The Schrödinger equation for the evolution operator reads:

$$ \frac{\partial U}{\partial t} = -\frac{i}{\hbar}HU $$

where for a time dependent Hamiltonian which need not commute with itself at different times we define

$$ U(t,t_0)=Te^{-\frac{i}{\hbar}\int_{t_0}^{t} H(t')dt'} $$

where $T$ is the time ordering operator.

If now one takes the conjugate transpose of the first equation:

$$ \frac{\partial U^\dagger}{\partial t} = +\frac{i}{\hbar}U^\dagger H \tag{1} $$

while if instead one looks at $U^\dagger$ (the conjugate transpose of the defined $U$), and takes its derivative we get:

$$ \frac{\partial U^\dagger}{\partial t} = +\frac{i}{\hbar}H U^\dagger \tag{2} $$ Obviously the two are equivalent only if the Hamiltonian commutes with itself at different times, in which case the time ordering is redundant and $[U,H]=0$.

If they don't commute which of the two is correct?

Notes to keep in mind:

  1. Though technically my first guess would be that the second line of thought is correct, I'm not sure about it since I'm not sure that $U^\dagger$ is just equivalent to taking $i\to-i$ in the above definition of $U$. Wouldn't we want $U^\dagger$ to be defined using time anti-ordering?
  2. If the first point is correct what does it imply about the time derivative? Does it imply that when taking the derivative $H$ should go to the right of $U^\dagger$?
  3. Note that putting $H$ to the right of $U^\dagger$, is appealing if one wants to obtain the the usual expressions for the Schrödinger equation, and the Liouville - Von Neumann equation in the interaction picture. If this is not the case one cannot for instance use the usual expressions for the interaction picture, while taking a time dependent Hamiltonian as the unperturbed Hamiltonian, as being done behind the scenes for instance here (in the last section).

3 Answers 3


Eq. (1) is correct. As OP already suspects, the Hermitian adjoint evolution operator $U^{\dagger}$ is associated with anti-time-order, so differentiation wrt. the final time brings the Hamiltonian $H$ down to the right of $U^{\dagger}$ in eq. (1), consistent with an anti-time-ordered late time. For more details, see also this related Phys.SE post.


The first one is correct. Note that $UU^\dagger=1$, and therefore $\dot UU^\dagger+U\dot U^\dagger=0$. Solving for $\dot U^\dagger$, $$ \dot U^\dagger=-U^\dagger\dot UU^\dagger=-U^\dagger (-iHU)U^\dagger=+i U^\dagger H $$

  • $\begingroup$ Thanks, this seems to make sense, as it results from first principles (the requirement that $U$ is unitary). However, how do you explain the presumed paradox? What is wrong in the other line of thought? $\endgroup$
    – Yair M
    Commented May 23, 2017 at 13:32

The usual way to treat a time-dependent evolution is to define a two parameter group of unitary evolutions $U(t,s)$ that satisfies

\begin{align}U(t,t)=1\;,\; U&(t,s)U(s,r)=U(t,r)\\i\partial_t U(t,s)&= H(t)U(t,s)\\ i\partial_s U(t,s)&= -U(t,s)H(s)\;.\end{align}

The formal solution (with $s$ instead of $t_0$ in the integral) is the one the OP wrote (and the system admits a unique solution whenever $t\mapsto H(t)\psi$ is strongly differentiable on a dense common core of all the $H(t)$).

Clearly, $$U(t,s)^*=U(s,t)=U(t,s)^{-1}\; .$$


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