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The LCAO Method

A simple technique for estimating the ground-state energy of a many-atom system involves using the variational principle with a linear combination of single-atom wavefunctions. While sound in principle, a calculation using this method on the hydrogen molecule ion converges very slowly on the correct binding energy, for the reason that single-atom wavefunctions are not sufficiently dense in the region of overlap.

Introduction

The Variational Principle

The variational principle is a simple and powerful method for establishing an upper bound on the ground-state energy of a quantum system. Given any wavefunction $\psi$, we may write $\psi$ as a linear combination of eigenfunctions $\psi _{i}$ of the Hamiltonian:

$\psi =\sum _{i}c_{i}\cdot \psi _{i}$

(where, to meet the requirement of normalization, we require $\sum _{i}c_{i}^{2}=1$). Then, if we take the inner product of the wavefunction with the Hamiltonian $H$, we find

$\langle \psi |H|\psi \rangle =\sum _{i}c_{i}^{2}\langle \psi _{i}|H|\psi _{i}\rangle$

Letting $E_{i}\equiv \langle \psi |H|\psi \rangle$ be the energy associated with each eigenfunction, and letting $\psi _{0}$ be the ground state, this becomes

$\langle \psi |H|\psi \rangle =\sum _{i}c_{i}^{2}E_{i}>\sum _{i}c_{i}^{2}E_{0}=E_{0}$

Or, more succinctly,

$\langle \psi |H|\psi \rangle >E_{0}$

If we have a large family of wavefunctions, we can calculate $\langle \psi |H|\psi \rangle$ for each one of them; with some amount of luck, one member of the family will be close to the actual ground-state wavefunction, and we'll have a good upper estimate for $E_{0}$.

The LCAO Technique

The LCAO (Linear Combination of Atomic Orbitals) technique uses the variational principle together with a specific family of wavefunctions to find ground-state energies. Specifically, for many-atom systems, each electron will be influenced by a potential

$V=\sum _{i}{\frac {Z_{i}e^{2}}{4\pi \epsilon _{0}r_{i}}}+(other\;electrons)$

Where $Z_{i}$ refers to the number of protons in each atom and $r_{i}$ refers to the distance between the electron and each atom. Regardless of other electrons, if we consider locations very close to a particular atomic center (say, $r_{i}\ll 1$ for a particular choice of $i$), we find that the non-relativistic Hamiltonian is

$H=-{\frac {\hbar ^{2}}{2m_{e}}}\nabla ^{2}-{\frac {Z_{i}e^{2}}{4\pi \epsilon _{0}}}\cdot {\frac {1}{r_{i}}}+{\mathcal {O}}(1)$

To most significant order, then, this looks exactly like the Hamiltonian for the Hydrogen atom with a charge $Ze$ instead of $1e$ at the center. As we come closer and closer, the agreement will become better and better as ${\frac {1}{r_{i}}}$ dominates over the rest of the potential terms in the Hamiltonian. If $\psi _{znlm}(r,\theta ,\phi )$ are the wavefunctions for Hydrogen with $z$ protons in the center nucleus, then we may well expect that a linear superposition of ground states, namely,

$\psi =\sum _{i}a_{i}\psi _{Z_{i}100}(r_{i},\theta _{i},\phi _{i})$

(for some constants $a_{i}$) will match the real ground-state wavefunction to most significant order.

What about higher-order corrections? While the linear combination of ground states is physically motivated, the LCAO technique for higher-order terms is more mathematically motivated. There will be some wave-function which is the ground state of the system; but any wave-function can be decomposed into sums of the eigenfunctions of the Hamiltonian for Hydrogen. Hence, no matter what the real ground-state wave-function looks like near the ith nucleus, we can decompose it into a sum of Hydrogen wave-functions. As this is true near any of the nuclei, we might expect that higher-order corrections to the potential will be reasonably matched by including higher-order wavefunctions of Hydrogen in our linear combination:

$\psi =\sum _{i,n,l,m}a_{inlm}\psi _{Z_{i}nlm}(r_{i},\theta _{i},\phi _{i})$

But in general, in the limit of including the entire set of eigenfunctions, we will match the ground-state wavefunction. Hence, if the infinite vector space of functions in this family is denoted by $\Psi$, we find that

$E_{0}=\min _{\psi \in \Psi }(\langle \psi |H|\psi \rangle )$

A Computation Using LCAO

Background

How well does this work in practice, however? In order to find out, we consider the simplest possible non-Hydrogen system, namely a molecular Hydrogen ion ($\mathrm {H} _{2}^{+}$). For this system, the Hamiltonian is

