Fix math mode rendering in browser

BasicGates/BasicGates.ipynb

@@ -214,7 +214,7 @@
     "\n",
     "**Input:** A qubit in state $|\\psi\\rangle = \\alpha |0\\rangle + \\beta |1\\rangle$.\n",
     "\n",
-    "**Goal:** Change the qubit state to $\\alpha |0\\rangle + \\color{red}i\\beta |1\\rangle$ (add a relative phase $i$ to $|1\\rangle$ component of the superposition).\n"
+    "**Goal:** Change the qubit state to $\\alpha |0\\rangle + {\\color{red}i}\\beta |1\\rangle$ (add a relative phase $i$ to $|1\\rangle$ component of the superposition).\n"
    ]
   },
   {
@@ -251,7 +251,7 @@
     "**Goal:**  Change the state of the qubit as follows:\n",
     "- If the qubit is in state $|0\\rangle$, don't change its state.\n",
     "- If the qubit is in state $|1\\rangle$, change its state to $e^{i\\alpha} |1\\rangle$.\n",
-    "- If the qubit is in superposition, change its state according to the effect on basis vectors: $\\beta |0\\rangle + \\color{red}{e^{i\\alpha}} \\gamma |1\\rangle$.\n"
+    "- If the qubit is in superposition, change its state according to the effect on basis vectors: $\\beta |0\\rangle + {\\color{red}{e^{i\\alpha}}} \\gamma |1\\rangle$.\n"
    ]
   },
   {
@@ -479,10 +479,10 @@
    "source": [
     "### Task 2.2. Two-qubit gate - 2\n",
     "\n",
-    "**Input:** Two unentangled qubits (stored in an array of length 2) in state $|+\\rangle \\otimes |+\\rangle = \\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle \\color{blue}+ |11\\rangle \\big)$.\n",
+    "**Input:** Two unentangled qubits (stored in an array of length 2) in state $|+\\rangle \\otimes |+\\rangle = \\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle {\\color{blue}+} |11\\rangle \\big)$.\n",
     "\n",
     "\n",
-    "**Goal:**  Change the two-qubit state to $\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle \\color{red}- |11\\rangle \\big)$.\n",
+    "**Goal:**  Change the two-qubit state to $\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle {\\color{red}-} |11\\rangle \\big)$.\n",
     "\n",
     "> Note that while the starting state can be represented as a tensor product of single-qubit states,\n",
     "> the resulting two-qubit state can not be represented in such a way."
@@ -514,10 +514,10 @@
    "source": [
     "### Task 2.3. Two-qubit gate - 3\n",
     "\n",
-    "**Input:** Two unentangled qubits (stored in an array of length 2) in an arbitrary two-qubit state $\\alpha |00\\rangle + \\color{blue}\\beta |01\\rangle + \\color{blue}\\gamma |10\\rangle + \\delta |11\\rangle$.\n",
+    "**Input:** Two unentangled qubits (stored in an array of length 2) in an arbitrary two-qubit state $\\alpha |00\\rangle + {\\color{blue}\\beta} |01\\rangle + {\\color{blue}\\gamma} |10\\rangle + \\delta |11\\rangle$.\n",
     "\n",
     "\n",
-    "**Goal:**  Change the two-qubit state to $\\alpha |00\\rangle + \\color{red}\\gamma |01\\rangle + \\color{red}\\beta |10\\rangle + \\delta |11\\rangle$.\n",
+    "**Goal:**  Change the two-qubit state to $\\alpha |00\\rangle + {\\color{red}\\gamma} |01\\rangle + {\\color{red}\\beta} |10\\rangle + \\delta |11\\rangle$.\n",
     "\n",
     "> This task can be solved using one intrinsic gate; as an exercise, try to express the solution using several (possibly controlled) Pauli gates."
    ]
@@ -549,9 +549,9 @@
     "### Task 2.4. Toffoli gate\n",
     "\n",
     "**Input:** Three qubits (stored in an array of length 3) in an arbitrary three-qubit state \n",
-    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + \\color{blue}\\eta|110\\rangle + \\color{blue}\\theta|111\\rangle$.\n",
+    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + {\\color{blue}\\eta}|110\\rangle + {\\color{blue}\\theta}|111\\rangle$.\n",
     "\n",
-    "**Goal:** Flip the state of the third qubit if the state of the first two is $|11\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + \\color{red}\\theta|110\\rangle + \\color{red}\\eta|111\\rangle$."
+    "**Goal:** Flip the state of the third qubit if the state of the first two is $|11\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + {\\color{red}\\theta}|110\\rangle + {\\color{red}\\eta}|111\\rangle$."
    ]
   },
   {
@@ -581,9 +581,9 @@
     "### Task 2.5. Fredkin gate\n",
     "\n",
     "**Input:** Three qubits (stored in an array of length 3) in an arbitrary three-qubit state \n",
-    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\color{blue}\\zeta|101\\rangle + \\color{blue}\\eta|110\\rangle + \\theta|111\\rangle$.\n",
+    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + {\\color{blue}\\zeta}|101\\rangle + {\\color{blue}\\eta}|110\\rangle + \\theta|111\\rangle$.\n",
     "\n",
-    "**Goal:** Swap the states of second and third qubit if and only if the state of the first qubit is $|1\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\color{red}\\eta|101\\rangle + \\color{red}\\zeta|110\\rangle + \\theta|111\\rangle$."
+    "**Goal:** Swap the states of second and third qubit if and only if the state of the first qubit is $|1\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + {\\color{red}\\eta}|101\\rangle + {\\color{red}\\zeta}|110\\rangle + \\theta|111\\rangle$."
    ]
   },
   {

