Verifying a simple QFT derivation in Peskin's

1. Sep 9, 2008

infiniteen

Hiya,

just stumbled upon this forum searching out 'Peskin errata' when trying to figure out a simple QFT calculation in the textbook. Apparently, there is no mention of the simple derivation that I'm struggling with, so there must be something wrong with my own working. I would really appreciate any help with this.

Anyway, here's the derivation -

basically, the I'd like to ask how (2.38) results from (2.35) and (2.37).

$$|\textbf{p}\rangle = \sqrt{2E_{\textbf{p}}}a^{t}_{\textbf{p}}|0\rangle$$
(2.35)​

$$U(\Lambda)|\textbf{p}\rangle = |\Lambda\textbf{p}\rangle$$
(2.37)​

$$U(\Lambda)a^{t}_{\textbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}}}a^{t}_{\Lambda\textbf{p}}$$
(2.38)​

It looks like something that you could hardly go wrong with, but I get the following instead:

$$U(\Lambda)a^{t}_{\textbf{p}} = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}}}a^{t}_{\Lambda\textbf{p}}$$

Thanks in advance for any help rendered :)

2. Sep 9, 2008

3. Sep 9, 2008

Avodyne

You can't actually get from (2.35) to (2.38), and P&S don't claim to. They are using the general framework of unitary transformations in QM. All QM statements (eg, commutation relations, eigenvalue equations, etc) are unchanged if every operator $A$ is replaced by $U\!\!AU^{-1}$, and every state $|\psi\rangle$ by $U|\psi\rangle$, where $U$ is a unitary operator.

4. Sep 9, 2008

koolmodee

But where does $$U(\Lambda)a^{t}_{\textbf{p}}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\textbf{p}}}{E_{\textbf{p}}} }a^{t}_{\Lambda\textbf{p}}$$ come from?

5. Sep 9, 2008

Avodyne

P&S are simply postulating the existence of such a unitary operator, since symmetries in QM are, in general, implemented by unitary operators. (Find a mind-numbingly detailed exposition of this, see Weinberg volume 1.)

6. Sep 10, 2008

haushofer

7. Sep 11, 2008

haushofer

8. Sep 11, 2008

Avodyne

The vacuum is rotation invariant and boost invariant; rotate it or boost it, and you get the same state back again. That's the meaning of $U(\Lambda)|0\rangle=|0\rangle$.

9. Sep 12, 2008

haushofer

Yes, I can understand that for inertial observers. But according to the Unruh-effect an accelerating observer will measure that the vacuum |0> as observed by inertial observers will be a heath bath with a certain temperature T. So, as far as I can understand, the statement

U|0> = |0>

only is true for those U which consists of Lorentz transformations transforming inertial observers to inertial observers.

Am I getting something wrong?

10. Sep 12, 2008

samalkhaiat

The existence of a unique, Poincare' invariant vacuum and its cyclicity is one of the postulates of the (special) relativistic quantum field theory.
I do understand your concerns regarding non-inertial observers and their choices of vacuum! However, I cannot produce a "short" and "easy" answer in here. A very readable account on the issues involved can be found in Sec. 3.3 of Birrell & Davies book "Quantum Fields in Curved Space".

regards

sam

11. Sep 12, 2008

strangerep

That's correct. The Hilbert/Fock space of standard QFT is constructed to carry a
representation of the Poincare group. If one then tries (naively) to represent a larger
group thereon (eg a group also containing accelerations) one gets a new Fock space which
is disjoint from the original. (Well, that's what the maths says anyway. Whether this is an
experimentally-real physical effect remains to be seen.)