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<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en">
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<head>
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<!-- 2021-01-31 Sun 08:40 -->
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<!-- 2021-01-31 Sun 09:26 -->
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<meta http-equiv="Content-Type" content="text/html;charset=utf-8" />
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<meta name="viewport" content="width=device-width, initial-scale=1" />
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<title>Quantum Monte Carlo</title>
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@ -329,152 +329,152 @@ for the JavaScript code in this tag.
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<h2>Table of Contents</h2>
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<div id="text-table-of-contents">
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<ul>
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<li><a href="#org0a42dc6">1. Introduction</a>
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<li><a href="#orgda3f50b">1. Introduction</a>
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<ul>
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<li><a href="#org6174a0a">1.1. Energy and local energy</a></li>
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<li><a href="#org036e966">1.1. Energy and local energy</a></li>
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</ul>
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</li>
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<li><a href="#orge35e3b0">2. Numerical evaluation of the energy of the hydrogen atom</a>
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<li><a href="#orgc590c55">2. Numerical evaluation of the energy of the hydrogen atom</a>
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<ul>
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<li><a href="#org536ab35">2.1. Local energy</a>
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<li><a href="#org18ff41d">2.1. Local energy</a>
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<ul>
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<li><a href="#orge1d3531">2.1.1. Exercise 1</a>
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<li><a href="#org683e4e2">2.1.1. Exercise 1</a>
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||||
<ul>
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||||
<li><a href="#orge29b2d5">2.1.1.1. Solution</a></li>
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||||
<li><a href="#org3f5fc9d">2.1.1.1. Solution</a></li>
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||||
</ul>
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||||
</li>
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||||
<li><a href="#org0082181">2.1.2. Exercise 2</a>
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<li><a href="#orgc407b41">2.1.2. Exercise 2</a>
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||||
<ul>
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||||
<li><a href="#orgcb5ffb6">2.1.2.1. Solution</a></li>
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||||
<li><a href="#org9104fc1">2.1.2.1. Solution</a></li>
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</ul>
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||||
</li>
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<li><a href="#org950428b">2.1.3. Exercise 3</a>
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<li><a href="#orgb70d960">2.1.3. Exercise 3</a>
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||||
<ul>
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<li><a href="#orgb3b7bff">2.1.3.1. Solution</a></li>
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<li><a href="#org77aed71">2.1.3.1. Solution</a></li>
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</ul>
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</li>
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<li><a href="#orgd1768b7">2.1.4. Exercise 4</a>
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<li><a href="#org7ae5fd2">2.1.4. Exercise 4</a>
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<ul>
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<li><a href="#orgca62f3b">2.1.4.1. Solution</a></li>
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<li><a href="#orgc9c223c">2.1.4.1. Solution</a></li>
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</ul>
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</li>
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<li><a href="#orgd9f5f66">2.1.5. Exercise 5</a>
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<li><a href="#org9fad9d8">2.1.5. Exercise 5</a>
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<ul>
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<li><a href="#org1e5b04a">2.1.5.1. Solution</a></li>
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<li><a href="#org6035a20">2.1.5.1. Solution</a></li>
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</ul>
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</li>
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</ul>
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</li>
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<li><a href="#org8d93b3a">2.2. Plot of the local energy along the \(x\) axis</a>
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<li><a href="#org541ad7b">2.2. Plot of the local energy along the \(x\) axis</a>
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<ul>
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<li><a href="#org73c7953">2.2.1. Exercise</a>
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<li><a href="#org6fe4366">2.2.1. Exercise</a>
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<ul>
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<li><a href="#org0fcc683">2.2.1.1. Solution</a></li>
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<li><a href="#org3f1245c">2.2.1.1. Solution</a></li>
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</ul>
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</li>
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</ul>
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</li>
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<li><a href="#org997c814">2.3. Numerical estimation of the energy</a>
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<li><a href="#org887fd3f">2.3. Numerical estimation of the energy</a>
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<ul>
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<li><a href="#org7a24d0a">2.3.1. Exercise</a>
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<li><a href="#org9274039">2.3.1. Exercise</a>
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<ul>
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<li><a href="#org44d22a3">2.3.1.1. Solution</a></li>
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<li><a href="#org451ad90">2.3.1.1. Solution</a></li>
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</ul>
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</li>
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</ul>
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</li>
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<li><a href="#org361114f">2.4. Variance of the local energy</a>
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<li><a href="#org276e07e">2.4. Variance of the local energy</a>
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<ul>
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<li><a href="#org2c40196">2.4.1. Exercise (optional)</a>
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<li><a href="#org43e53cc">2.4.1. Exercise (optional)</a>
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||||
<ul>
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<li><a href="#orgbecaeef">2.4.1.1. Solution</a></li>
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<li><a href="#org38d9972">2.4.1.1. Solution</a></li>
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</ul>
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</li>
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||||
<li><a href="#orgf1e8e3d">2.4.2. Exercise</a>
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<li><a href="#org733ec8a">2.4.2. Exercise</a>
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||||
<ul>
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<li><a href="#org8fc9a13">2.4.2.1. Solution</a></li>
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<li><a href="#org739fa49">2.4.2.1. Solution</a></li>
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</ul>
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</li>
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</ul>
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</li>
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</ul>
