Use the total circuit time divided by 10 instead.

This does not solve the problem completely. But it will avoid those mistakes that keep popping up when people try and compile one gate and send it to the solver. When there are more gates and a non-trivial circuit, the probability of this happening will be very low.

src/qutip_qip/device/processor.py

@@ -1196,19 +1196,23 @@ def run_state(
             # TODO, this can be simplified further, tlist in the solver only
             # determines the time step for intermediate result.
             tlist = self.get_full_tlist()
-        # set the max step size as 1/2 of the minimal gate time.
-        try:
-            half_gate_time = np.min(np.diff(self.get_full_tlist())) / 2
-        except ValueError:  # full_tlist not available e.g., no gate.
-            half_gate_time = 0.0
+        # Set the max step size as 1/10 of the total circuit time.
+        # A better solution is to use the gate, which
+        # is however, much harder to implement at this stage, see also
+        # https://github.com/qutip/qutip-qip/issues/184.
+        full_tlist = self.get_full_tlist()
+        if full_tlist is not None:
+            total_circuit_time = (full_tlist)[-1]
+        else:
+            total_circuit_time = 0.
         if is_qutip5:
             options = kwargs.get("options", qutip.SolverOptions())
             if options["max_step"] == 0.0:
-                options["max_step"] = half_gate_time
+                options["max_step"] = total_circuit_time / 10
         else:
             options = kwargs.get("options", qutip.Options())
             if options.max_step == 0.0:
-                options.max_step = half_gate_time
+                options.max_step = total_circuit_time / 10
         kwargs["options"] = options
         # choose solver:
         if solver == "mesolve":