path: root/content/know/concept/ritz-method/index.pdc

---
title: "Ritz method"
firstLetter: "R"
publishDate: 2022-09-18
categories:
- Physics
- Mathematics
- Perturbation
- Quantum mechanics
- Numerical methods

date: 2022-09-10T13:33:07+02:00
draft: false
markup: pandoc
---

# Ritz method

In various branches of physics,
the **Ritz method** is a technique to approximately find the lowest solutions to an eigenvalue problem.
Some call it the **Rayleigh-Ritz method**, the **Ritz-Galerkin method**,
or simply the **variational method**.



## Background

In the context of [variational calculus](/know/concept/calculus-of-variations/),
consider the following functional to be optimized:

$$\begin{aligned}
    R[u]
    = \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x}
\end{aligned}$$

Where $u(x) \in \mathbb{C}$ is the unknown function,
and $p(x), q(x) \in \mathbb{R}$ are given.
In addition, $S$ is the norm of $u$, which we demand be constant
with respect to a weight function $w(x) \in \mathbb{R}$:

$$\begin{aligned}
    S
    = \int_a^b w(x) \big|u(x)\big|^2 \dd{x}
\end{aligned}$$

To handle this normalization requirement,
we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $\lambda$,
and define the Lagrangian $\Lambda$ for the full constrained optimization problem as:

$$\begin{aligned}
    \Lambda
    \equiv \frac{1}{S} \bigg( \big( p |u_x|^2 - q |u|^2 \big) - \lambda \big( w |u|^2 \big) \bigg)
\end{aligned}$$

The resulting Euler-Lagrange equation is then calculated in the standard way, yielding:

$$\begin{aligned}
    0
    &= \pdv{\Lambda}{u^*} - \dv{x} \Big( \pdv{\Lambda}{u_x^*} \Big)
    \\
    &= - \frac{1}{S} \bigg( q u + \lambda w u + \dv{x} \big( p u_x \big) \bigg)
\end{aligned}$$

Which is clearly satisfied if and only if the following equation is fulfilled:

$$\begin{aligned}
    \dv{x} \big( p u_x \big) + q u
    = - \lambda w u
\end{aligned}$$

This has the familiar form of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/) (SLP),
with $\lambda$ representing an eigenvalue.
SLPs have useful properties, but before we can take advantage of those,
we need to handle an important detail: the boundary conditions (BCs) on $u$.
The above equation is only a valid SLP for certain BCs,
as seen in the derivation of Sturm-Liouville theory.

Let us return to the definition of $R[u]$,
and integrate it by parts:

$$\begin{aligned}
    R[u]
    &= \frac{1}{S} \int_a^b p u_x u_x^* - q u u^* \dd{x}
    \\
    &= \frac{1}{S} \Big[ p u_x u^* \Big]_a^b - \frac{1}{S} \int_a^b \dv{x} \Big(p u_x\Big) u^* + q u u^* \dd{x}
\end{aligned}$$

The boundary term vanishes for a subset of the BCs that make a valid SLP,
including Dirichlet BCs $u(a) = u(b) = 0$, Neumann BCs $u_x(a) = u_x(b) = 0$, and periodic BCs.
Therefore, we assume that this term does indeed vanish,
such that we can use Sturm-Liouville theory later:

$$\begin{aligned}
    R[u]
    &= - \frac{1}{S} \int_a^b \bigg( \dv{x} \Big(p u_x\Big) + q u \bigg) u^* \dd{x}
    \equiv - \frac{1}{S} \int_a^b u^* \hat{H} u \dd{x}
\end{aligned}$$

Where $\hat{H}$ is the self-adjoint Sturm-Liouville operator.
Because the constrained Euler-Lagrange equation is now an SLP,
we know that it has an infinite number of real discrete eigenvalues $\lambda_n$ with a lower bound,
corresponding to mutually orthogonal eigenfunctions $u_n(x)$.

