# Forced harmonic oscillator: exact solution and adiabatic approximation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.9.

The one-dimensional driven harmonic oscillator can be solved exactly both classically and in quantum mechanics. If the oscillator’s natural frequency is ${\omega}$, then with a driving force ${m\omega^{2}f\left(t\right)}$ where the function ${f}$ has dimensions of length and can be any function of time (with the condition that ${f\left(t\right)=0}$ for ${t\le0}$), then the total force is

$\displaystyle F\left(t\right)=m\ddot{x}\left(t\right)=m\omega^{2}\left(f\left(t\right)-x\left(t\right)\right) \ \ \ \ \ (1)$

The classical solution for ${x_{c}\left(t\right)}$ (subscript ‘c’ for ‘classical’) is given in Griffiths’s question as

$\displaystyle x_{c}=\omega\int_{0}^{t}f\left(t'\right)\sin\left[\omega\left(t-t'\right)\right]dt' \ \ \ \ \ (2)$

I’m not sure how to solve the original ODE to get this solution, but to show that it is a solution, we can work backwards. To do this requires finding the derivative of the integral, which is complicated by the presence of the limit of integration ${t}$ inside the sine function in the integrand. We can see what the derivative is in the more general case by using the definition of a derivative:

 $\displaystyle I\left(t\right)$ $\displaystyle \equiv$ $\displaystyle \int_{0}^{t}g\left(t-t'\right)dt'\ \ \ \ \ (3)$ $\displaystyle \dot{I}$ $\displaystyle =$ $\displaystyle \lim_{\Delta t\rightarrow0}\frac{1}{\Delta t}\left[\int_{0}^{t+\Delta t}g\left(t+\Delta t-t'\right)dt'-\int_{0}^{t}g\left(t-t'\right)dt'\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lim_{\Delta t\rightarrow0}\frac{1}{\Delta t}\left[\int_{0}^{t}\left(g\left(t+\Delta t-t'\right)-g\left(t-t'\right)\right)dt'+\int_{t}^{t+\Delta t}g\left(t+\Delta t-t'\right)dt'\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{t}\frac{dg\left(t-t'\right)}{dt}dt'+\lim_{\Delta t\rightarrow0}\frac{1}{\Delta t}\int_{t}^{t+\Delta t}g\left(t+\Delta t-t'\right)dt' \ \ \ \ \ (6)$

In the second term, as ${\Delta t\rightarrow0}$, the integral becomes

 $\displaystyle \lim_{\Delta t\rightarrow0}\frac{1}{\Delta t}\int_{t}^{t+\Delta t}g\left(t+\Delta t-t'\right)dt'$ $\displaystyle =$ $\displaystyle \lim_{\Delta t\rightarrow0}\frac{1}{\Delta t}\left[g\left(t+\Delta t-t\right)\left(t+\Delta t-t\right)\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g\left(0\right) \ \ \ \ \ (8)$

Therefore

$\displaystyle \frac{d}{dt}\int_{0}^{t}g\left(t-t'\right)dt'=\int_{0}^{t}\frac{dg\left(t-t'\right)}{dt}dt'+g\left(0\right) \ \ \ \ \ (9)$

Applying this to 2 we get

 $\displaystyle \dot{x}_{c}$ $\displaystyle =$ $\displaystyle \omega^{2}\int_{0}^{t}f\left(t'\right)\cos\left[\omega\left(t-t'\right)\right]dt'\ \ \ \ \ (10)$ $\displaystyle \ddot{x}_{c}$ $\displaystyle =$ $\displaystyle -\omega^{3}\int_{0}^{t}f\left(t'\right)\sin\left[\omega\left(t-t'\right)\right]dt'+\omega^{2}f\left(t\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega^{2}\left(f\left(t\right)-x_{c}\left(t\right)\right) \ \ \ \ \ (12)$

where in the last line, the function ${g}$ in 9 is

$\displaystyle g\left(t-t'\right)=f\left(t'\right)\cos\left[\omega\left(t-t'\right)\right] \ \ \ \ \ (13)$

so ${g\left(0\right)}$ occurs when ${t=t'}$, or

$\displaystyle g\left(0\right)=f\left(t\right)\cos\left(0\right)=f\left(t\right) \ \ \ \ \ (14)$

