$$f(u)=A_0e^{-u^2/2}, g(u)=2uA_1e^{-u^2/2}$$ Where both A's have a value so that$$\int\limits_{-\infty}^{\infty}f(u)du=\int\limits_{-\infty}^{\infty}g(u)du=1$$ Show that$$A_1=\frac{A_0}{\sqrt{2}}$$ From where I'm reading, it says: "Do not evaluate at A_0, but give a value for the particle'smean square position when in the second (quantum) state: $$=\int\limits_{-\infty}^{\infty}u^2g(u)^2du$$"

Question

$$f(u)=A_0e^{-u^2/2}, g(u)=2uA_1e^{-u^2/2}$$
Where both A's have a value so that$$\int\limits_{-\infty}^{\infty}f(u)du=\int\limits_{-\infty}^{\infty}g(u)du=1$$
Show that$$A_1=\frac{A_0}{\sqrt{2}}$$
From where I'm reading, it says: "Do not evaluate at A_0, but give a value for the particle'smean square position when in the second (quantum) state: $$<u^2>=\int\limits_{-\infty}^{\infty}u^2g(u)^2du$$"

experimentx · Answer

Normalize those functions ... using this property
|dw:1345989124482:dw|

anonymous · Answer

What does 'do not evaluate at A_0'mean?

experimentx · Answer

I don't know ... most probably ... you are asked to find the weighted averate.

experimentx · Answer

|dw:1345989312374:dw|
In this way you can avoid evaluating A_0