{
 "metadata": {
  "name": "commutators and sympy-Copy0"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#Commutator and expansion based computations with Python & Sympy"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "from sympy.physics.quantum import Commutator, Dagger, Operator\nfrom sympy import simplify, exp, series\ninit_printing()\nt = Symbol(\"t\")",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 229
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Here's a quick demo on how to do computations with commutators and expansions involving operators with Python and Sympy. Let's define a couple of operators first."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "L1= Operator(\"L1\")\nL2= Operator(\"L2\")\nL = L1+L2;",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 226
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Note that those operators by definition do **not** commute. "
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "Commutator(L1,L2)==0",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 234,
       "text": "False"
      }
     ],
     "prompt_number": 234
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Now let's move on to an operator defined by an expansion, the exponential operator. This is a well-defined operator as long as L is bounded. Most often it's denoted by $e^{Lt}$ and is really nothing more than \n\n$$e^{Lt}=\\sum_{k=0}^{\\infty}\\frac{(Lt)^k}{k!}$$\n\nLet's define a function that approximates this operator, up to a certain order."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "def expseries(L,t,k):\n    return series(exp(L*t),x=t,n=k);\n           ",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 235
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Now we can compute the expansions of those operators and combinations of them. For example, the first terms of $e^{(L_1+2L_2)t}$ are"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "expseries(L1+2*L2,t,3)\n\n\n\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$1 + t \\left(L_{1} + 2 L_{2}\\right) + t^{2} \\left(L_{1} L_{2} + \\frac{\\left(L_{1}\\right)^{2}}{2} + L_{2} L_{1} + 2 \\left(L_{2}\\right)^{2}\\right) + \\mathcal{O}\\left(t^{3}\\right)$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 199,
       "text": "                       \u239b          2                \u239e        \n                     2 \u239c        L\u2081                2\u239f    \u239b 3\u239e\n1 + t\u22c5(L\u2081 + 2\u22c5L\u2082) + t \u22c5\u239cL\u2081\u22c5L\u2082 + \u2500\u2500\u2500 + L\u2082\u22c5L\u2081 + 2\u22c5L\u2082 \u239f + O\u239dt \u23a0\n                       \u239d         2                 \u23a0        "
      }
     ],
     "prompt_number": 199
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "We can also compute the difference between different exponential operators. For example, it is a fact that we can approximate the operator $e^{Lt}=e^{(L_1+L_2)t}$ with $e^{L_1t}e^{L_2t}$ \n\nThis is called the Lie splitting and with Python & Sympy we can figure out the difference between those two, which is a new operator. "
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "expseries(L1+L2,t,3)-expseries(L1,t,3)*expseries(L2,t,3)\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$- \\left(1 + t L_{1} + \\frac{t^{2} \\left(L_{1}\\right)^{2}}{2} + \\mathcal{O}\\left(t^{3}\\right)\\right) \\left(1 + t L_{2} + \\frac{t^{2} \\left(L_{2}\\right)^{2}}{2} + \\mathcal{O}\\left(t^{3}\\right)\\right) + 1 + t \\left(L_{1} + L_{2}\\right) + t^{2} \\left(\\frac{L_{1} L_{2}}{2} + \\frac{\\left(L_{1}\\right)^{2}}{2} + \\frac{L_{2} L_{1}}{2} + \\frac{\\left(L_{2}\\right)^{2}}{2}\\right) + \\mathcal{O}\\left(t^{3}\\right)$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 214,
       "text": "  \u239b            2   2        \u239e \u239b            2   2        \u239e                     \n  \u239c           t \u22c5L\u2081     \u239b 3\u239e\u239f \u239c           t \u22c5L\u2082     \u239b 3\u239e\u239f                     \n- \u239c1 + t\u22c5L\u2081 + \u2500\u2500\u2500\u2500\u2500\u2500 + O\u239dt \u23a0\u239f\u22c5\u239c1 + t\u22c5L\u2082 + \u2500\u2500\u2500\u2500\u2500\u2500 + O\u239dt \u23a0\u239f + 1 + t\u22c5(L\u2081 + L\u2082) + \n  \u239d             2           \u23a0 \u239d             2           \u23a0                     \n\n   \u239b          2             2\u239e        \n 2 \u239cL\u2081\u22c5L\u2082   L\u2081    L\u2082\u22c5L\u2081   L\u2082 \u239f    \u239b 3\u239e\nt \u22c5\u239c\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u239f + O\u239dt \u23a0\n   \u239d  2      2      2      2 \u23a0        "
      }
     ],
     "prompt_number": 214
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Those are the error terms, but clearly we can have python do more simplifications for us. We use sympy's *simplify* for that."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "simplify(expseries(L1+L2,t,3)-expseries(L1,t,3)*expseries(L2,t,3))",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$\\frac{1}{2} \\left(t^{2} L_{2} L_{1} - t^{2} L_{1} L_{2} + \\mathcal{O}\\left(t^{3}\\right)\\right)$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 236,
