Homework Help: Ryder, QFT, page 57

1. Aug 9, 2007

Jimmy Snyder

1. The problem statement, all variables and given/known data
There is an unnumbered equation in the top half of the page:
$$(1 - i\mathbf{K\cdot\phi})(1 - iP\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + iP\cdot a) = 1 + [P_{\mu},P_{\nu}]a^{\mu}a^{\nu} + 2[P_{\mu}, K_i]a^{\mu}\phi_i + [K_i,K_j]\phi_i\phi_j$$

2. Relevant equations

3. The attempt at a solution
This makes no sense to me. For instance, the [P,P] and [K,K] terms work out to zero since each product that appears with a + sign also appears with a minus sign. For instance, suppressing the phi's:
$$[K_i,K_j] = K_1\cdot K_1 - K_1\cdot K_1 + K_1\cdot K_2 - K_2\cdot K_1 + K_1\cdot K_3 - K_3\cdot K_1$$
$$+ K_2\cdot K_1 - K_1\cdot K_2 + K_2\cdot K_2 - K_2\cdot K_2 + K_2\cdot K_3 - K_3\cdot K_2$$
$$+ K_3\cdot K_1 - K_1\cdot K_3 + K_3\cdot K_2 - K_2\cdot K_3 + K_3\cdot K_3 - K_3\cdot K_3 = 0$$
What am I missing?

Last edited: Aug 10, 2007
2. Aug 9, 2007

nrqed

yes, it's a confusing presentation!
As far as I can tell, you are right: the [K,K] and [P,P] terms add up to zero, leaving only the [K,P] terms. So boosting and translating do not commute becauseo fthat piece.

I have an older edition than yours so he may have change his discussion after this equation but his discussion is vague a bit. He should have rewritten the equation with only the [K,P] terms to make things more clear.

But I would say that I agree with you.

3. Aug 9, 2007

Jimmy Snyder

It's worse than that. I get:
$$(1 - i\mathbf{K\cdot\phi})(1 - iP\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + iP\cdot a) = 1 + P_{\mu}P_{\nu}a^{\mu}a^{\nu} + [P_{\mu}, K_i]a^{\mu}\phi_i + K_iK_j\phi_i\phi_j$$
Note that the PP and KK products are not commutators. If this is correct, then the entire paragraph becomes incomprehensible. As an aside, why doesn't the author mention the other combinations JP(-J)(-P) and KJ(-K)(-J)? Is he assuming that I would know he meant those too, or does he not mean them.

His main point is that the group structure is determined by the commutation relations. Is there a clearer explanation of this fact in another book? I looked in Kaku, Zee, and Srednicki, but did not find what I was looking for.

Last edited: Aug 10, 2007
4. Aug 9, 2007

meopemuk

It seems that you are trying to calculate

$$\sum_{ij}[K_i, K_j]$$

rather than

$$[K_i, K_j]$$

Eugene.

5. Aug 9, 2007

Jimmy Snyder

That's because I suppressed the phi's. The actual thing I am trying to calculate is:

$$[K_i, K_j]\phi_i\phi_j = \sum_{ij}[K_i, K_j]\phi_i\phi_j$$

6. Aug 9, 2007

meopemuk

Then you cannot "suppress the phi's". You can do that only if all $\phi_i$ are the same, i.e, independent on $i$. Then you can write

$$\phi_i = \phi$$

$$\sum_{ij}[K_i, K_j]\phi_i\phi_j = \Bigl( \sum_{ij}[K_i, K_j] \Bigr) \phi^2$$

$$\sum_{ij}[K_i, K_j]$$

In the general case when $\phi_i$ depend on $i$ this trick is not allowed.

Eugene.

7. Aug 9, 2007

Jimmy Snyder

Here is the equation without suppressing the phi's.
$$[K_i,K_j]\phi_i\phi_j = K_1\cdot K_1\phi_1\phi_1 - K_1\cdot K_1\phi_1\phi_1 + K_1\cdot K_2\phi_1\phi_2 - K_2\cdot K_1\phi_1\phi_2 + K_1\cdot K_3\phi_1\phi_3 - K_3\cdot K_1\phi_1\phi_3$$
$$+ K_2\cdot K_1\phi_1\phi_2 - K_1\cdot K_2\phi_1\phi_2 + K_2\cdot K_2\phi_2\phi_2 - K_2\cdot K_2\phi_2\phi_2 + K_2\cdot K_3\phi_2\phi_3 - K_3\cdot K_2\phi_2\phi_3$$
$$+ K_3\cdot K_1\phi_1\phi_3 - K_1\cdot K_3\phi_1\phi_3 + K_3\cdot K_2\phi_2\phi_3 - K_2\cdot K_3\phi_2\phi_3 + K_3\cdot K_3\phi_3\phi_3 - K_3\cdot K_3\phi_3\phi_3 = 0$$

8. Aug 9, 2007

meopemuk

You are right, I was wrong. Sorry for being so slow. Now I see that. In your original equation only [K,P] terms survive, which are proportional to P. (I believe that K are generators of boosts and P are generators of space-time translations).