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</li>
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<li><a href="#org7aa6a0a">3. Variational Monte Carlo</a>
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<li><a href="#org81347ce">3. Variational Monte Carlo</a>
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<ul>
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<li><a href="#org2087616">3.1. Computation of the statistical error</a>
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<li><a href="#orgd013ab9">3.1. Computation of the statistical error</a>
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<ul>
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<li><a href="#orgaeae4c2">3.1.1. Exercise</a>
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<li><a href="#org32ca3c3">3.1.1. Exercise</a>
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||||
<ul>
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||||
<li><a href="#orgdfc7b53">3.1.1.1. Solution</a></li>
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||||
<li><a href="#orgebeb146">3.1.1.1. Solution</a></li>
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</ul>
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</li>
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||||
</ul>
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</li>
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||||
<li><a href="#orge7d91c3">3.2. Uniform sampling in the box</a>
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<li><a href="#orgb2a9889">3.2. Uniform sampling in the box</a>
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<ul>
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||||
<li><a href="#org96d1bb2">3.2.1. Exercise</a>
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||||
<li><a href="#org4e6a1f7">3.2.1. Exercise</a>
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||||
<ul>
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||||
<li><a href="#orgc8c3acf">3.2.1.1. Solution</a></li>
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||||
<li><a href="#orgbffcbab">3.2.1.1. Solution</a></li>
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||||
</ul>
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</li>
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</ul>
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</li>
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<li><a href="#org0e66655">3.3. Metropolis sampling with \(\Psi^2\)</a>
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||||
<li><a href="#org0733fed">3.3. Metropolis sampling with \(\Psi^2\)</a>
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||||
<ul>
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||||
<li><a href="#org9692143">3.3.1. Exercise</a>
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||||
<li><a href="#org0c0af3c">3.3.1. Exercise</a>
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||||
<ul>
|
||||
<li><a href="#org88f43cd">3.3.1.1. Solution</a></li>
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||||
<li><a href="#org6c72dc3">3.3.1.1. Solution</a></li>
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||||
</ul>
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||||
</li>
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||||
</ul>
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||||
</li>
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||||
<li><a href="#org6c65f73">3.4. Gaussian random number generator</a></li>
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||||
<li><a href="#org70ec84a">3.5. Generalized Metropolis algorithm</a>
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||||
<li><a href="#orga0b0b8d">3.4. Gaussian random number generator</a></li>
|
||||
<li><a href="#org57a608a">3.5. Generalized Metropolis algorithm</a>
|
||||
<ul>
|
||||
<li><a href="#org6e34709">3.5.1. Exercise 1</a>
|
||||
<li><a href="#orgaa87988">3.5.1. Exercise 1</a>
|
||||
<ul>
|
||||
<li><a href="#orgcf485b0">3.5.1.1. Solution</a></li>
|
||||
<li><a href="#orgf391a29">3.5.1.1. Solution</a></li>
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||||
</ul>
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||||
</li>
|
||||
<li><a href="#orge3ee834">3.5.2. Exercise 2</a>
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||||
<li><a href="#org6abfe47">3.5.2. Exercise 2</a>
|
||||
<ul>
|
||||
<li><a href="#org2d51c0d">3.5.2.1. Solution</a></li>
|
||||
<li><a href="#org53b5631">3.5.2.1. Solution</a></li>
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||||
</ul>
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||||
</li>
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||||
</ul>
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||||
</li>
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||||
</ul>
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||||
</li>
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||||
<li><a href="#org58ca27b">4. Diffusion Monte Carlo</a>
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||||
<li><a href="#org5f1d81b">4. Diffusion Monte Carlo</a>
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||||
<ul>
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||||
<li><a href="#orgde86ff6">4.1. Schrödinger equation in imaginary time</a></li>
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||||
<li><a href="#org02ca0c2">4.2. Diffusion and branching</a></li>
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||||
<li><a href="#org1a44a46">4.3. Importance sampling</a>
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||||
<li><a href="#org95ab9cd">4.1. Schrödinger equation in imaginary time</a></li>
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||||
<li><a href="#org9fcbab9">4.2. Diffusion and branching</a></li>
|
||||
<li><a href="#org0aba1e8">4.3. Importance sampling</a>
|
||||
<ul>
|
||||
<li><a href="#org25d9eba">4.3.1. Appendix : Details of the Derivation</a></li>
|
||||
<li><a href="#orgcc09947">4.3.1. Appendix : Details of the Derivation</a></li>
|
||||
</ul>
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||||
</li>
|
||||
<li><a href="#org4206909">4.4. Fixed-node DMC energy</a></li>
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||||
<li><a href="#orgbc4844d">4.5. Pure Diffusion Monte Carlo (PDMC)</a></li>
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||||
<li><a href="#org0e0bf5c">4.6. Hydrogen atom</a>
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||||
<li><a href="#org2f26efe">4.4. Fixed-node DMC energy</a></li>
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||||
<li><a href="#orgfc317d3">4.5. Pure Diffusion Monte Carlo (PDMC)</a></li>
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||||
<li><a href="#orgf410952">4.6. Hydrogen atom</a>
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||||
<ul>
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||||
<li><a href="#org044109b">4.6.1. Exercise</a>
|
||||
<li><a href="#org5899d68">4.6.1. Exercise</a>
|
||||
<ul>
|
||||
<li><a href="#orgf0ab8a1">4.6.1.1. Solution</a></li>
|
||||
<li><a href="#orge0716bd">4.6.1.1. Solution</a></li>
|
||||
</ul>
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||||
</li>
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||||
</ul>
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||||
</li>
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||||
<li><a href="#orgce119bb">4.7. <span class="todo TODO">TODO</span> H<sub>2</sub></a></li>
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||||
<li><a href="#org5e287d8">4.7. <span class="todo TODO">TODO</span> H<sub>2</sub></a></li>
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||||
</ul>
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||||
</li>
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||||
<li><a href="#org02606e3">5. <span class="todo TODO">TODO</span> <code>[0/3]</code> Last things to do</a></li>
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||||
<li><a href="#org17d4d49">5. <span class="todo TODO">TODO</span> <code>[0/3]</code> Last things to do</a></li>
|
||||
</ul>
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||||
</div>
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||||
</div>
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||||
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||||
<div id="outline-container-org0a42dc6" class="outline-2">
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||||
<h2 id="org0a42dc6"><span class="section-number-2">1</span> Introduction</h2>
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||||
<div id="outline-container-orgda3f50b" class="outline-2">
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||||
<h2 id="orgda3f50b"><span class="section-number-2">1</span> Introduction</h2>
|
||||
<div class="outline-text-2" id="text-1">
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||||
<p>
|
||||
This website contains the QMC tutorial of the 2021 LTTC winter school
|
||||
@ -514,8 +514,8 @@ coordinates, etc).
