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For any homogenous proper orthochronous Lorentz transformation $$x\to\Lambda x$$ , there is a unitary linear operator $$U(\Lambda)$$ [Bogoliubov(1980),Srednicki(2007),Weinberg(1995)] which can be applied to either the state vector $$\left|\psi\right> \to U(\Lambda)\left|\psi\right>$$ of the field (S-picture) or the linear operators corresponding to the observables(H-picture) $$L \to U^{-1}(\Lambda)LU(\Lambda)$$ [Bogoliubov 1980]. Now,if we consider the transformation $$ x \to x'=\Lambda x \to x''=\tilde \Lambda x'$$ in the S-picture the transformation is following: $$\left|\psi\right> \to U(\Lambda)\left|\psi\right> \to U(\tilde \Lambda)U(\Lambda)\left|\psi\right>$$ and the operators corresponding to some physical quantity is fixed (assuming no interaction) whereas the H-Picture the transformation is the following: $$ L \to U^{-1}(\Lambda)LU(\Lambda) \to U^{-1}(\tilde \Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde \Lambda)$$ and the state vector remains fixed for all time (assuming no interaction). For S-picture the expectation value is: $$ \left<\psi\right|U^{-1}(\Lambda)U^{-1}(\tilde\Lambda)LU(\tilde\Lambda)U(\Lambda)\left|\psi\right>$$ And in the H-picture the value is : $$\left<\psi\right|U^{-1}(\tilde\Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde\Lambda)\left|\psi\right>$$ Now for the equivalence of both picture, we need the same expectation value of the observable in both picture which can be done by assuming the following: $$U(\Lambda)U(\tilde\Lambda) = U(\tilde\Lambda)U(\Lambda) \tag{1}\label{1}$$ My question is whether equation \eqref{1} is true and if not, what additional condition ,besides linearity and unitarity, enable $$U(\Lambda)$$ to give $$\left<\psi\right|U^{-1}(\tilde\Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde\Lambda)\left|\psi\right>=\left<\psi\right|U^{-1}(\Lambda)U^{-1}(\tilde\Lambda)LU(\tilde\Lambda)U(\Lambda)\left|\psi\right> \tag{2}$$ for any observable L

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  • $\begingroup$ Are you sure that's the right transformation for L? Shouldn't $U^{-1}$ be on the right and $U$ on the left? $\endgroup$
    – FrodCube
    Jul 12, 2021 at 10:08
  • $\begingroup$ I updated my answer, since I just want to make my point more clear that you look at an abelian subgroup of the Lorentz group i.e. the translations. $\endgroup$
    – SGG
    Jul 12, 2021 at 14:31

3 Answers 3

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Equation $(1)$ is false, because the Lorenz group is not abelian and there its representations are not abelian either (at least the one useful in QFT). There is no reason for $(2)$ to hold either.

This shows that what you wrote for the Heisenberg picture is not actually a representation of the Lorenz group. Instead, it should be : $$L\to U(\Lambda)LU(\Lambda)^{-1}$$ which behaves nicely under two successive Lorenz transformations : $$L \to U(\Lambda)LU^{-1}(\Lambda)\to U(\Lambda')U(\Lambda) LU(\Lambda)^{-1}U(\Lambda')^{-1} = U(\Lambda'\Lambda)LU(\Lambda'\Lambda)^{-1}$$

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  • $\begingroup$ When you are transforming the co-ordinate system, I think, you are changing both the state vector and the linear operator, whereas I used only the transformation of one, either state vector ( which I called S-picture) or the linear operators( which I called H-picture) and this I learnt from Bogoliubov's book. $\endgroup$
    – user306157
    Jul 12, 2021 at 13:41
  • $\begingroup$ I guess since the question concerns just the transformation between the Heisenberg and Schrödering picture equation (1) is true, since here one restricts to the abelian subgroup of time translations. $\endgroup$
    – SGG
    Jul 12, 2021 at 14:33
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I guess your transformation behaviour of $L$ is wrong. Given a transformation $U$ we have that it acts on a state $|\psi>$ as $U|\psi>$. From that one can deduce the transformation behaviour of an operator $A$: We have that $A|\psi>$ transforms as $UA|\psi>$ which is the same as $UAU^{-1}U|\psi>$ such that the action of the operator on the new transformed state $U|\psi>$ is given $UAU^{-1}$, which is defined as the transform operator by looking at the graph of the operator. Hence you obtain when deriving the expression of the expectation value you do not arrive at different orderings regarding the transformation parameter. You obtain the same ordering.

