06: This proof sucks

Exercise sheet 6/exercise_sheet_06.ipynb

@@ -222,15 +222,16 @@
     "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",
+    "Starting from the left-hand side, consider the same cluster from the same seed and unit vector $\\hat{n}$ but between different states $s$ and $s'$. Flipping between $s$ and $s'$ is done by the same cluster $C$, but from opposite sign. 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. This gives a ratio $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = \\frac{\\prod_{<i,j>^+} (1-p_{add}(s_i, s_j))}{\\prod_{<i,j>^-} (1-p_{add}(s_i, s_j))}$$ for pairs $<i,j>^\\pm$ with $s_i \\in C$ and $s_j$ neighbouring $s_i$ just outside the cluster with $\\pm(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}) > 0$. Substituting the formula for $p_{add}$ in, we find $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)} = \\frac{\\prod_{<i,j>^+} \\exp(-2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))}{\\prod_{<i,j>^-} \\exp(2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))} = \\prod_{<i,j>} \\exp(-2\\beta(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})) = \\exp(-2\\beta\\sum_{<i,j>} (s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n}))$$ with $<i,j>$ describing both types of pairs.\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",
     "$$\n",
-    "H_{XY}(s)-H_{XY}(s') = \\sum_{<i,j> \\in boundary} \\left[ s_i \\cdot s_j - s'_i \\cdot s_j \\right]\n",
-    "                     = \\sum_{<i,j> \\in boundary} 2(s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})\n",
+    "H_{XY}(s)-H_{XY}(s') = \\sum_{<i,j>} \\left[ s_i \\cdot s_j - s'_i \\cdot s_j \\right]\n",
+    "                     = 2\\sum_{<i,j>} (s_i \\cdot \\hat{n})(s_j \\cdot \\hat{n})\n",
     "$$\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"
+    "by summing over all pairs $<i,j>$ as above, as pairs not across the boundary do not give rise to an energy change. The last step is due to the flips from both signs to be accounted for.\n",
+    "This shows $$\\frac{P(s\\rightarrow s')}{P(s'\\rightarrow s)}=\\frac{\\pi(s')}{\\pi(s)},$$ thus that detailed balance holds."
    ]
   },
   {