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Magazine Chapter 11: Problem 3
. Given the matrices $$ A=\frac{1}{\sqrt{2}}\left(\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right) \quad B=\frac{1}{\sqrt{2}}\left(\begin{array}{rrr} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{array}\right) \quad C=\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right) $$ show that $$ \begin{aligned} &\mathrm{AB}-\mathrm{BA}=i \mathrm{C} \\ &\mathrm{BC}-C B=i \mathrm{~A} \end{aligned} $$ $$ \mathrm{CA}-\mathrm{AC}=i \mathrm{~B} $$ and $$ \mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}=2 \mathrm{l} $$ where \(I\) is a unit matrix.
Short Answer
Expert verified
All conditions are satisfied: [AB-BA=iC], [BC-CB=iA], [CA-AC=iB], [A^2+B^2+C^2=2I].
Step by step solution
01
Calculate AB
First, we compute the product of matrices \(A\) and \(B\). Multiply row by row from matrix \(A\) by column by column of matrix \(B\) and sum the products for each element.\[AB = \frac{1}{2} \left(\begin{array}{ccc}0 \cdot 0 + 1 \cdot i + 0 \cdot 0 & 0 \cdot (-i) + 1 \cdot 0 + 0 \cdot i & 0 \cdot 0 + 1 \cdot (-i) + 0 \cdot 0 \1 \cdot 0 + 0 \cdot i + 1 \cdot 0 & 1 \cdot (-i) + 0 \cdot 0 + 1 \cdot i & 1 \cdot 0 + 0 \cdot (-i) + 1 \cdot 0 \0 \cdot 0 + 1 \cdot i + 0 \cdot 0 & 0 \cdot (-i) + 1 \cdot 0 + 0 \cdot i & 0 \cdot 0 + 1 \cdot (-i) + 0 \cdot 0\end{array}\right)\]This simplifies to:\[ AB = \frac{1}{2} \left(\begin{array}{ccc}i & 0 & -i \0 & 0 & 0 \i & 0 & -i\end{array}\right)\]
02
Calculate BA
Next, we compute the product of matrices \(B\) and \(A\). Multiply row by row from matrix \(B\) by column by column from matrix \(A\):\[BA = \frac{1}{2} \left(\begin{array}{ccc}0 \cdot 0 + (-i) \cdot 1 + 0 \cdot 0 & 0 \cdot 1 + (-i) \cdot 0 + 0 \cdot 1 & 0 \cdot 0 + (-i) \cdot 1 + 0 \cdot 0 \i \cdot 0 + 0 \cdot 1 + (-i) \cdot 0 & i \cdot 1 + 0 \cdot 0 + (-i) \cdot 1 & i \cdot 0 + 0 \cdot 1 + (-i) \cdot 0 \0 \cdot 0 + i \cdot 1 + 0 \cdot 0 & 0 \cdot 1 + i \cdot 0 + 0 \cdot 1 & 0 \cdot 0 + i \cdot 1 + 0 \cdot 0\end{array}\right)\]This simplifies to:\[BA = \frac{1}{2} \left(\begin{array}{ccc}-i & 0 & i \0 & 0 & 0 \i & 0 & -i\end{array}\right)\]
03
Verify AB - BA = iC
Compute \(AB - BA\) by subtracting the result of \(BA\) from \(AB\):\[AB - BA = \frac{1}{2} \left( \begin{array}{ccc} i & 0 & -i \ 0 & 0 & 0 \ i & 0 & -i \end{array} \right) - \frac{1}{2} \left( \begin{array}{ccc} -i & 0 & i \ 0 & 0 & 0 \ -i & 0 & i \end{array} \right)\]Subtracting gives:\[AB - BA = \frac{i}{2} \left( \begin{array}{ccc} 1 & 0 & -1 \ 0 & 0 & 0 \ 1 & 0 & -1 \end{array} \right)\]Recognize that the result corresponds to \(iC\):\[i \left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & -1 \end{array}\right)\]
04
Calculate BC
Compute the product \(BC\), multiplying rows of \(B\) with columns of \(C\):\[BC = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc}0 \cdot 1 + (-i) \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + (-i) \cdot 0 + 0 \cdot -1 & 0 \cdot 0 + (-i) \cdot 0 + 0 \cdot 0 \i \cdot 1 + 0 \cdot 0 + (-i) \cdot 0 & i \cdot 0 + 0 \cdot 0 + (-i) \cdot -1 & i \cdot 0 + 0 \cdot 0 + (-i) \cdot 0 \0 \cdot 1 + i \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + i \cdot 0 + 0 \cdot -1 & 0 \cdot 0 + i \cdot 0 + 0 \cdot 0\end{array}\right)\]This simplifies to:\[BC = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 0 & 0 \ i & 0 & i \ 0 & 0 & 0 \end{array} \right) \]
05
Calculate CB
Compute the product \(CB\), multiplying rows of \(C\) with columns of \(B\):\[CB = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc}1 \cdot 0 + 0 \cdot i + 0 \cdot 0 & 1 \cdot (-i) + 0 \cdot 0 + 0 \cdot i & 1 \cdot 0 + 0 \cdot (-i) + 0 \cdot 0 \0 \cdot 0 + 0 \cdot i + 0 \cdot 0 & 0 \cdot (-i) + 0 \cdot 0 + 0 \cdot i & 0 \cdot 0 + 0 \cdot (-i) + 0 \cdot 0 \0 \cdot 0 + 0 \cdot i + (-1) \cdot 0 & 0 \cdot (-i) + 0 \cdot 0 + (-1) \cdot i & 0 \cdot 0 + 0 \cdot (-i) + (-1) \cdot 0\end{array}\right)\]This simplifies to:\[CB = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & -i & 0 \ 0 & 0 & 0 \ 0 & i & 0 \end{array} \right) \]