So this is the problem and not regarding to the difference of the Heisenberg and Schrödering picture.

The only difference between the Schrödering and Heisenberg picture is that all operators in the Heisenberg Picture have been evolved with the INVERSE time operator $U^{-1}$ (I guess this is the source of your confusion with the transformation of L) such that according to the above discussion we have that on a time-dependent state $|\psi(t)> \to U^{-1}(t)|\psi(t)>=|\psi(0)>$ so the new state is time-independent, while the operator $A$ now is given by $U^{-1}(t)AU(t)$.

UPDATE:

Since you just need to transform time to go from the Heisenberg picture to the Schrödering picture, you basically look at the subgroup of spacetime-translations, which is abelian. Hence, equation (1) holds for that subgroup (ofc not for the hole lorentz group).

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  • $\begingroup$ The transformation you said gives the same expectation value in all proper orthochronous lorentz frame which seems not physically true. The difference between both picture is: in the Schrödinger picture the state evolves over time and linear operators are kept fixed( if there is not external source) and in the Heisenberg picture the linear operators evolve over time whereas states are fixed $\endgroup$
    – user306157
    Jul 12, 2021 at 13:47
  • $\begingroup$ So you want to state that the expectation value in the schrödering picture is different from the one in the Heisenberg picture? $\endgroup$
    – SGG
    Jul 12, 2021 at 14:12
  • $\begingroup$ I updated my answer, since I just want to make my point more clear that you look at an abelian subgroup of the Lorentz group i.e. the translations. $\endgroup$
    – SGG
    Jul 12, 2021 at 14:30
  • $\begingroup$ I am not saying that both picture give different expectation value. I am saying that: if you are changing the operator $$ L\to ULU^{-1}$$ for the some physical quantity and the state vector $$\left|\psi\right> \to U(\Lambda)\left|\psi\right> $$ you have the same expectation value in all lorentz frame. This does not seem true to me. Since if the expectation value is $$P$$ for momentum in one frame, I think it will be $$\Lambda P$$ in the other frame. But your formula gives the expectation value $$P$$ in both frame $\endgroup$
    – user306157
    Jul 12, 2021 at 14:53
  • $\begingroup$ But it is true by $<\psi|L|\psi> \to <\psi|U^{-1} UL U^{-1}U|\psi>=<\psi|L|\psi>$. If you want to transform the states or either the operators, then it gives your result. However, the interpretation of the last thing is not going from the Heisenberg picture to the schrödinger or vice versa (i.e. this is boosting into a different frame). Going from the Heisenberg picture to the schrödering one is not boosting into a different frame. Here you transform BOTH the states AND the OPERATORS i.e. by the Transform taking the time dependence form the state to the operators. $\endgroup$
    – SGG
    Jul 12, 2021 at 15:10
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Shortening the notation the LHS of your formula (2) can be written as $$ \langle\psi,\tilde U^{-1}U^{-1}\,L\,U\,\tilde U\,\psi\rangle\,. $$ Because $U$ and $\tilde U$ are unitary this equals $$ \langle U\,\tilde U\psi,L\,U\,\tilde U\,\psi\rangle=\langle\psi,L\psi\rangle\,. $$ The last equals sign is true because $U\tilde U$ is unitary. Obviously we don't need assumption (1).

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  • $\begingroup$ You are just stating that $U\tilde U$ is a unitary operator. The OP's confusion comes from the fact that his transformation of operators in the Heisenberg picture is not a representation of the Lorenz group, since it does behave nicely under successive transformations. $\endgroup$ Jul 12, 2021 at 11:56
  • $\begingroup$ See the point. Thanks. $U$ must be a homomorphism also. $\endgroup$
    – Kurt G.
    Jul 12, 2021 at 12:51

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