# Harmonic oscillator: Hermite polynomials and orthogonality of eigenfunctions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercises 7.3.2 – 7.3.3.

The eigenfunctions of the harmonic oscillator are given by

$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

where ${H_{n}\left(u\right)}$ is a Hermite polynomial. The Hermite polynomials obey the recursion relation

$\displaystyle H_{n+1}(x)=2xH_{n}(x)-2nH_{n-1}(x) \ \ \ \ \ (2)$

The first few Hermite polynomials are given in Shankar’s equation 7.3.21, and we may use these to verify this relation for a couple of cases. Taking ${n=2}$ we have

 $\displaystyle H_{3}\left(x\right)$ $\displaystyle =$ $\displaystyle 2xH_{2}\left(x\right)-4H_{1}\left(x\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2x\left[-2\left(1-2x^{2}\right)\right]-4\left(2x\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -12x+8x^{3} \ \ \ \ \ (5)$

The last line agrees with ${H_{3}}$ as given in Shankar.

For ${n=3}$ we have

 $\displaystyle H_{4}\left(x\right)$ $\displaystyle =$ $\displaystyle 2xH_{3}\left(x\right)-6H_{2}\left(x\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2x\left[-12x+8x^{3}\right]-6\left[-2\left(1-2x^{2}\right)\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 12-48x^{2}+16y^{4} \ \ \ \ \ (8)$

which again agrees with Shankar’s equation.

We can see from the relation 2 that, given that ${H_{0}=1}$ and ${H_{1}=2x}$, all Hermite polynomials of even index contain only even powers of ${x}$, and all polynomials of odd index contain only odd powers of ${x}$. This means that all even Hermite polynomials are even functions of ${x}$, in the sense that ${H_{2n}\left(-x\right)=H_{2n}\left(x\right)}$, and all odd Hermite polynomials are odd functions of ${x}$, so that ${H_{2n+1}\left(-x\right)=-H_{2n+1}\left(x\right)}$.

If ${\psi\left(x\right)}$ is even and ${\phi\left(x\right)}$ is odd, then

$\displaystyle \psi\left(-x\right)\phi\left(-x\right)=-\psi\left(x\right)\phi\left(x\right) \ \ \ \ \ (9)$

That is, the product ${\psi\left(x\right)\phi\left(x\right)}$ is an odd function. Since the integral of any odd function over an interval symmetric about ${x=0}$ is zero, we have

$\displaystyle \int_{-\infty}^{\infty}\psi\left(x\right)\phi\left(x\right)dx=0 \ \ \ \ \ (10)$

Looking at the eigenfunctions 1, we see that the exponential factor is a Gaussian centred at ${x=0}$ and is therefore even, so that ${\psi_{n}}$ will be even or odd depending on whether ${n}$ is even or odd. In particular, the integral of any even ${\psi_{n}}$ multiplied by any odd ${\psi_{n}}$ over all ${x}$ will be zero.

To show that pairs of even functions are also orthogonal is a bit trickier, but we can do it in the simplest case, where we consider the functions ${\psi_{0}}$ and ${\psi_{2}}$.

 $\displaystyle \int_{-\infty}^{\infty}\psi_{0}\left(x\right)\psi_{2}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{8}}\int_{-\infty}^{\infty}H_{0}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)H_{2}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{8}}\int_{-\infty}^{\infty}\left(1\right)\left[-2\left(1-2\frac{m\omega}{\hbar}x^{2}\right)\right]e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2}}\left[\sqrt{\frac{\pi\hbar}{m\omega}}-\sqrt{\frac{\pi\hbar}{m\omega}}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

The two Gaussian integrals can be done using standard formulas as given in Shankar’s Appendix A.2. (I used Maple.)

# Harmonic oscillator – series solution revisited

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.1.

Shankar’s derivation of the eigenfunctions of the harmonic oscillator in the position basis is essentially the same as that in Griffiths, which we’ve covered before. The reader may wish to refresh their knowledge of this before reading the rest of this post.

To make the comparison we note that ${\epsilon}$ in Griffiths is ${2\varepsilon}$ in Shankar:

$\displaystyle \varepsilon\equiv\frac{E}{\hbar\omega} \ \ \ \ \ (1)$

The analysis begins with the Schrödinger equation for the harmonic oscillator, which is

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi \ \ \ \ \ (2)$

Making the substitution

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)$

we convert the equation to

$\displaystyle \psi^{\prime\prime}+\left(2\varepsilon-y^{2}\right)\psi=0 \ \ \ \ \ (4)$

where a prime indicates a derivative with respect to ${y}$.

As explained in the earlier post, we further convert this equation by defining another function ${u\left(y\right)}$ (Griffiths calls this function ${f\left(y\right)}$) as

$\displaystyle \psi(y)=e^{-y^{2}/2}u(y) \ \ \ \ \ (5)$

This results in a simpler differential equation for ${f}$:

$\displaystyle \frac{d^{2}u}{dy^{2}}-2y\frac{du}{dy}+(2\varepsilon-1)u=0 \ \ \ \ \ (6)$

We can solve this by proposing that ${u}$ is a power series in ${y}$:

$\displaystyle u\left(y\right)=\sum_{n=0}^{\infty}C_{n}y^{n} \ \ \ \ \ (7)$

This leads to the recursion relation for the coefficients ${C_{n}}$:

$\displaystyle C_{n+2}=C_{n}\frac{2n+1-2\varepsilon}{\left(n+1\right)\left(n+2\right)} \ \ \ \ \ (8)$

In order that ${u}$ is finite for large ${y}$, this series must terminate, which leads to the quantization condition for the energy:

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (9)$

Shankar poses as an exercise the question as to why we didn’t just try a series solution of 4, that is, we propose

$\displaystyle \psi\left(y\right)=\sum_{n=0}^{\infty}A_{n}y^{n} \ \ \ \ \ (10)$

for some other coefficients ${A_{n}}$. If we try this, there are three terms with different exponents for ${y}$ that result from plugging this into 4.

 $\displaystyle \psi^{\prime\prime}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}A_{n}n\left(n-1\right)y^{n-2}\ \ \ \ \ (11)$ $\displaystyle 2\varepsilon\psi$ $\displaystyle =$ $\displaystyle 2\varepsilon\sum_{n=0}^{\infty}A_{n}y^{n}\ \ \ \ \ (12)$ $\displaystyle -y^{2}\psi$ $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}A_{n}y^{n+2} \ \ \ \ \ (13)$

To compare the coefficients we reassign the summation ranges so that the powers of ${y}$ are the same in all three terms.

