is always zero.

(c) The quantum-mechanical expectation value of x^2 is given by <x^2> = ∫x^2|Ψ(x,t)|^2 dx. Plugging in the given wave function, we get <x^2> = (1/2a^2)+(1/2ω^2), as the integral of an even function over a symmetric interval is equal to twice the integral over half the interval.

(d) The quantum-mechanical expectation value of the square of momentum ^p2 is given by <^p^2> = ∫Ψ*(x,t)(-ħ^2d^2/dx^2)Ψ(x,t) dx. Plugging in the given wave function and simplifying, we get <^p^2> = ħ^2(a^2+ω^2)/2. This shows that the expectation value of ^p^2 is a constant, independent of time, which is a characteristic of a stationary state.