corrected use of psi and pi

learning/courses/basics-of-quantum-information/multiple-systems/classical-information.ipynb

@@ -489,10 +489,10 @@
     "$\\operatorname{Pr}(\\mathsf{X}=a) = 0,$ then we're dividing by zero and we obtain indeterminate form $\\frac{0}{0}.$\n",
     "This is not a problem, though, because if the probability associated with $a$ is zero, then we'll never obtain $a$ as an outcome of a measurement of $\\mathsf{X},$ so we don't need to be concerned with this possibility.\n",
     "\n",
-    "To express these formulas in terms of probability vectors, consider a probability vector $\\vert \\psi \\rangle$ describing a joint probabilistic state of $(\\mathsf{X},\\mathsf{Y}).$\n",
+    "To express these formulas in terms of probability vectors, consider a probability vector $\\vert \\pi \\rangle$ describing a joint probabilistic state of $(\\mathsf{X},\\mathsf{Y}).$\n",
     "\n",
     "$$\n",
-    "  \\vert\\psi\\rangle = \\sum_{(a,b)\\in\\Sigma\\times\\Gamma} p_{ab} \\vert ab\\rangle\n",
+    "  \\vert\\pi\\rangle = \\sum_{(a,b)\\in\\Sigma\\times\\Gamma} p_{ab} \\vert ab\\rangle\n",
     "$$\n",
     "\n",
     "Measuring $\\mathsf{X}$ alone yields each possible outcome $a\\in\\Sigma$ with probability\n",
@@ -510,20 +510,20 @@
     "Having obtained a particular outcome $a\\in\\Sigma$ of the measurement of $\\mathsf{X},$ the probabilistic state of $\\mathsf{Y}$ is updated according to the formula for conditional probabilities, so that it is represented by this probability vector:\n",
     "\n",
     "$$\n",
-    "  \\vert \\pi_a \\rangle\n",
+    "  \\vert \\psi_a \\rangle\n",
     "  = \\frac{\\sum_{b\\in\\Gamma}p_{ab}\\vert b\\rangle}{\\sum_{c\\in\\Gamma} p_{ac}}.\n",
     "$$\n",
     "\n",
     "In the event that the measurement of $\\mathsf{X}$ resulted in the classical state $a,$ we therefore update our description of the probabilistic state of the joint system $(\\mathsf{X},\\mathsf{Y})$ to\n",
-    "$\\vert a\\rangle \\otimes \\vert\\pi_a\\rangle.$\n",
+    "$\\vert a\\rangle \\otimes \\vert\\psi_a\\rangle.$\n",
     "\n",
-    "One way to think about this definition of $\\vert\\pi_a\\rangle$ is to see it as a *normalization* of the vector $\\sum_{b\\in\\Gamma} p_{ab} \\vert b\\rangle,$ where we divide by the sum of the entries in this vector to obtain a probability vector.\n",
+    "One way to think about this definition of $\\vert\\psi_a\\rangle$ is to see it as a *normalization* of the vector $\\sum_{b\\in\\Gamma} p_{ab} \\vert b\\rangle,$ where we divide by the sum of the entries in this vector to obtain a probability vector.\n",
     "This normalization effectively accounts for a conditioning on the event that the measurement of $\\mathsf{X}$ has resulted in the outcome $a.$\n",
     "\n",
     "For a specific example, suppose that classical state set of $\\mathsf{X}$ is $\\Sigma = \\{0,1\\},$ the classical state set of $\\mathsf{Y}$ is $\\Gamma = \\{1,2,3\\},$ and the probabilistic state of $(\\mathsf{X},\\mathsf{Y})$ is\n",
     "\n",
     "$$\n",
-    "  \\vert \\psi \\rangle\n",
+    "  \\vert \\pi \\rangle\n",
     "  = \\frac{1}{2}  \\vert 0,1 \\rangle\n",
     "  + \\frac{1}{12} \\vert 0,3 \\rangle\n",
     "  + \\frac{1}{12} \\vert 1,1 \\rangle\n",
@@ -533,10 +533,10 @@
     "\n",
     "Our goal will be to determine the probabilities of the two possible outcomes ($0$ and $1$), and to calculate what the resulting probabilistic state of $\\mathsf{Y}$ is for the two outcomes, assuming the system $\\mathsf{X}$ is measured.\n",
     "\n",
-    "Using the bilinearity of the tensor product, and specifically the fact that it is linear in the *second* argument, we may rewrite the vector $\\vert \\psi \\rangle$ as follows:\n",
+    "Using the bilinearity of the tensor product, and specifically the fact that it is linear in the *second* argument, we may rewrite the vector $\\vert \\pi \\rangle$ as follows:\n",
     "\n",
     "$$\n",
-    "  \\vert \\psi \\rangle\n",
+    "  \\vert \\pi \\rangle\n",
     "  = \\vert 0\\rangle \\otimes\n",
     "  \\biggl( \\frac{1}{2} \\vert 1 \\rangle + \\frac{1}{12} \\vert 3 \\rangle\\biggr)\n",
     "  + \\vert 1\\rangle \\otimes\n",
@@ -838,7 +838,7 @@
     "if $M$ is a probabilistic operation on $\\mathsf{X},$ $N$ is a probabilistic operation on $\\mathsf{Y},$ and the two operations are performed independently, then the resulting operation on the compound system $(\\mathsf{X},\\mathsf{Y})$ is the tensor product $M\\otimes N.$\n",
     "\n",
     "So, for both probabilistic states and probabilistic operations, *tensor products represent independence.*\n",
-    "If we have two systems $\\mathsf{X}$ and $\\mathsf{Y}$ that are independently in the probabilistic states $\\vert\\phi\\rangle$ and $\\vert\\pi\\rangle,$ then the compound system $(\\mathsf{X},\\mathsf{Y})$ is in the probabilistic state $\\vert\\phi\\rangle\\otimes\\vert\\pi\\rangle;$\n",
+    "If we have two systems $\\mathsf{X}$ and $\\mathsf{Y}$ that are independently in the probabilistic states $\\vert\\phi\\rangle$ and $\\vert\\psi\\rangle,$ then the compound system $(\\mathsf{X},\\mathsf{Y})$ is in the probabilistic state $\\vert\\phi\\rangle\\otimes\\vert\\psi\\rangle;$\n",
     "and if we apply probabilistic operations $M$ and $N$ to the two systems independently, then the resulting action on the compound system $(\\mathsf{X},\\mathsf{Y})$ is described by the operation $M\\otimes N.$\n",
     "\n",
     "Let's take a look at an example, which recalls a probabilistic operation on a single bit from the previous lesson:\n",