 $\displaystyle \psi^{\prime\prime}$ $\displaystyle =$ $\displaystyle \sum_{n=2}^{\infty}A_{n}n\left(n-1\right)y^{n-2}=\sum_{n=0}^{\infty}A_{n+2}\left(n+2\right)\left(n+1\right)y^{n}\ \ \ \ \ (14)$ $\displaystyle 2\varepsilon\psi$ $\displaystyle =$ $\displaystyle 2\varepsilon\sum_{n=0}^{\infty}A_{n}y^{n}\ \ \ \ \ (15)$ $\displaystyle -y^{2}\psi$ $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}A_{n}y^{n+2}=-\sum_{n=2}^{\infty}A_{n-2}y^{n} \ \ \ \ \ (16)$

Note that the top two sums start at ${n=0}$ while the last sum starts at ${n=2}$. To satisfy 4, the coefficient of each power of ${y}$ must be zero, that is

$\displaystyle A_{n+2}\left(n+2\right)\left(n+1\right)+2\varepsilon A_{n}-A_{n-2}=0 \ \ \ \ \ (17)$

There are two separate conditions here; one for even ${n}$ and the other for odd ${n}$. To get either sequence started, we need to specify the first two terms. For example, in the even sequence, we need to specify ${A_{0}}$ and ${A_{2}}$ which then allows calculation of ${A_{4}}$ (when ${n=2}$). We can then use ${A_{2}}$ and ${A_{4}}$ to get ${A_{6}}$ and so on. The general formula is

$\displaystyle A_{n+2}=\frac{A_{n-2}-2\varepsilon A_{n}}{\left(n+2\right)\left(n+1\right)} \ \ \ \ \ (18)$

# Hamiltonians for harmonic oscillators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.5; Exercises 2.5.2 – 2.5.3.

Here are a couple of examples of equations of motion using the Hamiltonian formalism. First, we look at the simple harmonic oscillator, in which we have a mass ${m}$ sliding on a frictionless horizontal surface. The mass is connected to a spring with constant ${k}$, with the other end of the spring connected to a fixed support.

The Hamiltonian is given by

$\displaystyle H\left(q,p\right)=\sum_{i}p_{i}\dot{q}_{i}-L\left(q,\dot{q}\right) \ \ \ \ \ (1)$

where the velocities ${\dot{q}_{i}}$ are expressed in terms of the positions ${q_{i}}$ and momenta ${p_{i}}$. In this case, we have, using the coordinate ${x}$ as the displacement from equilibrium

 $\displaystyle L\left(x,\dot{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2}\ \ \ \ \ (2)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{x}}=m\dot{x}\ \ \ \ \ (3)$ $\displaystyle \dot{x}$ $\displaystyle =$ $\displaystyle \frac{p}{m}\ \ \ \ \ (4)$ $\displaystyle L\left(x,\dot{x}\left(x,p\right)\right)$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\ \ \ \ \ (5)$ $\displaystyle H$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{m}-\left(\frac{p^{2}}{2m}-\frac{1}{2}kx^{2}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2} \ \ \ \ \ (7)$

We can now apply Hamilton’s canonical equations:

 $\displaystyle \frac{\partial H}{\partial p}$ $\displaystyle =$ $\displaystyle \dot{x}\ \ \ \ \ (8)$ $\displaystyle -\frac{\partial H}{\partial x}$ $\displaystyle =$ $\displaystyle \dot{p} \ \ \ \ \ (9)$

We get

 $\displaystyle \frac{\partial H}{\partial p}$ $\displaystyle =$ $\displaystyle \frac{p}{m}=\dot{x}\ \ \ \ \ (10)$ $\displaystyle -\frac{\partial H}{\partial x}$ $\displaystyle =$ $\displaystyle -kx=\dot{p} \ \ \ \ \ (11)$

We thus get a pair of first order ODEs which can be solved in the usual way, given ${x\left(0\right)}$ and ${p\left(0\right)}$. The second order ODE that we got by using the Lagrangian method can be obtained by differentiating the first equation and plugging it into the second:

 $\displaystyle \ddot{x}$ $\displaystyle =$ $\displaystyle \frac{\dot{p}}{m}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{k}{m}x \ \ \ \ \ (13)$

From 7 we see that, since in the absence of external force, the total energy ${H=T+V=E}$ is a constant,

$\displaystyle \frac{p^{2}}{2m}+\frac{1}{2}kx^{2}=E=\mbox{constant} \ \ \ \ \ (14)$

This can be written as the equation of an ellipse:

$\displaystyle \frac{p^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}=1 \ \ \ \ \ (15)$

where

 $\displaystyle a^{2}$ $\displaystyle =$ $\displaystyle \frac{2E}{k}\ \ \ \ \ (16)$ $\displaystyle b^{2}$ $\displaystyle =$ $\displaystyle 2mE \ \ \ \ \ (17)$

We can use the Hamiltonian formalism to get the equations of motion of the coupled harmonic oscillator. From our Lagrangian treatment, we had

$\displaystyle L=\frac{1}{2}m\left(\dot{x}_{1}^{2}+\dot{x}_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (18)$

Converting to coordinates and momenta, we have

 $\displaystyle p_{i}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{x}_{i}}=m\dot{x}_{i}\ \ \ \ \ (19)$ $\displaystyle \dot{x}_{i}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{m}\ \ \ \ \ (20)$ $\displaystyle H$ $\displaystyle =$ $\displaystyle \sum_{i}p_{i}\dot{x}_{i}-L\left(x,\dot{x}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{m}\left(p_{1}^{2}+p_{2}^{2}\right)-\left[\frac{1}{2m}m\left(p_{1}^{2}+p_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right)\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2m}\left(p_{1}^{2}+p_{2}^{2}\right)+k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (23)$

Applying the canonical equations gives

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \frac{p_{i}}{m}=\dot{x}_{i}\ \ \ \ \ (24)$ $\displaystyle -\frac{\partial H}{\partial x_{1}}$ $\displaystyle =$ $\displaystyle -2kx_{1}+kx_{2}=\dot{p}_{1}\ \ \ \ \ (25)$ $\displaystyle -\frac{\partial H}{\partial x_{2}}$ $\displaystyle =$ $\displaystyle -2kx_{2}+kx_{1}=\dot{p}_{2} \ \ \ \ \ (26)$