$H=-{\frac {\hbar ^{2}}{2m_{e}}}\nabla ^{2}-{\frac {e^{2}}{4\pi \epsilon _{0}}}\left({\frac {1}{r_{1}}}+{\frac {1}{r_{2}}}\right)$

where $r_{1}$ and $r_{2}$ are the distances from the electron to each proton. If we adopt natural units of length corresponding to

$L={\frac {4\pi \epsilon _{0}}{e^{2}}}\cdot {\frac {\hbar ^{2}}{2m_{e}}}$

(about 0.26 Angstrom) and natural units of energy corresponding to

$E=\left({\frac {e^{2}}{4\pi \epsilon _{0}}}\right)^{2}\cdot {\frac {2m_{e}}{\hbar }}$

(about 54.4 eV), the Hamiltonian reduces to the very simple form

$H=-\nabla ^{2}-{\frac {1}{r_{1}}}-{\frac {1}{r_{2}}}$

If we place both protons on the $z$-axis, separated by a distance $d$, we can write the Hamiltonian in cylindrical coordinates, where it becomes

$H=-{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial }{\partial r}}\right)-{\frac {\partial ^{2}}{\partial z^{2}}}-{\frac {1}{r^{2}}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}-{\frac {1}{\sqrt {r^{2}+(d/2-z)^{2}}}}-{\frac {1}{\sqrt {r^{2}+(d/2+z)^{2}}}}$

Written in this way, it becomes explicit that the ground-state energy depends on the separation distance $d$. Hence, without knowing the value of $d$ beforehand, we will have to calculate it as well:

$E_{0}=\min _{d\in \mathbf {R} ^{+};\;\psi \in \Psi }(\langle \psi |H(d)|\psi \rangle )$

To reduce the parameter space to a finite computational problem, we impose a cutoff on $n$, the highest order of the radial wave-functions of hydrogen which we consider. Additionally, we can exploit the symmetries of the Hamiltonian to note that the ground state must satisfy two conditions: $\psi _{0}(r,z,\phi )=\psi _{0}(r,z,0)$ (polar symmetry), and $\psi _{0}(r,z)=\psi _{0}(r,-z)$ (mirror symmetry). Variation from these conditions implies a less-than-minimal value of $-\nabla ^{2}$, which in turn implies a larger value for the energy of the wavefunction.

These two conditions lead to the following restrictions on the coefficients $a_{inlm}$:

$a_{in\ell {}m}={\begin{cases}0&{\mbox{if }}m\neq 0\\a_{(2-i)n\ell {}m}&{\mbox{if }}\ell \equiv 0{\pmod {2}}\\-a_{(2-i)n\ell {}m}&{\mbox{if }}\ell \equiv 1{\pmod {2}}\end{cases}}$

The latter two come from the fact that the Legendre polynomials in the wavefunctions of Hydrogen are odd if $\ell$ is odd, and they are even if $\ell$ is even. Hence, while we expect the lowest-order guess for the wavefunction to be of the form

$\psi _{a}=a_{1100}(\psi _{100}(r_{1},z_{1})+\psi _{100}(r_{2},z_{2}))$

The term involving the odd wavefunction $\psi _{210}$ will be antisymmetric:

$\psi _{b}=a_{1210}(\psi _{210}(r_{1},z_{1})-\psi _{210}(r_{2},z_{2}))$

Finally, one final computational aid may be realized. As the wavefunctions used are eigenfunctions of most of the Hamiltonian---except for the potential from the extra proton, we find that

$H\psi _{nlm}(r_{1},z_{1})=\left(E_{n}-{\frac {1}{r_{2}}}\right)\psi _{nlm}(r_{1},z_{1})$, and likewise,

$H\psi _{nlm}(r_{2},z_{2})=\left(E_{n}-{\frac {1}{r_{1}}}\right)\psi _{nlm}(r_{2},z_{2})$, where

$E_{n}=-{\frac {0.25}{n^{2}}}$ is the energy of the $n$th radial state of the Hydrogen atom (in natural units).

With these considerations in effect, we are now ready to perform the variational calculation.

