# Work Check: Wavefunction Normalisation

WWCY

## Homework Statement

Find relation between real normalisation constants ##B_1## and ##B_2## for the following wavefunction,
$$\Psi_k =\sum_{k=1,2} \frac{B_k}{\sqrt{4\sigma ^2 + 2it}} \exp (ip_k (x - \frac{p_k}{2}t) - \frac{(x - p_k t)^2}{4\sigma ^2 + 2it})$$

The working is rather long so thanks in advance for helping, it is greatly appreciated.

## The Attempt at a Solution

Normalisation requires ##\int_{\infty} dx |\Psi (x,0)|^2 = 1##, and
$$\Psi_k (x, 0) =\sum_{k=1,2} \frac{B_k}{2\sigma} \exp (ip_k (x) - \frac{(x)^2}{4\sigma ^2})$$
Knowing that,
$$|\Psi_k|^2 = |B_1 \psi_1|^2 + |B_2 \psi_2|^2 + (B_2 \psi_2)^* (B_1 \psi_1) + (B_2 \psi_2)(B_1 \psi_1)^*$$
$$|B_k \psi_k|^2 = \frac{B_k ^2}{4\sigma^2}\exp (-\frac{x^2}{2\sigma ^2})$$
$$(B_2 \psi_2)(B_1 \psi_1)^* = [\frac{B_1}{2\sigma} \exp (-ip_1 (x) - \frac{(x)^2}{4\sigma ^2})][\frac{B_2}{2\sigma} \exp (ip_2 (x) - \frac{(x)^2}{4\sigma ^2})]$$
$$= \frac{B_1 B_2 }{4\sigma^2} \exp [-i(p_1 - p_2)x - \frac{x^2}{2\sigma^2}]$$
Also,
$$(B_2 \psi_2)^* (B_1 \psi_1) = \frac{B_1 B_2 }{4\sigma^2} \exp [i(p_1 - p_2)x - \frac{x^2}{2\sigma^2}]$$
Before computing the integral, I simplified by letting ##\frac{B_1 B_2}{4\sigma ^2} = \phi##, ##\frac{1}{2\sigma ^2 } = g## and ##p_1 - p_2 = U##
This gave,
$$(B_2 \psi_2)(B_1 \psi_1)^* = \phi \exp (-gx^2 - iUx)$$
$$(B_2 \psi_2)^*(B_1 \psi_1) = \phi \exp (-gx^2 + iUx)$$
The expression for the integral is,
$$\int_{\infty} dx |\Psi (x,0)|^2 = 1 =\int_{\infty} dx [\frac{B_1 ^2 + B_2 ^2}{4\sigma^2}\exp (-gx^2) + \phi \exp (-gx^2 - iUx) + \phi \exp (-gx^2 + iUx)]$$
The integral of the first integrand can be computed with the Gaussian integral result which gives,
$$\frac{B_1 ^2 + B_2 ^2}{4\sigma ^2} \sqrt{\frac{\pi}{g}} = \frac{B_1 ^2 + B_2 ^2}{4\sigma ^2}\sqrt{2\pi \sigma ^2} = \frac{\sqrt{2\pi} (B_1 ^2 + B_2 ^2)}{4\sigma}$$
The integral of the second integrand, which is also equal to that of the third, was done by completing the square,
$$\int_{\infty} dx \phi \exp (-gx^2 - iUx) = \phi e^{\frac{-U^2}{4g}} \int_{\infty} dx \exp (-[\sqrt{g}x + \frac{iU}{2\sqrt{g}}]^2)$$
I used ##dy = \sqrt{g} dx## and rewrote this as,
$$\frac{\phi e^{\frac{-U^2}{4g}}}{\sqrt{g}} \int_{\infty} dx \exp (-y^2) = \frac{\phi e^{\frac{-U^2}{4g}}}{\sqrt{g}} \sqrt{\pi}$$
Substituting for ##g, U, \phi## then gives,
$$\frac{\sqrt{2\pi} B_1 B_2}{4\sigma} \exp (-\frac{(p_1 - p_2)^2 \sigma ^2}{2})$$

Finally, summing up all 3 integrals of all 3 terms gives,

$$1 =\frac{\sqrt{2\pi} (B_1 ^2 + B_2 ^2)}{4\sigma} + \frac{\sqrt{2\pi} B_1 B_2}{2\sigma} \exp (-\frac{(p_1 - p_2)^2 \sigma ^2}{2})$$

Gold Member
Shouldn't this be moved to the Advanced Physics Homework forum?

WWCY
Shouldn't this be moved to the Advanced Physics Homework forum?

I thought that this was pretty basic stuff (in the context of everything else), and decided to post it here. Apologies if I should have posted this in the other thread.

Gold Member
I thought that this was pretty basic stuff (in the context of everything else), and decided to post it here. Apologies if I should have posted this in the other thread.
Its not about you not posting in the correct forum. I think you might just get some better and faster answers in the advanced physics forum if what you're studying is wave functions and normalization.

WWCY
Could anybody take the time to assist? Many thanks!

Homework Helper
Dearly Missed

## Homework Statement

Find relation between real normalisation constants ##B_1## and ##B_2## for the following wavefunction,
$$\Psi_k =\sum_{k=1,2} \frac{B_k}{\sqrt{4\sigma ^2 + 2it}} \exp (ip_k (x - \frac{p_k}{2}t) - \frac{(x - p_k t)^2}{4\sigma ^2 + 2it})$$

The working is rather long so thanks in advance for helping, it is greatly appreciated.

## The Attempt at a Solution

Normalisation requires ##\int_{\infty} dx |\Psi (x,0)|^2 = 1##, and
$$\Psi_k (x, 0) =\sum_{k=1,2} \frac{B_k}{2\sigma} \exp (ip_k (x) - \frac{(x)^2}{4\sigma ^2})$$

***********************************************************************************

$$1 =\frac{\sqrt{2\pi} (B_1 ^2 + B_2 ^2)}{4\sigma} + \frac{\sqrt{2\pi} B_1 B_2}{2\sigma} \exp (-\frac{(p_1 - p_2)^2 \sigma ^2}{2})$$
Letting ##B_1## and ##B_2## be possibly complex, I used Maple and obtained
$$\int_R |\Psi(x,0)|^2 \, dx = \frac{\sqrt{2 \pi}}{4 \sigma} ( |B_1|^2 + |B_2|^2) + \frac{\sqrt{2 \pi}}{4 \sigma} \exp \left( -\frac{1}{2} \sigma^2 (p_1 - p_2)^2 \right) ( B_1 \bar{B_2} + B_2 \bar{B_1})$$
This agrees with your answer if we assume the ##B_i## are real.

WWCY
WWCY
Letting ##B_1## and ##B_2## be possibly complex, I used Maple and obtained
$$\int_R |\Psi(x,0)|^2 \, dx = \frac{\sqrt{2 \pi}}{4 \sigma} ( |B_1|^2 + |B_2|^2) + \frac{\sqrt{2 \pi}}{4 \sigma} \exp \left( -\frac{1}{2} \sigma^2 (p_1 - p_2)^2 \right) ( B_1 \bar{B_2} + B_2 \bar{B_1})$$
This agrees with your answer if we assume the ##B_i## are real.

Thank you!