I'm trying to understand the general way to treat a change of reference frame in quantum mechanics.

As I understand it, if I introduce the reference frame transform by a unitary transformation on the state vectors $$\left|\left.\psi'\right>\right. = U\left|\left.\psi\right>\right.\tag{1}$$ and I also make the transformation on the observables, $$\mathcal{O'} = U\mathcal{O} U^{\dagger},\tag{2}$$ the value of every quantity will be equivalently described in both reference frame $$\left<\left.\psi'\right|\right.\mathcal{O'} \left|\left.\psi'\right>\right. = \left<\left.\psi\right|\right.\mathcal{O} \left|\left.\psi\right>\right. .\tag{3}$$ So the Hamiltonian in the new reference frame should be given by $$H' = U H U^{\dagger}.\tag{4}$$

My problem is that in the old reference frame, Schrödinger equation says that the evolution is given by $$ H \left|\left.\psi\right>\right. = i \hbar\frac{d}{dt} \left|\left.\psi\right>\right.\tag{5} $$

Now expressed in the new reference frame, we have $$ U^{\dagger} H' U (U^{\dagger}\left|\left.\psi'\right>\right.) = i \hbar\frac{d}{dt}(U^{\dagger}\left|\left.\psi'\right>\right.)\\ U^{\dagger} H' \left|\left.\psi'\right>\right. = \left(i \hbar\frac{d}{dt}U^{\dagger}\right)\left|\left.\psi'\right>\right. + U^{\dagger}\left(i \hbar\frac{d}{dt}\left|\left.\psi'\right>\right.\right)\\ \left( H' - i \hbar U\frac{d}{dt}U^{\dagger}\right) \left|\left.\psi'\right>\right. = i \hbar\frac{d}{dt}\left|\left.\psi'\right>\right. \tag{6}$$

Which is not the form that I expected, that is $$H' \left|\left.\psi'\right>\right. = i \hbar\frac{d}{dt} \left|\left.\psi'\right>\right. .\tag{7}$$

So what is the meaning of this result? Is the Schrödinger equation modified when making a transformation that depends on time or is the new Hamiltonian really $$H'' = U H U^{\dagger} - i \hbar U\frac{d}{dt}U^{\dagger}~?\tag{8}$$ Is there a classical analogue for this result?

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    $\begingroup$ Using a time dependent $U$ is like changing into non-inertial coordinates: you get a "fictitious force" term in $H'$. $\endgroup$ Dec 25, 2016 at 22:33
  • 1
    $\begingroup$ @AccidentalFourierTransform In non-relativistic physics, changing from one inertial frame to another is a Galilean transformation. And Schrodinger equation is invariant under Galilean transformation. Do you mean the corresponding unitary operator $U$ is time-independent? $\endgroup$
    – SRS
    Dec 26, 2016 at 7:29
  • $\begingroup$ I do not have spare time to properly answer, but the result is correct and the same as in classical Hamitonian mechanics when dealing with canonical transformations which explicitly depend on time. Here the apparently anomalous term enters the computations when $U$ parametrically depends on time. The only case for Galileo group is when considering a boost transformation along an axis. $\endgroup$ Dec 26, 2016 at 8:56
  • $\begingroup$ @SRS I have now wrote an extended answer. BTW, no fictitious forces have to do with this issue... $\endgroup$ Dec 26, 2016 at 10:33

2 Answers 2


In QM there are so-called time-dependent dynamical symmetries: $t$-parametrized families of unitary operators $V(t)$ such that $$V(t)U_t = U_t V(0) \quad \forall t \in \mathbb R\:,\tag{1}$$ where $U_t = e^{-it H}$ is the temporal evolutor of the theory. The meaning of (1) is clear: if $$\mathbb R \ni t \mapsto \psi(t) := U_t \psi(0)$$ describes the evolution of a pure state in the Hilbert space $\cal H$ with initial condition $\psi(0)$, the map $$\mathbb R \ni t \mapsto \psi'(t) := V(t) \psi(t)$$ is still the story of a pure state of the system, the one with transformed initial condition $$\psi'(0) := V(0)\psi(0)\:.$$ When $V(t)=V(0)$ for all $t$ we have a much more usual dynamical symmetry associated with a conserved quantity by means of the quantum version of Noether theorem. Dynamical symmetries similarly define also conserved quantities but the procedure is a bit more involved.