Computational Results

Starting with $d=4$ and $a_{1100}=a_{2100}=1$, we may perform a gradient descent in parameter space to find the lowest energy value of the system. The results of this process are shown here:

Maximal $n$	$d$	$E_{0}$
1	1.32 Å	-1.76 eV
2	1.27 Å	-1.96 eV
3	1.25 Å	-2.07 eV
4	1.24 Å	-2.08 eV

In this table, $E_{0}$ refers to the binding energy of the system, namely, the energy of the electron and both protons relative to the energy of a free proton and a hydrogen atom. The column headed "Maximal $n$" refers to the largest $n$ for which $a_{inlm}$ was taken to be nonzero.

While these values seem to be converging rapidly, this similitude is misleading. The experimentally-determined value of $E_{0}$ is -2.65 eV, with a separation distance of 1.06Å. This is not an effect of relativistic dynamics (as the electron energy is much less than 511 keV) and is not an effect of fine structure or other corrections (which would modify the energy at the hundredth of a percent level). The full energy of the electron predicted from the last entry in the table is -27.2 eV. Comparatively, the full energy from experiment would be -29.6 eV.

Why is the LCAO method so terrible in this case? A naive guess would be because we have stopped at $n=4$. However, it is easy to see that adding further orbitals would result in mostly wasted efforts. The $n$th radial eigenfunction of Hydrogen depends on a characteristic radius $na$, where $a=2$ in the natural units we have assumed. However, as the optimal distance $d$ between the two protons is about 1.06 Å, or 4.0 in the natural units we have adopted, whereas the natural distance for the $n=5$ wavefunctions is 10, it is easy to see why adding them will have little additional effect. Most of the variation in the $n=5$ wavefunctions will happen at a radius beyond the distance between the two protons, whereas most of the interesting deviation in the ground-state wavefunction happens between the two protons. Hence, beyond $n=4$ the Hydrogen wavefunctions represent a particularly feeble basis with which to describe the ground-state wavefunction of the Hydrogen ion molecule, and so converge very poorly to the correct result.

The moral of this report is that, for calculating energies (or wavefunctions) based on the variational principle, the basis functions of the wavefunction family under consideration must quickly converge to arbitrary functions in the region of overlap between any two independent potentials in the Hamiltonian.

Analytical Imaginary Time Propagation at a Single Point

The Imaginary Time Propagation Method (ITP) has long been used as a way to find ground-state wavefunctions. However, if one is only interested in the ground-state energy, a few tricks may be employed to reduce the computational burden. In particular, a method is described whereby for $n$ particles with an arbitrary Coulombic potential, the analytical calculations may be done in a $n$-dimensional space instead of a $3n$-dimensional space.

Introduction to Imaginary Time Propagation

In general, finding the ground-state wavefunction directly from the Hamiltonian for a quantum system is a nightmare, with only a few known exactly-solvable systems. However, the Schrodinger equation yields to a cute mathematical trick, so that it is possible to approximate a ground state wavefunction to arbitrary accuracy. This trick, known as Imaginary Time Propagation (ITP), involves substituting imaginary time instead of real time into the time-dependent Schrodinger equation. This leads to the derivation of an imaginary-time operator which results in exponential decay of all states except for the ground state, which can then be applied to any initial trial wavefunction to compute an approximation to the actual ground state.

As mentioned, we begin with the time-dependent Schrodinger Equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=H\psi$

Equivalently, in terms of the derivative with respect to imaginary time $(i\cdot t)$, we find that

${\frac {\partial \psi }{\partial it}}=-H\psi /\hbar$

To see why this is useful, we first write $\psi$ as a sum of eigenfunctions of the Hamiltonian $H$, where each eigenfunction $\psi _{i}$ has corresponding energy $E_{i}$:

$\psi =\sum _{i}c_{i}\cdot \psi _{i}$

For each eigenfunction, the imaginary-time Schrodinger equation then becomes

${\frac {\partial \psi _{i}}{\partial it}}=-H\psi _{i}/\hbar =-E_{i}\psi _{i}/\hbar$

The solution to this expression is an exponential; if we define $\tau \equiv it/\hbar$, we find that

$\psi _{i}(\tau )=e^{-\tau E_{i}}\psi _{i}(0)=e^{-\tau H}\psi _{i}(0)$

Where $e^{-\tau H}$ is defined as usual by the Taylor expansion

$e^{-\tau H}=\sum _{n=0}^{\infty }{\frac {(-\tau )^{n}H^{n}}{n!}}$

Evidently, when propagated forward in imaginary time, each eigenfunction decays exponentially, with the rate of decay proportional to its energy. Indeed, all the states other than the ground state will die off exponentially quickly even compared to how quickly the ground state is vanishing:

