Normalization of ψ(x,0) - Can you tell me if my solution is correct?

[S_Flaherty]

We are given ψ(x,0) = A[ψ1(x)+ψ2(x)] and for the first part of my homework problem it asks us to normalize ψ(x,0) (it says find A).

What I did was ∫|ψ(x,0)|^2 dx = 1 = (|A|^2)∫(ψ1^2 + 2ψ1ψ2 + ψ2^2)dx and since
∫ψm(x)*ψn(x)dx = 0 when m≠n and it equals 1 when m=n I can make the integral equal to
(A^2)(1 + 2(0) + 1) = 1 so A^2 = 1/2, so A = 1/4.

Did I solve this correctly or did I completely misunderstand what I was doing?

 

[SammyS]

We are given ψ(x,0) = A[ψ1(x)+ψ2(x)] and for the first part of my homework problem it asks us to normalize ψ(x,0) (it says find A).

What I did was ∫|ψ(x,0)|^2 dx = 1 = (|A|^2)∫(ψ1^2 + 2ψ1ψ2 + ψ2^2)dx and since
∫ψm(x)*ψn(x)dx = 0 when m≠n and it equals 1 when m=n I can make the integral equal to
(A^2)(1 + 2(0) + 1) = 1 so A^2 = 1/2, so A = 1/4.

Did I solve this correctly or did I completely misunderstand what I was doing?

No.

It's correct up to \displaystyle {\text{A}\!^2}= \frac{1}{2}\ .

How do you solve for A after that?

 

[S_Flaherty]

No.

It's correct up to \displaystyle {\text{A}\!^2}= \frac{1}{2}\ .

How do you solve for A after that?

I just did the math wrong, I got A = 0.7071.