|
||||
</p>
|
||||
</div>
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||||
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||||
<div id="outline-container-org6174a0a" class="outline-3">
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||||
<h3 id="org6174a0a"><span class="section-number-3">1.1</span> Energy and local energy</h3>
|
||||
<div id="outline-container-org036e966" class="outline-3">
|
||||
<h3 id="org036e966"><span class="section-number-3">1.1</span> Energy and local energy</h3>
|
||||
<div class="outline-text-3" id="text-1-1">
|
||||
<p>
|
||||
For a given system with Hamiltonian \(\hat{H}\) and wave function \(\Psi\), we define the local energy as
|
||||
@ -593,8 +593,8 @@ $$ E ≈ \frac{1}{N<sub>\rm MC</sub>} ∑<sub>i=1</sub><sup>N<sub>\rm MC</
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||||
</div>
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||||
</div>
|
||||
|
||||
<div id="outline-container-orge35e3b0" class="outline-2">
|
||||
<h2 id="orge35e3b0"><span class="section-number-2">2</span> Numerical evaluation of the energy of the hydrogen atom</h2>
|
||||
<div id="outline-container-orgc590c55" class="outline-2">
|
||||
<h2 id="orgc590c55"><span class="section-number-2">2</span> Numerical evaluation of the energy of the hydrogen atom</h2>
|
||||
<div class="outline-text-2" id="text-2">
|
||||
<p>
|
||||
In this section, we consider the hydrogen atom with the following
|
||||
@ -623,8 +623,8 @@ To do that, we will compute the local energy and check whether it is constant.
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||||
</p>
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||||
</div>
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||||
|
||||
<div id="outline-container-org536ab35" class="outline-3">
|
||||
<h3 id="org536ab35"><span class="section-number-3">2.1</span> Local energy</h3>
|
||||
<div id="outline-container-org18ff41d" class="outline-3">
|
||||
<h3 id="org18ff41d"><span class="section-number-3">2.1</span> Local energy</h3>
|
||||
<div class="outline-text-3" id="text-2-1">
|
||||
<p>
|
||||
You will now program all quantities needed to compute the local energy of the H atom for the given wave function.
|
||||
@ -651,8 +651,8 @@ to catch the error.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orge1d3531" class="outline-4">
|
||||
<h4 id="orge1d3531"><span class="section-number-4">2.1.1</span> Exercise 1</h4>
|
||||
<div id="outline-container-org683e4e2" class="outline-4">
|
||||
<h4 id="org683e4e2"><span class="section-number-4">2.1.1</span> Exercise 1</h4>
|
||||
<div class="outline-text-4" id="text-2-1-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -696,8 +696,8 @@ and returns the potential.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orge29b2d5" class="outline-5">
|
||||
<h5 id="orge29b2d5"><span class="section-number-5">2.1.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org3f5fc9d" class="outline-5">
|
||||
<h5 id="org3f5fc9d"><span class="section-number-5">2.1.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-1-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -737,8 +737,8 @@ and returns the potential.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org0082181" class="outline-4">
|
||||
<h4 id="org0082181"><span class="section-number-4">2.1.2</span> Exercise 2</h4>
|
||||
<div id="outline-container-orgc407b41" class="outline-4">
|
||||
<h4 id="orgc407b41"><span class="section-number-4">2.1.2</span> Exercise 2</h4>
|
||||
<div class="outline-text-4" id="text-2-1-2">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -773,8 +773,8 @@ input arguments, and returns a scalar.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgcb5ffb6" class="outline-5">
|
||||
<h5 id="orgcb5ffb6"><span class="section-number-5">2.1.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org9104fc1" class="outline-5">
|
||||
<h5 id="org9104fc1"><span class="section-number-5">2.1.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-1-2-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -801,8 +801,8 @@ input arguments, and returns a scalar.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org950428b" class="outline-4">
|
||||
<h4 id="org950428b"><span class="section-number-4">2.1.3</span> Exercise 3</h4>
|
||||
<div id="outline-container-orgb70d960" class="outline-4">
|
||||
<h4 id="orgb70d960"><span class="section-number-4">2.1.3</span> Exercise 3</h4>
|
||||
<div class="outline-text-4" id="text-2-1-3">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -883,8 +883,8 @@ Therefore, the local kinetic energy is
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgb3b7bff" class="outline-5">
|
||||
<h5 id="orgb3b7bff"><span class="section-number-5">2.1.3.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org77aed71" class="outline-5">
|
||||
<h5 id="org77aed71"><span class="section-number-5">2.1.3.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-1-3-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -925,8 +925,8 @@ Therefore, the local kinetic energy is
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgd1768b7" class="outline-4">
|
||||
<h4 id="orgd1768b7"><span class="section-number-4">2.1.4</span> Exercise 4</h4>
|
||||
<div id="outline-container-org7ae5fd2" class="outline-4">
|
||||
<h4 id="org7ae5fd2"><span class="section-number-4">2.1.4</span> Exercise 4</h4>
|
||||
<div class="outline-text-4" id="text-2-1-4">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -969,8 +969,8 @@ local kinetic energy.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgca62f3b" class="outline-5">
|
||||
<h5 id="orgca62f3b"><span class="section-number-5">2.1.4.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-orgc9c223c" class="outline-5">
|
||||
<h5 id="orgc9c223c"><span class="section-number-5">2.1.4.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-1-4-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -1000,8 +1000,8 @@ local kinetic energy.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgd9f5f66" class="outline-4">
|
||||
<h4 id="orgd9f5f66"><span class="section-number-4">2.1.5</span> Exercise 5</h4>
|
||||
<div id="outline-container-org9fad9d8" class="outline-4">
|
||||
<h4 id="org9fad9d8"><span class="section-number-4">2.1.5</span> Exercise 5</h4>
|
||||
<div class="outline-text-4" id="text-2-1-5">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1011,8 +1011,8 @@ Find the theoretical value of \(a\) for which \(\Psi\) is an eigenfunction of \(
