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So I have read that if we translate a quantum scalar field, it transforms as

$$ U(\mathbf{a})^\dagger \hat{\phi}(\mathbf{x}) U(\mathbf{a}) = \hat{\phi}(\mathbf{x+a}). $$

where $U(\mathbf{a})$ is the translation operator. I do not understand why this is the case so I set about to prove it myself and I feel I have made some progress but any help would be much appreciated!

So, the field operator $\hat{\phi}(\mathbf{x}) $ comes from promoting the classical scalar field $ \phi(\mathbf{x}) $ to an operator. So my starting point is with the classical field. If I actively translate my classical field by a vector distance $ \mathbf{a} $, I find

$$ \phi(\mathbf{x}) \rightarrow \phi'(\mathbf{x})=\phi(\mathbf{x - a})$$

Now I will promote the fields $\phi(\mathbf{x})$ and $ \phi'(\mathbf{x})$ to operators with the interpretation that they produce a particle at the point $\mathbf{x}$, i.e.

$$ \hat{\phi}(\mathbf{x}) |0\rangle = | \mathbf{x} \rangle. $$

So

$$ \hat{\phi'}(\mathbf{x}) |0\rangle = \hat{\phi}(\mathbf{x-a}) |0\rangle \\ = |\mathbf{x-a}\rangle \\ =U(\mathbf{-a})|\mathbf{x}\rangle\\ =U(\mathbf{a})^\dagger\hat{\phi}(\mathbf{x}) |0\rangle.$$

So I have concluded that $ U(\mathbf{a})^\dagger\hat{\phi}(\mathbf{x}) = \hat{\phi}(\mathbf{x-a}) $ which doesn't agree with the first equation I wrote. I do not understand what the next step is, assuming what I have written is correct.

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  • $\begingroup$ You need two more ingredients: the vacuum is translation invariant and $U$ is hermitian, $U^\dagger U=1$. $\endgroup$
    – Wakabaloola
    Commented Feb 16, 2018 at 13:27
  • $\begingroup$ @Wakabaloola why you need hermitian? Unitary is enough $\endgroup$
    – Hkw
    Commented Feb 16, 2018 at 13:48
  • $\begingroup$ @Hare: apologies i meant unitary ($U^{\dagger}=U^{-1}$), certainly not hermitian $\endgroup$
    – Wakabaloola
    Commented Feb 16, 2018 at 14:24
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    $\begingroup$ You have not concluded that $U(\mathbf a)^\dagger\hat\phi(\mathbf x) = \hat\phi(\mathbf{x-a})$. All you can conclude for your calculation is that $U(\mathbf a)^\dagger\hat\phi(\mathbf x)$ is equal to $\hat\phi(\mathbf{x-a})$ when acting on $|0\rangle$. But it need not be the case that $U(\mathbf a)^\dagger\hat\phi(\mathbf x)$ is equal to $\hat\phi(\mathbf{x-a})$ when acting on other states. Indeed, they are not equal when acting on, say $|\mathbf y\rangle$. My advice is to repeat the argument but using $\hat\phi'(\mathbf x) |\mathbf y\rangle$ instead of $\hat\phi'(\mathbf x) |0\rangle$. $\endgroup$
    – AccidentalFourierTransform
    Commented Feb 16, 2018 at 16:48
  • $\begingroup$ Okay I will try that, thank you for your hint. What exactly is the interpretation of $ \hat{\phi}(\mathbf{x}) $ acting on $|\mathbf{y}\rangle$? Does it create a state $ |\mathbf{x,y}\rangle $? $\endgroup$
    – Hermitian_hermit
    Commented Feb 16, 2018 at 16:55

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