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Theme 3 — Postulates and Applications

Charged oscillator in an electric field

Keywords: harmonic oscillator · electric field · translation · polarisability · sudden approximation · wave-function overlap

Exercise 1 : Charged oscillator in an electric field

A particle of mass $m$ and charge $q$ is bound by a harmonic potential of angular frequency $\omega$ . From time $t=0$ onwards, it is subjected to a uniform, constant electric field $\mathcal E$ directed along $x$ . For $t>0$ , its Hamiltonian is

\hat H_{\mathcal E}=\frac{\hat P^2}{2m}+\frac12m\omega^2\hat X^2-q\mathcal E\hat X .

Let $\ell=\sqrt{\hbar/(m\omega)}$ , and let $\phi_n(x)$ denote the eigenfunctions of the oscillator without the field, with energies $\hbar\omega(n+\frac12)$ and $\phi_0(x)=\frac{1}{\pi^{1/4}\sqrt\ell}e^{-x^2/(2\ell^2)}$ . Recall the integral $\int_{\R}e^{-\alpha u^2}du=\sqrt{\pi/\alpha}$ for $\alpha>0$ .

Question 1

By completing the square, show that

\hat H_{\mathcal E}

can be written as the Hamiltonian of an oscillator centred at a position

d

, to be determined, up to an additive constant.

Solution

Write the potential as a square:

\frac12m\omega^2x^2-q\mathcal Ex=\frac12m\omega^2\Bigl(x^2-\frac{2q\mathcal E}{m\omega^2}x\Bigr)=\frac12m\omega^2(x-d)^2-\frac12m\omega^2d^2, \qquad d=\frac{q\mathcal E}{m\omega^2}.

Thus

\hat H_{\mathcal E}=\frac{\hat P^2}{2m}+\frac12m\omega^2(\hat X-d)^2-\frac{q^2\mathcal E^2}{2m\omega^2}.

The electric field does not change the shape of the potential: it shifts its minimum by $d$ , and lowers the bottom of the well by $q^2\mathcal E^2/(2m\omega^2)$ . Classically, $x=d$ is the equilibrium position at which the restoring force $-m\omega^2x$ balances the electric force $q\mathcal E$ .

Question 2

Hence deduce the spectrum and eigenfunctions of

\hat H_{\mathcal E}

Solution

In the stationary equation in the position-space representation,

-\frac{\hbar^2}{2m}\chi"(x)+\frac12m\omega^2(x-d)^2\chi(x)=\Bigl(E+\frac{q^2\mathcal E^2}{2m\omega^2}\Bigr)\chi(x),

the change of variable $u=x-d$ does not alter the second derivative and reduces the equation to that of the oscillator without the field. The solutions are therefore the translated functions $\chi_n(x)=\phi_n(x-d)$ , and the energies are

E_n^{(\mathcal E)}=\hbar\omega\Bigl(n+\frac12\Bigr)-\frac{q^2\mathcal E^2}{2m\omega^2}.

All the levels are lowered by the same amount and remain equally spaced. In the notation of Lesson 4, $\chi_n=T(d)\phi_n$ , where $T(d)=e^{-id\hat P/\hbar}$ is the translation operator.

Question 3

Calculate the mean position of the particle in the new ground state, and the mean dipole moment

\langle q\hat X\rangle

. Hence deduce the polarisability

\alpha_p

, defined by

\langle q\hat X\rangle=\alpha_p\mathcal E

Solution

The density

|\chi_0(x)|^2=|\phi_0(x-d)|^2

is a Gaussian centred at

d

, so

\langle\hat X\rangle=d

. The mean dipole moment is

\langle q\hat X\rangle=qd=\frac{q^2}{m\omega^2}\,\mathcal E, \qquad\alpha_p=\frac{q^2}{m\omega^2}.

