# Homework Help: Ryder, QFT 2nd Ed. Page 168

1. Aug 31, 2007

### Jimmy Snyder

1. The problem statement, all variables and given/known data
I have edited Ryder's text to emphasize the issue I am having. The actual text is approx. 40% down from the top of the page.
$$(\frac{2\alpha}{i})^{3/2}\int exp(\frac{i}{2\hbar}\mathbf{P\cdot x} + i\alpha \mathbf{x}^2)d\mathbf{x}$$
The integral may be evaluated by appealing to equation (5A.3) giving
$$exp(\frac{i\mathbf{P}^2(t_1 - t_0)}{8m\hbar})$$

2. Relevant equations
$$\alpha = \frac{m}{2h(t_1 - t_0)}$$
eqn (5A.3)
$$\int exp(-ax^2 + bx)dx = (\frac{\pi}{a})^{1/2}exp(\frac{b^2}{4a})$$

3. The attempt at a solution
The integral is in the form of eqn (5A.3) where
$$a = -i\alpha; b = \frac{i\mathbf{P}}{2\hbar}$$
and there are 3 dimensions. So
$$(\frac{2\alpha}{i})^{3/2}\int exp(\frac{i}{2\hbar}\mathbf{P\cdot x} + i\alpha \mathbf{x}^2)d\mathbf{x}$$
$$= (\frac{2\alpha}{i})^{3/2}(\frac{i\pi}{\alpha})^{3/2}exp(\frac{-i\mathbf{P}^2h(t_1 - t_0)}{8m\hbar^2})$$
$$= (2\pi)^{3/2}exp(\frac{-i\mathbf{P}^22\pi(t_1 - t_0)}{8m\hbar})$$
so there are a couple of embarassing factors of $2\pi$ hanging about.

2. Sep 1, 2007

### dextercioby

Where did you get that 2pi in the last equation from ?

3. Sep 1, 2007

### Jimmy Snyder

$$(\frac{2\alpha}{i})^{3/2}(\frac{i\pi}{\alpha})^{3/2} = (2\pi)^{3/2}$$

$$\frac{h}{\hbar} = 2\pi$$

Last edited: Sep 1, 2007
4. Sep 1, 2007

### Jimmy Snyder

Never mind, the issue is an error in the book. The book defines:
$$\alpha = \frac{m}{2h(t_1-t_0)}$$
and has the following factor in the equation:
$$(\frac{2\alpha}{i})^{3/2}$$
But if you define
$$\alpha = \frac{m}{2\hbar(t_1-t_0)}$$
then the correct factor is
$$(\frac{\alpha}{i\pi})^{3/2}$$
then everything works out.
I should have seen this earlier since this is the only way to justify the factor
$$e^{i\alpha\mathbf{x}^2}$$

I note another typo on the page, there is a missing minus sign in the exponent in equation (5.40). The minus sign is replaced when this equation is used in (5.42) on the next page. Also, even though it is not wrong, I find the choice of q to represent momentum in equation (5.41) to be weird.

Last edited: Sep 1, 2007
5. Sep 1, 2007

### dextercioby

I see, it was the second case. So you're claiming that Ryder's wrong, in the sense that his formula just b4 5.40 is wrong. Let's see. I'm getting something like

$$...\cdot \mbox{exp}\left(\frac{-8\pi^{2}i\left(t_{1}-t_{0}\right)}{mh}\right)$$

which is totally different than Ryder's or your result.

6. Sep 1, 2007

### Jimmy Snyder

Do you have a copy of the second edition at hand?

7. Sep 1, 2007

### dextercioby

Yes, i do. Why ?

8. Sep 1, 2007

### Jimmy Snyder

Note that in the equation at the bottom of page 167 there is a factor of
$$exp(\frac{im(\mathbf{x_0 - x_1})^2}{2\hbar(t_1 - t_0)})$$

On page 168, about 20% down the page there is an equation in which this has been converted to:

$$e^{i\alpha\mathbf{x}^2}$$
where
$$\alpha = \frac{m}{2h(t_1 - t_0)}$$
and
$$\mathbf{x = x_0 - x_1}$$

This is incorrectly converted. In order to fix the problem, either the exponent must be changed, or alpha must be changed. I have followed the second course and defined:

$$\alpha = \frac{m}{2\hbar(t_1 - t_0)}$$

This fixes the exponent, but then the factor of

$$(\frac{2\alpha}{i})^{3/2}$$
must become
$$(\frac{\alpha}{i\pi})^{3/2}$$

Once these changes are made, I believe everything else falls into place.

Last edited: Sep 1, 2007