- #1

WWCY

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## 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.

## Homework Equations

## 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})

$$