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( I understand my question is a bit vague. I will try to make it more precise at the end ).

Consider a vector field $\Phi^i$ in quantum mechanics. The crux of my question is, how does it transform?

In some texts, I have read that since it is a vector field, it will transform like a vector i.e. a ( matrix ) operator should hit it from the left-hand side to transform it into another vector. Something like this: $$ \phi^i(x) \rightarrow U(\epsilon) \phi^i(x) \tag{1}$$

In some texts, they say that since it is an operator, it needs to be sandwiched between an operator and its conjugate. Something like this: $$ \phi^i \rightarrow e^{-i \epsilon^A Q_A} \phi^i(x) e^{i \epsilon^A Q_A} \tag{2}$$

Which one is correct? Since I have gotten confused by reading different texts over a long period, I can't provide here the exact assumptions or keywords that different texts used in these two different cases. To make my question slightly more precise, I have seen these two transformations written in the same line recently in the textbook Supergravity ( by Freedman and Proeyen, 2012). Consider this equation from page 23 of that book:

$$ \phi^i(x) \rightarrow e^{-i \epsilon^A Q_A} \phi^i(x) e^{i \epsilon^A Q_A} = U(\epsilon) \phi^i(x) \tag{1.85}$$

On the face of it, this equation looks very confusing to me. I can't express the $e^{-i \epsilon^A Q_A} \phi^i(x) e^{i \epsilon^A Q_A}$ term as something acting on $\phi^i$ from left.

So here is my understanding:

In case of classical fields, equation (1) is the correct one. But when we are dealing with quantum fields, equation (2) is the correct one. In the textbook, however, the last notation ( $ U(\epsilon) \phi^i(x) $ ) is just symbolic. Perhaps they are reminding us how it would have looked in the case of classical field. For quantum field, only the sandwiched notation is the correct one.

Moreover, it is tacitly assumed that we are talking about operators in the Heisenberg picture. Right? Otherwise talking about the evolution of a quantum field/operator would not be sensible.

Is my understanding correct? If not, please tell me when to use which transformation.

Qmechanic
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baba26
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    It looks like the authors mean $e^{-i \epsilon^A Q_A} \phi^i(x) e^{i \epsilon^A Q_A} = \sum_j U(\epsilon)^i_{; j} \phi^j(x)$. Namely, the unitary transformation of each component of the vector $\phi^i$ [each is an operator] is equivalent to acting by $U(\epsilon)$ on the vector [as a column of such operators]. So, $e^{-i \epsilon^A Q_A}$ lives in the Hilbert space, whereas $U(\epsilon)$ lives in the spacetime [in this case]. In any case, your interpretation that the $U$ notation is just a reminder of the classical situation is unlikely to be correct. – Karen H. Aug 07 '23 at 02:16
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    see this answer of mine – ACuriousMind Aug 07 '23 at 22:59

1 Answers1

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In quantum mechanics, if we have a group $G$ acting on the Hilbert space through a unitary representation $U(g)$, all operators transform according to : $$\mathcal O\to U(g) \mathcal OU(g)^\dagger\tag{1}$$

Some operators have nice properties under this transformation. In the case of the rotation or Lorenz group, some of those would be called scalars, vectors, scalar fields, vector fields, etc. In very general terms, if $V$ is a representation of $G$ (taking a basis, say that $G$ acts by the matrices $T^i_{~~j}(g)$), then we say that a collection of operators $\mathcal O^i$ lives in the representation $V$ if we have, for all $g\in G$: $$U(g) \mathcal O^i U(g)^\dagger = \sum_j T^i_{~~j}(g) \mathcal O^j\tag{2}$$

For example, a vector operator is a set of operators ${ \mathcal O^i}$ such that under any rotation $R$, we have : $$U(R) {\mathcal O^i}U(R)^\dagger = R^i_{~~j}{\mathcal O^j}\tag 3$$ A vector field is a space dependent set of space-dependent operators $\mathcal O^i(x)$ such that under any rotation : $$U(R) \mathcal O^i(x)U(R)^\dagger = R^i_{~~j} \mathcal O^j(R^{-1}x) \tag{4}$$

In short, a vector field is an operator which transforms in a nice way, in that both transformation coincide, so you are free to use the expression that is most useful or most relevant in any given context.

SolubleFish
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  • Hey @SolubleFish, can I ask you a follow up question? I would like to see the derivations for the different ways operators transform. I've tried putting an operator in a bracket, transforming it and seeing what changes, but it's not working – Fernando Franco Félix Dec 22 '23 at 02:36
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    The fact that state transform as $|\psi\rangle \to U|\psi\rangle$ is a basic assumption of quantum mechanics that is probably covered in some way in every textbook on quantum mechanics. To make sure that matrix elements $\langle \psi|\mathcal O|\phi\rangle$ are unchanger, the operators need to transform according to eq. (1). The transformation law for operators in a representation (eq. (2)) is a definition, so strictly speaking it cannot be derived. – SolubleFish Jan 02 '24 at 18:03