06: Draft ugh

Exercise sheet 6/exercise_sheet_06.ipynb

@@ -20,13 +20,13 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 1,
    "id": "b388fbcf",
    "metadata": {},
    "outputs": [],
    "source": [
-    "NAME = \"\"\n",
-    "NAMES_OF_COLLABORATORS = \"\""
+    "NAME = \"Kees van Kempen\"\n",
+    "NAMES_OF_COLLABORATORS = \"Bart Steeman\""
    ]
   },
   {
@@ -62,7 +62,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 2,
    "id": "cb41d2a1",
    "metadata": {
     "deletable": false,
@@ -219,7 +219,18 @@
     }
    },
    "source": [
-    "YOUR ANSWER HERE"
+    "The condition for detailed balance is $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)}=\\frac{\\pi(s')}{\\pi(s)}.$$\n",
+    "Let's work to that expression by following the idea of the proof for the Ising model proof.\n",
+    "\n",
+    "Starting from the left-hand side, consider the same cluster from the same seed and unit vector $\\hat{n}$ but from different states $s$ and $s'$. Flipping between $s$ and $s'$ is done by the same cluster $C$. The probability of this happening is given by the boundaries of the clusters, namely that they are not added to the cluster, with probability $1 - p_{add}$ per boundary site. Again, like in the Ising model proof, we take the boundary of $C$ to be of $m + n$ length with $m$ the number of aligned sites and $n$ the number of anti-aligned sites. This gives a ratio $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = (1-p_{add})^{m-n}.$$\n",
+    "\n",
+    "For the right-hand side, $$\\frac{\\pi(s')}{\\pi(s)} = e^{\\beta \\left[H_{XY}(s)-H_{XY}(s')\\right]}$$ just from the definition of $\\pi$.\n",
+    "This difference in energy we can write as\n",
+    "$$\n",
+    "H_{XY}(s)-H_{XY}(s') = \\sum_{<i,j> \\in boundary} \\left[ s_i \\cdot s_j - s'_i \\cdot s_j \\right]\n",
+    "                     = \\sum_{<i,j> \\in boundary} 2(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})\n",
+    "$$\n",
+    "by summing over all pairs $i, j$ on the boundary of $C$, with $i \\in C$ and $j$ an outside neigbour, as pairs not across the boundary do not give rise to an energy change. The last step is due to"
    ]
   },
   {
@@ -245,7 +256,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 3,
    "id": "6e4a6d63",
    "metadata": {
     "deletable": false,
@@ -269,8 +280,17 @@
     "def xy_cluster_flip(state,seed,nhat,beta):\n",
     "    '''Perform a cluster move with specified seed and vector nhat on the state at temperature beta.'''\n",
     "    w = len(state)\n",
-    "    # YOUR CODE HERE\n",
-    "    raise NotImplementedError()\n",
+    "    spin = state[seed]\n",
+    "    state[seed] = state[seed] - 2*np.dot(state[seed], nhat)*nhat\n",
+    "    cluster_size = 1\n",
+    "    unvisited = deque([seed])   # use a deque to efficiently track the unvisited cluster sites\n",
+    "    while unvisited:   # while unvisited sites remain\n",
+    "        site = unvisited.pop()  # take one and remove from the unvisited list\n",
+    "        for nbr in neighboring_sites(site,w):\n",
+    "            if state[nbr] == spin and rng.uniform() < p_add:\n",
+    "                state[nbr] = -spin\n",
+    "                unvisited.appendleft(nbr)\n",
+    "                cluster_size += 1\n",
     "    return cluster_size\n",
     "\n",
     "def xy_cluster_move(state,beta):\n",
@@ -666,7 +686,7 @@
  ],
  "metadata": {
   "kernelspec": {
-   "display_name": "Python 3 (ipykernel)",
+   "display_name": "Python 3",
    "language": "python",
    "name": "python3"
   },