Displaced Oscillator

A displaced oscillator is a harmonic oscillator whose equilibrium position has been shifted by a linear term in the potential. The curvature is unchanged, so the energy spacing remains $\hbar\omega$, but all eigenfunctions are translated and all energies acquire a constant offset.

The simplest form is

$$\hat H_F = \frac{\hat p^2}{2m} +\frac12m\omega^2\hat x^2 -F\hat x,$$

where $F$ is a constant force. This model is the first exact example behind static forcing, displaced molecular potential surfaces, coherent-state preparation by sudden shifts, and linear couplings to oscillator coordinates.

Helpful background. Quantum Harmonic Oscillator to recognize oscillator energies, lengths, and eigenfunctions.

Completing The Square

Write the potential as

$$V_F(x) = \frac12m\omega^2x^2-Fx.$$

The shifted equilibrium is found by minimizing:

$$\frac{dV_F}{dx} = m\omega^2x-F=0.$$

Thus

$$x_F=\frac{F}{m\omega^2}.$$

Completing the square gives

$$V_F(x) = \frac12m\omega^2(x-x_F)^2 -\frac{F^2}{2m\omega^2}.$$

Therefore

$$\hat H_F = \frac{\hat p^2}{2m} +\frac12m\omega^2(\hat x-x_F)^2 -\frac{F^2}{2m\omega^2}.$$

The linear term has not changed the curvature $m\omega^2$. It has only moved the minimum and lowered the whole parabola by a constant.

Spectrum

Since the shifted coordinate

$$y=x-x_F$$

has the same oscillator curvature, the spectrum is

$$E_n(F) = \hbar\omega \left( n+\frac12 \right) -\frac{F^2}{2m\omega^2}, \qquad n=0,1,2,\ldots$$

The level spacing is unchanged:

$$E_{n+1}(F)-E_n(F)=\hbar\omega.$$

The constant offset matters when comparing two Hamiltonians with different forces or equilibrium positions. It does not change transition frequencies between adjacent levels of one fixed displaced oscillator.

Shifted Wavefunctions

Let $\psi_n^{(0)}(x)$ be the ordinary oscillator eigenfunction centered at the origin. The displaced-oscillator eigenfunctions are

$$\psi_n^{(F)}(x) = \psi_n^{(0)}(x-x_F).$$

In particular, the ground state is

$$\psi_0^{(F)}(x) = \left(\frac{1}{\pi\ell^2}\right)^{1/4} \exp\left[ -\frac{(x-x_F)^2}{2\ell^2} \right],$$

with the same oscillator length

$$\ell=\sqrt{\frac{\hbar}{m\omega}}.$$

The mean position in the shifted ground state is

$$\langle x\rangle_F=x_F,$$

while the position uncertainty is unchanged:

$$\Delta x=\frac{\ell}{\sqrt2}.$$

The oscillator has moved; it has not become wider or narrower.

Translation-Operator View

The unitary translation operator

$$\hat T(x_F) = \exp\left(-\frac{i}{\hbar}x_F\hat p\right)$$

acts in position representation as

$$(\hat T(x_F)\psi)(x)=\psi(x-x_F).$$

Thus the displaced eigenstates are translated ordinary eigenstates:

$$\lvert n;F\rangle = \hat T(x_F)\lvert n\rangle.$$

This also gives a quick way to check expectation values:

$$\langle n;F\vert\hat x\vert n;F\rangle=x_F, \qquad \langle n;F\vert\hat p\vert n;F\rangle=0.$$

The translation is spatial. It should not be confused with changing the oscillator frequency.

Ladder-Operator View

Using

$$\hat x = \frac{\ell}{\sqrt2} \left( \hat a+\hat a^\dagger \right),$$

the Hamiltonian becomes

$$\hat H_F = \hbar\omega \left( \hat a^\dagger\hat a+\frac12 \right) -\frac{F\ell}{\sqrt2} \left( \hat a+\hat a^\dagger \right).$$

Define

$$\eta = \frac{F\ell}{\sqrt2\,\hbar\omega} = \frac{x_F}{\sqrt2\,\ell},$$

and a shifted annihilation operator

$$\hat b=\hat a-\eta.$$

Since $\eta$ is a number,

$$[\hat b,\hat b^\dagger]=1.$$

Substitution gives

$$\hat H_F = \hbar\omega \left( \hat b^\dagger\hat b+\frac12 \right) -\hbar\omega\eta^2.$$

Because

$$\hbar\omega\eta^2 = \frac{F^2}{2m\omega^2},$$

this is the same spectrum found by completing the square.

The shifted ground state satisfies

$$\hat b\lvert0;F\rangle=0,$$

or

$$\hat a\lvert0;F\rangle=\eta\lvert0;F\rangle.$$

Relative to the original oscillator centered at zero, the new ground state is a coherent state with real amplitude $\eta$. Relative to the new Hamiltonian, it is simply the ground state.

Static Shift Versus Dynamical Drive

A static displacement changes the Hamiltonian from $\hat H_0$ to $\hat H_F$. If the force is present from the start, the stationary states are the shifted eigenstates above.

If the force is suddenly switched on while the system is in an eigenstate of the old oscillator, the state is generally not an eigenstate of the new Hamiltonian. The wave packet then oscillates around the new equilibrium. Starting from the old ground state produces coherent-state motion. Starting from an excited number state instead gives a displaced number state, with a moving center but a non-Gaussian density.