Again, by taking the derivative of the first line and substituting into the last two lines, we get back the previous equations of motion:

 $\displaystyle \ddot{x}_{1}$ $\displaystyle =$ $\displaystyle -2\frac{k}{m}x_{1}+\frac{k}{m}x_{2}\ \ \ \ \ (27)$ $\displaystyle \ddot{x}_{2}$ $\displaystyle =$ $\displaystyle \frac{k}{m}x_{1}-2\frac{k}{m}x_{2} \ \ \ \ \ (28)$

# Lagrangians for harmonic oscillators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.1; Exercises 2.1.1 – 2.1.2.

The Euler-Lagrange equations of motion, derived from the principle of least action are

$\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{i}}-\frac{\partial L}{\partial q_{i}}=0 \ \ \ \ \ (1)$

where ${q_{i}}$ and ${\dot{q_{i}}}$ are the generalized coordinates and velocities, respectively. Here are a couple of simple examples of how these equations can be used to derive equations of motion.

Example 1 The harmonic oscillator. We have a mass ${m}$ sliding on a frictionless horizontal surface with a spring of spring constant ${k}$ connected between one end of the mass and a fixed support. The horizontal displacement of the mass from its equilibrium position is given by ${x}$, with ${x<0}$ when the mass moves to the left, compressing the spring, and ${x>0}$ when it moves to the right, stretching the spring.

For systems where the potential energy ${V\left(q_{i}\right)}$ is independent of the velocities ${\dot{q}_{i}}$, the Lagrangian can be written as

$\displaystyle L=T-V \ \ \ \ \ (2)$

where ${T}$ is the kinetic energy. In the case of the mass

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}\ \ \ \ \ (3)$ $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{1}{2}kx^{2}\ \ \ \ \ (4)$ $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2} \ \ \ \ \ (5)$

The equation of motion is

 $\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{q}_{i}}-\frac{\partial L}{\partial q_{i}}$ $\displaystyle =$ $\displaystyle m\ddot{x}+kx=0\ \ \ \ \ (6)$ $\displaystyle m\ddot{x}$ $\displaystyle =$ $\displaystyle -kx \ \ \ \ \ (7)$

which is the familiar equation for the force on the mass equal to ${-kx}$.

Example 2 We can revisit the problem of two masses coupled by three springs, as described earlier. In this case, we have two coordinates ${x_{1}}$ and ${x_{2}}$. The total kinetic energy is

$\displaystyle T=\frac{1}{2}m\left(\dot{x}_{1}^{2}+\dot{x}_{2}^{2}\right) \ \ \ \ \ (8)$

The total potential energy is

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{1}{2}kx_{1}^{2}+\frac{1}{2}k\left(x_{2}-x_{1}\right)^{2}+\frac{1}{2}kx_{2}^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right) \ \ \ \ \ (10)$

The Lagrangian and equations of motion are then

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\left(\dot{x}_{1}^{2}+\dot{x}_{2}^{2}\right)-k\left(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}\right)\ \ \ \ \ (11)$ $\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{x}_{1}}-\frac{\partial L}{\partial x_{1}}$ $\displaystyle =$ $\displaystyle m\ddot{x}_{1}+2kx_{1}-kx_{2}=0\ \ \ \ \ (12)$ $\displaystyle \frac{d}{dt}\frac{\partial L}{\partial\dot{x}_{2}}-\frac{\partial L}{\partial x_{2}}$ $\displaystyle =$ $\displaystyle m\ddot{x}_{2}+2kx_{2}-kx_{1}=0 \ \ \ \ \ (13)$

This gives the same equations of motion we had earlier.

 $\displaystyle \ddot{x}_{1}$ $\displaystyle =$ $\displaystyle -2\frac{k}{m}x_{1}+\frac{k}{m}x_{2}\ \ \ \ \ (14)$ $\displaystyle \ddot{x}_{2}$ $\displaystyle =$ $\displaystyle \frac{k}{m}x_{1}-2\frac{k}{m}x_{2} \ \ \ \ \ (15)$

# Green’s functions; forced harmonic oscillator

To introduce the idea of Green’s functions, let’s return to the classical harmonic oscillator. Its equation of motion is

$\displaystyle m\ddot{x}+kx=0 \ \ \ \ \ (1)$

This has the general solution

$\displaystyle x\left(t\right)=A\sin\omega t+B\cos\omega t \ \ \ \ \ (2)$

where

$\displaystyle \omega=\sqrt{\frac{k}{m}} \ \ \ \ \ (3)$

and ${A}$ and ${B}$ are constants determined by the initial conditions.

Suppose now we add a forcing term to the oscillator. That is, in addition to the oscillator’s innate restoring force term ${kx}$, we drive the oscillator with an additional force ${F\left(t\right)}$. The equation of motion is now

$\displaystyle m\ddot{x}+kx=F\left(t\right) \ \ \ \ \ (4)$

Since this driving force can be anything, we obviously cannot solve this equation until we know ${F\left(t\right)}$.

Now suppose we had a function ${G\left(t-t^{\prime}\right)}$ which satisfied the equation

$\displaystyle m\frac{\partial^{2}G\left(t,t^{\prime}\right)}{\partial t^{2}}+kG\left(t,t^{\prime}\right)=\delta\left(t-t^{\prime}\right) \ \ \ \ \ (5)$

where ${\delta}$ is Dirac delta function. Then it turns out that the solution to 4 is

$\displaystyle x\left(t\right)=\int_{-\infty}^{\infty}G\left(t,t^{\prime}\right)F\left(t^{\prime}\right)dt^{\prime} \ \ \ \ \ (6)$

This can be verified by direct substitution:

 $\displaystyle m\ddot{x}+kx$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}m\frac{\partial^{2}G\left(t,t^{\prime}\right)}{\partial t^{2}}F\left(t^{\prime}\right)dt^{\prime}+\int_{-\infty}^{\infty}kG\left(t,t^{\prime}\right)F\left(t^{\prime}\right)dt^{\prime}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\left[m\frac{\partial^{2}G\left(t,t^{\prime}\right)}{\partial t^{2}}+kG\left(t,t^{\prime}\right)\right]F\left(t^{\prime}\right)dt^{\prime}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\delta\left(t-t^{\prime}\right)F\left(t^{\prime}\right)dt^{\prime}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle F\left(t\right) \ \ \ \ \ (10)$

The function ${G\left(t,t^{\prime}\right)}$ is called the Green’s function. If we know ${G}$, we can convert the original differential equation 4 into an ordinary integral 6 which is presumably easier to do.