Thus 2 is actually a solution of 1. The initial conditions are

 $\displaystyle x_{c}\left(0\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (15)$ $\displaystyle \dot{x}_{c}\left(0\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (16)$

Griffiths now asks us to show that the exact quantum mechanical solution to the Schrödinger equation is

$\displaystyle \Psi\left(x,t\right)=\psi_{n}\left(x-x_{c}\right)e^{\frac{i}{\hbar}\left[-\left(n+\frac{1}{2}\right)\hbar\omega t+m\dot{x}_{c}\left(x-\frac{x_{c}}{2}\right)+\frac{m\omega^{2}}{2}\int_{0}^{t}f\left(t'\right)x_{c}\left(t'\right)dt'\right]} \ \ \ \ \ (17)$

where ${\psi_{n}}$ is the eigenfunction for the unforced oscillator, with eigenvalue ${\left(n+\frac{1}{2}\right)\hbar\omega}$.

The Hamiltonian for the forced oscillator is

$\displaystyle H\left(t\right)=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2}m\omega^{2}x^{2}-m\omega^{2}xf\left(t\right) \ \ \ \ \ (18)$

We can show this by applying ${H}$ and ${i\hbar\frac{\partial}{\partial t}}$ to ${\Psi}$ and showing that they give the same result. We start with the time derivative, remembering that ${x_{c}}$ is a function of time. We’ll denote a derivative with respect to ${t}$ by a dot, and a derivative with respect to ${x}$ by a dash. We’ll also define

$\displaystyle \eta\equiv e^{\frac{i}{\hbar}\left[-\left(n+\frac{1}{2}\right)\hbar\omega t+m\dot{x}_{c}\left(x-\frac{x_{c}}{2}\right)+\frac{m\omega^{2}}{2}\int_{0}^{t}f\left(t'\right)x_{c}\left(t'\right)dt'\right]} \ \ \ \ \ (19)$

We get

 $\displaystyle i\hbar\frac{\partial\Psi}{\partial t}$ $\displaystyle =$ $\displaystyle -i\hbar\dot{x}_{c}\psi_{n}'\eta-\psi_{n}\eta\left[-\left(n+\frac{1}{2}\right)\hbar\omega+m\ddot{x}_{c}\left(x-\frac{x_{c}}{2}\right)-\frac{m\dot{x}_{c}^{2}}{2}+\frac{m\omega^{2}}{2}fx_{c}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\dot{x}_{c}\psi_{n}'\eta-\psi_{n}\eta\left[-\left(n+\frac{1}{2}\right)\hbar\omega+m\omega^{2}\left(f-x_{c}\right)\left(x-\frac{x_{c}}{2}\right)-\frac{m\dot{x}_{c}^{2}}{2}+\frac{m\omega^{2}}{2}fx_{c}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\dot{x}_{c}\psi_{n}'\eta-\psi_{n}\eta\left[-\left(n+\frac{1}{2}\right)\hbar\omega+m\omega^{2}\left(fx-xx_{c}+\frac{x_{c}^{2}}{2}\right)-\frac{m\dot{x}_{c}^{2}}{2}\right] \ \ \ \ \ (22)$

where we used 1 in the second line.

