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diff --git a/source/know/concept/ritz-method/index.md b/source/know/concept/ritz-method/index.md index 902b7cf..ef694da 100644 --- a/source/know/concept/ritz-method/index.md +++ b/source/know/concept/ritz-method/index.md @@ -25,25 +25,26 @@ consider the following functional to be optimized: $$\begin{aligned} R[u] - = \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x} + \equiv \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x} \end{aligned}$$ 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 +In addition, $$S$$ is the norm of $$u$$, which we take to be constant with respect to a weight function $$w(x) \in \mathbb{R}$$: $$\begin{aligned} S - = \int_a^b w(x) \big|u(x)\big|^2 \dd{x} + \equiv \int_a^b w(x) \big|u(x)\big|^2 \dd{x} \end{aligned}$$ -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: +This normalization requirement acts as a constraint +to the optimization problem for $$R[u]$$, +so we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $$\lambda$$, +and define the Lagrangian $$\mathcal{L}$$ for the full problem as: $$\begin{aligned} - \Lambda + \mathcal{L} \equiv \frac{1}{S} \bigg( \big( p |u_x|^2 - q |u|^2 \big) - \lambda \big( w |u|^2 \big) \bigg) \end{aligned}$$ @@ -51,7 +52,7 @@ The resulting Euler-Lagrange equation is then calculated in the standard way, yi $$\begin{aligned} 0 - &= \pdv{\Lambda}{u^*} - \dv{}{x}\Big( \pdv{\Lambda}{u_x^*} \Big) + &= \pdv{\mathcal{L}}{u^*} - \dv{}{x}\Big( \pdv{\mathcal{L}}{u_x^*} \Big) \\ &= - \frac{1}{S} \bigg( q u + \lambda w u + \dv{}{x}\big( p u_x \big) \bigg) \end{aligned}$$ @@ -69,15 +70,14 @@ 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 $$R[u]$$, +Let us return to the definition of $$R$$, and integrate it by parts: $$\begin{aligned} R[u] &= \frac{1}{S} \int_a^b p u_x u_x^* - q u u^* \dd{x} \\ - &= \frac{1}{S} \Big[ p u_x u^* \Big]_a^b - \frac{1}{S} \int_a^b \dv{}{x}\Big(p u_x\Big) u^* + q u u^* \dd{x} + &= \frac{1}{S} \Big[ p u_x u^* \Big]_a^b - \frac{1}{N} \int_a^b \dv{}{x}\Big(p u_x\Big) u^* + q u u^* \dd{x} \end{aligned}$$ The boundary term vanishes for a subset of the BCs that make a valid SLP, @@ -88,10 +88,11 @@ such that we can use Sturm-Liouville theory later: $$\begin{aligned} R[u] &= - \frac{1}{S} \int_a^b \bigg( \dv{}{x}\Big(p u_x\Big) + q u \bigg) u^* \dd{x} - \equiv - \frac{1}{S} \int_a^b u^* \hat{H} u \dd{x} + \\ + &\equiv - \frac{1}{S} \int_a^b u^* \hat{L} u \dd{x} \end{aligned}$$ -Where $$\hat{H}$$ is the self-adjoint Sturm-Liouville operator. +Where $$\hat{L}$$ 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 $$u_n(x)$$. @@ -102,16 +103,16 @@ and now insert one of the eigenfunctions $$u_n$$ into $$R$$: $$\begin{aligned} R[u_n] - &= - \frac{1}{S_n} \int_a^b u_n^* \hat{H} u_n \dd{x} - = \frac{1}{S_n} \int_a^b u_n^* \lambda_n w u_n \dd{x} + &= - \frac{1}{S_n} \int_a^b u_n^* \hat{L} u_n \dd{x} + \\ + &= \frac{1}{S_n} \int_a^b \lambda_n w |u_n|^2 \dd{x} \\ - &= \frac{1}{S_n} \lambda_n \int_a^b w |u_n|^2 \dd{x} - = \frac{S_n}{S_n} \lambda_n + &= \frac{S_n}{S_n} \lambda_n \end{aligned}$$ Where $$S_n$$ is the normalization of $$u_n$$. -In other words, when given $$u_n$$, -the functional $$R$$ yields the corresponding eigenvalue $$\lambda_n$$: +In other words, when given $$u_n$$ as input, +the functional $$R$$ returns the corresponding eigenvalue $$\lambda_n$$: $$\begin{aligned} \boxed{ @@ -121,6 +122,11 @@ $$\begin{aligned} \end{aligned}$$ This powerful