The catch, of course, is that finding the Green’s function is seldom easy and can often turn out to be just as difficult as solving the differential equation directly. For the harmonic oscillator, however, finding the Green’s function is possible, so here we go.

To make things definite, we’ll take ${F\left(t\right)=\delta\left(t\right)}$ in 4. This corresponds to an impulse force applied to the oscillator at ${t=0}$. We’ll also assume that prior to ${t=0}$, the oscillator was at rest, so ${x=0}$ for ${t<0}$. The forced oscillator equation of motion is thus

$\displaystyle m\ddot{x}+kx=\delta\left(t\right) \ \ \ \ \ (11)$

For ${t>0}$, ${F\left(t\right)=0}$, so the solution in this region must reduce to the unforced oscillator 2. To determine ${A}$ and ${B}$, we need to impose boundary conditions at ${t=0}$, so we’ll need to investigate what happens here in a bit more detail. The process is reminiscent of the calculations we did when analyzing the delta function potential in quantum mechanics.

Since all the action occurs around ${t=0}$, we’ll integrate 11 over a small interval ${t\in\left[-\epsilon,\epsilon\right]}$:

$\displaystyle \int_{-\epsilon}^{\epsilon}m\ddot{x}\left(t\right)dt+\int_{-\epsilon}^{\epsilon}kx\left(t\right)dt=\int_{-\epsilon}^{\epsilon}\delta\left(t\right)dt \ \ \ \ \ (12)$

Term by term, we get

 $\displaystyle \int_{-\epsilon}^{\epsilon}m\ddot{x}\left(t\right)dt$ $\displaystyle =$ $\displaystyle m\left.\dot{x}\left(t\right)\right|_{-\epsilon}^{\epsilon}\ \ \ \ \ (13)$ $\displaystyle \left|\int_{-\epsilon}^{\epsilon}kx\left(t\right)dt\right|$ $\displaystyle <$ $\displaystyle 2\epsilon k\max\left|x\left(t\right)\right|\ \ \ \ \ (14)$ $\displaystyle \int_{-\epsilon}^{\epsilon}\delta\left(t\right)dt$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

In the limit ${\epsilon\rightarrow0}$, the second line tends to zero, since we’re assuming that the maximum displacement of the oscillator is finite. Therefore, we must have

$\displaystyle \lim_{\epsilon\rightarrow0}m\left.\dot{x}\left(t\right)\right|_{-\epsilon}^{\epsilon}=m\dot{x}_{0^{+}}-m\dot{x}_{0^{-}}=1 \ \ \ \ \ (16)$

That is, the velocity of the oscillator is discontinuous at ${t=0}$. Since the oscillator is at rest for ${t<0}$, ${\dot{x}_{0^{-}}=0}$ so

$\displaystyle \dot{x}_{0^{+}}=\frac{1}{m} \ \ \ \ \ (17)$

[By the way, a discontinuous velocity implies an infinite acceleration, but that’s what we would expect given that the force applied at ${t=0}$ is a delta function, which is infinite.]

The displacement ${x}$ is continuous across ${t=0}$, since the velocity is finite and in the limit ${\epsilon\rightarrow0}$

$\displaystyle \left|\int_{-\epsilon}^{\epsilon}\dot{x}\left(t\right)dt\right|<2\epsilon\max\left|\dot{x}\right|\rightarrow0 \ \ \ \ \ (18)$

Therefore, since ${x=0}$ for ${t<0}$ and ${x=A\sin\omega t+B\cos\omega t}$ for ${t>0}$, we must have ${x\left(0\right)=0}$ so ${B=0}$. From 17 we have

 $\displaystyle \dot{x}\left(0^{+}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{m}=A\omega\ \ \ \ \ (19)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{m\omega} \ \ \ \ \ (20)$

Thus the general solution to the forced oscillator with a delta function forcing term is

$\displaystyle x\left(t\right)=\begin{cases} 0 & t<0\\ \frac{1}{m\omega}\sin\omega t & t>0 \end{cases} \ \ \ \ \ (21)$

If we shift the origin from ${t=0}$ to ${t=t^{\prime}}$, the equation of motion 11 becomes

$\displaystyle m\ddot{x}+kx=\delta\left(t-t^{\prime}\right) \ \ \ \ \ (22)$

and the solution becomes

$\displaystyle x\left(t,t^{\prime}\right)=\begin{cases} 0 & tt^{\prime} \end{cases} \ \ \ \ \ (23)$

Note that 22 has exactly the same form as the Green’s function equation 5, so we have actually found the Green’s function for the harmonic oscillator. This means that a general solution to the forced oscillator is given by

 $\displaystyle x\left(t\right)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x\left(t,t^{\prime}\right)F\left(t^{\prime}\right)dt^{\prime}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{m\omega}\int_{-\infty}^{t}\sin\omega\left(t-t^{\prime}\right)F\left(t^{\prime}\right)dt^{\prime} \ \ \ \ \ (25)$

Note that the upper limit of the integral becomes ${t}$ in the last line, since ${x\left(t,t^{\prime}\right)=0}$ for ${t^{\prime}>t}$.

A physical interpretation of this solution is that we’re adding up the contributions from the forcing function that occurred before the specified time ${t}$. Clearly what the forcing function ${F\left(t^{\prime}\right)}$ does in the future (that is, for ${t^{\prime}>t}$) can’t affect the position ${x\left(t\right)}$ at the current time ${t}$, but all values of ${F\left(t^{\prime}\right)}$ in the past do affect the oscillator’s current position.

A Green’s function that is zero before a certain point, as our oscillator Green’s function is, is called a retarded Green’s function.

# Klein-Gordon solutions from harmonic oscillator

Michael E. Peskin & Daniel V. Schroeder, An Introduction to Quantum Field Theory, (Perseus Books, 1995) – Chapter 2.

