# Translations of field operators in QFT

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A question in the book QFT of Srednicki: This concerns the relativistic QFT generalization

$$\tag{2.21} {{e}^{-i\hat{P}x/\hbar}}\psi (0){{e}^{i\hat{P}x/\hbar}}~=~\psi (x)$$

of the formula

$$\tag{2.20} {{e}^{i\hat{H}t/\hbar}}\psi (\vec{x},0){{e}^{-i\hat{H}t/\hbar}}~=~\psi (\vec{x},t)$$

from non-relativistic QM.

How can we prove this two formula just use Poincare algebra? $\psi (x)$ is an field operator. About the second formula we can use Schrodinger equation to prove. I think there is a proof just using the Poincare algebra. I just have no idea about this.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user henry
asked Nov 4, 2012

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In the answer of Argopulos, one must assume that the Taylor series converges. This is never done, and maybe physicists don't care.

From a mathematical perspective, (2.20) is in fact the definition of the dynamics, and the commutator relation follows simply by differentiating both sides. No special assumption is needed to get this.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user Arnold Neumaier
answered Nov 4, 2012 by (15,787 points)
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Symmetry group of the space time$^1$ on which QFT is defined is usually required to have a representation on the space of states.

Quantum mechanics is just QFT in one dimensions. The spacetime in this case is the time line $\mathbb R$. Fields are $X(t)$, and $P(t)$. Symmetry group is group of translations $t\rightarrow t+b$ of $\mathbb R$. Infinitesimal form of (Hermitian) time translation operator is Hamiltonian $H=i\partial/\partial t$. One requires that Hilbert space of states have a representation of this group, which in Heisenberg picture is given as $$exp(iHt)X(0)exp(-iHt)=X(t),\:exp(iHt)P(0)exp(-iHt)=P(t)$$ The case of QFT on Minkowski space is similar. We again require the symmetry group (which is Poincare group in this case) to have a representation on the space of states. In particular for the subgroup of Poincare group generated by four translation operators $P_0,P_1,P_2,P_3$ we require $$exp(iP_0x^0)\psi(0)exp(-iP_0x^0)=\psi(x^0,0,0,0)$$ $$exp(iP_1x^1)\psi(0)exp(-iP_1x^1)=\psi(0,x^1,0,0)$$ $$exp(iP_2x^2)\psi(0)exp(-iP_2x^2)=\psi(0,0,x^2,0)$$ $$exp(iP_3x^3)\psi(0)exp(-iP_3x^3)=\psi(0,0,0,x^3)$$ Since $P_\mu$'s commute with each other these conditions can be collectively written as : $$exp(iPx)\psi(0)exp(-iPx)=\psi(x)$$ where $P=(P^\mu), \: P^\mu=\eta^{\mu\nu}P_\nu$ and $\eta=diag(1,-1,-1,-1)$. Sredniki's sign convention is different i think.

1. For example if space time is a manifold with metric $\eta$ then symmetry group is group of transfomations which preserve $\eta$.
This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user10001
answered Nov 4, 2012 by (635 points)
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Taylor expand the right-hand side, and you will see that it matches, order by order in $x$, with the left-hand side. You just need to know the definition of $\hat{P}$, i.e. $[\hat{P},\hat{\psi}(x)]=i \hat{\psi}'(x)$, where the prime denotes derivative with respect to the coordinate associated to $\hat{P}$. Of course, you need to remember that the momenta all commute among each other (and in fact derivatives do commute). Back to your original intuition, you need to know in practice only the abelian subalgebra of the Poincaré algebra.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user Argopulos
answered Nov 4, 2012 by (100 points)
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The formula $e^{-i\hat{P}a/\hbar}\hat\psi(x)e^{i\hat{P}a/\hbar}=\hat\psi(x+a)$ is just the finite transformation version of infinitesimal transformation version formula $[\hat{P},\hat{\psi}(x)]=i\hbar\partial_x\hat{\psi}(x)$ as mentioned by DaniH and Argopulos above. $e^{-i\hat{P}a/\hbar}$ is a finite transformation operator acting on the Hilbert space, while $x\rightarrow x+a$ is the finite transformation acting on the field space. The infinitesimal transformations or the generator of the Lie group is given by $\hat P$ and $i\hbar\partial_x$ accordingly.

Finite transformation can be derived from infinitesimal transformation as followed:

$e^{-i\hat{P}a/\hbar}\hat\psi(x)e^{i\hat{P}a/\hbar}=e^{-i(a/\hbar)[\hat{P},\cdot]}\hat\psi(x)=e^{-i(a/\hbar)(i\hbar\partial_x)}\hat\psi(x)=e^{a\partial_x}\hat\psi(x)=\hat\psi(x+a)$.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user Tengen
answered Nov 5, 2012 by (105 points)
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I have been puzzled by this for a long time, but I am studying Conformal Field Theory recently, and it seems that there is an answer. If we consider then conformal algebra in d>=3, there are such commutation relations:

where the first one means the generator of dilation transformation commutes with all the generators of Lorenz transformations, if our quantum fields is still given by the irreducible representation of Lorentz group, the Schur's lemma tells us $\tilde{\Delta}$ is proportional to identity, then $\kappa_\mu$ which represents the generators of spacetime translation and SCT is 0 by the second commutation relation! Since $\kappa_\mu$ vanish, its exponential is equal to one, that's exactly what you asked in the question.

answered May 26, 2021 by anonymous
edited May 26, 2021 by Dilaton

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