# Why is the ground state energy of a linearly perturbed quantum oscillator always lower than its harmonic counterpart?

I am concerned with a QHO that is linearly perturbed in $$x$$, i.e. $$H = \hbar \omega \left(\hat{n} + \frac{1}{2}\right) + \lambda \underbrace{\left(\hat{b}+ \hat{b}^\dagger \right)}_{\propto \hat{x}}.$$ For $$\lambda = 0$$, the ground state energy is of course $$E_0 = \frac{\hbar \omega}{2}$$. For non-zero $$\lambda$$, we can absorb the $$\lambda \hat{x}$$ perturbation into the $$\hat{x}^2$$ term of the unperturbed Hamiltonian, leading to an energy shift $$\propto -\lambda^2$$; see e.g. here for details.

In particular, this means that the ground state energy of the perturbed system is always lower than $$\frac{\hbar \omega}{2}$$.

Mathematically, sure; trivial, actually. But physically, this goes against my intuition—how does shifting an axis, or putting the potential on a slope, always decrease the ground state energy, regardless of which way I shift it (i.e. regardless of the sign of $$\lambda$$)?

Specifically, in the context of the Holstein model or a (basic) Franck-Condon excitation, this implies that exciting an electron makes vibrational states accessible with a lower energy than was accessible when the electron was not excited. It seems paradoxical that the vibrational ground state energy would always, for any parameters, be lower in the excited-electron manifold. Are other cases (where e.g. the lowest-energy vibrational state is actually associated with the lowest-energy electronic state) described by including higher-order perturbation terms?

• Feb 26, 2020 at 2:15
• I agree that that question is relevant, in a certain sense it's a generalization of this question. However, I think that march's answer below gives a different perspective than the answers given there, which is what I was hoping for. Feb 26, 2020 at 2:29

Physically, the perturbation is a constant force pointing in one direction or the other. This is because if $$V(x) = \alpha x$$, then the force is given by $$F_x=-\frac{d}{dx}V(x) = -\alpha.$$ One of the effects of a constant force on a harmonic oscillator is to shift the equilibrium position in the direction of the force. This is also the direction of decreasing $$V(x)$$, and so one consequence of this shift is that the total energy at the minimum is also decreased.

Now, this decrease in minimum energy (corresponding to the ground-state energy being smaller than $$\hbar\omega/2$$) is irrelevant, because the spacing between harmonic oscillator levels doesn't change: it's still $$\hbar\omega$$. Therefore, we still need to add the same amount of energy to add a vibrational excitation.

• Another similar (classical) argument: with an additional $\alpha x$ energy, the energy at $x = 0$ stays the same. But because the harmonic potential is flat at $x=0$, it is always possible to reduce slightly the energy by moving in the direction in which $\alpha x$ decreases, and since $\alpha x$ is the dominant term around $x=0$, this will lead to a decrease of energy. Feb 25, 2020 at 23:48
• @QuantumApple. I actually think that your argument is not another argument: instead, it's a necessary part that I left out. I thought about adding the fact that the energy decreases because the bottom of the potential is flat, and so the linear term wins for small deviations from $x=0$, but I didn't! Feb 25, 2020 at 23:52
• I agree that there is nothing inherently quantum about this! After working through the classical problem, I found this point which was crucial to my understanding: After applying a force $-a$, the classical "ground" state is shifted to the phase space point $(x=x_0 - \frac{a}{2k}, p=0)$ (e.g. via classical Hamilton's function). The energy of this state is $-\frac{a^2}{4k}$. The work that is performed by the external force is precisely $\int_{x_0}^{x_0 - \frac{a}{2k}} kx \mathrm{d}x = \frac{a^2}{4k}$—in other words, the energy is decreased (classically) because the external force performs work. Feb 26, 2020 at 1:59
• OK, corrected version, since I can't edit the other comment: The classical lowest-energy phase space point is $(x = x_0 - \frac{a}{k}, p=0)$. The energy of this point is $-\frac{a^2}{2k}$. The work performed by the external force is $\int_{x_0}^{x_0 - \frac{a}{k}} k(x - x_0)\mathrm{d} x = \frac{a^2}{2k}$. Also, this decrease in the ground state energy does actually make a difference when there is a transition between the two settings, e.g. in a Franck-Condon excitation. Then this ground-state shift enables a vibrational relaxation from the unperturbed ground state into the perturbed one. Feb 26, 2020 at 2:21

I'm going to expand on march's idea, and specifically my own comments underneath his answer, which don't represent the whole picture. This should all be basic textbook stuff, but that's basically what it was supposed to end up as!