result was not at all clear from $$R$$'s initial definition. +Note that some authors use the opposite sign for $$\lambda$$ in their SLP definition, +in which case this result can still be obtained +simply by also defining $$R$$ with the opposite sign. +This sign choice is consistent with quantum mechanics, +with the Hamiltonian $$\hat{H} = - \hat{L}$$. @@ -137,81 +143,79 @@ $$\begin{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: +Next, by definition: $$\begin{aligned} \boxed{ R[u] - = - \frac{\displaystyle\int u^* \hat{H} u \dd{x}}{\displaystyle\int u^* w u \dd{x}} + = - \frac{\displaystyle\int u^* \hat{L} u \dd{x}}{\displaystyle\int u^* w u \dd{x}} } \end{aligned}$$ -This quantity is known as the **Rayleigh quotient**. +This quantity is known as the **Rayleigh quotient**, +and again beware of the sign in its definition; see the remark above. Inserting our ansatz $$u$$, -and using that the true $$u_n$$ have corresponding eigenvalues $$\lambda_n$$: +and using that the true $$u_n$$ have corresponding eigenvalues $$\lambda_n$$, +we have: $$\begin{aligned} R[u] - &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \: \hat{H} \Big\{ u_0 + \sum_n c_n u_n \Big\} \dd{x}} + &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \: \hat{L} \Big\{ u_0 + \sum_n c_n u_n \Big\} \dd{x}} {\displaystyle\int w \Big( u_0 + \sum_n c_n u_n \Big) \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \dd{x}} \\ - &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( \!-\! \lambda_0 w u_0 - \sum_n c_n \lambda_n w u_n \Big) \dd{x}} + &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) + \Big( \!-\! \lambda_0 w u_0 - \sum_n c_n \lambda_n w u_n \Big) \dd{x}} {\displaystyle\int w \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( u_0 + \sum_n c_n u_n \Big) \dd{x}} \end{aligned}$$ For convenience, we switch to [Dirac notation](/know/concept/dirac-notation/) -before evaluating further. +before evaluating further: $$\begin{aligned} - R - &= \frac{\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) \cdot \Big( \lambda_0 \Ket{w u_0} + \sum_n c_n \lambda_n \Ket{w u_n} \Big)} - {\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) \cdot \Big( \Ket{w u_0} + \sum_n c_n \Ket{w u_n} \Big)} + R[u] + &= \frac{\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) + \Big( \lambda_0 \Ket{w u_0} + \sum_n c_n \lambda_n \Ket{w u_n} \Big)} + {\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) \Big( \Ket{w u_0} + \sum_n c_n \Ket{w u_n} \Big)} \\ - &= \frac{\displaystyle \lambda_0 \Inprod{u_0}{w u_0} + \lambda_0 \sum_{n = 1}^\infty c_n^* \Inprod{u_n}{w u_0} - + \sum_{n = 1}^\infty c_n \lambda_n \Inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \lambda_n \Inprod{u_m}{w u_n}} - {\displaystyle \Inprod{u_0}{w u_0} + \sum_{n = 1}^\infty c_n^* \Inprod{u_n}{w u_0} - + \sum_{n = 1}^\infty c_n \Inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \Inprod{u_m}{w u_n}} + &= \frac{\displaystyle \lambda_0 \inprod{u_0}{w u_0} + \lambda_0 \sum_{n} c_n^* \inprod{u_n}{w u_0} + + \sum_{n} c_n \lambda_n \inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \lambda_n \inprod{u_m}{w u_n}} + {\displaystyle \inprod{u_0}{w u_0} + \sum_{n} c_n^* \inprod{u_n}{w u_0} + + \sum_{n} c_n \inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \inprod{u_m}{w u_n}} \end{aligned}$$ -Using orthogonality $$\Inprod{u_m}{w u_n} = S_n \delta_{mn}$$, +Using orthogonality $$\inprod{u_m}{w u_n} = S_n \delta_{mn}$$, and the fact that $$n \neq 0$$ by