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org1e5b04a" class="outline-5">
|
||||
<h5 id="org1e5b04a"><span class="section-number-5">2.1.5.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org6035a20" class="outline-5">
|
||||
<h5 id="org6035a20"><span class="section-number-5">2.1.5.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-1-5-1">
|
||||
\begin{eqnarray*}
|
||||
E &=& \frac{\hat{H} \Psi}{\Psi} = - \frac{1}{2} \frac{\Delta \Psi}{\Psi} -
|
||||
@ -1032,8 +1032,8 @@ equal to -0.5 atomic units.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org8d93b3a" class="outline-3">
|
||||
<h3 id="org8d93b3a"><span class="section-number-3">2.2</span> Plot of the local energy along the \(x\) axis</h3>
|
||||
<div id="outline-container-org541ad7b" class="outline-3">
|
||||
<h3 id="org541ad7b"><span class="section-number-3">2.2</span> Plot of the local energy along the \(x\) axis</h3>
|
||||
<div class="outline-text-3" id="text-2-2">
|
||||
<div class="note">
|
||||
<p>
|
||||
@ -1044,8 +1044,8 @@ choose a grid which does not contain the origin.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org73c7953" class="outline-4">
|
||||
<h4 id="org73c7953"><span class="section-number-4">2.2.1</span> Exercise</h4>
|
||||
<div id="outline-container-org6fe4366" class="outline-4">
|
||||
<h4 id="org6fe4366"><span class="section-number-4">2.2.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-2-2-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1128,8 +1128,8 @@ plot './data' index 0 using 1:2 with lines title 'a=0.1', \
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org0fcc683" class="outline-5">
|
||||
<h5 id="org0fcc683"><span class="section-number-5">2.2.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org3f1245c" class="outline-5">
|
||||
<h5 id="org3f1245c"><span class="section-number-5">2.2.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-2-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -1204,8 +1204,8 @@ plt.savefig(<span style="color: #8b2252;">"plot_py.png"</span>)
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org997c814" class="outline-3">
|
||||
<h3 id="org997c814"><span class="section-number-3">2.3</span> Numerical estimation of the energy</h3>
|
||||
<div id="outline-container-org887fd3f" class="outline-3">
|
||||
<h3 id="org887fd3f"><span class="section-number-3">2.3</span> Numerical estimation of the energy</h3>
|
||||
<div class="outline-text-3" id="text-2-3">
|
||||
<p>
|
||||
If the space is discretized in small volume elements \(\mathbf{r}_i\)
|
||||
@ -1235,8 +1235,8 @@ The energy is biased because:
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-org7a24d0a" class="outline-4">
|
||||
<h4 id="org7a24d0a"><span class="section-number-4">2.3.1</span> Exercise</h4>
|
||||
<div id="outline-container-org9274039" class="outline-4">
|
||||
<h4 id="org9274039"><span class="section-number-4">2.3.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-2-3-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1305,8 +1305,8 @@ To compile the Fortran and run it:
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org44d22a3" class="outline-5">
|
||||
<h5 id="org44d22a3"><span class="section-number-5">2.3.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org451ad90" class="outline-5">
|
||||
<h5 id="org451ad90"><span class="section-number-5">2.3.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-3-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -1421,8 +1421,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org361114f" class="outline-3">
|
||||
<h3 id="org361114f"><span class="section-number-3">2.4</span> Variance of the local energy</h3>
|
||||
<div id="outline-container-org276e07e" class="outline-3">
|
||||
<h3 id="org276e07e"><span class="section-number-3">2.4</span> Variance of the local energy</h3>
|
||||
<div class="outline-text-3" id="text-2-4">
|
||||
<p>
|
||||
The variance of the local energy is a functional of \(\Psi\)
|
||||
@ -1449,8 +1449,8 @@ energy can be used as a measure of the quality of a wave function.
|
||||
</p>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org2c40196" class="outline-4">
|
||||
<h4 id="org2c40196"><span class="section-number-4">2.4.1</span> Exercise (optional)</h4>
|
||||
<div id="outline-container-org43e53cc" class="outline-4">
|
||||
<h4 id="org43e53cc"><span class="section-number-4">2.4.1</span> Exercise (optional)</h4>
|
||||
<div class="outline-text-4" id="text-2-4-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1461,8 +1461,8 @@ Prove that :
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgbecaeef" class="outline-5">
|
||||
<h5 id="orgbecaeef"><span class="section-number-5">2.4.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org38d9972" class="outline-5">
|
||||
<h5 id="org38d9972"><span class="section-number-5">2.4.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-4-1-1">
|
||||
<p>
|
||||
\(\bar{E} = \langle E \rangle\) is a constant, so \(\langle \bar{E}
|
||||
@ -1481,8 +1481,8 @@ Prove that :
|
||||
</div>
|
||||
</div>
|
||||
</div>
|
||||
<div id="outline-container-orgf1e8e3d" class="outline-4">
|
||||
<h4 id="orgf1e8e3d"><span class="section-number-4">2.4.2</span> Exercise</h4>
|
||||
<div id="outline-container-org733ec8a" class="outline-4">
|
||||
<h4 id="org733ec8a"><span class="section-number-4">2.4.2</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-2-4-2">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1556,8 +1556,8 @@ To compile and run:
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org8fc9a13" class="outline-5">
|
||||
<h5 id="org8fc9a13"><span class="section-number-5">2.4.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org739fa49" class="outline-5">
|
||||
<h5 id="org739fa49"><span class="section-number-5">2.4.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-2-4-2-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -1694,8 +1694,8 @@ a = 2.0000000000000000 E = -8.0869806678448772E-002 s2 = 1.8068814
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org7aa6a0a" class="outline-2">
|
||||
<h2 id="org7aa6a0a"><span class="section-number-2">3</span> Variational Monte Carlo</h2>
|
||||
<div id="outline-container-org81347ce" class="outline-2">
|
||||
<h2 id="org81347ce"><span class="section-number-2">3</span> Variational Monte Carlo</h2>
|
||||
<div class="outline-text-2" id="text-3">
|
||||
<p>
|
||||
Numerical integration with deterministic methods is very efficient
|
||||
@ -1711,8 +1711,8 @@ interval.
