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If H is a Hermitian operator, then its eigenvalues are real. Is the converse true?
micromass said:No. Take as counterexample the matrix [tex]\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right).[/tex]
However, if you know that the operator is normal (that is: if [itex]AA^*=A^*A[/itex]), then it is true that it is Hermitian iff the eigenvalues are real.
Hermitian operators are mathematical entities used in quantum mechanics to represent physical observables, such as position, momentum, and energy. They are self-adjoint operators that have the property of being equal to their own conjugate transpose.
Hermitian operators have several important properties that make them useful in quantum mechanics. They have real eigenvalues, which correspond to measurable quantities in experiments. They also have orthogonal eigenvectors, which can be used to construct a complete basis for representing any quantum state.
In quantum mechanics, physical observables are represented by Hermitian operators. The measurement of an observable corresponds to finding the eigenvalue of the associated Hermitian operator. This allows us to make predictions about the outcomes of measurements on quantum systems.
No, not all operators in quantum mechanics are Hermitian. However, any operator can be decomposed into a sum of a Hermitian operator and an anti-Hermitian operator. The Hermitian part represents the observable quantity, while the anti-Hermitian part has no physical meaning.
Unitary operators are a special type of Hermitian operator that represent transformations that preserve the length of a quantum state vector. This means that unitary operators are responsible for the time evolution of quantum systems. Additionally, Hermitian operators can be diagonalized using unitary operators.