Ryder QFT Page 57 Homework: Solving Equation

[Jimmy Snyder]

Homework Statement

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

Homework Equations

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?

 

[nrqed]

Homework Statement

There is an unnumbered equation in the top half of the page:
(1 - i\mathbf{K\cdot\phi})(1 - P\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + P\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

Homework Equations

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] = K1\cdot K1 - K1\cdot K1 + K1\cdot K2 - K2\cdot K1 + K1\cdot K3 - K3\cdot K1
+ K2\cdot K1 - K1\cdot K2 + K2\cdot K2 - K2\cdot K2 + K2\cdot K3 - K3\cdot K2
+ K3\cdot K1 - K1\cdot K3 + K3\cdot K2 - K2\cdot K3 + K3\cdot K3 - K3\cdot K3 = 0
What am I missing?

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.

 

[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.

 

[meopemuk]

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?

It seems that you are trying to calculate

\sum_{ij}[K_i, K_j]

rather than

[K_i, K_j]

Eugene.

 

[Jimmy Snyder]

It seems that you are trying to calculate

\sum_{ij}[K_i, K_j]

rather than

[K_i, K_j]

Eugene.

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

 

[meopemuk]

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

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

and use your result for

\sum_{ij}[K_i, K_j]

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

Eugene.

 

[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

 

[meopemuk]

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

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.

 

[olgranpappy]

It's worse than that. I get:
(1 - i\mathbf{K\cdot\phi})(1 - P\cdot a)(1 + i\mathbf{K\cdot\phi})(1 + P\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

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:

<br /> 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\;.<br />

(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.

 

[Jimmy Snyder]

(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.)

Oops, my bad, and your nice catch. I edited my original post to read 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} + 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

 

[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.

 

[Jimmy Snyder]

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.

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:

Hence the structure of the group is known when the commutation relations between the generators are known.

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?

 

[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:
<br /> e^{A}e^{B}=e^{A+B+\frac{1}{2}[A,B]+\ldots}\;.<br />
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
<br /> e^{i\vec G\cdot \vec \alpha}e^{i\vec G\cdot \vec \beta}<br /> =e^{i\vec G\cdot(\vec\alpha+\vec\beta-1/2\vec\alpha\times\vec\beta+\ldots)}\;.<br />

 

[meopemuk]

Hence the structure of the group is known when the commutation relations between the generators are known

Check out books on Lie algebras and Lie groups.

Eugene.

 

[Jimmy Snyder]

Thanks everyone. I will look into these matters.

 

[dextercioby]

e^{-i\mathbf{K}\cdot\mathbf{\phi}} e^{-i\mathbf{P}\cdot\mathbf{a}} e^{i\mathbf{K}\cdot\mathbf{\phi}} e^{i\mathbf{P}\cdot\mathbf{a}} =\left(1-i\mathbf{K}\cdot\mathbf{\phi}\right)<br /> \left(1-i\mathbf{P}\cdot\mathbf{a}\right)\left(1+i\mathbf{K}\cdot\mathbf{\phi}\right)<br /> \left(1+i\mathbf{P}\cdot\mathbf{a}\right)=1+\left[P_{\mu},P_{\nu}\right]a^{\mu}a^{\nu}<br /> +2\left[P_{\mu},K_{i}\right]a^{\mu}\phi_{i}+\left[K_{i},K_{j}}\right]\phi_{i}\phi_{j}

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.

 

[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.

 

[Jimmy Snyder]

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.

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.

 

[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...?

 

[olgranpappy]

And the whole enchilada is seen to be simply

1+\left[P_{\mu},K_{i}\right]a^{\mu}\phi_{i}...

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

 

[olgranpappy]

It's not true...

no. It's true all right.

 

[Jimmy Snyder]

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...?