$$V(x)$$ has two components: $$V(x) = \frac{1}{2} kx^2 + \lambda x$$. If we plot these two simple functions, it becomes evident that for any $$\lambda$$ (or $$a$$), there is a region close to the origin in which the linear term is larger than the quadratic term (this is basically QuantumApple's point):

Heuristically, the new ground state will be located where the linear potential maximally counteracts the quadratic potential, which is indicated by the arrow in the diagram. If we consider the blue curve to represent the negative of the perturbation, $$-\lambda x$$, then we can rephrase the true extremality condition in terms of our heuristic as $$\frac{\mathrm{d}}{\mathrm{d}x} \left[\frac{1}{2} kx^2 - (-\lambda x)\right] = 0$$. This point of maximal counteraction is located at $$x=-\frac{\lambda}{k}$$. This result is, of course, the same result as obtained by minimizing the classical Hamilton function, or shifting the axis in the Hamiltonian/Hamilton function as $$x \mapsto x + \frac{\lambda}{k}$$ by completing the square.

Whatever the sign of $$\lambda$$, the near-0 region where this holds true will always exist, either left or right of the origin.

Now, this is already (IMO) a more intuitive picture, and additionally it should be valid in either the QM or classical case. However, by going fully classical, it is possible to take another perspective:

Assuming we know, one way or the other, that there is a new minimum at $$x=-\frac{\lambda}{k} = -\frac{a}{k}$$, we can look at the balance of forces and the energy/work that gets us there. First assume the classical HO (e.g. as a mass on a spring) is initially at its equilibrium position $$x=0$$, and at rest, $$p=0$$. This is the classical "ground" state.

Now we introduce the linear perturbation in the energy, which is classically equivalent to a constant force $$F = -\frac{\mathrm{d}}{\mathrm{d}x} ax = -a$$. This force shall displace the spring-loaded mass from $$x=0$$ to our newfound minimum at $$x=-\frac{a}{k}$$. While doing so, it performs work equal to $$\int_{x=0}^{-\frac{a}{k}} F(x) \mathrm{d} x = - \int_{x=0}^{-\frac{a}{k}} a \mathrm{d} x = \frac{a^2}{k}.$$ However, if we look at the total energy that the mass should have at its new energy minimum, it should be at $$H(x,p) = -\frac{a^2}{2k}$$, i.e. only half this energy. What's going on?

Well, the external force $$F = -a$$ has to counteract the "internal" force of the spring (equivalently: the total force acting on $$m$$ is the sum of the two forces). Counteracting the harmonic force requires work equal to: $$\int_{x=0}^{-\frac{a}{k}} \left( -kx\right) \mathrm{d} x = -\frac{a^2}{2k}.$$ Note that this counteraction principle is exactly what I described in terms of potentials above.

Therefore, the net work that is performed in moving the particle from its old to its new "ground" state position is equal to $$\frac{a^2}{k} -\frac{a^2}{2k} = \frac{a^2}{2k}$$. Correspondingly, the total internal energy change should be (due to conservation of energy), and is, $$-\frac{a^2}{2k}$$—which evidently arises from the fact that the work that can be performed by the external, constant force exceeds the counteracting harmonic force (i.e. $$\frac{a^2}{k} > \frac{a^2}{2k}$$)!

May this be helpful to some soul in the future.

The key point is that the perturbation is completely off diagonal in the energy basis. (An energy-diagonal perturbation can shift the ground-state energy up or down without restriction.) If you then calculate the expectation value of the energy in the unperturbed ground state, $$\langle\psi^{(0)}_{g}|H|\psi^{(0)}_{g}\rangle$$, it is unchanged.

That expectation value is a weighted sum of the energies of some collection of perturbed eigenstates. This weighted sum cannot be smaller than the smallest perturbed eigenvalue. That means that, among the modified energy eigenstates, there must be at least one with a lower energy than the unperturbed ground state. In other words, the ground state energy goes down.

Another way to think of this to remember that any perturbation that mixes two (unperturbed) energy eigenstates actually pushes them apart in energy. Since the initial ground state can only be mixed with higher-energy states, it's energy is always pushed downward.

• My inner mathematician likes this argument because it has proof-like character, although I wouldn't say it speaks to my non-mathematical everyday intuition. Concerning level repulsion: This system actually has the interesting twist, as march noted, that the level spacing stays the same, so that principle only half works here as far as I can tell. Also, I don't quite understand the way you phrased your very last sentence...? Feb 26, 2020 at 1:29