The Variational Principle

I’m currently working on writing an animation package for visualizing quantum circuits. I had hoped it would have been finished sooner so I could complete Part 2 of visualizing entanglement. Until then, I’d like to talk about something I learned in a lecture that may be one of the most useful results in quantum mechanics, while also being embarrassingly simple. It’s called the Variational Principle and is the basis for how variational quantum algorithms work, specifically the variational quantum eigensolver (VQE).

I’ll boldly state the result here and build up to it as we go on. Pick any normalized wave function $\psi$ whatsoever. Then

$$E_{\text{gs}}\le ⟨\psi |\hat{H}|\psi ⟩\equiv ⟨\hat{H}⟩$$

which says the expectation value of the Hamiltonian $\hat{H}$ with any arbitrary wave function $\psi$ will always overestimate the ground state energy of $\hat{H}$. Of course, if $\psi$ just happens to be the ground state wave function, then the above equation is an equivalence. If $\psi$ happens to be any of the excited states, then of course $E_{\text{gs}}\le ⟨\hat{H}⟩$. The point is that this overestimation holds for any $\psi$ whatsoever.

The proof is actually quite simple.

Proof: Since the (unknown) eigenstates of $\hat{H}$ are orthonormal and linearly independent (i.e. $⟨{\psi }_i|{\psi }_j⟩={\delta }_{ij}$) for any two eigenstates, they form a complete set. Therefore, we can express any arbitrary $\psi$ as a linear combination of them:

$$\psi =\sum _n{c}_n{\psi }_n,\text{with }\hat{H}{\psi }_n=E_n{\psi }_n,$$

Since $\psi$ is normalized,

$$1=⟨\psi |\psi ⟩=⟨\sum _m{c}_m{\psi }_m|\sum _n{c}_n{\psi }_n⟩$$ $$=\sum _m\sum _n{c}_m^{\ast }{c}_n⟨{\psi }_m|{\psi }_n⟩=\sum _n|c_n{|}^2.$$

(because any two eigenstates are orthonormal: $⟨{\psi }_m|{\psi }_n⟩={\delta }_{mn}$).

Calculating $⟨\hat{H}⟩$,

$$⟨\hat{H}⟩=⟨\sum _m{c}_m{\psi }_m|\hat{H}|\sum _n{c}_n{\psi }_n⟩=⟨c_m{\psi }_m|\hat{H}\sum _n{c}_n{\psi }_n⟩$$ $$=\sum _m\sum _n{c}_m^{\ast }{E}_n{c}_n⟨{\psi }_m|{\psi }_n⟩=\sum _n{E}_n|c_n{|}^2.$$

But the ground state energy $E_{\text{gs}}$ is, by definition, the lowest eigenvalue, $E_{\text{gs}}\le E_n\mathrm{\forall }n$. Hence,

$$⟨\hat{H}⟩=\sum _n{E}_n|c_n{|}^2\ge E_{\text{gs}}\sum _n|c_n{|}^2=E_{\text{gs}}$$

$$⟨\hat{H}⟩\ge E_{\text{gs}}.$$

Helium

I think the best example to showcase the power of the Variational Principle is calculating the ground state energy of Helium.

Despite Helium being the second simplest elemental Hamiltonian, consisting of only two electrons in orbit around a nucleus containing two protons (with some neutrons), its Schrödinger equation is actually unsolvable.

$$\text{The Helium Atom}.$$

Like any many-body system, its differential equation is impossible to solve exactly. When astrodynamicists calculate trajectories of satellites traveling to, say, Jupiter, they would have to account for the gravitational attraction of the Sun, Earth, and Jupiter at a minimum to be exact. However, there exists no solvable equation to account for the pull of a three-body system. They instead discretize the system, calculating the Earth-satellite system when the satellite is near Earth, and the Jupiter-satellite system when the satellite is near Jupiter. (Helium electrons are always close together, unfortunately, so you can’t really do this.) Elsewhere, they employ numerical techniques and possibly some perturbation theory.

The Hamiltonian for Helium (ignoring fine structure) is given by

$$\hat{H}=-\frac{{\hslash }^2}{2m}({\mathrm{\nabla }}_1^2+{\mathrm{\nabla }}_2^2)-\frac{e^2}{4\pi {ϵ}_0}(\frac{2}{r_1}+\frac{2}{r_2}-\frac{1}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}).$$

The ground state of Helium has been measured experimentally to be $$E_{\text{gs}}\approx -78.975\text{ eV (experimental)}.$$

This is the quantity we’ll do our best to reproduce mathematically.

