Ryder, QFT 2nd Ed. Page 168

[Jimmy Snyder]

Homework Statement

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
$$exp(\frac{i\mathbf{P}^2(t_1 - t_0)}{8m\hbar})$$

Homework Equations

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

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.

 

[dextercioby]

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

 

[Jimmy Snyder]

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

Are you asking about:
$$(\frac{2\alpha}{i})^{3/2}(\frac{i\pi}{\alpha})^{3/2} = (2\pi)^{3/2}$$

or about:
$$\frac{h}{\hbar} = 2\pi$$

 

[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.

 

[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.

 

[Jimmy Snyder]

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

 

[dextercioby]

Yes, i do. Why ?

 

[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.