FaradayCage said:Given some state [itex]\left|\psi\right\rangle[/itex], and two operators [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex], how do you prove that if [itex]\langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{B}| \psi\rangle[/itex] then [itex]\hat{A} = \hat{B}[/itex] ?
FaradayCage said:Given some state [itex]\left|\psi\right\rangle[/itex], and two operators [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex], how do you prove that if [itex]\langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{B}| \psi\rangle[/itex] then [itex]\hat{A} = \hat{B}[/itex] ?
The concept of "Proof: Operators with same expectation value" refers to the mathematical proof that two operators in quantum mechanics can have the same expectation value without being identical, as long as they have the same eigenvalues and corresponding eigenvectors.
An operator in quantum mechanics is a mathematical representation of a physical observable, such as position, momentum, or energy. Operators act on quantum states to produce a measurable value, known as an eigenvalue.
The expectation value for an operator is calculated by taking the inner product of the operator with the quantum state and then normalizing by the inner product of the state with itself. This can be written as 〈ψ|A|ψ〉, where ψ represents the quantum state and A represents the operator.
Yes, two operators can have the same expectation value for all quantum states if they have the same eigenvalues and corresponding eigenvectors. This can be proven mathematically using the concept of completeness, which states that any quantum state can be expressed as a linear combination of eigenvectors of a given operator.
The concept of "Proof: Operators with same expectation value" is fundamental in understanding and predicting the behavior of quantum systems, such as atoms, molecules, and subatomic particles. It is also used in the development of quantum algorithms for quantum computers and in various fields of quantum technology, such as quantum cryptography and quantum sensing.