Another way of deriving the Klein-Gordon field and its conjugate momentum density is by drawing an analogy with the harmonic oscillator. The idea is to start with the classical Klein-Gordon field ${\phi\left(\mathbf{x},t\right)}$ and expand it by using a Fourier transform:

$\displaystyle \phi\left(\mathbf{x},t\right)=\frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\phi\left(\mathbf{p},t\right) \ \ \ \ \ (1)$

Since ${\phi\left(\mathbf{x},t\right)}$ is a classical field, it must be real, which we can ensure by requiring that ${\phi^*\left(\mathbf{p},t\right)=\phi\left(-\mathbf{p},t\right)}$. Plugging this into the Klein-Gordon equation gives

 $\displaystyle \partial_{\mu}\partial^{\mu}\phi\left(\mathbf{x},t\right)+m^{2}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;\left[\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}+m^{2}\right]e^{i\mathbf{p\cdot x}}\phi\left(\mathbf{p},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\left[\frac{\partial^{2}}{\partial t^{2}}+\left|\mathbf{p}\right|^{2}+m^{2}\right]\phi\left(\mathbf{p},t\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

Since this must be true for all ${\phi\left(\mathbf{p},t\right)}$, the integrand must be zero so we get

$\displaystyle \left[\frac{\partial^{2}}{\partial t^{2}}+\left|\mathbf{p}\right|^{2}+m^{2}\right]\phi\left(\mathbf{p},t\right)=0 \ \ \ \ \ (5)$

The classical equation of motion for a harmonic oscillator in one dimension is

 $\displaystyle F$ $\displaystyle =$ $\displaystyle -kx\ \ \ \ \ (6)$ $\displaystyle m_{0}\frac{d^{2}x}{dt^{2}}+kx$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

The solution is an oscillator with frequency

$\displaystyle \omega=\sqrt{\frac{k}{m_{0}}} \ \ \ \ \ (8)$

The corresponding Hamiltonian is

$\displaystyle H=\frac{p^{2}}{2m_{0}}+\frac{1}{2}kx^{2} \ \ \ \ \ (9)$

Comparing this with 5, we see that the Klein-Gordon equation in momentum space has the same form as a harmonic oscillator with ${x}$ replaced by ${\mathbf{p}}$, ${k=\left|\mathbf{p}\right|^{2}+m^{2}}$ and ${m_{0}=1}$, so its solution is an oscillator with frequency

$\displaystyle \omega_{\mathbf{p}}=\sqrt{\left|\mathbf{p}\right|^{2}+m^{2}} \ \ \ \ \ (10)$

In the algebraic solution of the harmonic oscillator in non-relativistic quantum mechanics, we introduced the raising and lowering operators (renaming ${a_{+}=a^{\dagger}}$ and ${a_{-}=a}$ to be consistent with P&S’s notation):

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m_{0}\omega}}\left[-ip+m_{0}\omega x\right]\ \ \ \ \ (11)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m_{0}\omega}}\left[ip+m_{0}\omega x\right] \ \ \ \ \ (12)$

In natural units ${\hbar=1}$ and with ${m_{0}=1}$ this gives

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left[-ip+\omega x\right]\ \ \ \ \ (13)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left[ip+\omega x\right] \ \ \ \ \ (14)$

which can be inverted to give

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega}}\left(a+a^{\dagger}\right)\ \ \ \ \ (15)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle -i\sqrt{\frac{\omega}{2}}\left(a-a^{\dagger}\right) \ \ \ \ \ (16)$

Using the commutation relation

$\displaystyle \left[x,p\right]=i \ \ \ \ \ (17)$

we get

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (18)$

and the Hamiltonian comes out to

$\displaystyle H_{SHO}=\omega\left(a^{\dagger}a+\frac{1}{2}\right) \ \ \ \ \ (19)$

The effects of ${a}$ and ${a^{\dagger}}$ are to lower and raise a state by an energy quantum ${\omega}$. The wave functions are

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (20)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (21)$

To apply all this to the quantum Klein-Gordon equation, we want to interpret ${x}$ in 15 as a quantum field ${\phi\left(\mathbf{p}\right)}$, and leave ${p}$ in 16 as it is. The field ${\phi\left(\mathbf{x}\right)}$ in position space is the integral 1 over ${\phi\left(\mathbf{p}\right)}$ in momentum space, with each momentum ${\mathbf{p}}$ contributing its own raising and lowering operators ${a_{\mathbf{p}}}$ and ${a_{-\mathbf{p}}^{\dagger}}$ from 13 and 14. That is, we get

 $\displaystyle \phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (22)$ $\displaystyle \pi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\right)^{3}}\int d^{3}p\;e^{i\mathbf{p\cdot x}}\sqrt{\frac{\omega_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}-a_{-\mathbf{p}}^{\dagger}\right) \ \ \ \ \ (23)$

We can invert these equations to get expressions for ${a}$ and ${a^{\dagger}}$. We multiply both sides by ${e^{-i\mathbf{p^{\prime}\cdot x}}}$ and integrate over ${d^{3}x}$:

 $\displaystyle \int d^{3}xe^{-i\mathbf{p^{\prime}\cdot x}}\phi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle \int d^{3}p\left(\frac{1}{\left(2\pi\right)^{3}}\int d^{3}xe^{i\mathbf{\left(p-p^{\prime}\right)\cdot x}}\right)\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}p\delta\left(\mathbf{p-p}^{\prime}\right)\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\left(a_{\mathbf{p}}+a_{-\mathbf{p}}^{\dagger}\right)\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\omega_{\mathbf{p}^{\prime}}}}\left(a_{\mathbf{p}^{\prime}}+a_{-\mathbf{p}^{\prime}}^{\dagger}\right)\ \ \ \ \ (26)$ $\displaystyle \int d^{3}xe^{-i\mathbf{p^{\prime}\cdot x}}\pi\left(\mathbf{x}\right)$ $\displaystyle =$ $\displaystyle -i\sqrt{\frac{\omega_{\mathbf{p}^{\prime}}}{2}}\left(a_{\mathbf{p}^{\prime}}-a_{-\mathbf{p}^{\prime}}^{\dagger}\right) \ \ \ \ \ (27)$

Dropping the primes on ${\mathbf{p}^{\prime}}$ and solving for ${a}$ and ${a^{\dagger}}$ we get (since ${\omega_{\mathbf{p}}=\omega_{-\mathbf{p}}}$):

 $\displaystyle a_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \int d^{3}x\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}e^{-i\mathbf{p\cdot x}}\left[i\pi\left(\mathbf{x}\right)+\omega_{\mathbf{p}}\phi\left(\mathbf{x}\right)\right]\ \ \ \ \ (28)$ $\displaystyle a_{\mathbf{p}}^{\dagger}$ $\displaystyle =$ $\displaystyle \int d^{3}x\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}e^{i\mathbf{p\cdot x}}\left[-i\pi\left(\mathbf{x}\right)+\omega_{\mathbf{p}}\phi\left(\mathbf{x}\right)\right] \ \ \ \ \ (29)$

# Creation and annihilation operators: commutators and anticommutators

References: Amitabha Lahiri & P. B. Pal, A First Book of Quantum Field Theory, Second Edition (Alpha Science International, 2004) – Chapter 1, Problems 1.1 – 1.2.

