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hokhani
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- How momentum is Hermitian
Momentum operator is ##p=-i\frac{d}{dx}## and its adjoint is ##p^\dagger=i\frac{d}{dx}##. So, ##p^\dagger=-p##. How is the momentum Hermitian?
No, this is not correct. The correct equation is ##p^\dagger = (- i)^* \hbar \left( \frac{d}{dx} \right)^\dagger##. Then:hokhani said:Momentum operator is ##p=-i\frac{d}{dx}## and its adjoint is ##p^\dagger=i\frac{d}{dx}##.
Which means that ##p^\dagger = (- i)^* \hbar \left( \frac{d}{dx} \right)^\dagger = i \hbar \left(- \frac{d}{dx} \right) = - i \hbar \frac{d}{dx} = p##.Gaussian97 said:You can prove indeed that ##\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}##
I can't prove it. Could you please help me with that?Gaussian97 said:You can prove indeed that ##\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}##
Post #3 is a proof of it. If the presence of the factor of ##i\hbar## in post #3 confuses you, just eliminate it; then you have a straightforward proof. The key to the proof is the sign flip that comes with the integration by parts.hokhani said:I can't prove it. Could you please help me with that?
Note, btw, that the proof in post #3 is only valid for functions that vanish at the boundary, i.e., at infinity, so the boundary term in the integration by parts goes away. The usual argument is that any function that can actually be physically realized will have this property; this is what mathematicians often call (somewhat disdainfully) a physicist's level of rigor.PeterDonis said:Post #3 is a proof of it.
The Hermitian momentum operator is a mathematical operator used in quantum mechanics to describe the momentum of a particle. It is represented by the symbol p and is defined as the negative gradient of the wave function.
The Hermitian momentum operator is related to the classical momentum through the correspondence principle, which states that in the classical limit, quantum mechanics should reproduce classical mechanics. In this case, the Hermitian momentum operator reduces to the classical momentum, p = mv.
The Hermitian property of the momentum operator means that it is a self-adjoint operator, meaning that its eigenvalues are real and its eigenvectors are orthogonal. This property is important in quantum mechanics because it ensures that the measurement of the momentum will always yield a real value.
The Hermitian momentum operator is used in the Schrödinger equation to describe the time evolution of a quantum system. It appears in the kinetic energy term of the equation and is used to calculate the momentum of the particle at a given point in time.
Yes, the Hermitian momentum operator can be used to measure the momentum of a particle in quantum mechanics. When the momentum operator acts on the wave function of a particle, it yields the eigenvalue of the momentum, which can be interpreted as the momentum of the particle at that point in time.