Normalization Constant for Gaussian

[atomicpedals]

Homework Statement

Find the normalization constant N for the Gaussian wave packet
$$\psi (x) = N e^{-(x-x_{0})^{2}/2 K^{2}}$$

Homework Equations

$$1 = \int |\psi (x)|^{2} dx$$

The Attempt at a Solution

$$1 = \int |\psi (x)|^{2} dx = N^{2} \int e^{-(x-x_{0})^{2}/K^{2}} dx$$
Substitute $y=(x-x_{0})$
$$N^{2} \int e^{-y^{2}/K^{2}} dy$$
Substitute again $z = y/|K|$
$$N^{2} \int e^{-z^{2}} dz = N^{2} x_{0} K \sqrt{\pi}$$
$$N= ( \frac{1}{K x_{0} \sqrt{\pi}})^{1/2}$$
Where my question lies is with the $x_{0}$ in N. Should that be there?

 

[atomicpedals]

Actually... that first substitution may be flawed.

 

[atomicpedals]

Ok, I think I see where I went wrong. The $x_{0}$ doesn't belong in the final answer.

 

[BruceW]

yes, you've figured it out.

P.S. on this line, on the left hand side, there should be K since you have changed dy for dz:
$$N^{2} \int e^{-z^{2}} dz = N^{2} x_{0} K \sqrt{\pi}$$
But after that you've remembered the K, so I guess you just forgot to type the K here, but you understand the right answer.

 

[atomicpedals]

Yeah, I forgot about the K, so what I should end up with is:
$$N^{2} K \int e^{-z^{2}} dz = N^{2} K \sqrt{\pi}$$

 

[BruceW]

yep, looks right to me :)