Daily Archives: Fri, 5 June 2015

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.