|
||||
</p>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org2087616" class="outline-3">
|
||||
<h3 id="org2087616"><span class="section-number-3">3.1</span> Computation of the statistical error</h3>
|
||||
<div id="outline-container-orgd013ab9" class="outline-3">
|
||||
<h3 id="orgd013ab9"><span class="section-number-3">3.1</span> Computation of the statistical error</h3>
|
||||
<div class="outline-text-3" id="text-3-1">
|
||||
<p>
|
||||
To compute the statistical error, you need to perform \(M\)
|
||||
@ -1752,8 +1752,8 @@ And the confidence interval is given by
|
||||
</p>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgaeae4c2" class="outline-4">
|
||||
<h4 id="orgaeae4c2"><span class="section-number-4">3.1.1</span> Exercise</h4>
|
||||
<div id="outline-container-org32ca3c3" class="outline-4">
|
||||
<h4 id="org32ca3c3"><span class="section-number-4">3.1.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-3-1-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -1791,8 +1791,8 @@ input array.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgdfc7b53" class="outline-5">
|
||||
<h5 id="orgdfc7b53"><span class="section-number-5">3.1.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-orgebeb146" class="outline-5">
|
||||
<h5 id="orgebeb146"><span class="section-number-5">3.1.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-3-1-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -1851,8 +1851,8 @@ input array.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orge7d91c3" class="outline-3">
|
||||
<h3 id="orge7d91c3"><span class="section-number-3">3.2</span> Uniform sampling in the box</h3>
|
||||
<div id="outline-container-orgb2a9889" class="outline-3">
|
||||
<h3 id="orgb2a9889"><span class="section-number-3">3.2</span> Uniform sampling in the box</h3>
|
||||
<div class="outline-text-3" id="text-3-2">
|
||||
<p>
|
||||
We will now perform our first Monte Carlo calculation to compute the
|
||||
@ -1872,15 +1872,15 @@ Clearly, the square of the wave function is a good choice of probability density
|
||||
</p>
|
||||
|
||||
\begin{eqnarray*}
|
||||
E & = & \frac{\int E_L(\mathbf{r})\frac{|\Psi(\mathbf{r})|^2}{p(\mathbf{r})}p(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{|\Psi(\mathbf{r})|^2 }{p(\mathbf{r})}p(\mathbf{r})d\mathbf{r}}\,.
|
||||
E & = & \frac{\int E_L(\mathbf{r})\frac{|\Psi(\mathbf{r})|^2}{P(\mathbf{r})}P(\mathbf{r})\, \,d\mathbf{r}}{\int \frac{|\Psi(\mathbf{r})|^2 }{P(\mathbf{r})}P(\mathbf{r})d\mathbf{r}}\,.
|
||||
\end{eqnarray*}
|
||||
|
||||
<p>
|
||||
Here, we will sample a uniform probability \(p(\mathbf{r})\) in a cube of volume \(L^3\) centered at the origin:
|
||||
Here, we will sample a uniform probability \(P(\mathbf{r})\) in a cube of volume \(L^3\) centered at the origin:
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[ p(\mathbf{r}) = \frac{1}{L^3}\,, \]
|
||||
\[ P(\mathbf{r}) = \frac{1}{L^3}\,, \]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
@ -1913,8 +1913,8 @@ compute the statistical error.
|
||||
</p>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org96d1bb2" class="outline-4">
|
||||
<h4 id="org96d1bb2"><span class="section-number-4">3.2.1</span> Exercise</h4>
|
||||
<div id="outline-container-org4e6a1f7" class="outline-4">
|
||||
<h4 id="org4e6a1f7"><span class="section-number-4">3.2.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-3-2-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -2014,8 +2014,8 @@ well as the index of the current step.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgc8c3acf" class="outline-5">
|
||||
<h5 id="orgc8c3acf"><span class="section-number-5">3.2.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-orgbffcbab" class="outline-5">
|
||||
<h5 id="orgbffcbab"><span class="section-number-5">3.2.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-3-2-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -2129,8 +2129,8 @@ E = -0.49518773675598715 +/- 5.2391494923686175E-004
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org0e66655" class="outline-3">
|
||||
<h3 id="org0e66655"><span class="section-number-3">3.3</span> Metropolis sampling with \(\Psi^2\)</h3>
|
||||
<div id="outline-container-org0733fed" class="outline-3">
|
||||
<h3 id="org0733fed"><span class="section-number-3">3.3</span> Metropolis sampling with \(\Psi^2\)</h3>
|
||||
<div class="outline-text-3" id="text-3-3">
|
||||
<p>
|
||||
We will now use the square of the wave function to sample random
|
||||
@ -2155,29 +2155,75 @@ sampling:
|
||||
<p>
|
||||
To sample a chosen probability density, an efficient method is the
|
||||
<a href="https://en.wikipedia.org/wiki/Metropolis%E2%80%93Hastings_algorithm">Metropolis-Hastings sampling algorithm</a>. Starting from a random
|
||||
initial position \(\mathbf{r}_0\), we will realize a random walk as follows:
|
||||
initial position \(\mathbf{r}_0\), we will realize a random walk:
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[ \mathbf{r}_0 \rightarrow \mathbf{r}_1 \rightarrow \mathbf{r}_2 \ldots \mathbf{r}_{N_{\rm MC}}\,, \]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
according to the following algorithm.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
At every step, we propose a new move according to a transition probability \(T(\mathbf{r}_{n+1},\mathbf{r}_n)\) of our choice.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
For simplicity, let us move the electron in a 3-dimensional box of side \(2\delta L\) centered at the current position
|
||||
of the electron:
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \mathbf{u}
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta L \, \mathbf{u}
|
||||
\]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
where \(\delta t\) is a fixed constant (the so-called <i>time-step</i>), and
|
||||
where \(\delta L\) is a fixed constant, and
|
||||
\(\mathbf{u}\) is a uniform random number in a 3-dimensional box
|
||||
\((-1,-1,-1) \le \mathbf{u} \le (1,1,1)\). We will then add the
|
||||
\((-1,-1,-1) \le \mathbf{u} \le (1,1,1)\).
|
||||
</p>
|
||||
|
||||
<p>
|
||||
After having moved the electron, add the
|
||||
accept/reject step that guarantees that the distribution of the
|
||||
\(\mathbf{r}_n\) is \(\Psi^2\):
|
||||
\(\mathbf{r}_n\) is \(\Psi^2\). This amounts to accepting the move with
|
||||
probability
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
A{\mathbf{r}_{n+1},\mathbf{r}_n) = \min\left(1,\frac{T(\mathbf{r}_{n},\mathbf{r}_{n+1}) P(\mathbf{r}_{n+1})}{T(\mathbf{r}_{n+1},\mathbf{r}_n)P(\mathbf{r}_{n})}\right)\,,
|
||||
\]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
which, for our choice of transition probability, becomes
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
A{\mathbf{r}_{n+1},\mathbf{r}_n) = \min\left(1,\frac{P(\mathbf{r}_{n+1})}{P(\mathbf{r}_{n})}\right)= \min\left(1,\frac{\Psi(\mathbf{r}_{n+1})^2}{\Psi(\mathbf{r}_{n})^2}
|
||||
\]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
Explain why the transition probability cancels out in the expression of \(A\). Also note that we do not need to compute the norm of the wave function!
