# What is the proper translation of a field operator?

I am trying to write the correct expression for a translated quantum field operator. There appear to be conflicting expressions given in this PSE post, and this one. In the former linked PSE post, the following is stated:

$$\hat\phi(x+a)=U(a)^\dagger \, \hat\phi(x)\,U(a)\tag{1}$$

where $$U(a)=\exp(i a\cdot \hat P)$$ [Edit: corrected sign as per comment by @CosmasZachos] is the unitary translation operator, and $$P_\mu=-i\partial_\mu$$. In explicit-component notation, this is

$$U(a)=\exp\left(i\left(-a^0 H + \vec a \cdot \vec P \right)\right)\tag{2}$$

The second post however states that

$$\hat\phi(\vec x + \vec a)=U(\vec a)\, \hat\phi(\vec x)\,U(\vec a)^\dagger \tag{3}$$

My question is, which is right?

Here is my attempt at answering it. Suppose $$|x\rangle = \hat\phi (x)|0\rangle$$ is the state with one $$\phi$$-particle at point $$x$$. Suppose that these states form a suitable orthogonal basis for the single-particle Hilbert space in our theory. The full Hilbert space is the Fock-space built out of these.

$$|f\rangle = \sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,|x_1,x_2,\ldots,x_i\rangle \tag{4}$$

Let's act with $$\hat\phi(x+a)$$ on this state. In the following equations, $$\color{red}{\textrm{red}}$$ indicates important changes relative to the previous line.

\begin{align} \hat\phi(x+a)|f\rangle &= \hat\phi(x+a)\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,|x_1,x_2,\ldots,x_i\rangle\tag{5}\\ &=\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,\color{red}{\hat\phi (x+a)}|x_1,x_2,\ldots,x_i\rangle\tag{6}\\ &=\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,|\color{red}{x+a},x_1,x_2,\ldots,x_i\rangle\tag{7}\\ &=\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,\color{red}{U(a)|x,x_1-a,x_2-a,\ldots,x_i-a\rangle}\tag{8}\\ &=\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,U(a)\color{red}{\phi(x)|x_1-a,x_2-a,\ldots,x_i-a\rangle}\tag{9}\\ &=\sum_{i=\textrm{# of particles}}\int \prod_i d^4x_i\,f_i(\{x_i\})\,U(a)\phi(x)\color{red}{U(-a)|x_1,x_2,\ldots,x_i\rangle}\tag{10}\\ &=U(a)\hat\phi(x)U(-a) |f\rangle\\ &=U(a)\hat\phi(x)U(a)^\dagger |f\rangle\tag{11}\\ &\\ \implies & \boxed{\hat\phi(x+a) = U(a)\hat\phi(x)U(a)^\dagger}\tag{12} \end{align}

I have been a bit loose with notation but I hope my logic is clear. Please let me know if anything is vague or incorrect.

• @CosmasZachos For a simple real/complex function, isn't $f(x+a)=e^{a\partial_x} f(x)$? Commented Sep 21, 2021 at 0:47
• @CosmasZachos I see how I messed up the signs though, we should have $U(a)=\exp (a\partial)=\exp (iaP)$, since $P=-i\partial$. Commented Sep 21, 2021 at 0:49

## 1 Answer

In classical QM,

$$\partial_x = i \hbar^{-1} p.$$

However, the time derivative is (According to the Schrodinger equation)

$$\partial_t = -i \hbar^{-1} H.$$

Whilst spatial translations are generated by the momentum, time shifts are actually generated by minus the energy (or depending on your definitions, spatial translations are generated by minus the momentum and time shifts by just the energy).

It's no wonder the same thing goes on in field theory:

$$U(a) = e^{i a^{\mu} P_{\mu}} = e^{i a^0 H + i \vec{a} \cdot (- \vec{P}))}.$$

Again, the generators are the momentum and minus the energy.

To go from here to finite transformations is trivial: for functions we have $$f(x - a) = e^{- a \partial_x} f(x)$$ (note that $$f(x - a)$$ and not $$f(x + a)$$ is the result of shifting $$f$$ to the right in $$x$$), similarly $$f(t - a) = e^{- a \partial_t} f(t).$$

For operators we must have $$\phi(t, x - a) = e^{i a \hbar^{-1} p_x} \phi(t, x) e^{-i a \hbar^{-1} p_x};$$ $$\phi(t - a, x) = e^{- i a \hbar^{-1} H} \phi(t, x) e^{i a \hbar^{-1} H}.$$

Looks like the first post gets it right (since the extra sign comes from considering $$\phi(x + a)$$ instead of $$\phi(x - a)$$).

• Ok so for single-valued functions, you're defining $U(a)=\exp(-a\partial_x)$, which is opposite to how I defined $U(a)$ (which is fine). Your last two equations, for the translation of a field operator, agree with the one I derived. Do you agree with the rationale I gave in my post? Or do you have a different derivation? Commented Sep 21, 2021 at 16:13
• @ArturodonJuan honestly I don’t understand your rationale. It sounds like you’re lost in signs while I am not lost, so perhaps you could use my derivation to figure out which parts of your question are right and which are wrong? Commented Sep 21, 2021 at 21:05
• I was only clarifying a subtle difference in signs residing in your answer, compared to my OP. If $a$ is spacelike (say, along the $x$-direction), your expression for $U(a)$ is the following: $U(a)=\exp (ia^{\mu} P_{\mu})=\exp (-ia^x P_x)=\exp (-a^{x}\partial_x)$. This is in contrast with my OP, where I defined $U(a)=\exp (a\partial_x)$. Luckily this subtle difference in signs doesn't show up in your expression for $\phi(t,x-a)$. You didn't provide a derivation for the transformation for operators under translation $\phi(t,x-a)$. You simply stated "we must have". Commented Sep 22, 2021 at 17:39
• I have provided equation-tags in my OP so you can point out any individual line. Commented Sep 22, 2021 at 17:43
• @ArturodonJuan forget about $U$ for a moment. How do $\partial_{\mu}$ generate spacetime translations? $f(x^{\mu} - a^{\mu}) = e^{- a^{\mu} \partial_{\mu}} f(x^{\mu})$, this is just Taylor expansion. Plug the generators of translations appropriately signed as in my post instead of $\partial_{\mu}$ and you will get your answer. You can define $U$ however you like to simplify some of the formulas in that answer you obtained. Commented Sep 22, 2021 at 21:30