definition, we find: $$\begin{aligned} - R + R[u] &= \frac{\displaystyle \lambda_0 S_0 + \lambda_0 \sum_n c_n^* S_n \delta_{n0} + \sum_n c_n \lambda_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* \lambda_n S_n \delta_{mn}} {\displaystyle S_0 + \sum_n c_n^* S_n \delta_{n0} + \sum_n c_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* S_n \delta_{mn}} \\ - &= \frac{\displaystyle \lambda_0 S_0 + 0 + 0 + \sum_{n} c_n c_n^* \lambda_n S_n} - {\displaystyle S_0 + 0 + 0 + \sum_{n} c_n c_n^* S_n} - = \frac{\displaystyle \lambda_0 S_0 + \sum_{n} |c_n|^2 \lambda_n S_n} + &= \frac{\displaystyle \lambda_0 S_0 + \sum_{n} |c_n|^2 \lambda_n S_n} {\displaystyle S_0 + \sum_{n} |c_n|^2 S_n} \end{aligned}$$ It is always possible to choose our normalizations such that $$S_n = S$$ for all $$u_n$$, leaving: $$\begin{aligned} - R - &= \frac{\displaystyle \lambda_0 S + \sum_{n} |c_n|^2 \lambda_n S} - {\displaystyle S + \sum_{n} |c_n|^2 S} - = \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_n} + R[u] + &= \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_n} {\displaystyle 1 + \sum_{n} |c_n|^2} \end{aligned}$$ And finally, after rearranging the numerator, we arrive at the following relation: $$\begin{aligned} - R + R[u] &= \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_0 + \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)} {\displaystyle 1 + \sum_{n} |c_n|^2} - = \lambda_0 + \frac{\displaystyle \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)} + \\ + &= \lambda_0 + \frac{\displaystyle \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)} {\displaystyle 1 + \sum_{n} |c_n|^2} \end{aligned}$$ -Thus, if we improve our guess $$u$$, +Thus, if we improve our guess $$u$$ (i.e. reduce $$|c_n|$$), 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: @@ -228,19 +232,21 @@ In the context of quantum mechanics, this is not surprising, since any superposition of multiple states is guaranteed to have a higher energy than the ground state. -Note that the convergence to $$\lambda_0$$ goes as $$|c_n|^2$$, +As our guess $$u$$ is improved, $$\lambda_0$$ converges 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 $$\lambda_0$$. +so even a fairly bad ansatz $$u$$ gives 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. +since the results hold for both continuous functions $$u(x)$$ +and discrete vectors $$\vb{u}$$, +as long as the operator $$\hat{L}$$ is self-adjoint. Suppose we express our guess $$\Ket{u}$$ as a linear combination -of *known* basis vectors $$\Ket{f_n}$$ with weights $$a_n \in \mathbb{C}$$: +of *known* basis vectors $$\Ket{f_n}$$ with weights $$a_n \in \mathbb{C}$$, +where $$\Ket{f_n}$$ are not necessarily eigenvectors of $$\hat{L}$$: $$\begin{aligned} \Ket{u} @@ -250,11 +256,11 @@ $$\begin{aligned} \end{aligned}$$ For numerical tractability, we truncate the sum at $$N$$ terms, -and for generality, we allow $$\Ket{f_n}$$ to be non-orthogonal, +and for generality we allow $$\Ket{f_n}$$ to be non-orthogonal, as described by an *overlap matrix* with elements $$S_{mn}$$: $$\begin{aligned} - \Inprod{f_m}{w f_n} = S_{m n} + \inprod{f_m}{w f_n} = S_{m n} \end{aligned}$$ From the discussion above, @@ -262,11 +268,10 @@ we know that the ground-state eigenvalue $$\lambda_0$$ is estimated by: $$\begin{aligned} \lambda_0 - \approx \lambda - = R[u] - = \frac{\inprod{u}{\hat{H} u}}{\Inprod{u}{w u}} - = \frac{\displaystyle \sum_{m n} a_m^* a_n \inprod{f_m}{\hat{H} f_n}}{\displaystyle \sum_{m