BasicGates/Workbook_BasicGates.ipynb

@@ -622,7 +622,7 @@
     "\n",
     "**Input:** A qubit in state $|\\psi\\rangle = \\alpha |0\\rangle + \\beta |1\\rangle$.\n",
     "\n",
-    "**Goal:** Change the qubit state to $\\alpha |0\\rangle + \\color{red}i\\beta |1\\rangle$ (add a relative phase $i$ to $|1\\rangle$ component of the superposition).\n"
+    "**Goal:** Change the qubit state to $\\alpha |0\\rangle + {\\color{red}i}\\beta |1\\rangle$ (add a relative phase $i$ to $|1\\rangle$ component of the superposition).\n"
    ]
   },
   {
@@ -745,7 +745,7 @@
     "**Goal:**  Change the state of the qubit as follows:\n",
     "- If the qubit is in state $|0\\rangle$, don't change its state.\n",
     "- If the qubit is in state $|1\\rangle$, change its state to $e^{i\\alpha} |1\\rangle$.\n",
-    "- If the qubit is in superposition, change its state according to the effect on basis vectors: $\\beta |0\\rangle + \\color{red}{e^{i\\alpha}} \\gamma |1\\rangle$."
+    "- If the qubit is in superposition, change its state according to the effect on basis vectors: $\\beta |0\\rangle + {\\color{red}{e^{i\\alpha}}} \\gamma |1\\rangle$."
    ]
   },
   {
@@ -794,14 +794,14 @@
     "  =\n",
     "\\begin{bmatrix}\n",
     "   1.\\beta + 0.\\gamma\\\\\n",
-    "   0.\\beta + \\color{red}{e^{i\\alpha}}\\gamma\n",
+    "   0.\\beta + {\\color{red}{e^{i\\alpha}}}\\gamma\n",
     "  \\end{bmatrix} \n",
     "  =\n",
     " \\begin{bmatrix}\n",
     "   \\beta\\\\\n",
-    "   \\color{red}{e^{i\\alpha}}\\gamma\n",
+    "   {\\color{red}{e^{i\\alpha}}}\\gamma\n",
     "  \\end{bmatrix}  \n",
-    "  = \\beta |0\\rangle + \\color{red}{e^{i\\alpha}} \\gamma |1\\rangle\n",
+    "  = \\beta |0\\rangle + {\\color{red}{e^{i\\alpha}}} \\gamma |1\\rangle\n",
     "$$"
    ]
   },
@@ -1166,10 +1166,10 @@
    "source": [
     "## Task 2.2. Two-qubit gate - 2\n",
     "\n",
-    "**Input:** Two unentangled qubits (stored in an array of length 2) in state $|+\\rangle \\otimes |+\\rangle = \\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle \\color{blue}+ |11\\rangle \\big)$.\n",
+    "**Input:** Two unentangled qubits (stored in an array of length 2) in state $|+\\rangle \\otimes |+\\rangle = \\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle {\\color{blue}+} |11\\rangle \\big)$.\n",
     "\n",
     "\n",
-    "**Goal:**  Change the two-qubit state to $\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle \\color{red}- |11\\rangle \\big)$."
+    "**Goal:**  Change the two-qubit state to $\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle {\\color{red}-} |11\\rangle \\big)$."
    ]
   },
   {
@@ -1278,7 +1278,7 @@
     "   1\\cdot\\color{red}{e^{i\\alpha}}\\\\\n",
     "\\end{bmatrix}\n",
     "=\n",
-    "\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle \\color{red}- |11\\rangle \\big)\n",
+    "\\frac{1}{2} \\big( |00\\rangle + |01\\rangle + |10\\rangle {\\color{red}-} |11\\rangle \\big)\n",
     "$$"
    ]
   },
@@ -1337,10 +1337,10 @@
    "source": [
     "## Task 2.3. Two-qubit gate - 3\n",
     "\n",
-    "**Input:** Two unentangled qubits (stored in an array of length 2) in an arbitrary two-qubit state $\\alpha |00\\rangle + \\color{blue}\\beta |01\\rangle + \\color{blue}\\gamma |10\\rangle + \\delta |11\\rangle$.\n",
+    "**Input:** Two unentangled qubits (stored in an array of length 2) in an arbitrary two-qubit state $\\alpha |00\\rangle + {\\color{blue}\\beta} |01\\rangle + {\\color{blue}\\gamma} |10\\rangle + \\delta |11\\rangle$.\n",
     "\n",
     "\n",
-    "**Goal:**  Change the two-qubit state to $\\alpha |00\\rangle + \\color{red}\\gamma |01\\rangle + \\color{red}\\beta |10\\rangle + \\delta |11\\rangle$.\n",