Eugene.

9. Aug 10, 2007

olgranpappy

From looking at the LHS of the equation you gave, it appears that your RHS is wrong: the (P.a)(P.a) term should have a minus sign and the [P,K] term is missing a factor of 'i'. Expanding the LHS to second order in the operators, I find:

$$1+K^iK^j\phi_i\phi_j-P^\mu P^\nu a_\mu a_\nu + i[K^i,P^\mu]\phi_i a_\mu\;.$$

(is the LHS really supposed to have (1-P.a)? Is there a missing 'i'?... If that P.a term on the LHS you gave is missing an 'i' then let a-->ia and my result equals your RHS.)

I think that's correct. I don't have Ryder at my house, but I'll take a look at it in the morning (probably more like the afternoon, actually) and give my 2 cents then. Cheers.

Last edited: Aug 10, 2007
10. Aug 10, 2007

Jimmy Snyder

$$(1 - i\mathbf{K\cdot\phi})(1 - iP\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + iP\cdot a) = 1 + [P_{\mu},P_{\nu}]a^{\mu}a^{\nu} + 2[P_{\mu}, K_i]a^{\mu}\phi_i + [K_i,K_j]\phi_i\phi_j$$
and my post #3 as follows:
$$(1 - i\mathbf{K\cdot\phi})(1 - iP\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + iP\cdot a) = 1 + P_{\mu}P_{\nu}a^{\mu}a^{\nu} + [P_{\mu}, K_i]a^{\mu}\phi_i + K_iK_j\phi_i\phi_j$$

11. Aug 10, 2007

olgranpappy

Yeah, it looks like that equation in Ryder is just wrong... I searched the web for an errata to the text, but I couldn't find one.

I sent an email to Ryder... we'll see what he says.

12. Aug 10, 2007

Jimmy Snyder

Actually, I had emailed him a few weeks ago just to ask if there was an errata page on the web. He hasn't replied. I think he is retired. But if the equation is wrong, then what of the paragraph. It says:
That rings true even if the equation is wrong. In fact, I think it's true of all groups. But what is the right way to show it?

13. Aug 10, 2007

olgranpappy

Well, I'm not sure exactly. But, for example, you can think about SU(2) as an example. Most quantum mechanics books (Messiah, e.g.) show how one can start from just the commutation relations between the generators and end up with explicit expressions for rotation matrices (representations of elements of the group). I.e., from the generators and their commutation relations we can figure out the elements of the group (which I think might only be true of groups whose elements are continuously connected with the identity element... I forget... I'd check a book on group theory if I were you).

Also, to motivate the fact that knowledge of the commutation relations is enough to figure out the structure of the group, consider the Baker-Campbell-Hausdorff Theorem:
$$e^{A}e^{B}=e^{A+B+\frac{1}{2}[A,B]+\ldots}\;.$$
where the \ldots involve commutators of commutators, etc.\

If A and B are generators, then e^A and e^B are elements of the group and thus also e^Ae^B is an element of the group (since it's a group). But what element, exactly? If we know the commutation relations then we can, in principle figure it out.

For example, if I consider SU(2) I can write
$$e^{i\vec G\cdot \vec \alpha}e^{i\vec G\cdot \vec \beta} =e^{i\vec G\cdot(\vec\alpha+\vec\beta-1/2\vec\alpha\times\vec\beta+\ldots)}\;.$$

14. Aug 11, 2007

meopemuk

Check out books on Lie algebras and Lie groups.

Eugene.

15. Aug 11, 2007

Jimmy Snyder

Thanks everyone. I will look into these matters.

16. Aug 11, 2007

dextercioby

What Ryder wrote is wrong. To second order in the operators the second member of this equality is different from the third one. The thing that could still be true is the equality between the first and the third. That's wrong too, all 3 terms involving commutators should be divided by 2. For the calculation you have to use 1 blank page, 1 pen and the Baker - Campbell - Hausdorff identity several times and always stopping at the second order in the operators.

Last edited: Aug 12, 2007
17. Aug 11, 2007

dextercioby

And the whole enchilada is seen to be simply

$$1+\left[P_{\mu},K_{i}\right]a^{\mu}\phi_{i}$$, because the other 2 terms are trivially 0 because of the fact that antisymmetric 2-nd rank (Lorentz and cartesian) tensors are contracted with symmetric ones.

Last edited: Aug 12, 2007
18. Aug 11, 2007

Jimmy Snyder

It seems almost trivial to me that the left hand equality is correct. If it is, and if the extreme ends of the equation are equal to each other, then all three must be equal and we are back where we started.

19. Aug 12, 2007

dextercioby

It's not true. You have to evaluate the whole exponent of "e" and then expand to 2-nd order in the operators. Since you're dealing with possible noncommuting operators the product of 4 exponentials, when expanded to 2-nd order will definitely differ from the product of linearizations. And one more thing. Why would you resort to linearization of the exponentials, when you need the 2-nd order in the operators...?

20. Aug 12, 2007

olgranpappy

yes, the whole wrong equation reduces to a simpler wrong equation. That is true.