${\frac {\psi _{i}(\tau )}{\psi _{0}(\tau )}}\propto e^{-\tau (E_{i}-E_{0})}$

If we apply the solution for the eigenfunctions to the original wave function, we find

$\psi (\tau )=\sum _{i}c_{i}\cdot \psi _{i}(\tau )=\sum _{i}e^{-\tau E_{i}}c_{i}\psi _{i}(0)$

Hence, in the limit of imaginary time propagation, the coefficients for the scattering part of the wavefunction approach zero and those for bound states approach infinity. Note, however, that the relative proportion of the ground state wave function in the overall wave function always increases as a function of imaginary time because $E_{i}>E_{0}$:

$\lim _{\tau \to \infty }{\frac {\langle \psi (\tau )|c_{0}\psi _{0}(\tau )\rangle }{\langle \psi (\tau )|\psi (\tau )\rangle }}=\lim _{\tau \to \infty }{\frac {c_{0}^{2}e^{-2\tau E_{0}}}{c_{0}^{2}e^{-2\tau E_{0}}+\sum _{i=1}^{\infty }c_{i}^{2}e^{-2\tau E_{i}}}}=1$

Thus, starting with any wavefunction $\psi$, we find that in the limit of large $\tau$, the wavefunction will become proportional to the ground-state wavefunction:

$\psi (\tau )=c_{0}e^{-\tau H}\psi _{0}(0)+{\mathcal {O}}(e^{-\tau (E_{1}-E_{0})})$

And, in particular, we may compute the ground-state wavefunction exactly, as long as the overlap $c_{0}$ between our initial wavefunction and the ground-state wavefunction is nonzero:

$\psi _{0}(0)=\lim _{\tau \to \infty }{\frac {\psi (\tau )}{\sqrt {\langle \psi (\tau )|\psi (\tau )\rangle }}}$

The Ground-State Energy

ITP at a Single Point

Full ITP normally involves numerically propagating a trial wavefunction for many time steps over a large volume in parameter space. If we are only interested in finding the ground-state energy of a system, this seems particularly wasteful. For an eigenfunction $\psi _{i}$ of the Hamiltonian, it is true at any point in parameter space that $H\psi _{0}=E_{0}\psi _{0}$. This suggests a simple method for ITP at a single point. Suppose that in natural units our Hamiltonian for $n$ particles is of the form

$H=V-\partial _{i}^{2}$

Where, as usual, $\partial _{i}^{2}$ is the sum over all (rectilinear) space dimensions of second derivatives:

$\partial _{i}={\frac {\partial }{\partial x^{i}}}$ and $\partial _{i}^{2}=\sum _{i}{\frac {\partial ^{2}}{\partial (x^{i})^{2}}}$

As rectilinear partial derivatives commute with themselves as well as partials with respect to imaginary time ($\partial _{0}$), we may then write down the equation for the imaginary time evolution of the derivatives of $\psi$:

$\partial _{0}\cdot (\partial _{j_{1}}\cdot \ldots \cdot \partial _{j_{n}})\psi =-\partial _{i}^{2}\cdot (\partial _{j_{1}}\cdot \ldots \cdot \partial _{j_{n}})\psi +(\partial _{j_{1}}\cdot \ldots \cdot \partial _{j_{n}})(V\psi )$

Now, if we start with the initial value of $\psi$ at a point and some large number of its derivatives, we can evolve both $\psi$ and its derivatives according to the imaginary-time Schrodinger equation. After some appropriate amount of time, we can then extract the ground-state energy just by applying $H$ to $\psi$ at the particular point we have chosen.

However, this is not especially better than a brute-force solution. Even though derivatives commute, if there are $n$ dimensions, there are ${\frac {(m+1)!}{(n-1)!(m-n+2)!}}\propto (m+1)^{n-1}$ distinct ways to take $m$ derivatives. While we would not have to worry about spatial boundary conditions, we would have to worry about "boundary" conditions on the derivatives which we choose not to propagate. Hence, we have simply mapped the complexity of the problem in a virtually one-to-one manner into a different (but no more useful) form.