+    "**Goal:**  Change the two-qubit state to $\\alpha |00\\rangle + {\\color{red}\\gamma} |01\\rangle + {\\color{red}\\beta} |10\\rangle + \\delta |11\\rangle$.\n",
     "\n",
     "> This task can be solved using one intrinsic gate; as an exercise, try to express the solution using several (possibly controlled) Pauli gates."
    ]
@@ -1423,7 +1423,7 @@
     "   \\delta\\\\\n",
     "\\end{bmatrix}\n",
     "=\n",
-    "|00\\rangle + \\color{red}\\gamma |01\\rangle + \\color{red}\\beta |10\\rangle + \\delta |11\\rangle\n",
+    "|00\\rangle + {\\color{red}\\gamma} |01\\rangle + {\\color{red}\\beta} |10\\rangle + \\delta |11\\rangle\n",
     "$$"
    ]
   },
@@ -1533,9 +1533,9 @@
     "## Task 2.4. Toffoli gate\n",
     "\n",
     "**Input:** Three qubits (stored in an array of length 3) in an arbitrary three-qubit state \n",
-    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + \\color{blue}\\eta|110\\rangle + \\color{blue}\\theta|111\\rangle$.\n",
+    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + {\\color{blue}\\eta}|110\\rangle + {\\color{blue}\\theta}|111\\rangle$.\n",
     "\n",
-    "**Goal:** Flip the state of the third qubit if the state of the first two is $|11\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + \\color{red}\\theta|110\\rangle + \\color{red}\\eta|111\\rangle$.\n",
+    "**Goal:** Flip the state of the third qubit if the state of the first two is $|11\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + {\\color{red}\\theta}|110\\rangle + {\\color{red}\\eta}|111\\rangle$.\n",
     "\n",
     "### Solution\n",
     "\n",
@@ -1629,7 +1629,7 @@
     "   \\color{red}\\eta\\\\       \n",
     "\\end{bmatrix}\n",
     "=\n",
-    "\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + \\color{red}\\theta|110\\rangle + \\color{red}\\eta|111\\rangle\n",
+    "\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\zeta|101\\rangle + {\\color{red}\\theta}|110\\rangle + {\\color{red}\\eta}|111\\rangle\n",
     "$$"
    ]
   },
@@ -1660,9 +1660,9 @@
     "## Task 2.5. Fredkin gate\n",
     "\n",
     "**Input:** Three qubits (stored in an array of length 3) in an arbitrary three-qubit state \n",
-    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\color{blue}\\zeta|101\\rangle + \\color{blue}\\eta|110\\rangle + \\theta|111\\rangle$.\n",
+    "$\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + {\\color{blue}\\zeta}|101\\rangle + {\\color{blue}\\eta}|110\\rangle + \\theta|111\\rangle$.\n",
     "\n",
-    "**Goal:** Swap the states of second and third qubit if and only if the state of the first qubit is $|1\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\color{red}\\eta|101\\rangle + \\color{red}\\zeta|110\\rangle + \\theta|111\\rangle$.\n",
+    "**Goal:** Swap the states of second and third qubit if and only if the state of the first qubit is $|1\\rangle$, i.e., change the three-qubit state to $\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + {\\color{red}\\eta}|101\\rangle + {\\color{red}\\zeta}|110\\rangle + \\theta|111\\rangle$.\n",
     "\n",
     "\n",
     "### Solution\n",
@@ -1758,7 +1758,7 @@
     "   \\theta\\\\       \n",
     "\\end{bmatrix}\n",
     "=\n",
-    "\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + \\color{red}\\eta|101\\rangle + \\color{red}\\zeta|110\\rangle + \\theta|111\\rangle\n",
+    "\\alpha |000\\rangle + \\beta |001\\rangle + \\gamma |010\\rangle + \\delta |011\\rangle + \\epsilon |100\\rangle + {\\color{red}\\eta}|101\\rangle + {\\color{red}\\zeta}|110\\rangle + \\theta|111\\rangle\n",
     "$$"
    ]
   },
@@ -1800,7 +1800,7 @@
    "file_extension": ".qs",
    "mimetype": "text/x-qsharp",
    "name": "qsharp",
-   "version": "0.14"
+   "version": "0.24"
   },
   "widgets": {
    "application/vnd.jupyter.widget-state+json": {