If the force is changed slowly enough and the relevant gap remains open, the state can follow the instantaneous shifted eigenstate adiabatically. Time-dependent driving and adiabaticity are dynamical questions; the static page here supplies the exactly solvable Hamiltonian used in those later analyses.

Physical Examples

For a charged oscillator in a uniform static electric field, $F=qE$. The equilibrium shift is

$$x_F=\frac{qE}{m\omega^2},$$

and the energy shift is

$$\Delta E = -\frac{q^2E^2}{2m\omega^2}.$$

This is the oscillator version of an induced-dipole energy. It is a simple model for polarizability, not a full atomic Stark-effect calculation.

In molecular physics, two electronic configurations may have approximately harmonic nuclear potentials with different equilibrium positions. Vibrational wavefunctions on one surface then overlap shifted wavefunctions on another surface. Those overlaps are the beginning of Franck–Condon physics; the detailed spectroscopy belongs in the atoms, molecules, and light volume.

In oscillator-bath and polaron models, linear couplings to oscillator coordinates are often removed by completing the square or by a displacement transformation. Polarons Preview applies the same idea mode by mode, where recoil, convergence, and the bare-particle overlap add new physics.

Common Mistakes

• Treating the linear term as if it changed the level spacing.
• Forgetting the constant energy shift $-F^2/(2m\omega^2)$.
• Shifting the coordinate but forgetting to shift the wavefunction argument.
• Confusing the shifted ground state with an excited state of the shifted Hamiltonian.
• Saying “coherent state” without specifying whether the reference Hamiltonian is the old or shifted oscillator.
• Applying the static displaced-oscillator spectrum directly to a time-dependent driven oscillator.
• Ignoring the sign convention: $V(x)=m\omega^2x^2/2-Fx$ shifts the minimum to $+F/(m\omega^2)$.

Exercises

1. Complete the square for $V(x)=m\omega^2x^2/2-Fx$ and identify the shifted minimum.

Solution

Set

$$x_F=\frac{F}{m\omega^2}.$$

Then

$$\frac12m\omega^2(x-x_F)^2 = \frac12m\omega^2x^2 -m\omega^2x_Fx +\frac12m\omega^2x_F^2.$$

Since $m\omega^2x_F=F$,

$$\frac12m\omega^2x^2-Fx = \frac12m\omega^2(x-x_F)^2 -\frac12m\omega^2x_F^2.$$

Thus

$$V(x) = \frac12m\omega^2(x-x_F)^2 -\frac{F^2}{2m\omega^2}.$$

2. Find the spectrum of $\hat H_F$.

Solution

After completing the square, $\hat H_F$ is an ordinary oscillator in the coordinate $x-x_F$ plus a constant:

$$\hat H_F = \frac{\hat p^2}{2m} +\frac12m\omega^2(\hat x-x_F)^2 -\frac{F^2}{2m\omega^2}.$$

Therefore

$$E_n(F) = \hbar\omega \left( n+\frac12 \right) -\frac{F^2}{2m\omega^2}.$$

3. Show that the shifted ground-state wavefunction has the same uncertainty $\Delta x$ as the unshifted ground state.

Solution

The shifted ground state is

$$\psi_0^{(F)}(x) = \psi_0^{(0)}(x-x_F).$$

A translation changes the mean position but not the width. Explicitly,

$$\langle x\rangle_F=x_F,$$

and

$$\langle (x-x_F)^2\rangle_F=\frac{\ell^2}{2}.$$

Thus

$$\Delta x=\frac{\ell}{\sqrt2},$$

the same as for the unshifted oscillator.

4. An oscillator starts in the ground state $\lvert0\rangle$ of $\hat H_0$. A constant force $F$ is switched on suddenly, so the state is unchanged during the switch. Find the probabilities of measuring the new energies $E_n(F)$ and the mean energy immediately afterward. What is the probability of the new ground state when $x_F=\sqrt2\,\ell$?

Solution

The old vacuum obeys $\hat a\lvert0\rangle=0$, so the shifted annihilation operator gives

$$\hat b\lvert0\rangle=-\eta\lvert0\rangle, \qquad \eta=\frac{x_F}{\sqrt2\,\ell}.$$

It is therefore a coherent state with amplitude $-\eta$ in the new oscillator basis. Using the number expansion of a coherent state,

$$\lvert0\rangle =e^{-\eta^2/2}\sum_{n=0}^{\infty} \frac{(-\eta)^n}{\sqrt{n!}}\lvert n;F\rangle, \qquad P_F(n)=e^{-\eta^2}\frac{\eta^{2n}}{n!}.$$

This Poisson distribution has mean occupation $\eta^2$. Keeping the constant offset in the new energies gives

$$\langle H_F\rangle =\hbar\omega\left(\eta^2+\frac12\right) -\hbar\omega\eta^2 =\frac12\hbar\omega.$$

This also follows directly from $\langle0|\hat H_F|0\rangle=\langle0|\hat H_0|0\rangle-F\langle0|\hat x|0\rangle$. The unchanged mean energy does not imply that the state is the new ground state: its mean excitation energy above $E_0(F)$ is $\hbar\omega\eta^2$. At $x_F=\sqrt2\,\ell$, $\eta=1$ and $P_F(0)=e^{-1}\approx0.368$. A sudden change of equilibrium does not instantly translate the wavefunction to the new minimum.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. R. Klauder and B.-S. Skagerstam, Coherent States, World Scientific, 1985.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• B. Zwiebach, “Quantum Dynamics”, 8.05 Quantum Physics II, MIT OpenCourseWare, Fall 2013, §4.4.