Normalization of Slit #2 Wave Amplitude

[Tyst]

Homework Statement

In a double-slit experiment, the slits are on the y-axis and the electrons are detected
on a vertical screen. When only slit #1 is open, the amplitude of the wave which
gets through is
$$\psi(y,t) = A \exp^{-y^2} \exp^{-i((ky-\omega t)}$$

when only slit #2 is open, the amplitude of the wave which gets through is
$$\psi(y,t) = A \exp^{-y^2} \exp^{-i(k+\pi)y-\omega t)}$$

(a) Normalize 1 and 2.

Homework Equations

Normalization Condition
$$\int_{-\infty}^{\infty}\psi*\psi dy = 1$$

The Attempt at a Solution

1.
$$1=\int_{-\infty}^{\infty} A (\exp^{-y^2} \exp^{-i((ky-\omega t)}) (\exp^{-y^2} \exp^{i((ky-\omega t)}) dy$$
$$1=A \int_{-\infty}^{\infty} \exp^{-2y^2} dy$$
$$1/A= \sqrt{\pi/2}$$ From integral tables

That's the solution i came up with for the first normalisation
Is the second the same?

 

[Marco_84]

Homework Statement

In a double-slit experiment, the slits are on the y-axis and the electrons are detected
on a vertical screen. When only slit #1 is open, the amplitude of the wave which
gets through is
$$\psi(y,t) = A \exp^{-y^2} \exp^{-i((ky-\omega t)}$$

when only slit #2 is open, the amplitude of the wave which gets through is
$$\psi(y,t) = A \exp^{-y^2} \exp^{-i(k+\pi)y-\omega t)}$$

(a) Normalize 1 and 2.

Homework Equations

Normalization Condition
$$\int_{-\infty}^{\infty}\psi*\psi dy = 1$$

The Attempt at a Solution

1.
$$1=\int_{-\infty}^{\infty} A (\exp^{-y^2} \exp^{-i((ky-\omega t)}) (\exp^{-y^2} \exp^{i((ky-\omega t)}) dy$$
$$1=A \int_{-\infty}^{\infty} \exp^{-2y^2} dy$$
$$1/A= \sqrt{\pi/2}$$ From integral tables

That's the solution i came up with for the first normalisation
Is the second the same?

i think u forgot to put the square over the A

cioa

 

[Tyst]

Ha! So i did, thanks.