# Why is the generalized momentum replaced by the momentum operator but not the ordinary momentum?

I was trying to understand the derivation of the Hamiltonian for a charged particle in an electromagnetic field. https://en.wikipedia.org/wiki/Hamiltonian_mechanics#Charged_particle_in_an_electromagnetic_field

This Hamiltonian is used frequently in quantum mechanics. In quantum mechanics usually the momentum is replaced by the momentum operator. But in case of this Hamiltonian the generalized momentum is replaced by the momentum operator.

I'm very surprised about that because the generalized momentum isn't actually the momentum of the particle. When I try to calculate the expectation value of a state using this momentum operator, do I actually calculate the expectation value of the momentum of the particle then?

• Generalized momentum is conserved as a result of invariance to spatial translations. Momentum operator commutes with Hamiltonian if the latter is translationally invariant. If this is the case, the eigenvalues of the momentum operator are conserved. Thats the connection. See Weinberg's lectures on quantum mechanics
– Cryo
Jun 30, 2019 at 23:22
• Possible duplicates: physics.stackexchange.com/q/45796/2451 , physics.stackexchange.com/q/104178/2451 and links therein. Jul 1, 2019 at 6:34

• The canonical momentum $$\mathbf{p}$$, which is the conjugate variable for position, defined with $$p_i=\partial L/\partial \dot{x_i}$$. In classical mechanics, it is an abstract quantity that indicates a phase space coordinate. In quantum mechanics, the equivalent (though not identical, as elucidated here) definition is more helpfully phrased as "canonical momentum is the generator of translation;" this statement becomes somewhat more digestible when one considers a derivation (of the variety shown here) of its form in the position representation, $$-\mathrm{i}\hbar\nabla$$ . Notice that there is no immediately obvious connection between this definition and the time over which a body translates (i.e. the speed of the body).
• The kinetic momentum $$\vec{\pi}=m\dot{\mathbf{x}}$$, which is intuitively visualized a scalar multiple of the rate of change of position in classical mechanics and the rate of change of position expectation value in quantum mechanics. Its form as an operator in quantum mechanics is not immediately obvious from this description, but we can define the kinetic energy operator as $$\hat{T}=|\vec{\pi}|^2/2m$$; you may appeal to your classical intuition to convince yourself that this is the right definition of $$\hat{T}$$.
• In a number of cases, the potential is independent of $$\dot{\mathbf{x}}$$. Then, the Lagrangian is $$L=T(\dot{\mathbf{x}})-V(\mathbf{x})$$. The canonical momentum is given by $$p_i=\partial L/\partial\dot{x}_i=\partial T/\partial\dot{x}_i=m\dot{x}_i$$. This is the implicit assumption when one uses $$-\mathrm{i}\hbar\nabla$$ as the momentum operator in the Schrödinger equation or when calculating expectation values; the intent is to refer to kinetic momentum, but they happen to be equal.
• In the case of a charge in an electromagnetic field, the assumption that the potential is independent of $$\dot{\mathbf{x}}$$ is incorrect, so the canonical and kinetic momentum are no longer equal. Specifically, if one writes down the Lagrangian$$L=\frac{1}{2}m\dot{\mathbf{x}}^2-q(\phi-\dot{\mathbf{x}}\cdot\mathbf{A}),$$ which, when inserted in the Euler-Lagrange equation, gives the Lorentz force equation $$m\ddot{\mathbf{x}}=q(\mathbf{E}+\dot{\mathbf{x}}\times\mathbf{B})$$, and differentiates that with respect to $$\dot{x}_i$$, the result is the canonical momentum $$p_i=m\dot{x}_i+qA_i$$, but the kinetic momentum is unchanged, $$\vec{\pi}=m\dot{\mathbf{x}}$$. In such cases, one would eventually use the kinetic momentum, expressed as $$\vec{\pi}=\mathbf{p}-q\mathbf{A}$$, when calculating expectation values or solving the Schrödinger equation, but the Hamiltonian$$H=\frac{1}{2m}|\mathbf{p}-q\mathbf{A}|^2+q\phi$$is actually arrived at using $$H=\dot{\mathbf{x}}\cdot\mathbf{p}-L$$. Moreover, if you try to find the expectation value of the canonical momentum for a valid state, you may end up with a gauge-dependent answer, which is not problematic because the canonical momentum is not an observable quantity, unlike the kinetic momentum (the expectation value of which will be gauge-invariant).