Solving the time-dependent Schrödinger equation with a time delta potential

Question

Suppose I want to model a system described by a Hamiltonian $H_0$ to which I give a quick kick at time $t = 0$. I would use the time-dependent Hamiltonian $$ \mathcal{H}(t) = H_0 + \bar{V} \delta (t)$$ with $\bar{V}$ an operator.

I started by trying to solve analytically the Schrödinger equation with $H_0 = p^2/2m$, but I'm getting a bit stuck. Indeed, it is quite simple to see that the wave function is discontinuous at $t = 0$ ; but this discontinuity itself depends on the value of the wave function at $t=0$.

Is this type of problem even solvable ?

I have the impression that this is a "standard problem" but I have not found any discussion in the literature.

Answer 1

Solution using a basis set expansion

Let $H_0|n\rangle=E_n|n\rangle$ and lets make the following ansatz for our time dependent state, $$ |\psi(t)\rangle = \sum_n c_n(t) \exp\left(-\frac{i}{\hbar }E_nt\right)|n\rangle . $$ Inserting this into the time dependent Schrödinger equation $i\hbar \frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangle$ with $H(t)=H_0 +\delta(t)V$ leads after cancelation of some terms to

$$ i\hbar \sum_n \dot c_n(t) \exp\left(-\frac{i}{\hbar }E_nt\right)|n\rangle = \sum_l c_l(t) \exp\left(-\frac{i}{\hbar }E_lt\right)V\delta (t)|l\rangle $$ Projection unto state $\langle k|$ yields with the definition $V_{kl}= \langle k|V|l\rangle$ and using the orthonormality of states $\langle k|n\rangle = \delta_{kn} $ $$ i\hbar \dot c_k(t) \exp\left(-\frac{i}{\hbar }E_kt\right)= \sum_l V_{kl} c_l(t) \exp\left(-\frac{i}{\hbar }E_lt\right)\delta(t) $$ Now we multiply with $$-\frac{i}{\hbar}\exp\left(\frac{i}{\hbar }E_kt\right) $$ and define $\tilde V_{kl}(t)=V_{kl}\exp\left(-\frac{i}{\hbar }(E_l-E_k)t\right)$ which leads to $$ \dot c_k = -\frac{i}{\hbar } \sum_l \tilde V_{kl}(t) c_l(t)\delta(t) $$ This can be written as vector equation $$ \mathbf{ \dot c} = -\frac{i}{\hbar } \mathbf{\tilde V(t)c(t)}\delta(t) $$ We can now integrate the equation, $$\begin{aligned} \int_0^t dt' \mathbf{ \dot c} &= -\frac{i}{\hbar } \int^t_0 dt' \mathbf{\tilde V(t')c(t')}\delta(t')\\ \mathbf{c(t)} - \mathbf{c(0)} &= -\frac{i}{\hbar } \mathbf{\tilde V(0)c(0)}\\ \mathbf{c(t)} &= \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\right )\mathbf{c(0)} \end{aligned}$$ This equations holds for $t \geq 0$. If we assume that our system propagates freely before $t=0$, we can introduce the Heaviside theta function to obtain $$ \mathbf{c}(t) = \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\theta(t)\right )\mathbf{c(0)} $$

Solution without basis set expansion

We could also do the same without expansion in an explicit basis(which is just a bad habit of mine due to my Comp. Chem. background ...)

For that define the state in the interaction picture as $$\begin{aligned} |\psi(t)\rangle = \exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \\ |\psi(0)\rangle = |\psi_I(0)\rangle \end{aligned}$$

Now insert this into the TDSE, $$\begin{aligned} i\hbar \frac{d}{dt}|\psi(t)\rangle &= (H_0 + V\delta(t) )|\psi(t)\rangle \\ i\hbar \frac{d}{dt} \left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right) &= (H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right) \\ i\hbar \left( -\frac{i}{\hbar}H_0 \exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle +\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &= (H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ H_0\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right)|\psi_I(t)\rangle \right)+i\hbar\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &=(H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ \text{Cancel }H_0 \text{ term on both sides}\\ i\hbar\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &= V\delta(t)\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ \text{Multiply with }\exp\left(\frac{i}{\hbar}H_0 t\right)\text{ from the left}\\ i\hbar |\frac{d}{dt}\psi_I(t)\rangle &= \exp\left(\frac{i}{\hbar}H_0 t\right)V\delta(t)\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \\ \int^t_0 dt' |\frac{d}{dt'}\psi_I(t')\rangle & =-\frac{i}{\hbar} \int^t_0 dt' \exp\left(\frac{i}{\hbar}H_0 t'\right)V\delta(t')\exp\left({-\frac{i}{\hbar}}H_0 t'\right) |\psi_I(t')\rangle \\ |\psi_I(t)\rangle - |\psi_I(0)\rangle &= -\frac{i}{\hbar} \exp\left(\frac{i}{\hbar}H_0 \cdot 0\right)V\exp\left({-\frac{i}{\hbar}}H_0 \cdot 0\right) |\psi_I(0)\rangle \\ |\psi_I(t)\rangle &= |\psi_I(0)\rangle -\frac{i}{\hbar}V|\psi_I(0)\rangle\\ |\psi_I(t)\rangle &= (\mathbf{1} -\frac{i}{\hbar}V)|\psi_I(0)\rangle\\ \exp\left({\frac{i}{\hbar}}H_0 t\right) |\psi(t)\rangle &= (\mathbf{1} -\frac{i}{\hbar}V)|\psi(0)\rangle\\ |\psi(t)\rangle &=\exp\left(-{\frac{i}{\hbar}}H_0 t\right)(\mathbf{1} -\frac{i}{\hbar}V)|\psi(0)\rangle\\ \end{aligned}$$

This is equivalent to my first calculation. The line $$ \mathbf{c}(t) = \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\right )\mathbf{c(0)} $$ is equivalent to $$ |\psi_I(t)\rangle = (\mathbf{1} -\frac{i}{\hbar}V)|\psi_I(0)\rangle\\ $$

Answer 2

I've been thinking about this problem again and I want to propose an other answer, which I think is more correct than the previous one. We have to be more careful when we integrate $\delta (t)$ from $0$ to $t > 0$. First, because $|\Psi (t) \rangle$ is discontinuous at $t=0$ we will consider the initial state to be taken at $t = 0^-$. Then if we define $$\int_{0^-}^{t > 0} dt' \delta(t') |\Psi (t') \rangle = |\Psi (0^-) \rangle$$ we recover the previous answer. But this definition is problematic because it gives us a non-unitary time evolution operator. Another evident/intuitive solution of $i\hbar \frac{d}{dt} |\Psi_I (t) \rangle = V \delta(t) |\Psi_I (t) \rangle $ is $$ |\Psi_I (t) \rangle = e^{-iV/\hbar} |\Psi_I (0^-) \rangle$$ for $t>0$ ; the dynamics is now unitary as expected.

Now why this second solution would be more sensible than the first one (if we forget the unitarity) ? Because this is exactly what we expect if we regularize the Dirac with a constant and narrow potential of norm $V/2\varepsilon$ and length $2\varepsilon$. Indeed, in this case

$$|\Psi_I (\varepsilon) \rangle = e^{-iV/\hbar} |\Psi_I (-\varepsilon) \rangle$$

In the limit $\varepsilon \rightarrow 0$ we recover

$$ |\Psi_I (0^+) \rangle = e^{-iV/\hbar} |\Psi_I (0^-) \rangle$$