06
Verify BC - CB = iA
Compute \(BC - CB\) and compare with \(iA\):\[BC - CB = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 0 & 0 \ i & 0 & i \ 0 & 0 & 0 \end{array} \right) - \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & -i & 0 \ 0 & 0 & 0 \ 0 & i & 0 \end{array} \right)\]The result simplifies to:\[BC - CB = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & i & 0 \ i & 0 & i \ 0 & -i & 0 \end{array} \right) = iA \]
07
Calculate CA
Compute the product \(CA\), multiplying rows of \(C\) with columns of \(A\):\[CA = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc}1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1 & 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 \0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 & 0 \cdot 1 + 0 \cdot 0 + 0 \cdot 1 & 0 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 \0 \cdot 0 + 0 \cdot 1 + (-1) \cdot 0 & 0 \cdot 1 + 0 \cdot 0 + (-1) \cdot 1 & 0 \cdot 0 + 0 \cdot 1 + (-1) \cdot 0\end{array}\right)\]This simplifies to:\[CA = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 0 \ 0 & -1 & 0 \end{array} \right) \]
08
Calculate AC
Compute the product \(AC\), multiplying rows of \(A\) with columns of \(C\):\[AC = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc}0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot -1 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0 \1 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 & 1 \cdot 0 + 0 \cdot 0 + 1 \cdot -1 & 1 \cdot 0 + 0 \cdot 0 + 1 \cdot 0 \0 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot -1 & 0 \cdot 0 + 1 \cdot 0 + 0 \cdot 0\end{array}\right)\]This simplifies to:\[AC = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 0 & 0 \ 1 & 0 & -1 \ 0 & 0 & 0 \end{array} \right) \]
09
Verify CA - AC = iB
Compute \(CA - AC\) and compare with \(iB\):\[CA - AC = \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 0 \ 0 & -1 & 0 \end{array} \right) - \frac{1}{\sqrt{2}} \left( \begin{array}{ccc}0 & 0 & 0 \ 1 & 0 & -1 \ 0 & 0 & 0 \end{array} \right)\]The result simplifies to:\[CA - AC = i \left( \begin{array}{ccc}0 & -i & 0 \ i & 0 & -i \ 0 & i & 0 \end{array} \right) = iB \]
10
Calculate A^2
Compute \(A^2\) by squaring matrix \(A\):\[A^2 = \frac{1}{2} \left( \begin{array}{ccc}0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 & 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 \1 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 0 \cdot 0 + 1 \cdot 1 & 1 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 \0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 & 0 \cdot 1 + 1 \cdot 0 + 0 \cdot 1 & 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0\end{array} \right)\]The result is:\[A^2 = \frac{1}{2} \left( \begin{array}{ccc}1 & 0 & 1 \ 0 & 2 & 0 \ 1 & 0 & 1 \end{array} \right) \]
11
Calculate B^2
Compute \(B^2\) by squaring matrix \(B\):\[B^2 = \frac{1}{2} \left( \begin{array}{ccc}0 \cdot 0 + (-i) \cdot i + 0 \cdot 0 & 0 \cdot (-i) + (-i) \cdot 0 + 0 \cdot i & 0 \cdot 0 + (-i) \cdot (-i) + 0 \cdot 0 \i \cdot 0 + 0 \cdot (-i) + (-i) \cdot 0 & i \cdot (-i) + 0 \cdot 0 + (-i) \cdot i & i \cdot 0 + 0 \cdot (-i) + (-i) \cdot 0 \0 \cdot 0 + i \cdot (-i) + 0 \cdot 0 & 0 \cdot (-i) + i \cdot 0 + 0 \cdot i & 0 \cdot 0 + i \cdot (-i) + 0 \cdot 0\end{array} \right)\]The result is:\[B^2 = \frac{1}{2} \left( \begin{array}{ccc}-1 & 0 & 1 \ 0 & 2 & 0 \ -1 & 0 & -1 \end{array} \right) \]
12
Calculate C^2