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The algorithm is summarized as follows:
|
||||
</p>
|
||||
|
||||
<ol class="org-ol">
|
||||
<li>Compute \(\Psi\) at a new position \(\mathbf{r'} = \mathbf{r}_n +
|
||||
\delta t\, \mathbf{u}\)</li>
|
||||
<li>Compute the ratio \(R = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}\)</li>
|
||||
\delta L\, \mathbf{u}\)</li>
|
||||
<li>Compute the ratio \(A = \frac{\left[\Psi(\mathbf{r'})\right]^2}{\left[\Psi(\mathbf{r}_{n})\right]^2}\)</li>
|
||||
<li>Draw a uniform random number \(v \in [0,1]\)</li>
|
||||
<li>if \(v \le R\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
|
||||
<li>if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
|
||||
<li>else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)</li>
|
||||
<li>evaluate the local energy at \(\mathbf{r}_{n+1}\)</li>
|
||||
</ol>
|
||||
@ -2195,30 +2241,30 @@ All samples should be kept, from both accepted and rejected moves.
|
||||
</div>
|
||||
|
||||
<p>
|
||||
If the time step is infinitely small, the ratio will be very close
|
||||
to one and all the steps will be accepted. But the trajectory will
|
||||
be infinitely too short to have statistical significance.
|
||||
If the box is infinitely small, the ratio will be very close
|
||||
to one and all the steps will be accepted. However, the moves will be
|
||||
very correlated and you will visit the configurational space very slowly.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
On the other hand, as the time step increases, the number of
|
||||
On the other hand, if you propose too large moves, the number of
|
||||
accepted steps will decrease because the ratios might become
|
||||
small. If the number of accepted steps is close to zero, then the
|
||||
space is not well sampled either.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The time step should be adjusted so that it is as large as
|
||||
The size of the move should be adjusted so that it is as large as
|
||||
possible, keeping the number of accepted steps not too small. To
|
||||
achieve that, we define the acceptance rate as the number of
|
||||
accepted steps over the total number of steps. Adjusting the time
|
||||
step such that the acceptance rate is close to 0.5 is a good compromise.
|
||||
step such that the acceptance rate is close to 0.5 is a good compromise for the current problem.
|
||||
</p>
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-org9692143" class="outline-4">
|
||||
<h4 id="org9692143"><span class="section-number-4">3.3.1</span> Exercise</h4>
|
||||
<div id="outline-container-org0c0af3c" class="outline-4">
|
||||
<h4 id="org0c0af3c"><span class="section-number-4">3.3.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-3-3-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -2325,8 +2371,8 @@ Can you observe a reduction in the statistical error?
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org88f43cd" class="outline-5">
|
||||
<h5 id="org88f43cd"><span class="section-number-5">3.3.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org6c72dc3" class="outline-5">
|
||||
<h5 id="org6c72dc3"><span class="section-number-5">3.3.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-3-3-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -2471,8 +2517,8 @@ A = 0.51695266666666673 +/- 4.0445505648997396E-004
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org6c65f73" class="outline-3">
|
||||
<h3 id="org6c65f73"><span class="section-number-3">3.4</span> Gaussian random number generator</h3>
|
||||
<div id="outline-container-orga0b0b8d" class="outline-3">
|
||||
<h3 id="orga0b0b8d"><span class="section-number-3">3.4</span> Gaussian random number generator</h3>
|
||||
<div class="outline-text-3" id="text-3-4">
|
||||
<p>
|
||||
To obtain Gaussian-distributed random numbers, you can apply the
|
||||
@ -2534,14 +2580,17 @@ In Python, you can use the <a href="https://numpy.org/doc/stable/reference/rando
|
||||
</p>
|
||||
</div>
|
||||
</div>
|
||||
<div id="outline-container-org70ec84a" class="outline-3">
|
||||
<h3 id="org70ec84a"><span class="section-number-3">3.5</span> Generalized Metropolis algorithm</h3>
|
||||
|
||||
<div id="outline-container-org57a608a" class="outline-3">
|
||||
<h3 id="org57a608a"><span class="section-number-3">3.5</span> Generalized Metropolis algorithm</h3>
|
||||
<div class="outline-text-3" id="text-3-5">
|
||||
<p>
|
||||
One can use more efficient numerical schemes to move the electrons,
|
||||
but the Metropolis accepation step has to be adapted accordingly:
|
||||
the acceptance
|
||||
probability \(A\) is chosen so that it is consistent with the
|
||||
One can use more efficient numerical schemes to move the electrons by choosing a smarter expression for the transition probability.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The Metropolis acceptance step has to be adapted accordingly to ensure that the detailed balance condition is satisfied. This means that
|
||||
the acceptance probability \(A\) is chosen so that it is consistent with the
|
||||
probability of leaving \(\mathbf{r}_n\) and the probability of
|
||||
entering \(\mathbf{r}_{n+1}\):
|
||||
</p>
|
||||
@ -2565,7 +2614,7 @@ numbers. Hence, the transition probability was
|
||||
<p>
|
||||
\[
|
||||
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
|
||||
\text{constant}
|
||||
\text{constant}\,,
|
||||
\]
|
||||
</p>
|
||||
|
||||
@ -2584,7 +2633,7 @@ choose to draw Gaussian random numbers with zero mean and variance
|
||||
\[
|
||||
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
|
||||
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
|
||||
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]
|
||||
\mathbf{r}_{n+1} - \mathbf{r}_{n} \right)^2}{2\delta t} \right]\,.
|
||||
\]
|
||||
</p>
|
||||
|
||||
@ -2597,23 +2646,23 @@ acceptance ratios and improve the sampling.
|
||||
</p>
|
||||
|
||||
<p>
|
||||
To do this, we can add the drift vector
|
||||
To do this, we can use the gradient of the probability density
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}.