       "text": " 2          2          \u239b 3\u239e\nt \u22c5L\u2082\u22c5L\u2081 - t \u22c5L\u2081\u22c5L\u2082 + O\u239dt \u23a0\n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n             2             "
      }
     ],
     "prompt_number": 236
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Now things look a lot better! Note that the term in there is actually the commutator of $L_1$ and $L_2$. It's easy to isolate that term in python with an *expand* and *coeff*."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "expand(simplify(expseries(L1+L2,t,3)-expseries(L1,t,3)*expseries(L2,t,3))).coeff(t,2)\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$- \\frac{L_{1} L_{2}}{2} + \\frac{L_{2} L_{1}}{2}$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 237,
       "text": "  L\u2081\u22c5L\u2082   L\u2082\u22c5L\u2081\n- \u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\n    2       2  "
      }
     ],
     "prompt_number": 237
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "We can use the same idea to compute the errors between arbitrarily complex splittings.\n\nFor example, we can also approximate $e^{Lt}$ with $e^{L_1t/2}e^{L_2t}e^{L_1t/2}$. This is called the Strang splitting. Again we try to compute the error term."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "simplify(expseries(L1+L2,t,3)-expseries(L1/2,t,3)*expseries(L2,t,3)*expseries(L1,t,3))",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$\\frac{1}{8} \\left(- 4 t L_{1} - 4 t^{2} L_{2} L_{1} - 5 t^{2} \\left(L_{1}\\right)^{2} + \\mathcal{O}\\left(t^{3}\\right)\\right)$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 217,
       "text": "             2            2   2    \u239b 3\u239e\n-4\u22c5t\u22c5L\u2081 - 4\u22c5t \u22c5L\u2082\u22c5L\u2081 - 5\u22c5t \u22c5L\u2081  + O\u239dt \u23a0\n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n                   8                   "
      }
     ],
     "prompt_number": 217
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "In this case, the error term contains first order terms. Since we *know* that Strang is more accurate than Lie, it must be the case that we don't have enough terms in the expansion. A lot of multiplications will be hidden under the following *simplify*."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "simplify(expseries(L1+L2,t,4)-expseries(L1/2,t,4)*expseries(L2,t,4)*expseries(L1/2,t,4))",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$\\frac{1}{24} \\left(- 2 t^{3} \\left(L_{2}\\right)^{2} L_{1} + t^{3} L_{2} \\left(L_{1}\\right)^{2} + 4 t^{3} L_{2} L_{1} L_{2} + t^{3} \\left(L_{1}\\right)^{2} L_{2} - 2 t^{3} L_{1} \\left(L_{2}\\right)^{2} - 2 t^{3} L_{1} L_{2} L_{1} + \\mathcal{O}\\left(t^{4}\\right)\\right)$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 222,
       "text": "     3   2       3      2      3             3   2         3      2      3    \n- 2\u22c5t \u22c5L\u2082 \u22c5L\u2081 + t \u22c5L\u2082\u22c5L\u2081  + 4\u22c5t \u22c5L\u2082\u22c5L\u2081\u22c5L\u2082 + t \u22c5L\u2081 \u22c5L\u2082 - 2\u22c5t \u22c5L\u2081\u22c5L\u2082  - 2\u22c5t \u22c5L\u2081\u22c5\n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n                                             24                               \n\n         \u239b 4\u239e\nL\u2082\u22c5L\u2081 + O\u239dt \u23a0\n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\n             "
      }
     ],
     "prompt_number": 222
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "Once again, we can collect the important terms by *expand* and *coeff*."
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "expand(Out[222]).coeff(t,3)",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": "$$- \\frac{L_{1} L_{2}}{12} L_{1} - \\frac{L_{1} \\left(L_{2}\\right)^{2}}{12} + \\frac{\\left(L_{1}\\right)^{2} L_{2}}{24} + \\frac{L_{2} L_{1}}{6} L_{2} + \\frac{L_{2} \\left(L_{1}\\right)^{2}}{24} - \\frac{\\left(L_{2}\\right)^{2} L_{1}}{12}$$",
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 224,
       "text": "                  2     2                      2     2   \n  L\u2081\u22c5L\u2082\u22c5L\u2081   L\u2081\u22c5L\u2082    L\u2081 \u22c5L\u2082   L\u2082\u22c5L\u2081\u22c5L\u2082   L\u2082\u22c5L\u2081    L\u2082 \u22c5L\u2081\n- \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500 - \u2500\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500 + \u2500\u2500\u2500\u2500\u2500\u2500 - \u2500\u2500\u2500\u2500\u2500\u2500\n     12        12       24        6         24       12  "
      }
     ],
     "prompt_number": 224
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "",
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}