# 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.)

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