n} a_m^* a_n \Inprod{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}} + \approx R[u] + = - \frac{\inprod{u}{\hat{L} u}}{\inprod{u}{w u}} + = - \frac{\displaystyle \sum_{m n} a_m^* a_n \inprod{f_m}{\hat{L} f_n}}{\displaystyle \sum_{m n} a_m^* a_n \inprod{f_m}{w f_n}} + \equiv - \frac{\displaystyle \sum_{m n} a_m^* a_n L_{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]$$, @@ -274,25 +279,27 @@ so we vary $$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)} + = - \pdv{R}{a_k^*} + &= \frac{\displaystyle \Big( \sum_{n} a_n L_{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^* L_{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)}{\Inprod{u}{w u}} - = \frac{\displaystyle \sum_{n} a_n \big(H_{k n} - \lambda S_{k n}\big)}{\Inprod{u}{w u}} + &= \frac{\displaystyle \Big( \sum_{n} a_n L_{k n} \Big) - R[u] \Big( \sum_{n} a_n S_{k n}\Big)} + {\displaystyle \sum_{m n} a_n a_m^* S_{mn}} + \\ + &= \sum_{n} a_n \frac{\big(L_{k n} - \lambda S_{k n}\big)}{\inprod{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) + = \sum_{n = 0}^{N - 1} a_n \big(L_{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}$$: +with $$M_{k n} \equiv L_{k n} - \lambda S_{k n}$$: $$\begin{aligned} \begin{bmatrix} @@ -311,53 +318,47 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -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 $$\lambda$$: +This looks like an eigenvalue problem for $$\lambda$$, +so we demand that its determinant vanishes: $$\begin{aligned} 0 - = \det\!\Big[ \overline{M} \Big] - = \det\!\Big[ \overline{H} - \lambda \overline{S} \Big] + = \det\!\Big[ \bar{M} \Big] + = \det\!\Big[ \bar{L} - \lambda \bar{S} \Big] \end{aligned}$$ 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). +which are exact eigenvalues of $$\bar{L}$$, +and estimated eigenvalues of $$\hat{L}$$ +(recall that $$\bar{L}$$ is $$\hat{L}$$ 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. +gives the optimal weights $$a_n$$ 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 $$\hat{L}$$, +although in practice the results become less accurate the higher we go. +If we only care about the ground state, +then we already know $$\lambda$$ from $$R[u]$$, +so we just need to solve the matrix equation for $$a_n$$. -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. +You may find this result unsurprising: +it makes some intuitive sense that approximating $$\hat{L}$$ +in a limited basis would yield a matrix $$\bar{L}$$ giving rough eigenvalues. The point of this discussion is to rigorously show the validity of this approach. -If we only care about the ground state, -then we already know $$\lambda$$ from $$R[u]$$, -so all we need to do is solve the above matrix equation for $$a_n$$. -Keep in mind that $$\overline{M}$$ is singular, -and $$a_n$$ are only defined up to a constant factor. - Nowadays, there exist many other methods to calculate eigenvalues -of complicated operators $$\hat{H}$$, +of complicated operators $$\hat{L}$$, but an attractive feature of the Ritz method is that it is single-step, whereas its competitors tend to be iterative. -That said, the Ritz method cannot recover from a poorly chosen basis. +That said, this method cannot recover from a poorly chosen basis $$\{\Ket{f_n}\}$$. + +Indeed, 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 improves the results, +but at a computational cost; +it is usually more efficient to carefully choose *which* $$\ket{f_n}$$ to use, +rather than just *how many*. |