As a bit of background to the quantum field theoretic use of creation and annihilation operators we’ll look again at the harmonic oscillator. The creation and annihilation operators (called raising and lowering operators by Griffiths) are defined in terms of the position and momentum operators as

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[-ip+m\omega x\right]\ \ \ \ \ (1)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[ip+m\omega x\right] \ \ \ \ \ (2)$

From the commutator ${\left[x,p\right]=i\hbar}$ we can work out

 $\displaystyle \left[a,a^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2\hbar m\omega}\left(-im\omega\left[x,p\right]\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (4)$

The annihilation operator ${a}$ acting on the vacuum or ground state ${\left|0\right\rangle }$ gives 0, and the creation operator ${a^{\dagger}}$ produces a state ${a^{\dagger}\left|0\right\rangle =\left|1\right\rangle }$ with energy eigenvalue ${\frac{3}{2}\hbar\omega}$. Successive applications of ${a^{\dagger}}$ produce states with higher energy, where each quantum of energy is ${\hbar\omega}$.

Normalization

Given that the ground state is normalized so that ${\left\langle \left.0\right|0\right\rangle =1}$, we can find the factor required to normalize higher states so that ${\left\langle \left.n\right|n\right\rangle =1}$. Consider ${n=2}$. We have

$\displaystyle a^{\dagger}a^{\dagger}\left|0\right\rangle =A\left|2\right\rangle \ \ \ \ \ (5)$

where ${A}$ is to be determined. We have

 $\displaystyle \left\langle 0\left|aaa^{\dagger}a^{\dagger}\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a\left(1+a^{\dagger}a\right)a^{\dagger}\right|0\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|aa^{\dagger}\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}aa^{\dagger}\right|0\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|\left(1+a^{\dagger}a\right)\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}\left(1+a^{\dagger}a\right)\right|0\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle 0\left|aa^{\dagger}\right|0\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle 0\left|\left(1+a^{\dagger}a\right)\right|0\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \left.0\right|0\right\rangle +\left\langle \left.0\right|0\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{A^{2}}\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}} \ \ \ \ \ (14)$

For ${n=3}$ we get ${\left\langle 0\left|aaaa^{\dagger}a^{\dagger}a^{\dagger}\right|0\right\rangle }$. We need to commute each ${a}$ through the ${a^{\dagger}}$ operators to its right. The first ${a}$ will generate the factor ${\left(1+a^{\dagger}a\right)}$ 3 times as it commutes with each ${a^{\dagger}}$ operator. Each of these terms will be ${\left\langle 0\left|aaa^{\dagger}a^{\dagger}\right|0\right\rangle }$ and we already know that this term produces a factor of 2. Therefore

$\displaystyle \left\langle 0\left|aaaa^{\dagger}a^{\dagger}a^{\dagger}\right|0\right\rangle =3\times2=6 \ \ \ \ \ (15)$

We can extend this result to the general case:

$\displaystyle \left\langle 0\left|a^{n}\left(a^{\dagger}\right)^{n}\right|0\right\rangle =n! \ \ \ \ \ (16)$

The normalization must then be

$\displaystyle \left|n\right\rangle =\frac{1}{\sqrt{n!}}\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (17)$

Number operator

We’ve met the number operator ${N}$ in the field case, but there is an analogous operator for the harmonic oscillator. We have

$\displaystyle N\equiv a^{\dagger}a \ \ \ \ \ (18)$

As with the field case, we can work out its commutators:

 $\displaystyle \left[N,a^{\dagger}\right]$ $\displaystyle =$ $\displaystyle a^{\dagger}aa^{\dagger}-a^{\dagger}a^{\dagger}a\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}a^{\dagger}a+a^{\dagger}-a^{\dagger}a^{\dagger}a\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}\ \ \ \ \ (21)$ $\displaystyle \left[N,a\right]$ $\displaystyle =$ $\displaystyle a^{\dagger}aa-aa^{\dagger}a\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}aa-a+a^{\dagger}aa\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a \ \ \ \ \ (24)$

Applying this to ${\left|n\right\rangle }$ we get

$\displaystyle N\left|n\right\rangle =\frac{1}{\sqrt{n!}}N\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (25)$

We get

 $\displaystyle N\left(a^{\dagger}\right)^{n}$ $\displaystyle =$ $\displaystyle \left[a^{\dagger}+a^{\dagger}N\right]\left(a^{\dagger}\right)^{n-1}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{2}\left(1+N\right)\left(a^{\dagger}\right)^{n-2}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ldots\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{n}N\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\left(a^{\dagger}\right)^{n}+\left(a^{\dagger}\right)^{n}a^{\dagger}a \ \ \ \ \ (30)$

When operating on ${\left|0\right\rangle }$, the last term gives 0, so

$\displaystyle N\left|n\right\rangle =\frac{n}{\sqrt{n!}}\left(a^{\dagger}\right)^{n}\left|0\right\rangle \ \ \ \ \ (31)$

Multiple oscillators

If we now have a system of ${N}$ non-interacting harmonic oscillators with equal masses and frequencies ${\omega_{i}}$, ${i=1,\ldots,N}$, the Hamiltonian is

$\displaystyle H=\frac{1}{2m}\sum_{i}\left(p_{i}^{2}+m^{2}\omega_{i}^{2}x_{i}^{2}\right) \ \ \ \ \ (32)$

Since the oscillators are not coupled, the creation and annihilation operators for different operators all commute, so that

$\displaystyle \left[a_{i},a_{j}^{\dagger}\right]=\delta_{ij} \ \ \ \ \ (33)$

so the normalized state where oscillator ${i}$ is in the ${n_{i}}$th excited state is

$\displaystyle \left|n_{1}n_{2}\ldots n_{N}\right\rangle =\prod_{i=1}^{N}\frac{\left(a_{i}^{\dagger}\right)^{n_{i}}}{\sqrt{n_{i}!}}\left|0\right\rangle \ \ \ \ \ (34)$

The number operator in this case is

$\displaystyle \mathcal{N}=\sum_{i=1}^{N}\left(a_{i}^{\dagger}a_{i}\right) \ \ \ \ \ (35)$

This works because the commutation relation 33 allows each term ${a_{i}^{\dagger}a_{i}}$ in the sum to pick out the number of quanta of oscillator ${i}$.

