# Mandl and Shaw 2.5

1. Apr 28, 2008

### jdstokes

[SOLVED] Mandl and Shaw 2.5

The question is to show that the unitary transformation corresponding to spacetime translation $\delta_\alpha$ of a scalar field is $U = e^{-(\mathrm{i}/\hbar) \delta_\alpha P^\alpha }$ where $P^\alpha$ is the energy-momentum 4-vector of the field.

$\varphi (x) \mapsto \varphi'(x') = \varphi(x_\alpha - \delta_\alpha) = U\varphi(x)U^\dag$.

Essentially this boils down to showing that

$\varphi(x_\alpha-\delta_\alpha) = U \varphi(x_\alpha)U^\dag$.

I'm sure I need to use the identity

$[P^\alpha, U] = -\mathrm{i}\hbar\frac{\partial U}{\partial x_\alpha}$,

but I'm not sure how to contort it into a form that will give me what I want.

2. Apr 28, 2008

### George Jones

Staff Emeritus
I think the idea is to use

$$U \psi \left(x\right) = \left[ U , \psi \left(x\right) \right] + \psi \left(x\right) U$$

to write

$$U \psi \left(x\right) U^{\dagger} = \left[ U , \psi \left(x\right) \right] U^\dagger + \psi \left(x\right).$$

Then use

$$\left[P^a , \psi \left(x\right) \right] =-i \hbar \frac{\partial \psi}{\partial x^a} \left(x\right)$$

and the power series expansion of $U$ to work out

$$\left[ U , \psi \left(x\right) \right].$$

Last edited: Apr 28, 2008
3. Apr 28, 2008

### jdstokes

Interesting idea but I'm not sure if it can be made to work,

$[P^\alpha,\varphi] = -\mathrm{i}\hbar\partial^\alpha \varphi \implies [P^\alpha P^\beta,\varphi] = -\mathrm{i}\hbar(P^\alpha \partial^\beta\varphi + \partial^\alpha \varphi P^\beta)$.

Higher terms in the expansion of $e^{-\mathrm{i}\hbar \delta_\alpha P^\alpha}$ contain bigger and bigger versions of this.

4. Apr 29, 2008

### jdstokes

Scrap that last post, the point is when the displacement is infinitesimal, higher order terms don't contribute. Finite displacements can be obtained by infinitely compounding the infinitesimal operator.

5. Apr 29, 2008

### George Jones

Staff Emeritus
I think that by playing with commutators, the full Taylor series expansion of $\varphi(x_\alpha-\delta_\alpha)$, can be obtained, not just the first two terms of the Taylor series expansion.

In the middle, things probably get somewhat messy, though.