Solving Mandl & Shaw 2.5: Show U = e^(-i/h)δαPα

[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.

 

[George Jones]

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

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

 

[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.

 

[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.

 

[George Jones]

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.

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.