Understanding the Hermiticity of the Momentum Operator in Quantum Mechanics

In quantum mechanics, an operator is considered Hermitian (or self-adjoint) if it satisfies the condition:

(langle psi ,hat{A} ,phi rangle langle hat{A} ,psi ,phi rangle)
for all states (left|psirightrangle) and (left|phirightrangle) in the Hilbert space, where (hat{A}) is the operator in question.

Introduction to Hermitian Operators in Quantum Mechanics

Hermitian operators play a crucial role in quantum mechanics because they guarantee that the eigenvalues of the operator, which correspond to measurable quantities, are real. This property is fundamental for the formulation of physical theories in quantum mechanics, as it ensures that observable quantities have sensible and well-defined values.

The Momentum Operator in Quantum Mechanics

The momentum operator, (hat{p}), in one-dimensional quantum mechanics is defined as:

(hat{p} -i hbar frac{d}{dx})
where (hbar) is the reduced Planck constant.

Proving the Hermiticity of the Momentum Operator

To prove that the momentum operator is Hermitian, we need to show that:

(langle psi ,hat{p} ,phi rangle langle hat{p} ,psi ,phi rangle)
for all wave functions (psi(x)) and (phi(x)).

Calculating the Left-Hand Side

The left-hand side is given by:

[langle psi ,hat{p} ,phi rangle int_{-infty}^{infty} psi(x) ,hat{p} ,phi(x) ,dx int_{-infty}^{infty} psi(x) left(-i hbar frac{d}{dx} phi(x)right) dx]

Applying integration by parts:

[ -i hbar left[psi(x) phi(x)right]_{-infty}^{infty} i hbar int_{-infty}^{infty} phi(x) frac{d}{dx} psi(x) ,dx]

Assuming that (psi) and (phi) vanish at infinity, the boundary term vanishes:

[langle psi ,hat{p} ,phi rangle i hbar int_{-infty}^{infty} phi(x) frac{d}{dx} psi(x) ,dx]

Calculating the Right-Hand Side

The right-hand side is given by:

[langle hat{p} ,psi ,phi rangle int_{-infty}^{infty} hat{p} ,psi(x) ,phi(x) ,dx int_{-infty}^{infty} left(-i hbar frac{d}{dx} psi(x)right) phi(x) ,dx]

This simplifies to:

[ -i hbar int_{-infty}^{infty} phi(x) frac{d}{dx} psi(x) ,dx]

Equating the Two Sides

Equating the two sides:

[i hbar int_{-infty}^{infty} phi(x) frac{d}{dx} psi(x) ,dx -i hbar int_{-infty}^{infty} phi(x) frac{d}{dx} psi(x) ,dx]

Rearranging terms and recognizing that both expressions are equal, we conclude:

[langle psi ,hat{p} ,phi rangle langle hat{p} ,psi ,phi rangle]

Thus, the momentum operator (hat{p}) is indeed Hermitian.

Summary

The momentum operator is Hermitian because it satisfies the equality condition for inner products which holds true under appropriate boundary conditions, typically that wave functions vanish at infinity. This property is essential in quantum mechanics as it ensures that the eigenvalues of the operator, which correspond to measurable quantities, are real.