The polarisability is the same as in classical mechanics, and does not depend on $\hbar$ . This is a special property of the harmonic oscillator, for which the centre of the wave packet obeys the classical laws exactly. This model of a harmonically bound electron underlies the classical description of the refractive index of transparent media.

Question 4

For

t<0

, there is no field, and the particle is in the ground state

\phi_0

. The field is switched on suddenly at

t=0

, over a time much shorter than

1/\omega

, so that the wave function has no time to change while the field is being switched on. Calculate the probability of finding the particle in the ground state of the new Hamiltonian immediately after the field is switched on.

Solution

Immediately after the field is switched on, the state is still

\phi_0

. According to the Born rule, the probability of finding the particle in the ground state

\chi_0

of the new Hamiltonian is

|\braket{\chi_0}{\phi_0}|^2

. Since both functions are real,

\braket{\chi_0}{\phi_0}=\int_{\R}\phi_0(x-d)\,\phi_0(x)\,dx=\frac{1}{\sqrt\pi\,\ell}\int_{\R}\exp\Bigl[-\frac{(x-d)^2+x^2}{2\ell^2}\Bigr]dx .

We complete the square in the exponent: $(x-d)^2+x^2=2\bigl(x-\frac d2\bigr)^2+\frac{d^2}{2}$ . With $u=x-d/2$ ,

\braket{\chi_0}{\phi_0}=\frac{1}{\sqrt\pi\,\ell}\,e^{-d^2/(4\ell^2)}\int_{\R}e^{-u^2/\ell^2}du=\frac{1}{\sqrt\pi\,\ell}\,e^{-d^2/(4\ell^2)}\sqrt\pi\,\ell=e^{-d^2/(4\ell^2)}.

The required probability is therefore

P_0=e^{-d^2/(2\ell^2)}.

It is close to one if the displacement $d$ is small compared with the width $\ell$ of the ground state: the new ground state then closely resembles the old one. Conversely, if $d\gg\ell$ , the two Gaussians hardly overlap, and the particle is very likely to be found in an excited state of the new Hamiltonian.

Question 5

Calculate the expectation value

\langle\hat H_{\mathcal E}\rangle

immediately after the field is switched on, and compare it with the energy of the new ground state. Then determine the evolution of

\langle\hat X\rangle(t)

for

t>0

, and interpret it.

Solution

In the state

\phi_0

, we have

\langle\hat X\rangle=0

by parity, and the expectation value of the energy of the oscillator without the field is

\hbar\omega/2

. Thus

\langle\hat H_{\mathcal E}\rangle=\frac{\hbar\omega}{2}-q\mathcal E\cdot0=\frac{\hbar\omega}{2}.

The energy of the new ground state is $\frac{\hbar\omega}{2}-\frac12m\omega^2d^2$ : the particle therefore has a mean excess energy $\frac12m\omega^2d^2$ relative to the new ground state. This is exactly the potential energy of a classical particle displaced by $d$ from its new equilibrium position. This excess is distributed over the excited states, consistently with the probability $P_0<1$ .

For the evolution, the potential is quadratic, and Ehrenfest's theorem gives exactly classical equations:

\frac{d\langle\hat X\rangle}{dt}=\frac{\langle\hat P\rangle}{m},\qquad\frac{d\langle\hat P\rangle}{dt}=-m\omega^2\langle\hat X\rangle+q\mathcal E=-m\omega^2\bigl(\langle\hat X\rangle-d\bigr).

With $\langle\hat X\rangle(0)=0$ and $\langle\hat P\rangle(0)=0$ , the solution is

\langle\hat X\rangle(t)=d\,\bigl(1-\cos\omega t\bigr).

The centre of the wave packet oscillates between $0$ and $2d$ about the new equilibrium position $d$ , exactly as would a mass attached to a spring and suddenly subjected to a constant force. Since all the Bohr frequencies of the oscillator are multiples of $\omega$ , this oscillation continues indefinitely without damping in the absence of coupling to the environment.