tutorials/ComplexArithmetic/ComplexArithmetic.ipynb

@@ -68,10 +68,10 @@
     "\n",
     "As we said before, $i$ can't be a real number. In that case, we'll call it an **imaginary unit**. However, there is no reason for us to define it as acting any different from any other number, other than the fact that $i^2 = -1$:\n",
     "\n",
-    "$$i + i = 2i \\\\\n",
-    "i - i = 0 \\\\\n",
-    "-1 \\cdot i = -i \\\\\n",
-    "(-i)^{2} = -1$$\n",
+    "$$i + i = 2i$$\n",
+    "$$i - i = 0$$\n",
+    "$$-1 \\cdot i = -i$$\n",
+    "$$(-i)^{2} = -1$$\n",
     "\n",
     "We'll call the number $i$ and its real multiples **imaginary numbers**.\n",
     "\n",
@@ -251,8 +251,8 @@
     "\n",
     "Another property of the conjugate is that it distributes over both complex addition and complex multiplication:\n",
     "\n",
-    "$$\\overline{x + y} = \\overline{x} + \\overline{y} \\\\\n",
-    "\\overline{x \\cdot y} = \\overline{x} \\cdot \\overline{y}$$"
+    "$$\\overline{x + y} = \\overline{x} + \\overline{y}$$\n",
+    "$$\\overline{x \\cdot y} = \\overline{x} \\cdot \\overline{y}$$"
    ]
   },
   {

tutorials/LinearAlgebra/LinearAlgebra.ipynb

@@ -405,8 +405,8 @@
     "\n",
     "Another, equivalent definition highlights what makes this an interesting property. For any matrices $B$ and $C$ of compatible sizes:\n",
     "\n",
-    "$$A^{-1}(AB) = A(A^{-1}B) = B \\\\\n",
-    "(CA)A^{-1} = (CA^{-1})A = C$$\n",
+    "$$A^{-1}(AB) = A(A^{-1}B) = B$$\n",
+    "$$(CA)A^{-1} = (CA^{-1})A = C$$\n",
     "\n",
     "A square matrix has a property called the **determinant**, with the determinant of matrix $A$ being written as $|A|$. A matrix is invertible if and only if its determinant isn't equal to $0$.\n",
     "\n",
@@ -948,8 +948,8 @@
     "    A_{n-1,0} \\cdot \\color{blue} {\\begin{bmatrix}B_{0,0} & \\dotsb & B_{0,l-1} \\\\ \\vdots & \\ddots & \\vdots \\\\ B_{k-1,0} & \\dotsb & B_{k-1,l-1} \\end{bmatrix}} & \\dotsb &\n",
     "    A_{n-1,m-1} \\cdot \\color{red} {\\begin{bmatrix}B_{0,0} & \\dotsb & B_{0,l-1} \\\\ \\vdots & \\ddots & \\vdots \\\\ B_{k-1,0} & \\dotsb & B_{k-1,l-1} \\end{bmatrix}}\n",
     "\\end{bmatrix}\n",
-    "= \\\\\n",
-    "=\n",
+    "=$$\n",
+    "$$=\n",
     "\\begin{bmatrix}\n",
     "    A_{0,0} \\cdot \\color{red} {B_{0,0}} & \\dotsb & A_{0,0} \\cdot \\color{red} {B_{0,l-1}} & \\dotsb & A_{0,m-1} \\cdot \\color{blue} {B_{0,0}} & \\dotsb & A_{0,m-1} \\cdot \\color{blue} {B_{0,l-1}} \\\\\n",
     "    \\vdots & \\ddots & \\vdots & \\dotsb & \\vdots & \\ddots & \\vdots \\\\\n",