# Why is Coupled Cluster not variational?

It has been noted in several sources (e.g. J. Romero et al. Strategies for quantum computing molecular energies using the unitary coupled cluster ansatz. arXiv:1701.02691 [quant-ph]) that one of the disadvantages of the Coupled Cluster method for the description of electron correlation is that, since $$e^T$$ in $$E_\mathrm{CC} = \langle0|e^{-T}He^T|0\rangle$$ isn't unitary, the method isn't variational (the energy obtained by solving CC equations isn't a rigorous upper bound to the exact energy).

Could someone please explain to me the connection between non-unitarity of $$e^T$$ and the method not being variational.

# Variational coupled cluster (vCC)

First, you can make a variational CC with a non-unitary T operator in the following way.

Recall the variational principle (taken from Wikipedia):

$$\varepsilon[\Psi] = \frac{\langle\Psi| \hat{H} |\Psi\rangle}{\langle\Psi|\Psi\rangle}$$ The variational principle states that

• $$\varepsilon \geq E_{0}$$, where $$E_{0}$$ is the lowest energy eigenstate (ground state) of the hamiltonian
• $$\varepsilon = E_{0}$$ if and only if $$\Psi$$ is exactly equal to the wave function of the ground state of the studied system.

So all you have to do to make a variational CC is set $$|\Psi\rangle = e^{\hat{T}}|0\rangle$$ in the above expression. This is called "variational coupled cluster" or "vCC", because it literally uses the "variational principle" shown above.

# Unitary coupled cluster (uCC)

Unitary transformations preserve eigenvalues, so if $$\hat{U}$$ is a unitary transformation (unlike $$e^{\hat{T}}$$), then $$UHU^\dagger$$ has the same eigenvalues as $$H$$. Since the energies are eigenvalues of $$H$$, unitary coupled cluster can recover the correct energies with the unitary operator $$U = e^{\hat{T} - \hat{T}^{\dagger}}$$. For a long time people thought vCC and uCC were just two ways of writing the same thing, but this was shown to be wrong very recently (G. Harsha et al. On the difference between variational and unitary coupled cluster theories. arXiv:1711.00579 [cond-mat.str-el]; J. Chem. Phys. 2018, 148 (4), 044107.).

# Something that a lot of people don't know

Regular CC is actually variational in the full correlation limit. This means that for $$\ce{Li2}$$ for example, CCSDTQPH is variational. This is because $$\ce{Li2}$$ has $$6$$ electrons, so CCSDTQPH is FCI, and FCI is variational. CCSDTQP and CCSDTQ and even CCSDT(Q) are not likely to overshoot the variational limit either since they are very good approximations of CCSDTQPH for $$\ce{Li2}$$.

• Thanks! Despite not being variational, does regular CC have any advantages over other types of CC? – GingerBadger Jul 9 '18 at 11:18
• In regular CC the wavefunction is not normalised, how can it obey the variational principle if you don't renormalise? The last paragraph also neglects that this can only be true at the complete basis set limit. – Martin - マーチン Jul 9 '18 at 17:04
• @Martin-マーチン: You are wrong. CC(N) for N electrons is variational within any basis set. Just like CI(N), also known as FCI. – user1271772 Jul 10 '18 at 21:05
• That is why I suggested a Q&A style. Just ask a question like "Why are CC(N) and CI(N) variational in any basis set?" and post the corresponding answer there. Then post the link as a comment to this answer; and you'll earn quite a bit of reputation along the way. And if anything gets moved to chat, then the discussion would or could continue there. But for the benefit of many more people, I would prefer the Q&A style. – Martin - マーチン Jul 11 '18 at 10:34
• Do we have to do the Q/A style again where I ask "What is the point of regular CC?" The point is that CC(2) = CCSD contains a bit of the higher-order excitations (T,Q,P, etc.) thanks to the $e^T$ operator. Therefore CC(N) converges MUCH faster with respect to N than CI(N). This is given in the table comparing CC(N) to CI(N) in that paper I gave you. Can you see it? I do not have journal access but I remember the table. That paper is one of the most important papers in CC, which is why it has more than 500 citations. – user1271772 Jul 11 '18 at 11:17