To understand the significance of this result,
suppose we have solved the SLP,
and now insert one of the eigenfunctions $u_n$ into $R$:

$$\begin{aligned}
    R[u_n]
    &= - \frac{1}{S_n} \int_a^b u_n^* \hat{H} u_n \dd{x}
    = \frac{1}{S_n} \int_a^b u_n^* \lambda_n w u_n \dd{x}
    \\
    &= \frac{1}{S_n} \lambda_n \int_a^b w |u_n|^2 \dd{x}
    = \frac{S_n}{S_n} \lambda_n
\end{aligned}$$

Where $S_n$ is the normalization of $u_n$.
In other words, when given $u_n$,
the functional $R$ yields the corresponding eigenvalue $\lambda_n$:

$$\begin{aligned}
    \boxed{
        R[u_n]
        = \lambda_n
    }
\end{aligned}$$

This powerful result was not at all clear from $R$'s initial definition.



## Justification

But what if we do not know the eigenfunctions? Is $R$ still useful?
Yes, as we shall see. Suppose we make an educated guess $u(x)$
for the ground state (i.e. lowest-eigenvalue) solution $u_0(x)$:

$$\begin{aligned}
    u(x)
    = u_0(x) + \sum_{n = 1}^\infty c_n u_n(x)
\end{aligned}$$

Here, we are using the fact that the eigenfunctions of an SLP form a complete set,
so our (known) guess $u$ can be expanded in the true (unknown) eigenfunctions $u_n$.
We are assuming that $u$ is already quite close to its target $u_0$,
such that the (unknown) expansion coefficients $c_n$ are small;
specifically $|c_n|^2 \ll 1$.
Let us start from what we know:

$$\begin{aligned}
    \boxed{
        R[u]
        = - \frac{\displaystyle\int u^* \hat{H} u \dd{x}}{\displaystyle\int u^* w u \dd{x}}
    }
\end{aligned}$$

This quantity is known as the **Rayleigh quotient**.
Inserting our ansatz $u$,
and using that the true $u_n$ have corresponding eigenvalues $\lambda_n$:

$$\begin{aligned}
    R[u]
    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \: \hat{H} \Big\{ u_0 + \sum_n c_n u_n \Big\} \dd{x}}
    {\displaystyle\int w \Big( u_0 + \sum_n c_n u_n \Big) \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \dd{x}}
    \\
    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( \!-\! \lambda_0 w u_0 - \sum_n c_n \lambda_n w u_n \Big) \dd{x}}
    {\displaystyle\int w  \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( u_0 + \sum_n c_n u_n \Big) \dd{x}}
\end{aligned}$$

For convenience, we switch to [Dirac notation](/know/concept/dirac-notation/)
before evaluating further.

$$\begin{aligned}
    R
    &= \frac{\displaystyle \Big( \bra{u_0} + \sum_n c_n^* \bra{u_n} \Big) \cdot \Big( \lambda_0 \ket{w u_0} + \sum_n c_n \lambda_n \ket{w u_n} \Big)}
    {\displaystyle \Big( \bra{u_0} + \sum_n c_n^* \bra{u_n} \Big) \cdot \Big( \ket{w u_0} + \sum_n c_n \ket{w u_n} \Big)}
    \\
    &= \frac{\displaystyle \lambda_0 \braket{u_0}{w u_0} + \lambda_0 \sum_{n = 1}^\infty c_n^* \braket{u_n}{w u_0}
    + \sum_{n = 1}^\infty c_n \lambda_n \braket{u_0}{w u_n} + \sum_{m n} c_n c_m^* \lambda_n \braket{u_m}{w u_n}}
    {\displaystyle \braket{u_0}{w u_0} + \sum_{n = 1}^\infty c_n^* \braket{u_n}{w u_0}
    + \sum_{n = 1}^\infty c_n \braket{u_0}{w u_n} + \sum_{m n} c_n c_m^* \braket{u_m}{w u_n}}
\end{aligned}$$