The issue with trying to solve this Hamiltonian comes from the electron-electron repulsion potential,

$$V_{ee}=\frac{e^2}{4\pi {ϵ}_0}\frac{1}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}.$$

We could employ time-independent perturbation theory using $\widehat{H^{\prime }}=V_{ee}$. However, because the $V_{ee}$ is not a small perturbation, this approximation would be quite far off.

If we just ignore $V_{ee}$, the Helium Hamiltonian simplifies into two hydrogenic Hamiltonians

$$\hat{H}=-(\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}_1^2-\frac{e^2}{4\pi {ϵ}_0}\frac{2}{r_1})-(\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}_2^2-\frac{e^2}{4\pi {ϵ}_0}\frac{2}{r_2})$$.

with a nuclear charge of $2e$ instead of $e$. The exact solution is just the product of hydrogenic wave functions:

$${\psi }_0(\overrightarrow{r_1},\overrightarrow{r_2})\equiv {\psi }_{100}(\overrightarrow{r_1}){\psi }_{100}(\overrightarrow{r_2})=\frac{8}{\pi a^3}{e}^{-2(r_1+r_2)/a},$$

and the energy is $8E_1\approx -109$ eV, which is very far from $-78.975$ eV. Let’s instead use the Variational Principle with the hydrogenic ${\psi }_0$ as the ansatz wave function.

As we’ll see soon, the closer the ansatz is to the actual ground state wave function, the better an approximation the Variational Principle gives. So this ansatz make sense since it’s an eigenfunction for most of the Hamiltonian:

$$\hat{H}{\psi }_0=(8E_1+V_{ee}){\psi }_0.$$

Hydrogenic Ansatz

The Variational Principle tells us $E_{\text{gs}}\le ⟨\hat{H}⟩$. Hence, with the hydrogenic ansatz, we are tasked with solving

$$⟨\hat{H}⟩=8E_1+⟨V_{ee}⟩.$$

where $$⟨V_{ee}⟩=⟨{\psi }_0|\frac{e^2}{4\pi {ϵ}_0}\frac{1}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}|{\psi }_0⟩.$$ $$=\left(\frac{e^2}{4\pi {ϵ}_0}\right){\left(\frac{8}{\pi a^3}\right)}^2\int \frac{e^{-4(r_1+r_2)/a}}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}{d}^3\overrightarrow{r_1}{d}^3\overrightarrow{r_2}.$$

Solving the $\overrightarrow{r_2}$ integral first, we align $\overrightarrow{r_1}$ to be along the polar axis in the $\overrightarrow{r_2}$ frame.

By the law of cosines, $|\overrightarrow{r_1}-\overrightarrow{r_2}|=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_2}$. Hence,

$$I_2\equiv \int \frac{e^{-4r_2/a}}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}{d}^3{\overrightarrow{r}}_2=\int \frac{e^{-4r_2/a}}{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_2}}{r}_2^{2}d{\theta }_{2}d{\phi }_2$$

The integrand has no ${\phi }_2$ dependence, so the its contribution is just $2\pi$. The ${\theta }_2$ integral can be calculated,

$${\int }_0^{\pi }\frac{\sin {\theta }_2}{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_2}}d{\theta }_2=\frac{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_2}}{r_1{r}_2}{|}_0^{\pi }$$ $$=\frac{1}{r_1{r}_2}[(r_1+r_2)-|r_1-r_2|]=\{\begin{array}{l}2/r_1,r_2<r_1,\\ 2/r_2,r_2>r_1\end{array}.$$

Therefore,

$$I_2=4\pi (\frac{1}{r_1}{\int }_0^{r_1}{e}^{-4r_2/a}{r}_2^{2}d{r}_2+{\int }_{r_1}^{\mathrm{\infty }}{e}^{-4r_2/a}{r}_{2}d{r}_2)$$ $$=\frac{\pi a^3}{8r_1}[1-(1+\frac{2r_1}{a})e^{-4r_1/a}].$$

The $r_1$ part of the integral is still left. Finally solving for $V_{ee}$,

$$⟨V_{ee}⟩=\left(\frac{e^2}{4\pi {ϵ}_0}\right)\left(\frac{8}{\pi a^3}\right)\int [1-(1+\frac{2r_1}{a})e^{-4r_1/a}]e^{-4r_1/a}{r}_1\sin {\theta }_{1}d{r}_{1}d{\theta }_{1}d{\phi }_1$$

$$=\left(\frac{e^2}{4\pi {ϵ}_0}\right)\left(\frac{8}{\pi a^3}\right)\cdot 4\pi {\int }_0^{\mathrm{\infty }}[r{e}^{-4r/a}-(r+\frac{2r^2}{a})e^{-8r/a}]dr$$

$$=\frac{5}{4a}\left(\frac{e^2}{4\pi {ϵ}_0}\right)=-\frac{5}{2}{E}_1=34\text{ eV}.$$

And hence,

$$⟨\hat{H}⟩=-109\text{ eV}+34\text{ eV}=-75\text{ eV}.$$

Given the experimental value is $\approx -79$ eV, we’re already only 5% off from solving an unsolvable equation. Let’s keep going!

