What is the solution to Zwiebach eqn (17.44) on page 395?

[Jimmy Snyder]

[SOLVED] Zwiebach eqn (17.44) page 395

Homework Statement

Equation (17.44) is:
$$e^{-i\ell x_0/R}pe^{i\ell x_0/R} = p + \frac{\ell}{R}$$

Homework Equations

Equation (17.43)
$$\ell \in \mathbb{Z}$$

The Attempt at a Solution

Equation (17.44) is easy to derive if $\frac{\ell}{R}$ is small. but since $\ell$ can be any integer, there doesn't seem to be any way to insure that. Did the author simply forget to mention some conditions necessary for it to be true?

 

[nrqed]

Homework Statement

Equation (17.44) is:
$$e^{-i\ell x_0/R}pe^{i\ell x_0/R} = p + \frac{\ell}{R}$$

Homework Equations

Equation (17.43)
$$\ell \in \mathbb{Z}$$

The Attempt at a Solution

Equation (17.44) is easy to derive if $\frac{\ell}{R}$ is small. but since $\ell$ can be any integer, there doesn't seem to be any way to insure that. Did the author simply forget to mention some conditions necessary for it to be true?

It's true to all orders. Just use $[ x_0^n, p] = i n x_0^{n-1}$.

 

[Jimmy Snyder]

It's true to all orders. Just use $[ x_0^n, p] = i n x_0^{n-1}$.

Thanks for taking a look at this nrqed. If $\frac{\ell}{R}$ were small, then I would expand the exponentials to linear terms. But it isn't small, so it would seem that I need to keep all terms. But if I consider up to quadratic terms I get:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(x_0^2p + x_0px_0 - px_0^2) + \mathcal{O}({\frac{\ell^3}{R^3}})[/itex]<br /> Using your suggestion this becomes:<br /> [tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(2ix_0 + x_0px_0) + \mathcal{O}(\frac{\ell^3}{R^3})[/itex]<br /> This doesn't seem to improve things. Is there a simpler approach that I am missing?[/tex][/tex]

 

[nrqed]

Thanks for taking a look at this nrqed. If $\frac{\ell}{R}$ were small, then I would expand the exponentials to linear terms. But it isn't small, so it would seem that I need to keep all terms. But if I consider up to quadratic terms I get:
[tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(x_0^2p + x_0Px_0 - px_0^2) + \mathcal{O}({\frac{\ell^3}{R^3}})[/itex]<br /> Using your suggestion this becomes:<br /> [tex]p + \frac{\ell}{R} + \frac{\ell^2}{2!R^2}(2ix_0 + x_0Px_0) + \mathcal{O}(\frac{\ell^3}{R^3})[/itex]<br /> This doesn't seem to improve things. Is there a simpler approach that I am missing?[/tex][/tex]

[tex]$$<br /> Hi Jimmy,<br /> <br /> I am in a hurry because I am teaching a class in 15 minutes and have stuff to do. I will get back to you early this afternoon.$$[/tex]

 

[Jimmy Snyder]

I will get back to you early this afternoon.

Take your time. When you get back to it, you will find that I erroniously put upper case P's in my equations. They should be lower case. I fixed it in my post.

Edit - With your help, I have figured this out. Thanks nrqed.

 

[nrqed]

Take your time. When you get back to it, you will find that I erroniously put upper case P's in my equations. They should be lower case. I fixed it in my post.

Edit - With your help, I have figured this out. Thanks nrqed.

I just got back from my classes. I am glad it worked out! As you see, it works to all orders.

Glad I could help!


Patrick

 

[Jimmy Snyder]

Just for the record, here is the solution:
$$e^{-ilx_0/R}pe^{ilx_0/R}$$

$$= (\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^np)e^{ilx_0/R}$$

Now using nrqed's suggestion:

$$= (\Sigma\frac{1}{n!}(\frac{-il}{R})^npx_0^n + \Sigma_0\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}$$

$$= (p\Sigma\frac{1}{n!}(\frac{-il}{R})^nx_0^n + \Sigma_1\frac{1}{n!}(\frac{-il}{R})^ninx_0^{n-1})e^{ilx_0/R}$$

$$= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_1\frac{1}{(n-1)!}(\frac{-il}{R})^{n-1}x_0^{n-1})e^{ilx_0/R}$$

$$= (pe^{-ilx_0/R} + \frac{\ell}{R}\Sigma_0\frac{1}{n!}(\frac{-il}{R})^n}x_0^n)e^{ilx_0/R}$$

$$= (pe^{-ilx_0/R} + \frac{\ell}{R}e^{-ilx_0/R})e^{ilx_0/R}$$

$$= p + \frac{\ell}{R}$$