There are at least two important cases of time-dependent dynamical symmetries in physics, the boost transformation representing the change of inertial reference frame both in classical and in relativistic (quantum) physics. The structure is however the same $$V_v(t) = e^{iv K(t)}\quad t \in \mathbb R$$ where $v \in \mathbb R$ is the parameter of the one-parameter group at fixed $t$ of respectively the Galileo or Poincaré group unitarily represented. For Galileo group (there is such a generator for every direction of the $3$-space) $$K(t) = MX - tP$$ where $M$ is the mass of the system and $X$ the position of the center of mass. (It is possible to prove that this generator together with the other generators $9$ generators of the Lie algebra of projective Galileo group are essentially self-adjoint on a dense common invariant domain, but I will not discuss these more mathematical issues here.)

Now let us consider the time evolution of a pure state represented by the unit vector $\psi(t)$ subjected to a boost transformation (so viewed as the initial vector in the reference frame). The new evolutor is $U'_t$ and we have $$U'_t \psi'(0) = \psi'(t)$$ together with $$\psi'(t) = V(t) \psi(t)$$ so that $$U'_t \psi'(0) = \psi'(t) = V(t) U_t \psi(t) = V(t) U_t V(0)^{-1} \psi'(0)\:.$$ Since $\psi'(0)$ is arbitrary, we conclude that $$U'_t = V(t) U_t V(0)^{-1}\:. \tag{2}$$ If, at least formally, taking the (strong operator topology) $t$-derivative at $t=0$ (and this makes sense on the above-mentioned domain for Galileo group) we find $$H' = V(0)HV(0)^{-1} +i \frac{\partial V}{\partial t}|_{t=0} V(0)^{-1}\:. \tag{3}$$ If explicitly computing the derivative we actually find $H'=H$. This is because, from (1), (2) can be recast into $$U'_t = U_t V(0) V(0)^{-1} = U_t\:. \tag{2'}$$ which immediately yields $H'=H$ without exploiting the awkward identity (3).

Similar results are valid in classical Hamiltonian formalism when dealing with canonical transformations explicitly depending on time, for instance used to canonically represent the action of Galileo group. It is related to the fact that the Hamiltonian function is not a scalar differently from the Lagrangian function in classical mechanics and this fact happens when the change of coordinates depends on time.

(1) has another for when written in terms of Hamiltonian you can easily establish using it together the identity (Stone theorem) $$\frac{dU_t}{dt} = -i H U_t = -i U_t H\:.$$ I mean the identity $$H = V(t) H V(t)^{-1} + i \frac{dV(t)}{dt} V(t)^{-1}\:.\tag{4}$$ which can be rephrased into the form you found out yourself, $$V(t) H V(t)^{-1} = H - i \frac{dV(t)}{dt} V(t)^{-1}\:.$$ The first term on the right-hand side of (4) is the transformation law of observables under the boost symmetry $V(t)$, which is valid for instance for the momentum and the position operator: $$X \to X' = V(t) X V(t)^{-1} = X + tvI, \quad P \to P' = V(t) P V(t)^{-1} + MvI\:. \tag{5}$$ This notion of transformation is based on the idea that if one simultaneously transforms states and observables, then expectation values do not change. The point is that when applying this idea to the Hamiltonian observable under the boost symmetry, the transformed Hamiltonian ceases to be the Hamiltonian of the system since (1) holds in place of $V(t) U_t V(t)^{-1} = U_t$ (wrong) as instead happens for the other symmetries of Galileo group: for an isolated system, $H$ is a scalar under translations and rotations but not under boosts.


Your equation (4) is only valid if $U$ is time-independent. If $U$ becomes time-dependent then, you should use equation (8) for transformation of Hamiltonian.


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