Anticommutators

Now suppose that instead of the commutation relations 33 we have anticommutation relations as follows:

 $\displaystyle \left\{ a_{i},a_{j}^{\dagger}\right\}$ $\displaystyle \equiv$ $\displaystyle a_{i}a_{j}+a_{j}a_{i}=\delta_{ij}\ \ \ \ \ (36)$ $\displaystyle \left\{ a_{i}^{\dagger},a_{j}^{\dagger}\right\}$ $\displaystyle =$ $\displaystyle \left\{ a_{i},a_{j}\right\} =0 \ \ \ \ \ (37)$

If we start with the vacuum state ${\left|0\right\rangle }$ and require ${a_{i}^{\dagger}\left|0\right\rangle =\left|0\ldots1_{i}\ldots0\right\rangle }$ (that is, ${a_{i}^{\dagger}}$ creates one quantum in category ${i}$), then if we try to create another quantum in the same state, we get

 $\displaystyle \left\langle 0\left|a_{i}a_{i}a_{i}^{\dagger}a_{i}^{\dagger}\right|0\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}\left(1-a_{i}^{\dagger}a_{i}\right)a_{i}^{\dagger}\right|0\right\rangle \ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}a_{i}a_{i}^{\dagger}\right|0\right\rangle \ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}\left(1-a_{i}^{\dagger}a_{i}\right)\right|0\right\rangle \ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle -\left\langle 0\left|a_{i}a_{i}^{\dagger}\right|0\right\rangle +\left\langle 0\left|a_{i}a_{i}^{\dagger}a_{i}^{\dagger}a_{i}\right|0\right\rangle \ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (42)$

Thus, attempting to create two quanta in the same state produces zero, so at most one quantum can occupy each state. The commutator case 33 thus behaves like bosons and the anticommutator case like fermions.

# Harmonic oscillator: statistics

Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers

Required physics: Schrödinger equation, probability density

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.9.

Suppose a particle is in the quantum state

$\displaystyle \Psi\left(x,t\right)=Ae^{-amx^{2}/\hbar}e^{-iat} \ \ \ \ \ (1)$

where ${A}$ is the normalization constant and ${a}$ is a constant with dimensions of 1/time. We can find ${A}$ from normalization:

 $\displaystyle \int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\int_{-\infty}^{\infty}e^{-2amx^{2}/\hbar}dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\sqrt{\frac{\pi\hbar}{2ma}}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \left(\frac{2ma}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (5)$

The spatial component of the wave function is

$\displaystyle \psi\left(x\right)=\left(\frac{2ma}{\pi\hbar}\right)^{1/4}e^{-amx^{2}/\hbar} \ \ \ \ \ (6)$

and it must satisfy the time-independent Schrödinger equation in one dimension

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$ $\displaystyle =$ $\displaystyle E\psi(x) \ \ \ \ \ (7)$

The energy ${E}$ can be found from the time equation:

$\displaystyle i\hbar\frac{\partial\Xi}{\partial t}=E\Xi \ \ \ \ \ (8)$

where

$\displaystyle \Xi\left(t\right)=e^{-iat} \ \ \ \ \ (9)$

Therefore

$\displaystyle E=\hbar a \ \ \ \ \ (10)$

From 7 we have

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}$ $\displaystyle =$ $\displaystyle \left(\frac{2ma}{\pi\hbar}\right)^{1/4}a\left(\hbar-2amx^{2}\right)e^{-amx^{2}/\hbar}\ \ \ \ \ (11)$ $\displaystyle V\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{E\psi(x)+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}}{\psi\left(x\right)}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2ma^{2}x^{2} \ \ \ \ \ (13)$

This is the harmonic oscillator potential, and the wave function is actually the ground state of that potential.

We can work out a few average values:

$\displaystyle \left\langle x\right\rangle =0 \ \ \ \ \ (14)$

since ${\psi\left(x\right)}$ is even.

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi^{2}dx=\frac{\hbar}{4am}\ \ \ \ \ (15)$ $\displaystyle \left\langle p\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-\infty}^{\infty}\psi\frac{\partial\psi}{\partial x}dx=0\ \ \ \ \ (16)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-\infty}^{\infty}\psi\frac{\partial^{2}\psi}{\partial x^{2}}dx=\hbar ma \ \ \ \ \ (17)$

The standard deviations are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\frac{1}{2}\sqrt{\frac{\hbar}{ma}}\ \ \ \ \ (18)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\hbar ma} \ \ \ \ \ (19)$

and the uncertainty principle is

$\displaystyle \sigma_{x}\sigma_{p}=\frac{\hbar}{2} \ \ \ \ \ (20)$

so in this case, the uncertainty is the minimum possible.

# Adiabatic approximation: higher order corrections

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.10.

In deriving the adiabatic theorem, Griffiths (in his section 10.1) shows that the solution to the time-dependent Schrödinger equation can be written as

$\displaystyle \Psi\left(x,t\right)=\sum_{n}c_{n}\left(t\right)\psi_{n}\left(x,t\right)e^{i\theta_{n}\left(t\right)} \ \ \ \ \ (1)$

where the ${\psi_{n}}$ form an orthonormal set of functions that are eigenfunctions of the Hamiltonian at a particular instant of time, and ${\theta_{n}}$ is the dynamic phase. The coefficients ${c_{n}}$ are the usual weighting factors, and they depend only on time.