Using orthogonality $\braket{u_m}{w u_n} = S_n \delta_{mn}$,
and the fact that $n \neq 0$ by definition, we find:

$$\begin{aligned}
    R
    &= \frac{\displaystyle \lambda_0 S_0 + \lambda_0 \sum_n c_n^* S_n \delta_{n0}
    + \sum_n c_n \lambda_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* \lambda_n S_n \delta_{mn}}
    {\displaystyle S_0 + \sum_n c_n^* S_n \delta_{n0} + \sum_n c_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* S_n \delta_{mn}}
    \\
    &= \frac{\displaystyle \lambda_0 S_0 + 0 + 0 + \sum_{n} c_n c_n^* \lambda_n S_n}
    {\displaystyle S_0 + 0 + 0 + \sum_{n} c_n c_n^* S_n}
    = \frac{\displaystyle \lambda_0 S_0 + \sum_{n} |c_n|^2 \lambda_n S_n}
    {\displaystyle S_0 + \sum_{n} |c_n|^2 S_n}
\end{aligned}$$

It is always possible to choose our normalizations such that $S_n = S$ for all $u_n$, leaving:

$$\begin{aligned}
    R
    &= \frac{\displaystyle \lambda_0 S + \sum_{n} |c_n|^2 \lambda_n S}
    {\displaystyle S + \sum_{n} |c_n|^2 S}
    = \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_n}
    {\displaystyle 1 + \sum_{n} |c_n|^2}
\end{aligned}$$

And finally, after rearranging the numerator, we arrive at the following relation:

$$\begin{aligned}
    R
    &= \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_0 + \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
    {\displaystyle 1 + \sum_{n} |c_n|^2}
    = \lambda_0 + \frac{\displaystyle \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
    {\displaystyle 1 + \sum_{n} |c_n|^2}
\end{aligned}$$

Thus, if we improve our guess $u$,
then $R[u]$ approaches the true eigenvalue $\lambda_0$.
For numerically finding $u_0$ and $\lambda_0$, this gives us a clear goal: minimize $R$, because:

$$\begin{aligned}
    \boxed{
        R[u]
        = \lambda_0 + \frac{\displaystyle \sum_{n = 1}^\infty |c_n|^2 (\lambda_n - \lambda_0)}
        {\displaystyle 1 + \sum_{n = 1}^\infty |c_n|^2}
        \ge \lambda_0
    }
\end{aligned}$$

Note that the convergence to $\lambda_0$ goes as $|c_n|^2$,
while $u$ converges to $u_0$ as $|c_n|$ by definition,
so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$.



## The method

In the following, we stick to Dirac notation,
since the results hold for both continuous functions $u(x)$ and discrete vectors $\vb{u}$,
as long as the operator $\hat{H}$ is self-adjoint.
Suppose we express our guess $\ket{u}$ as a linear combination
of *known* basis vectors $\ket{f_n}$ with weights $a_n \in \mathbb{C}$:

$$\begin{aligned}
    \ket{u}
    = \sum_{n = 0}^\infty c_n \ket{u_n}
    = \sum_{n = 0}^\infty a_n \ket{f_n}
    \approx \sum_{n = 0}^{N - 1} a_n \ket{f_n}
\end{aligned}$$

For numerical tractability, we truncate the sum at $N$ terms,
and for generality, we allow $\ket{f_n}$ to be non-orthogonal,
as described by an *overlap matrix* with elements $S_{mn}$:

$$\begin{aligned}
    \braket{f_m}{w f_n} = S_{m n}
\end{aligned}$$

From the discussion above,
we know that the ground-state eigenvalue $\lambda_0$ is estimated by:

$$\begin{aligned}
    \lambda_0
    \approx \lambda
    = R[u]
    = \frac{\braket*{u}{\hat{H} u}}{\braket{u}{w u}}
    = \frac{\displaystyle \sum_{m n} a_m^* a_n \braket*{f_m}{\hat{H} f_n}}{\displaystyle \sum_{m n} a_m^* a_n \braket{f_m}{w f_n}}
    \equiv \frac{\displaystyle \sum_{m n} a_m^* a_n H_{m n}}{\displaystyle \sum_{m n} a_m^* a_n S_{mn}}
\end{aligned}$$