Now we apply ${H}$ to ${\Psi\left(x,t\right)}$. However, the unforced eigenfunction in 17 is given as a function of ${x-x_{c}}$, not ${x}$, so the Hamiltonian that gives the standard harmonic oscillator eigenvalues is

 $\displaystyle H_{0}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial\left(x-x_{c}\right)^{2}}+\frac{1}{2}m\omega^{2}\left(x-x_{c}\right)^{2}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2}m\omega^{2}\left(x^{2}-2xx_{c}+x_{c}^{2}\right) \ \ \ \ \ (24)$

so our forced Hamiltonian is

$\displaystyle H=H_{0}+\frac{1}{2}m\omega^{2}\left(2xx_{c}-x_{c}^{2}\right)-m\omega^{2}xf \ \ \ \ \ (25)$

Applying this to 17 requires finding the second ${x}$ derivative:

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial\Psi}{\partial x}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\eta\left(\psi_{n}'+\psi_{n}\frac{i}{\hbar}m\dot{x}_{c}\right)\ \ \ \ \ (26)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\eta\left(\psi_{n}^{\prime\prime}+2\psi_{n}'\left(\frac{i}{\hbar}m\dot{x}_{c}\right)-\frac{m^{2}\dot{x}_{c}^{2}}{\hbar^{2}}\psi_{n}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta\left(-\frac{\hbar^{2}}{2m}\psi_{n}^{\prime\prime}-i\hbar\psi_{n}'\dot{x}_{c}+\frac{m\dot{x}_{c}^{2}}{2}\psi_{n}\right) \ \ \ \ \ (28)$

Putting it together, we get

 $\displaystyle H\Psi$ $\displaystyle =$ $\displaystyle \eta\left(-\frac{\hbar^{2}}{2m}\psi_{n}^{\prime\prime}-i\hbar\psi_{n}'\dot{x}_{c}+\frac{m\dot{x}_{c}^{2}}{2}\psi_{n}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\frac{1}{2}m\omega^{2}\left(x^{2}-2xx_{c}+x_{c}^{2}\right)+\frac{1}{2}m\omega^{2}\left(2xx_{c}-x_{c}^{2}\right)-m\omega^{2}xf\right]\eta\psi_{n}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \eta\left[H_{0}+\frac{m\dot{x}_{c}^{2}}{2}+\frac{1}{2}m\omega^{2}\left(2xx_{c}-x_{c}^{2}\right)-m\omega^{2}xf\right]\psi_{n}-i\hbar\eta\dot{x}_{c}\psi_{n}'\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\dot{x}_{c}\psi_{n}'\eta-\psi_{n}\eta\left[-\left(n+\frac{1}{2}\right)\hbar\omega+m\omega^{2}\left(fx-xx_{c}+\frac{x_{c}^{2}}{2}\right)-\frac{m\dot{x}_{c}^{2}}{2}\right] \ \ \ \ \ (31)$

which is identical to 22, so 17 is indeed a solution of the Schrödinger equation.

Using a similar argument, we can find the eigenfunctions and eigenvalues of the full Hamiltonian. Griffiths gives the eigenfunctions as

$\displaystyle \psi_{n}\left(x,t\right)=\psi_{n}\left(x-f\right) \ \ \ \ \ (32)$

where ${\psi_{n}\left(x-f\right)}$ is the unforced eigenfunction evaluated at position ${x-f}$. We can define an unforced Hamiltonian as above:

 $\displaystyle H_{f}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial\left(x-f\right)^{2}}+\frac{1}{2}m\omega^{2}\left(x-f\right)^{2}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2}m\omega^{2}\left(x^{2}-2xf+f^{2}\right) \ \ \ \ \ (34)$

The full Hamiltonian becomes

 $\displaystyle H$ $\displaystyle =$ $\displaystyle H_{f}+\frac{1}{2}m\omega^{2}\left(2xf-f^{2}\right)-m\omega^{2}xf\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle H_{f}-\frac{1}{2}m\omega^{2}f^{2} \ \ \ \ \ (36)$

Applying this to ${\psi_{n}\left(x-f\right)}$ we get

 $\displaystyle H\psi_{n}\left(x-f\right)$ $\displaystyle =$ $\displaystyle H_{f}\psi_{n}\left(x-f\right)-\frac{1}{2}m\omega^{2}f^{2}\psi_{n}\left(x-f\right)\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\left(n+\frac{1}{2}\right)\hbar\omega-\frac{1}{2}m\omega^{2}f^{2}\right]\psi_{n}\left(x-f\right) \ \ \ \ \ (38)$

which shows that ${\psi_{n}\left(x-f\right)}$ is an eigenfunction and the eigenvalues are

$\displaystyle E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{1}{2}m\omega^{2}f^{2} \ \ \ \ \ (39)$

Note that both the eigenfunctions and eigenvalues are time-dependent, through the parameter ${f}$.