To beat this approximation, we have to find a better ansatz wave function. The closer the ansatz is to the intrinsic Helium ground state, the better.

Helium is a system wherein electrons ($-e$) not only get pulled by the nuclear charge $(Z=+2e)$, but also feel a repulsion between each other. This repulsion acts against the inward pull from the nucleus. In this manner, we can think of each electron partially “shielding” the nucleus from the other by, making the net effective nuclear charge $Z$ slightly less than 2. This suggests a ansatz of the form

$${\psi }_1(\overrightarrow{r_1},\overrightarrow{r_2})\equiv \frac{Z^3}{\pi a^3}{e}^{-Z(r_1+r_2)/a},$$

where we treat $Z$ as a variational parameter, rather than equating it to 2. This equation is an eigenstate of a hydrogenic Hamiltonian, only with $Z$ this time instead of 2. Thus, $\hat{H}$ is evidently

$$\hat{H}=-\frac{{\hslash }^2}{2m}({\mathrm{\nabla }}_1^2+{\mathrm{\nabla }}_2^2)-\frac{e^2}{4\pi {ϵ}_0}(\frac{Z}{r_1}+\frac{Z}{r_2})$$

$$+\frac{e^2}{4\pi {ϵ}_0}(\frac{Z-2}{r_1}+\frac{Z-2}{r_2}+\frac{1}{|\overrightarrow{r_1}-\overrightarrow{r_2}|}).$$

We’ll solve for $⟨\hat{H}⟩$ similarly which will give us an expression involving an arbitary Z.

$$⟨\hat{H}⟩=2Z^2{E}_1+2(Z-2)\left(\frac{e^2}{4\pi {ϵ}_0}\right)⟨\frac{1}{r}⟩+⟨V_{ee}⟩.$$

For hydrogenic wave functions with nuclear charge $Z$ , the expectation value $⟨\frac{1}{r}⟩=\frac{Z}{a}$ ($a$ being the Bohr radius). The expectation value $⟨V_{ee}⟩$ is the exact same as before, just with $Z$ instead of 2. Because the Bohr radius also scaled inversely with $Z$, we also adjust $a\mapsto \frac{2}{Z}a$.

$$⟨V_{ee}⟩=\frac{5Z}{8a}\left(\frac{e^2}{4\pi {ϵ}_0}\right)=-\frac{5Z}{4}{E}_1.$$

Hence,

$$⟨\hat{H}⟩=[2Z^2-4Z(Z-2)-\frac{5}{4}Z]E_1=[-2Z^2+\frac{27}{4}Z]E_1.$$

We’re getting close now. Because $E_{\text{gs}}\le ⟨\hat{H}⟩$, that is, the value above exceeds $E_{\text{gs}}$ for any value of $Z$, the closest value to $E_{\text{gs}}$ occurs when $⟨\hat{H}⟩$ is minimized.

$$\frac{d}{dZ}⟨\hat{H}⟩=[-4Z+\frac{27}{4}]E_1=0.$$

It directly follows that

$$Z=\frac{27}{16}=1.6875.$$

So the electrons in Helium shield the nucleus by roughly 16%. Plugging this value of $Z$ back into $⟨\hat{H}⟩$,

$$⟨\hat{H}⟩=\frac{1}{2}{\left(\frac{3}{2}\right)}^6{E}_1=-77.5\text{ eV}.$$

This is within 2% of Helium’s actual ground state.

While this was a bit of math, it does showcase the power of the Variational Principle and how easy it is to use, albeit the complicated integrals. This procedure is the basis for quantum algorithms like VQE, which quantum chemists use to approximate the ground state energies for any system. VQE generates its ansatzs using parameterized quantum circuits. In this way, the Variational Principle turns into a machine learning problem. By continuously adjusting the parameters of the quantum circuit using gradient descent on the system’s Hilbert space, more accurate ansatzs are formed, which yield better ground state energy approximations.

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