# Integral of Schrodinger equation for a time-dependent Hamiltonian

I am given the following Hamiltonian, $H=H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega_1^2x^2$ for $t<0$ and $H=H_2=\frac{p^2}{2m}+\frac{1}{2}m\omega_2^2x^2$ for $t\geq 0$. Now I want to integrate my Schrodinger equation as $$\int_{t_1}^{t_2} \frac{d \psi}{|\psi\rangle}=\frac{-i}{\hbar}\int_{t_1}^{t_2}\hat{H}dt=\frac{-i}{\hbar}\int_{t_1}^{0} H_1 dt+\frac{-i}{\hbar}\int_{0}^{t_2} H_2 dt$$ for $t_1<0$ and $t_2>0$. Can I do such splitting of integral? Also, as in the bounds $(t_1,0)$, $H_1$ is time independent and so is $H_2$ in $(0,t_2)$.

• How do you divide with a ket? – Qmechanic Jun 8 '18 at 8:52
• So there is nothing stopping you from splitting an integral. The question here is whether the quantity that you wrote above is meaningful and well defined. Could you please justify your expression $\hat{H}$ is an operator so how are you integrating $\hat{H}$dt? – ohneVal Jun 8 '18 at 9:06
• I may have messed up with the notations but conceptually, I am trying to follow what I read in Sakurai, Chap 2, Page 72. – Naman Agarwal Jun 8 '18 at 9:10
• Related post by OP: physics.stackexchange.com/q/410366/2451 – Qmechanic Jun 8 '18 at 9:20

You can split the time interval as you say: the problem then becomes "solve the time-dependent Schrödinger equation with time-independent Hamiltonian $H_1$ from $t_1$ to $0$, and then solve it with $H_2$ from $0$ to $t_2$". However, you cannot just write the solution as an integral in this way, it is not a simple quadrature, and as the earlier comments said, you can't just divide by $\psi$ (and leave the Hamiltonian operators on the right with nothing to operate on). Formally you are solving a differential equation, with given initial conditions (the initial wavefunction) $\psi(t_1)$. The final wavefunction from the first stage becomes the initial wavefunction for the second stage. Both stages can be easily solved.
• I agree that I have messed up the notations. So in short I can write the time evolution operator for my state $|\psi(t_2)\rangle=U(t_2,t_1)|\psi(t_1)\rangle$, where $U(t_2,t_1)=exp\left [ \frac{-iH_1(-t_1)}{\hbar} + \frac{-iH_2(t_2)}{\hbar}\right ]$ – Naman Agarwal Jun 8 '18 at 9:21
• Better to write it as $U_2(t_2-0)U_1(0-t_1)\psi(t_1)$ where $U_1(t)=\exp(-i H_1 t/\hbar)$ and $U_2(t)=\exp(-i H_2 t/\hbar)$. I know that it looks almost the same, but it expresses the fact that $U_1$ acts first, for time $t_1$, and $U_2$ acts second, for time $t_2$. Writing it your way, you are just adding the two hamiltonian terms together. In the general case, two operators might not commute, so the exponential of their sum would not be the same as the product of their exponentials. – user197851 Jun 8 '18 at 9:40
• But in this case, since your Hamiltonians commute, your expression for the propagator is correct (i.e. the exponent is the time integral of the Hamiltonian over the whole interval). I was just trying to make clear that this gets more complicated when $H_1$ and $H_2$ do not commute. – user197851 Jun 8 '18 at 10:02