So far, everything has been exact, but we can now apply the adiabatic theorem to the case where the forcing function ${f}$ varies slowly with time. First, we can return to the classical result 2, and rewrite it as

 $\displaystyle x_{c}\left(t\right)$ $\displaystyle =$ $\displaystyle \omega\int_{0}^{t}f\left(t'\right)\sin\left[\omega\left(t-t'\right)\right]dt'\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega\int_{0}^{t}f\left(t'\right)\frac{1}{\omega}\frac{d}{dt'}\cos\left[\omega\left(t-t'\right)\right]dt'\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.f\left(t'\right)\cos\left[\omega\left(t-t'\right)\right]\right|_{t'=0}^{t'=t}-\int_{0}^{t}\cos\left[\omega\left(t-t'\right)\right]\frac{d}{dt'}f\left(t'\right)dt'\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle f\left(t\right)-\int_{0}^{t}\cos\left[\omega\left(t-t'\right)\right]\frac{d}{dt'}f\left(t'\right)dt' \ \ \ \ \ (43)$

where we used ${f\left(0\right)=0}$ in the last line.

Now if ${f}$ varies slowly compared to the natural frequency ${\omega}$ we can take its derivative outside the integral to get an approximation:

 $\displaystyle x_{c}\left(t\right)$ $\displaystyle \approx$ $\displaystyle f\left(t\right)-\dot{f}\left(t\right)\int_{0}^{t}\cos\left[\omega\left(t-t'\right)\right]dt'\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle f\left(t\right)+\frac{\dot{f}\left(t\right)}{\omega}\sin\left[\omega\left(t-t'\right)\right] \ \ \ \ \ (45)$

If

$\displaystyle \left|\dot{f}\left(t\right)\right|\ll\omega\left|f\left(t\right)\right| \ \ \ \ \ (46)$

then we neglect the second term to get the classical adiabatic approximation

$\displaystyle x_{c}\left(t\right)\approx f\left(t\right) \ \ \ \ \ (47)$

We can use this approximation to get an adiabatic approximation for the quantum wave function 17:

 $\displaystyle \Psi\left(x,t\right)$ $\displaystyle \approx$ $\displaystyle \psi_{n}\left(x-f\right)e^{\frac{i}{\hbar}\left[-\left(n+\frac{1}{2}\right)\hbar\omega t+m\dot{f}\left(x-\frac{f}{2}\right)+\frac{m\omega^{2}}{2}\int_{0}^{t}f^{2}\left(t'\right)dt'\right]}\ \ \ \ \ (48)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi_{n}\left(x-f\right)e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)} \ \ \ \ \ (49)$

where

 $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2\hbar}\int_{0}^{t}f^{2}\left(t'\right)dt'-\left(n+\frac{1}{2}\right)\omega t\ \ \ \ \ (50)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\dot{f}}{\hbar}\left(x-\frac{f}{2}\right) \ \ \ \ \ (51)$

Here the phase factors are ${\theta}$ (the dynamic phase) and ${\gamma}$ (the geometric phase). From 39 we see that

$\displaystyle \theta_{n}\left(t\right)=-\frac{1}{\hbar}\int_{0}^{t}E_{n}\left(t'\right)dt' \ \ \ \ \ (52)$

which agrees with its earlier definition.

If the eigenfunctions are real, then the geometric phase should be zero. This isn’t strictly true here, but we’re assuming that ${\dot{f}}$ is small, so ${\gamma_{n}}$ should be close to zero.