Later in the derivation, he arrives at a differential equation for the ${c_{m}}$:

$\displaystyle \dot{c}_{m}\left(t\right)=-\sum_{j}c_{j}\left\langle \psi_{m}\left|\dot{\psi}_{j}\right.\right\rangle e^{i\left(\theta_{j}-\theta_{m}\right)} \ \ \ \ \ (2)$

In the adiabatic approximation, this equation has the approximate solution

 $\displaystyle c_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle c_{m}\left(0\right)e^{i\gamma_{m}\left(t\right)}\ \ \ \ \ (3)$ $\displaystyle \gamma_{m}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{m}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{m}\left(t'\right)\right.\right\rangle dt' \ \ \ \ \ (4)$

where ${\gamma_{m}}$ is the geometric phase. In particular, if the system starts in a definite eigenstate ${\psi_{n}}$ then ${c_{m}\left(0\right)=\delta_{nm}}$ so

$\displaystyle c_{m}\left(t\right)=\delta_{nm}e^{i\gamma_{n}\left(t\right)} \ \ \ \ \ (5)$

with the result that the overall solution becomes

$\displaystyle \Psi_{n}\left(x,t\right)=\psi_{n}\left(x,t\right)e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)} \ \ \ \ \ (6)$

that is, the system stays in the ${n^{th}}$ state over time, although its phase can change.

We can extend the adiabatic approximation recursively by using the first approximation 5 to generate the next approximation. We can do this by inserting 5 into 2 and then solving the resulting differential equation. The sum in 2 is reduced to a single term where ${j=n}$, the eigenstate in which the system starts at ${t=0}$.

 $\displaystyle \dot{c}_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle -e^{i\gamma_{n}\left(t\right)}\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}\ \ \ \ \ (7)$ $\displaystyle c_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle c_{m}\left(0\right)-\int_{0}^{t}e^{i\gamma_{n}\left(t\right)}\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}dt' \ \ \ \ \ (8)$

This correction to the basic adiabatic approximation now has the ability to predict transitions from the initial state ${\psi_{n}}$ to other states ${\psi_{m}}$ where ${m\ne n}$. We can apply this to the forced oscillator, where we found that in the adiabatic approximation

 $\displaystyle \psi_{n}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \psi_{n}\left(x-f\right)\ \ \ \ \ (9)$ $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2\hbar}\int_{0}^{t}f^{2}\left(t'\right)dt'-\left(n+\frac{1}{2}\right)\omega t\ \ \ \ \ (10)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\dot{f}}{\hbar}\left(x-\frac{f}{2}\right)\approx0 \ \ \ \ \ (11)$

Here, ${m\omega^{2}f\left(t\right)}$ is the forcing term, and the adiabatic approximation is obtained by assuming that ${f}$ changes very slowly, or to be precise:

$\displaystyle \left|\dot{f}\left(t\right)\right|\ll\omega\left|f\left(t\right)\right| \ \ \ \ \ (12)$

To work out the correction, we need to find ${\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle }$ in 8. We can do this using the raising and lowering operators for the harmonic oscillator. In particular, the momentum operator can be written in terms of them as

$\displaystyle p=i\sqrt{\frac{\hbar m\omega}{2}}\left(a_{+}-a_{-}\right) \ \ \ \ \ (13)$

Also, recall that the effects of ${a_{\pm}}$ are

 $\displaystyle a_{+}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\psi_{n}\ \ \ \ \ (14)$ $\displaystyle a_{-}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n}\psi_{n-1} \ \ \ \ \ (15)$

How does this help us? We need to find the derivative ${\partial\psi_{n}\left(x-f\right)/\partial t'}$, so we get, defining ${z\equiv x-f}$:

 $\displaystyle \frac{\partial\psi_{n}\left(x-f\right)}{\partial t'}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi_{n}\left(z\right)}{\partial z}\frac{\partial z}{\partial t'}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\psi_{n}\left(z\right)}{\partial z}\dot{f}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\psi_{n}}{\partial x}\dot{f} \ \ \ \ \ (18)$

where the last line follows because ${z=x-f}$ and ${f}$ doesn’t depend on ${x}$.

Now the momentum operator is

$\displaystyle p=\frac{\hbar}{i}\frac{\partial}{\partial x} \ \ \ \ \ (19)$

so our derivative is

 $\displaystyle \frac{\partial\psi_{n}\left(x-f\right)}{\partial t'}$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\dot{f}p\psi_{n}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\left(a_{+}-a_{-}\right)\psi_{n}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\left(\sqrt{n+1}\psi_{n+1}-\sqrt{n}\psi_{n-1}\right) \ \ \ \ \ (22)$

Using the orthonormality of the ${\psi_{m}}$ we have

 $\displaystyle \left\langle \psi_{n+1}\left|\dot{\psi}_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n+1}\ \ \ \ \ (23)$ $\displaystyle \left\langle \psi_{n-1}\left|\dot{\psi}_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle -\dot{f}\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n} \ \ \ \ \ (24)$

with all other matrix elements equal to zero.

Returning to 8 we can work out the phase terms from 10 and 11.

 $\displaystyle \gamma_{n}$ $\displaystyle \approx$ $\displaystyle 0\ \ \ \ \ (25)$ $\displaystyle \theta_{n}-\theta_{n+1}$ $\displaystyle =$ $\displaystyle \omega t\ \ \ \ \ (26)$ $\displaystyle \theta_{n}-\theta_{n-1}$ $\displaystyle =$ $\displaystyle -\omega t \ \ \ \ \ (27)$

Therefore, since ${c_{n+1}\left(0\right)=c_{n-1}\left(0\right)=0}$,

 $\displaystyle c_{n+1}\left(t\right)$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n+1}\int_{0}^{t}\dot{f}e^{i\omega t'}dt'\ \ \ \ \ (28)$ $\displaystyle c_{n-1}\left(t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}\sqrt{n}\int_{0}^{t}\dot{f}e^{-i\omega t'}dt' \ \ \ \ \ (29)$

[These answers aren’t the same as those given in Griffiths’s question (although the square moduli are the same) but I can’t see anything wrong with my derivation. Comments welcome.]

Note that

$\displaystyle \left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle =0 \ \ \ \ \ (30)$

so 8 predicts that ${c_{n}\left(t\right)=c_{n}\left(0\right)=1}$, thus the sum of the square moduli of the ${c_{m}}$s is greater than 1. However, these values for the ${c_{m}}$s are correct only to first order in ${\dot{f}}$. To get the second order corrections, we’d need to insert 28 and 29 back into 2 and integrate again to get new values for the ${c_{m}}$s, which would give ${c_{n}\left(t\right)<1}$ for ${t>0}$. The process can be continued as long as we like, giving an adiabatic series.

# 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.