And we also know that our goal is to minimize $R[u]$,
so we vary $a_k^*$ to find its extremum:

$$\begin{aligned}
    0
    = \pdv{R}{a_k^*}
    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) \Big( \sum_{m n} a_n a_m^* S_{mn} \Big)
    - \Big( \sum_{n} a_n S_{k n} \Big) \Big( \sum_{m n} a_n a_m^* H_{mn} \Big)}
    {\Big( \displaystyle \sum_{m n} a_n a_m^* S_{mn} \Big)^2}
    \\
    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) - R[u] \Big( \sum_{n} a_n S_{k n}\Big)}{\braket{u}{w u}}
    = \frac{\displaystyle \sum_{n} a_n \big(H_{k n} - \lambda S_{k n}\big)}{\braket{u}{w u}}
\end{aligned}$$

Clearly, this is only satisfied if the following holds for all $k = 0, 1, ..., N\!-\!1$:

$$\begin{aligned}
    0
    = \sum_{n = 0}^{N - 1} a_n \big(H_{k n} - \lambda S_{k n}\big)
\end{aligned}$$

For illustrative purposes,
we can write this as a matrix equation
with $M_{k n} \equiv H_{k n} - \lambda S_{k n}$:

$$\begin{aligned}
    \begin{bmatrix}
        M_{0,0} & M_{0,1} & \cdots & M_{0,N-1} \\
        M_{1,0} & \ddots & & \vdots \\
        \vdots & & \ddots & \vdots \\
        M_{N-1,0} & \cdots & \cdots & M_{N-1,N-1}
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
        a_0 \\ a_1 \\ \vdots \\ a_{N-1}
    \end{bmatrix}
    =
    \begin{bmatrix}
        0 \\ 0 \\ \vdots \\ 0
    \end{bmatrix}
\end{aligned}$$

Note that this looks like an eigenvalue problem for $\lambda$.
Indeed, demanding that $\overline{M}$ cannot simply be inverted
(i.e. the solution is non-trivial)
yields a characteristic polynomial for $\lambda$:

$$\begin{aligned}
    0
    = \det\!\Big[ \overline{M} \Big]
    = \det\!\Big[ \overline{H} - \lambda \overline{S} \Big]
\end{aligned}$$

This gives a set of $\lambda$,
which are the exact eigenvalues of $\overline{H}$,
and the estimated eigenvalues of $\hat{H}$
(recall that $\overline{H}$ is $\hat{H}$ expressed in a truncated basis).
The eigenvector $\big[ a_0, a_1, ..., a_{N-1} \big]$ of the lowest $\lambda$
gives the optimal weights to approximate $\ket{u_0}$ in the basis $\{\ket{f_n}\}$.
Likewise, the higher $\lambda$'s eigenvectors approximate
excited (i.e. non-ground) eigenstates of $\hat{H}$,
although in practice the results are less accurate the higher we go.

The overall accuracy is determined by how good our truncated basis is,
i.e. how large a subspace it spans
of the [Hilbert space](/know/concept/hilbert-space/) in which the true $\ket{u_0}$ resides.
Clearly, adding more basis vectors will improve the results,
at the cost of computation.
For example, if $\hat{H}$ represents a helium atom,
a good choice for $\{\ket{f_n}\}$ would be hydrogen orbitals,
since those are qualitatively similar.

You may find this result unsurprising;
it makes some intuitive sense that approximating $\hat{H}$
in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues.
The point of this discussion is to rigorously show
the validity of this approach.

Nowadays, there exist many other methods to calculate eigenvalues
of complicated operators $\hat{H}$,
but an attractive feature of the Ritz method is that it is single-step,
whereas its competitors tend to be iterative.



## References
1.  G.B. Arfken, H.J. Weber,
    *Mathematical methods for physicists*, 6th edition, 2005,
    Elsevier.
2.  O. Bang,
    *Applied mathematics for physicists: lecture notes*, 2019,
    unpublished.