|
||||
\frac{\nabla [ \Psi^2 ]}{\Psi^2} = 2 \frac{\nabla \Psi}{\Psi}\,,
|
||||
\]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The numerical scheme becomes a drifted diffusion:
|
||||
and add the so-called drift vector, so that the numerical scheme becomes a drifted diffusion:
|
||||
</p>
|
||||
|
||||
<p>
|
||||
\[
|
||||
\mathbf{r}_{n+1} = \mathbf{r}_{n} + \delta t\, \frac{\nabla
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi
|
||||
\Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi \,,
|
||||
\]
|
||||
</p>
|
||||
|
||||
@ -2628,14 +2677,40 @@ The transition probability becomes:
|
||||
T(\mathbf{r}_{n} \rightarrow \mathbf{r}_{n+1}) =
|
||||
\frac{1}{(2\pi\,\delta t)^{3/2}} \exp \left[ - \frac{\left(
|
||||
\mathbf{r}_{n+1} - \mathbf{r}_{n} - \frac{\nabla
|
||||
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]
|
||||
\Psi(\mathbf{r}_n)}{\Psi(\mathbf{r}_n)} \right)^2}{2\,\delta t} \right]\,.
|
||||
\]
|
||||
</p>
|
||||
|
||||
<p>
|
||||
The algorithm of the previous exercise is only slighlty modified summarized:
|
||||
</p>
|
||||
|
||||
<ol class="org-ol">
|
||||
<li>For the starting position compute \(\Psi\) and the drif-vector \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\)</li>
|
||||
<li><p>
|
||||
Compute a new position \(\mathbf{r'} = \mathbf{r}_n +
|
||||
\delta t\, \frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})} + \chi\)
|
||||
</p>
|
||||
|
||||
<p>
|
||||
Evaluate \(\Psi\) and \(\frac{\nabla \Psi(\mathbf{r})}{\Psi(\mathbf{r})}\) at the new position
|
||||
</p></li>
|
||||
<li>Compute the ratio $A = \frac{T(\mathbf{r}<sub>n+1</sub> → \mathbf{r}<sub>n</sub>) P(\mathbf{r}<sub>n+1</sub>)}</li>
|
||||
</ol>
|
||||
<p>
|
||||
{T(\mathbf{r}<sub>n</sub> → \mathbf{r}<sub>n+1</sub>) P(\mathbf{r}<sub>n</sub>)}$
|
||||
</p>
|
||||
<ol class="org-ol">
|
||||
<li>Draw a uniform random number \(v \in [0,1]\)</li>
|
||||
<li>if \(v \le A\), accept the move : set \(\mathbf{r}_{n+1} = \mathbf{r'}\)</li>
|
||||
<li>else, reject the move : set \(\mathbf{r}_{n+1} = \mathbf{r}_n\)</li>
|
||||
<li>evaluate the local energy at \(\mathbf{r}_{n+1}\)</li>
|
||||
</ol>
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-org6e34709" class="outline-4">
|
||||
<h4 id="org6e34709"><span class="section-number-4">3.5.1</span> Exercise 1</h4>
|
||||
<div id="outline-container-orgaa87988" class="outline-4">
|
||||
<h4 id="orgaa87988"><span class="section-number-4">3.5.1</span> Exercise 1</h4>
|
||||
<div class="outline-text-4" id="text-3-5-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -2669,8 +2744,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgcf485b0" class="outline-5">
|
||||
<h5 id="orgcf485b0"><span class="section-number-5">3.5.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-orgf391a29" class="outline-5">
|
||||
<h5 id="orgf391a29"><span class="section-number-5">3.5.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-3-5-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -2703,8 +2778,8 @@ Write a function to compute the drift vector \(\frac{\nabla \Psi(\mathbf{r})}{\P
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orge3ee834" class="outline-4">
|
||||
<h4 id="orge3ee834"><span class="section-number-4">3.5.2</span> Exercise 2</h4>
|
||||
<div id="outline-container-org6abfe47" class="outline-4">
|
||||
<h4 id="org6abfe47"><span class="section-number-4">3.5.2</span> Exercise 2</h4>
|
||||
<div class="outline-text-4" id="text-3-5-2">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -2798,8 +2873,8 @@ Modify the previous program to introduce the drifted diffusion scheme.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org2d51c0d" class="outline-5">
|
||||
<h5 id="org2d51c0d"><span class="section-number-5">3.5.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-org53b5631" class="outline-5">
|
||||
<h5 id="org53b5631"><span class="section-number-5">3.5.2.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-3-5-2-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -2985,12 +3060,12 @@ A = 0.78839866666666658 +/- 3.2503783452043152E-004
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org58ca27b" class="outline-2">
|
||||
<h2 id="org58ca27b"><span class="section-number-2">4</span> Diffusion Monte Carlo   <span class="tag"><span class="solution">solution</span></span></h2>
|
||||
<div id="outline-container-org5f1d81b" class="outline-2">
|
||||
<h2 id="org5f1d81b"><span class="section-number-2">4</span> Diffusion Monte Carlo   <span class="tag"><span class="solution">solution</span></span></h2>
|
||||
<div class="outline-text-2" id="text-4">
|
||||
</div>
|
||||
<div id="outline-container-orgde86ff6" class="outline-3">
|
||||
<h3 id="orgde86ff6"><span class="section-number-3">4.1</span> Schrödinger equation in imaginary time</h3>
|
||||
<div id="outline-container-org95ab9cd" class="outline-3">
|
||||
<h3 id="org95ab9cd"><span class="section-number-3">4.1</span> Schrödinger equation in imaginary time</h3>
|
||||
<div class="outline-text-3" id="text-4-1">
|
||||
<p>
|
||||
Consider the time-dependent Schrödinger equation:
|
||||
@ -3049,8 +3124,8 @@ system.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org02ca0c2" class="outline-3">
|
||||
<h3 id="org02ca0c2"><span class="section-number-3">4.2</span> Diffusion and branching</h3>
|
||||
<div id="outline-container-org9fcbab9" class="outline-3">
|
||||
<h3 id="org9fcbab9"><span class="section-number-3">4.2</span> Diffusion and branching</h3>
|
||||
<div class="outline-text-3" id="text-4-2">
|
||||
<p>
|
||||
The <a href="https://en.wikipedia.org/wiki/Diffusion_equation">diffusion equation</a> of particles is given by
|
||||
@ -3104,8 +3179,8 @@ the combination of a diffusion process and a branching process.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org1a44a46" class="outline-3">
|
||||
<h3 id="org1a44a46"><span class="section-number-3">4.3</span> Importance sampling</h3>
|
||||
<div id="outline-container-org0aba1e8" class="outline-3">
|
||||
<h3 id="org0aba1e8"><span class="section-number-3">4.3</span> Importance sampling</h3>
|
||||
<div class="outline-text-3" id="text-4-3">
|
||||
<p>
|
||||
In a molecular system, the potential is far from being constant,
|
||||
@ -3162,8 +3237,8 @@ error known as the <i>fixed node error</i>.