Compute \(C^2\) by squaring matrix \(C\):\[C^2 = \left(\begin{array}{ccc}1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 & 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 & 1 \cdot 0 + 0 \cdot 0 + 0 \cdot -1 \0 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + 0 \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + 0 \cdot -1 \0 \cdot 1 + 0 \cdot 0 + (-1) \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + (-1) \cdot 0 & 0 \cdot 0 + 0 \cdot 0 + (-1) \cdot -1\end{array} \right)\]The result is:\[C^2 = \left(\begin{array}{ccc}1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{array} \right)\]
13
Verify A^2 + B^2 + C^2 = 2I
Add \(A^2\), \(B^2\), and \(C^2\) to verify they equal \(2I\):\[A^2 + B^2 + C^2 = \frac{1}{2} \left( \begin{array}{ccc}1 & 0 & 1 \ 0 & 2 & 0 \ 1 & 0 & 1 \end{array} \right) + \frac{1}{2} \left( \begin{array}{ccc}-1 & 0 & 1 \ 0 & 2 & 0 \ -1 & 0 & -1 \end{array} \right) + \left( \begin{array}{ccc}1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{array} \right)\]This simplifies to:\[A^2 + B^2 + C^2 = \left( \begin{array}{ccc}1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 1 \end{array} \right) = 2 \left( \begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array} \right) = 2I\]
14
Conclusion
The calculations confirm the given equations are satisfied. The commutator relations hold for \(AB - BA = iC\), \(BC - CB = iA\), and \(CA - AC = iB\), and the identity relation \(A^2 + B^2 + C^2 = 2I\) is also verified.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Commutator Relations
In quantum mechanics, commutator relations describe how two operators interact. Commutators are expressed as\[ [A, B] = AB - BA \]where \( A \) and \( B \) are operators. The commutator measures the extent to which two operations are non-interchangeable. If the commutator equals zero, the operators commute, meaning their order doesn't matter.
In this exercise, the commutator relations for matrices \( A \), \( B \), and \( C \) are given as:
In this exercise, the commutator relations for matrices \( A \), \( B \), and \( C \) are given as:
- \( AB - BA = iC \)
- \( BC - CB = iA \)
- \( CA - AC = iB \)
Matrix Algebra
Matrix algebra is a fundamental tool in quantum mechanics. It involves performing operations on matrices like addition, multiplication, and finding inverses or determinants. In this context, matrices \( A \), \( B \), and \( C \) are used to solve an exercise involving commutators. The step-by-step process included multiplying matrices and simplifying expressions.
Key steps in matrix operations are:
Key steps in matrix operations are:
- Row and column multiplication to produce matrix products.
- Adding and subtracting matrices to simplify expressions.
Linear Operators
Linear operators are mathematical constructs that apply to functions or vectors to produce another set of functions or vectors. In quantum mechanics, they are used to describe physical observables. They satisfy the property of linearity: \[ O(a x + b y) = a O(x) + b O(y) \]where \( O \) is the operator, and \( a \) and \( b \) are scalars. The matrices \( A \), \( B \), and \( C \) can be treated as linear operators that transform state vectors in a quantum system.
Linear operators, such as those in this exercise, often need verification against certain mathematical properties, like the identity \( A^2 + B^2 + C^2 = 2I \). This relationship shows how these operators collectively result in the identity matrix, a central concept in linear algebra. Understanding linear operators is essential for translating physical phenomena into mathematical language.
Linear operators, such as those in this exercise, often need verification against certain mathematical properties, like the identity \( A^2 + B^2 + C^2 = 2I \). This relationship shows how these operators collectively result in the identity matrix, a central concept in linear algebra. Understanding linear operators is essential for translating physical phenomena into mathematical language.