Of course. Now I get:
(1 - i\mathbf{K\cdot\phi} -\frac{1}{2}K_iK_j\phi_i\phi_j)(1 - i\mathbf{P\cdot a}-\frac{1}{2}{P_{\mu}P_{\nu}a^{\mu}a^{\nu})(1 + i\mathbf{K\cdot\phi}-\frac{1}{2}K_iK_j\phi_i\phi_j)(1 + i\mathbf{P\cdot a}-\frac{1}{2}P_{\mu}P_{\nu}a^{\mu}a^{\nu}) = 1 + [P_{\mu}, K_i]a^{\mu}\phi_i
Since we all agree that the [K,K] and [P,P] sums are zero, the only point of contention is the factor of 2. At least the paragraph makes sense now. Don't I need to verify similar equations for the other products in the group though? PJ(-P)(-J) and KJ(-K)(-J).

Now I see that none of the properties of K and P were used in the derivation, so the equation for PJ and for KJ will have the same form. Indeed, this probably proves the fact for all continuous groups.

 

[dextercioby]

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

What do you mean by <simpler wrong equation> ?

 

[olgranpappy]

What do you mean by <simpler wrong equation> ?

I mean that the correct expression for the far RHS of the equation we are discussing is
<br /> 1-K_iK_j\phi^i\phi^j-P_\mu P_\nu a^\mu a^\nu+[P_\mu , K_i]a^\mu \phi^i<br />[EDIT: Actually, sorry, I'm wrong. See below]

 

[olgranpappy]

oh, never mind. I just realized what I was doing wrong... what I wrote above is wrong. I forgot to put the 1/2 in the second term of the expansion of the exponential... duh...

Here is the work (keeping terms up to second order only at each step and writing k.f for K_i\phi^i and writing p.a for P_\mu a^\mu to simplify the writing):

(1 - ik.f - k.fk.f/2)( 1 - ip.a - p.ap.a/2)(1 + ik.f - k.fk.f/2)(1 + ip.a - p.ap.a/2)
=
(1-ip.a -p.ap.a/2 -ik.f -k.fp.a -k.fk.f/2)(1 +ip.a-p.ap.a/2+ik.f-k.fp.a -k.fk.f/2)
=
(1 + ip.a - p.ap.a/2 + ik.f - k.fp.a - k.fk.f/2
- ip.a + p.ap.a +p.ak.f
-p.ap.a/2
-ik.f + k.fp.a +k.fk.f
-k.fp.a
-k.fk.f/2 )

Then canceling a bunch of terms gives:

(1
+p.ak.f
-k.fp.a)I.e.
<br /> 1+[P_\mu , K_i]a^\mu \phi^i<br />

 

[Jimmy Snyder]

Thanks everyone, for your help on this.

 

[olgranpappy]

P.S. I got an email back from Ryder. He seemed pleased that we had alerted him to an error in his book, and he confirmed that, indeed, there is an extra factor of two and what we eventually found in this thread is correct.

Also, he admitted that it is quite misleading to have shown the exponentials expanded only to first order after the first equal sign since they should all be expanded to second order (as dex pointed out) and then combinded (keeping all terms to second order) if one wants to arrive at he correct answer (involving only the P-K commutator, and without the incorrect factor of 2).

Cheers.

 

[Jimmy Snyder]

P.S. I got an email back from Ryder. He seemed pleased that we had alerted him to an error in his book.

Thanks for the extra effort olgranpappy. I have often sent e-mails to authors concerning errata and they are always pleased. I had sent one to Professor Ryder concerning a different matter and he never responded. Did you ask if there was an on-line errata page? Can you PM me with his e-mail address? I don't think there's a problem sharing it retail, but it might not be a good idea to share it wholesale as it might bring him even more junk mail than he gets now.

I have made great progress in the past few days. The math in Chapter 3 seems more straightforward than Chapter 2. I don't fully understand all of the concepts though. I sure could use a tutor.

 

[olgranpappy]

Thanks for the extra effort olgranpappy. I have often sent e-mails to authors concerning errata and they are always pleased. I had sent one to Professor Ryder concerning a different matter and he never responded. Did you ask if there was an on-line errata page?...

He said there was now list of errata online, unfortunately.