A Better Approach

For a better approach, we recall the expression for the ITP of $\psi$, decomposed into eigenfunctions:

$\psi (\tau )=\sum _{i}e^{-\tau H}\cdot c_{i}\psi _{i}(0)=e^{-\tau H}\sum _{i}c_{i}\cdot \psi _{i}(0)=e^{-\tau H}\psi (0)$

Evidently, if we can calculate the effect of the operator $e^{-\tau H}$ on $\psi (0)$, we can immediately recover the ground-state energy:

$E_{0}={\frac {H\psi _{0}}{\psi _{0}}}=\lim _{\tau \to \infty }{\frac {He^{-\tau H}\psi }{e^{-\tau H}\psi }}$

Then, we simply have to calculate a number of terms in the expansion of $e^{-\tau H}$. How many? The answer naturally depends on the initial wave function $\psi$ which we choose and the desired accuracy which we want. Suppose, for instance, we pick a wavefunction which is some (unknown) superposition of bound states and we desire an accuracy of one part in $10^{9}$. We expect that (unless we get extremely unlucky with our choice) we will find

${\frac {|\langle \psi _{i}|\psi \rangle |}{|\langle \psi _{0}|\psi \rangle |}}\ll 10^{5}$

This would suggest that we would like $e^{-\tau (E_{1}-E_{0})}<10^{-15}$, or, in other words, $\tau (E_{1}-E_{0})>\ln(10^{15})\approx 35$. In that case, how fast does the series for $e^{-\tau H}$ converge? If we note that $(H^{n}\psi _{i})/\psi _{i}=E_{i}^{n}$, then

${\frac {\tau ^{n}(H^{n})\psi _{i}}{\psi _{i}}}=\tau ^{n}|E_{i}|^{n}\approx \left({\frac {35E_{i}}{E_{1}-E_{0}}}\right)^{n}$

Thus, for our desired accuracy, we want

${\frac {1}{n!}}\left({\frac {35E_{i}}{E_{1}-E_{0}}}\right)^{n}<10^{-15}$ for $i>0$.

We expect this to be maximum for $i=1$, if the states are ordered in terms of increasing energy. For Hydrogen, for instance, where $3E_{1}=(E_{1}-E_{0})$, this implies that about 56 terms in the expansion are necessary. For Helium, where $3(E_{0}-E_{1})\approx E_{1}$, this would imply the calculation of 320 terms. If one had an estimate of $E_{1}$, one could naturally take away $E_{1}/2$ from the entire energy scale, dramatically reducing the number of terms necessary (38 terms for Hydrogen, 173 terms for Helium).

This, at least, sets a sensible upper bound on the size of the calculation. For any one-dimensional calculation, this method is clearly a fine choice---as each derivative in the Hamiltonian can only be taken one way, calculating each successive term in the expansion only requires that one derivative be taken. However, one-dimensional calculations are hardly impressive. In fact, up to the present time, we have gained no advantages by evaluating $e^{-\tau H}\psi$ at a single point. Since a new Laplacian is applied for each term, we end up taking $2m$ derivatives to calculate $m$ terms, resulting in a computational burden of about $(2m)^{n-1}$ distinct derivatives at the final step.

While one could pick a simple initial wavefunction with simple derivatives (such as $\psi (0)=e^{-r}$), this does not help matters. Observe, for example,

$H^{2}\psi =(-\nabla ^{2}+V)(-\nabla ^{2}+V)\psi =\nabla ^{4}\psi +V^{2}\psi -V\nabla ^{2}\psi -\nabla ^{2}(V\psi )$

The last term is the trouble, for it means that no matter how simple our wavefunction is, we still end up calculating derivatives of $V$ and, by extension, $V^{2}$, $V^{3}$, and so on.

Coulombic Potentials

Nonetheless, with certain potentials, the derivatives of $V^{n}$ can still end up taking extremely simple forms. One case in point is Coulombic potentials, where

$V={\frac {1}{2}}\sum _{i\neq j}{\frac {c_{ij}}{|{\vec {r}}_{1}-{\vec {r}}_{2}|}}$

Most obviously, $\nabla ^{2}V=0$ (except at certain points) which reduces the number of distinct derivatives one must calculate. However, there's a more subtle consideration which is much more important. Consider a two-particle potential, with one particle at the origin: $V={\frac {1}{r}}=(x^{2}+y^{2}+z^{2})^{-1/2}$

If we evaluate $e^{-\tau H}\psi$ at the point $(x,y,z)=(0,0,1)$, then any elements of our answer which involve positive powers of $x$ or $y$ will evaluate to zero, and we can skip calculating them entirely. Indeed, as it turns out, we can skip calculating derivatives with respect to $x$ entirely until the very end of our calculation, because they all take a very simple form. In particular, we can make a simple table of even derivatives, noting that all odd derivatives will evaluate to zero at $x=0$:

