# Is the Momentum Operator Hermitian? A Proof

• B
• hokhani
In summary, the momentum operator ##p=-i\frac{d}{dx}## and its adjoint ##p^\dagger = i\frac{d}{dx}## are equal to each other, as shown by the proof in post #3. This proves that the momentum operator is Hermitian, as required by the definition of an autoadjoint operator.
hokhani
TL;DR Summary
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?

Remember, the exact definition of an autoadjoint operator is that ##\left<\psi\right|\hat{p}\left|\phi\right>=\left<\phi\right|\hat{p}\left|\psi\right>^*##
You can check that with such a definition the momentum is really autoadjoint.
You can prove indeed that ##\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}##

hokhani and vanhees71
write this\begin{align*}

(f, p g) = -i \hbar \int_D f^*(x) \frac{\partial g}{\partial x}(x)dx &= i\hbar \int_D \frac{\partial f^*}{\partial x}(x) g(x) dx - {\underbrace{\left[ i \hbar f^*(x) g(x) \right]}_{\overset{!}{=} \, 0}}^{\partial D} \\

&= \int_D \left( - i \hbar \frac{\partial f}{\partial x}(x) \right)^* g(x) dx \\

&= (p f, g)

\end{align*}

Clear277, hokhani and vanhees71
hokhani said:
Momentum operator is ##p=-i\frac{d}{dx}## and its adjoint is ##p^\dagger=i\frac{d}{dx}##.
No, this is not correct. The correct equation is ##p^\dagger = (- i)^* \hbar \left( \frac{d}{dx} \right)^\dagger##. Then:

Gaussian97 said:
You can prove indeed that ##\left(\frac{d}{dx}\right)^\dagger = -\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##.

hokhani and vanhees71
Gaussian97 said:
You can prove indeed that ##\left(\frac{d}{dx}\right)^\dagger = -\frac{d}{dx}##

please let ##L = \dfrac{d}{dx}## and re-write ##(f, Lg)## in the form ##(Mf, g)## and then tell what is ##M##

hokhani and vanhees71
hokhani said:
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
PeterDonis said:
Post #3 is a proof of it.
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.

hokhani and vanhees71

## 1. What is the Hermitian momentum operator?

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.

## 2. How is the Hermitian momentum operator related to the classical momentum?

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.

## 3. What is the significance of the Hermitian property of the momentum operator?

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.

## 4. How is the Hermitian momentum operator used in the Schrödinger equation?

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.

## 5. Can the Hermitian momentum operator be used to measure the momentum of a particle?

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.

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