# Unitary operator + Lorents transformations (question from Peskin)

## Main Question or Discussion Point

Hi. I am trying to understand a statement from Peskin and Schroeder at page 59 they write;

"The one particle states
$$|\vec p ,s \rangle \equiv \sqrt{2E_{\vec p}}a_{\vec p}^{s \dagger} |0\rangle$$
are defined so that their inner product
$$\langle \vec p, r| \vec q,s\rangle = 2 \vec E_\vec{p} (2\pi)^3 \delta^{(3)}(\vec p - \vec q) \delta^{rs}$$
is Lorentz invariant. This implies that the operator $U(\Lambda)$ that implements Lorentz transformations on hte states of the Hilbert space is unitary, even tough for boosts $\Lambda_{1/2}$ is not unitary."

Then they draw the conclusion from the above equations that

$$U(\Lambda)a_\vec{p}^s U^{-1}(\Lambda) = \sqrt{ \frac{ E_{\Lambda \vec{p} } }{E_{\vec p} }} a_{\Lambda \vec p}^s.$$

So my question is; how do they see that $U(\Lambda)$ must be unitary? And how do they conclude with the last equation? :)

## Answers and Replies

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Bill_K
center o bass, I don't have Peskin & Schroeder, but this argument is a routine part of QFT, so I think I can explain it. The task is to go from a field ψ(x) in position space to a Fourier-transformed field a(k) in momentum space, and then quantize. The trouble is, we want to wind up with the usual commutation relations [a(k), a*(k')] = δ3(k - k'), which are not Lorentz invariant. So a normalization factor √2E must be inserted.

The formula starts off with a straight four-dimensional Fourier transform, with a delta function restriction to the mass shell:

ψ(x) = (2π)-3/2∫d4k δ(k2 - m2) b(k) eik·x

which is relativistically invariant. Doing out the k0 integral gives you

ψ(x) = (2π)-3/2∫d3k/(2k0) b(k) eik·x

which is still relativistically invariant (even though it doesn't look like it) because d3k/(2k0) is the invariant volume element on the mass hyperboloid.

However the states created/destroyed by b(k) are also relativistically invariant, i.e. they're normalized to

<k|k'> = 2k0 δ3(k - k')

and the b's therefore obey the rather strange looking commutation relations

[b(k), b*(k')] = 2k0 δ3(k - k')

To get the usual commutator you have to replace b(k) by a(k):

b(k) = √(2k0) a(k)

What your quote is saying is that the states created by the b's form a Lorentz invariant set. Their inner product is Lorentz invariant, their commutators are Lorentz invariant, and therefore the operators U(Λ) that take you from one of these states to another is unitary, i.e. norm-preserving. Likewise for the b's: U(Λ)b(k)U(Λ)-1 = b(Λk). But when this is written in terms of the a's you need to put in the factor of √(2k0) explicitly, as in the last equation you quote.