3
$\begingroup$

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).
$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 May 23 '17 at 13:32
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.