${\frac {\partial ^{2}}{\partial x^{2}}}V^{n}{\big |}_{x=0}=(n)(n+2)x^{2}V^{n+4}-nV^{n+2}{\big |}_{x=0}=-nV^{n+2}(x=0)$ ${\frac {\partial ^{4}}{\partial x^{4}}}V^{n}{\big |}_{x=0}=(n)(n+2)(n+4)(n+6)x^{4}V^{n+8}-6n(n+2)(n+4)x^{2}V^{n+6}+3n(n+2)V^{n+4}{\big |}_{x=0}=3n(n+2)V^{n+4}(x=0)$

$\vdots$

${\frac {\partial ^{2k}}{\partial x^{2k}}}V^{n}{\big |}_{x=0}={\frac {(n+2(k-1))!!}{(n-2)!!}}C(k)V^{n+2k}(x=0)$

Where $C(k)$ is a simple function purely depending on $k$: $C(k)={\frac {\partial ^{2k}}{\partial x^{2k}}}\left({\frac {1}{\sqrt {x^{2}+1}}}\right){\Big |}_{x=0}$, and where $m!!$ is the semifactorial: $m!!=(m)(m-2)(m-4)\ldots (4)(2)$ for even $m$, and $m!!=(m)(m-2)(m-4)\ldots (3)(1)$ for odd integers.

Evidently, we can apply this to derivatives with respect to $y$, where we find

${\frac {\partial ^{2k+2\ell }}{\partial x^{2k}y^{2\ell }}}V^{n}{\big |}_{x=0}={\frac {(n+2(k+\ell -1))!!}{(n-2)!!}}C(k)C(\ell )V^{n+2k+2l}(x,y=0)$

Or, in general, if we need to make up $m$ total derivatives this way, we find

$\sum _{k+\ell =m}{\frac {m!}{k!\ell !}}{\frac {\partial ^{2k+2\ell }}{\partial x^{2k}y^{2\ell }}}V^{n}{\big |}_{x,y=0}={\frac {(n+2(k+\ell -1))!!}{(n-2)!!}}D(m)V^{n+2k+2l}(x=0)$

where

$D(m)=\sum _{k+\ell =m}{\frac {m!}{k!\ell !}}C(k)C(\ell )$

If, for whatever reason, we are examining the two-particle system in higher-dimensional spacetime, then further coefficients in the style of $D(m)$ are no harder to calculate:

$E(m)=\sum _{j+k+\ell =m}{\frac {m!}{j!k!\ell !}}C(j)C(k)C(l)=\sum _{j}{\frac {m!}{j!(m-j)!}}C(j)\sum _{k+\ell =m-j}D(k+\ell )$

So that calculations of this type are linear in terms of the number of dimensions involved.

Hence, we may treat this as a purely one-dimensional problem, taking only derivatives with respect to $z$, and then only at the very end taking derivatives with respect to $x$ and $y$ as necessary to round out the Laplacians. That is to say, the effect of adding extra dimensions is a purely combinatorical factor multiplying the final result. Now, of course, this should not be impressive yet---after all, we started with a 1D-potential, namely $V=1/r$.

However, with more particles, this trick---namely, that the derivatives all take the same general form---becomes astoundingly useful. We cannot expect to place all the particles on top of each other, obviously. Nonetheless, in problems where the particle locations are not constrained, we choose to evaluate $e^{-\tau H}\psi$ at a location in parameter space where all the particles are lined up in a straight line along the z-axis. That is to say, $x_{i}=y_{i}=0$ for each particle, so that $(x_{i}-x_{j})=0$ and $y_{i}-y_{j}=0$ for any pair of particles. Then, we find that the only nontrivial derivatives in the expansion of $e^{-\tau H}$---i.e., terms which are not purely powers of the distances between particles---result from derivatives with respect to the particle locations on the z-axis. At the end, we may apply the effects of the extra dimensions all at once---as they cannot result in new terms, and as they only result in multiplication by a simply-calculable number, their computational cost becomes linear in the number of nontrivial derivatives already calculated.

Consequently, we have reduced a $3n$-dimensional problem to a $n$-dimensional problem. Hydrogen, as before, remains a 1D-problem; Helium becomes a 2D problem; Lithium becomes a 3D problem, and so on and so forth.

Future Work

In the previous section, we only considered fixing the locations of particles in two dimensions each. We consider it likely that by (for example) fixing particles to be at a constant spacing along the $z-$axis, it may be possible write all the derivatives along the $z-$ axis in terms of each other---resulting in the derivative of only one dimension being unique. This would be truly satisfying---as energy is a single-dimensional quantity---we would then be at a minimum of work required to calculate it.