|
||||
</p>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org25d9eba" class="outline-4">
|
||||
<h4 id="org25d9eba"><span class="section-number-4">4.3.1</span> Appendix : Details of the Derivation</h4>
|
||||
<div id="outline-container-orgcc09947" class="outline-4">
|
||||
<h4 id="orgcc09947"><span class="section-number-4">4.3.1</span> Appendix : Details of the Derivation</h4>
|
||||
<div class="outline-text-4" id="text-4-3-1">
|
||||
<p>
|
||||
\[
|
||||
@ -3225,8 +3300,8 @@ Defining \(\Pi(\mathbf{r},t) = \psi(\mathbf{r},\tau)
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-org4206909" class="outline-3">
|
||||
<h3 id="org4206909"><span class="section-number-3">4.4</span> Fixed-node DMC energy</h3>
|
||||
<div id="outline-container-org2f26efe" class="outline-3">
|
||||
<h3 id="org2f26efe"><span class="section-number-3">4.4</span> Fixed-node DMC energy</h3>
|
||||
<div class="outline-text-3" id="text-4-4">
|
||||
<p>
|
||||
Now that we have a process to sample \(\Pi(\mathbf{r},\tau) =
|
||||
@ -3278,8 +3353,8 @@ energies computed with the trial wave function.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgbc4844d" class="outline-3">
|
||||
<h3 id="orgbc4844d"><span class="section-number-3">4.5</span> Pure Diffusion Monte Carlo (PDMC)</h3>
|
||||
<div id="outline-container-orgfc317d3" class="outline-3">
|
||||
<h3 id="orgfc317d3"><span class="section-number-3">4.5</span> Pure Diffusion Monte Carlo (PDMC)</h3>
|
||||
<div class="outline-text-3" id="text-4-5">
|
||||
<p>
|
||||
Instead of having a variable number of particles to simulate the
|
||||
@ -3331,13 +3406,13 @@ code, so this is what we will do in the next section.
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org0e0bf5c" class="outline-3">
|
||||
<h3 id="org0e0bf5c"><span class="section-number-3">4.6</span> Hydrogen atom</h3>
|
||||
<div id="outline-container-orgf410952" class="outline-3">
|
||||
<h3 id="orgf410952"><span class="section-number-3">4.6</span> Hydrogen atom</h3>
|
||||
<div class="outline-text-3" id="text-4-6">
|
||||
</div>
|
||||
|
||||
<div id="outline-container-org044109b" class="outline-4">
|
||||
<h4 id="org044109b"><span class="section-number-4">4.6.1</span> Exercise</h4>
|
||||
<div id="outline-container-org5899d68" class="outline-4">
|
||||
<h4 id="org5899d68"><span class="section-number-4">4.6.1</span> Exercise</h4>
|
||||
<div class="outline-text-4" id="text-4-6-1">
|
||||
<div class="exercise">
|
||||
<p>
|
||||
@ -3436,8 +3511,8 @@ energy of H for any value of \(a\).
|
||||
</div>
|
||||
</div>
|
||||
|
||||
<div id="outline-container-orgf0ab8a1" class="outline-5">
|
||||
<h5 id="orgf0ab8a1"><span class="section-number-5">4.6.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div id="outline-container-orge0716bd" class="outline-5">
|
||||
<h5 id="orge0716bd"><span class="section-number-5">4.6.1.1</span> Solution   <span class="tag"><span class="solution">solution</span></span></h5>
|
||||
<div class="outline-text-5" id="text-4-6-1-1">
|
||||
<p>
|
||||
<b>Python</b>
|
||||
@ -3653,8 +3728,8 @@ A = 0.98788066666666663 +/- 7.2889356133441110E-005
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-orgce119bb" class="outline-3">
|
||||
<h3 id="orgce119bb"><span class="section-number-3">4.7</span> <span class="todo TODO">TODO</span> H<sub>2</sub></h3>
|
||||
<div id="outline-container-org5e287d8" class="outline-3">
|
||||
<h3 id="org5e287d8"><span class="section-number-3">4.7</span> <span class="todo TODO">TODO</span> H<sub>2</sub></h3>
|
||||
<div class="outline-text-3" id="text-4-7">
|
||||
<p>
|
||||
We will now consider the H<sub>2</sub> molecule in a minimal basis composed of the
|
||||
@ -3675,8 +3750,8 @@ the nuclei.
|
||||
</div>
|
||||
|
||||
|
||||
<div id="outline-container-org02606e3" class="outline-2">
|
||||
<h2 id="org02606e3"><span class="section-number-2">5</span> <span class="todo TODO">TODO</span> <code>[0/3]</code> Last things to do</h2>
|
||||
<div id="outline-container-org17d4d49" class="outline-2">
|
||||
<h2 id="org17d4d49"><span class="section-number-2">5</span> <span class="todo TODO">TODO</span> <code>[0/3]</code> Last things to do</h2>
|
||||
<div class="outline-text-2" id="text-5">
|
||||
<ul class="org-ul">
|
||||
<li class="off"><code>[ ]</code> Give some hints of how much time is required for each section</li>
|
||||
@ -3692,7 +3767,7 @@ the H\(_2\) molecule at $R$=1.4010 bohr. Answer: 0.17406 a.u.</li>
|
||||
</div>
|
||||
<div id="postamble" class="status">
|
||||
<p class="author">Author: Anthony Scemama, Claudia Filippi</p>
|
||||
<p class="date">Created: 2021-01-31 Sun 08:40</p>
|
||||
<p class="date">Created: 2021-01-31 Sun 09:26</p>
|
||||
<p class="validation"><a href="http://validator.w3.org/check?uri=referer">Validate</a></p>
|
||||
</div>
|
||||
</body>
|
||||
|
||||
Loading…
Reference in New Issue
Block a user