Question

Prove that if \(\mathbf{A}\) and \(\mathbf{B}\) are hermitian, then \(i[\mathbf{A}, \mathbf{B}]\) is also hermitian.

Short answer

Based on the given problem, we used the definition of the commutator and the properties of the Hermitian conjugate to prove the equality and thus show that \(i[A, B]\) is Hermitian.

Step-by-step solution

Compute the Commutator

Express \(i[A, B]\) using the commutator: \(i[A, B] = i(AB - BA)\)

Compute the Hermitian Conjugate (Transpose)

Transform \(i[A, B]\) to its Hermitian conjugate: \((i[A, B])^H = -i(AB-BA)^H\)

Apply the Hermitian Properties

Apply the Hermitian properties \(\mathbf{A}^H=\mathbf{A}\) and \(\mathbf{B}^H=\mathbf{B}\) to rewrite the expression: \(-i(AB-BA)^H = -i(B^HA^H - A^HB^H) = -i(BA - AB) = i(AB - BA) = i[A, B]\)

Key concepts

Commutator of Operators

Understanding the commutator of operators is essential in quantum mechanics, as it measures the degree to which two observable properties (represented by operators) can be measured independently of each other. For Hermitian operators \textbf{A} and \textbf{B}, which represent observable quantities, their commutator is defined as \[ [A, B] = AB - BA \].

If \textbf{A} and \textbf{B} do not commute, this implies that the observables cannot be simultaneously determined with arbitrary precision due to the Heisenberg uncertainty principle. In the exercise, the commutator translates into the operation \(i[A, B]\), which represents a new operator that is also required to be Hermitian for the proof. When we compute the commutator for Hermitian operators, we are looking for its characteristic behavior under Hermitian conjugation, which will determine if the resultant operator is indeed Hermitian.

Hermitian Conjugate

The Hermitian conjugate, also known as the adjoint of an operator, is a fundamental concept in quantum mechanics. It is denoted by A^H for an operator A. For a matrix representation, obtaining the Hermitian conjugate involves taking the transpose of the operator's matrix and then taking the complex conjugate of each element. Mathematically, if \( A = (a_ij) \) then \( A^H = (\bar{a}_{ji}) \).

For our exercise, the Hermitian conjugate of the operator \(i[A, B]\) is calculated, resulting in \( -i(AB-BA)^H \). When considering Hermitian operators \textbf{A} and \textbf{B}, one uses the fact that \(A^H = A\) and \(B^H = B\) to show that the Hermitian conjugate of the commutator returns to itself times a complex unit, which is the key step in proving it is a Hermitian operator.

Hermitian Properties

Hermitian operators have several vital properties that make them integral in quantum mechanics. One such property is that they are equal to their own Hermitian conjugate, which means that \(A^H = A\) for a Hermitian operator \textbf{A}. This is significant, as it implies that the operator corresponds to a measurable physical quantity with real eigenvalues. Furthermore, the expectation values taken with respect to eigenstates of a Hermitian operator are also real.

In the context of our exercise, applying these Hermitian properties allows us to show that \( (i[A, B])^H = i[A, B] \), completing the proof that \(i[A, B]\) is indeed a Hermitian operator. The step-by-step solution takes advantage of these properties when it transforms the operator \(i[A, B]\) under Hermitian conjugation and demonstrates that it is invariant up to a complex unit, which is indicative of Hermitian operators.