What does the canonical momentum $\textbf{p}=m\textbf{v}+e\textbf{A}$ mean? Is it just momentum accounting for electromagnetic effects?

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    $\begingroup$ To clarify, are you looking for physical intuition? For example, the magnitude of the standard, non-relativistic momentum of a particle tells you "how difficult it would be to stop the particle" in a sense that can be made precise using the concept of impulse. Are you looking for something analogous to this but for canonical momentum as opposed to, say, a discussion of how it mathematically arises from a Lagrangian for electrodynamics? $\endgroup$ – joshphysics Jul 24 '13 at 22:27
  • $\begingroup$ Intuitively, is it the entire momentum of asystem? So magnetic fields have momentum? $\endgroup$ – cpc333 Jul 25 '13 at 1:53

In Lagrangian mechanics, "momentum" is just a conserved quantity, and is the derivative of the Lagrangian with respect to velocity ($\frac{dL}{d\dot{q}}$). For the case of a point charge traveling through a uniform magnetic field $\mathbf{B}$, $\mathbf{p} = m \mathbf{v}$ simply isn't conserved anymore, as the charge travels in a circular path due to the magnetic field, causing its momentum to constantly change direction. A quantity known as the canonical momentum, $\mathbf{P} = m\mathbf{v} + e \mathbf{A}$ ends up being conserved throughout the charged particle's trajectory. (Setting the total time derivative of the canonical momentum equal to zero simply results in $m \mathbf{a} = e \mathbf{v} \times \mathbf{B}$, which is just the expression for magnetic force.) In short, the canonical momentum is simply "the quantity that is conserved" in electromagnetic interactions, while the kinetic momentum is just the product of mass and velocity.

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    $\begingroup$ It is not true that the canonical momentum is "conserved in electromagnetic interactions." It satisfies the Euler-Lagrange equation of motion, $\frac{d}{dt}\frac{dL}{d\dot{q}} - \frac{dL}{dq}$. For instance, in a uniform magnetic field in the Landau gauge, $\mathbf{A} = -By\mathbf{\hat{x}}$ and the Lagrangian is $L = \frac{1}{2}mv^2 - e\Phi(x) + e \mathbf{v}.\mathbf{A}$, so $\frac{dP_y}{dt} = e \dot{x} B \neq 0$. $\endgroup$ – Ted Pudlik Jul 2 '14 at 19:55
  • $\begingroup$ Ted, Does your critique invalidate the answer? If so what is the correct answer to the original question? $\endgroup$ – James Bowery Nov 2 '14 at 13:09
  • $\begingroup$ @JamesBowery See my answer, I also think this answer is problematic. $\endgroup$ – an offer can't refuse Nov 6 '14 at 14:54

The canonical momentum $p$ is just a conjugate variable of position in classical mechanics, for we have the relation $p=\frac{\partial L}{\partial \dot{r}}$. When making the transition to quantum mechanics: we need substitute $p$ by an operator $-ih\nabla$ in the Hamiltonian; similarly, we need substitute $r$ by $i\hbar \nabla_p$ in momentum representation.

The kinetic momentum is named because it represent velocity of the particle in classical mechanics or when we are talking about the quantum mechanical expectation values:

$$\frac{d\langle\vec r\rangle}{d t}=\frac{\langle\vec{P}\rangle}{m}$$

Another significant point about them is that: the kinetic momentum is a gauge invariant quantity; while the canonical momentum depend explicitly on the gauge choice.

Neither the canonical $\hat p=-i\hbar\nabla$ nor the kinetic momentum $\hat{P}=-i\hbar\nabla-q\vec{A}$ is a conserved quantity in the general case.

Consider the Hamiltonian in electromagnetic field:$$H=\frac{1}{2m}(\hat p- q \vec{A})^2+q \varphi$$, one can check that: $$\frac{d\langle \vec P \rangle}{dt}=q\langle \vec E\rangle+\frac{q}{2m}\langle(\hat p\times \vec B-\vec B\times \hat p)\rangle-\frac{q^2}{m}\langle\vec{A}\times\vec{B}\rangle$$. and $$\frac{d\langle \vec p \rangle}{dt}=-q\langle\nabla\varphi\rangle+ \frac{q}{2m}\sum_j\langle(\nabla A_j) p_j+p_j(\nabla A_j)-2qA_j\nabla A_j\rangle$$

Where $$\vec E=\nabla \varphi-\partial\vec A/\partial t$$ $$\vec B=\nabla\times\vec A$$

So you can see in general they are not conserved. Even in the very special case as @Frederic BrĂ¼nner point out: $\vec A$ is position independent.

So forget about the conservation of both, they may be conserved only in some very special cases.

  • $\begingroup$ I cannot believe that this is the first answer not talking about conservation in some form or another. Good one! $\endgroup$ – ACuriousMind Nov 6 '14 at 16:13
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    $\begingroup$ Re the latest edit: Quantization does not require you to "replace" $p$ by $\nabla$, that is only in one particular representation. In general, $x$ and $p$ are abstract operators on equal footing, and it can be shown that there's only "one" kind of representing them on a particular space, and that is by making one the derivative w.r.t. the other. $\endgroup$ – ACuriousMind Nov 6 '14 at 17:22
  • $\begingroup$ @ACuriousMind Yes, you are right. We don't have to substitute $p$. I edit the answer $\endgroup$ – an offer can't refuse Nov 6 '14 at 17:56

Imagine this situation:

enter image description here

at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity.

at time t=T, we make the current to be I, thus we have a $ \mathbf{B}$ field, and $ \mathbf{A}$ field.

during this process, $ \mathbf{A}$ is build up from zero to some value, therefore we have induced electric field $ \mathbf{E}= - \frac{ \partial \mathbf{A}}{\partial t}$

$\Delta (m\mathbf{v})=\int \mathbf{F} dt = \int q \mathbf{E} dt =\int q \frac{ - \partial \mathbf{A}}{\partial t} dt $

assume this process happened very fast, the particle almost stays at the same position, $ \frac{ \partial \mathbf{A}}{\partial t} =\frac{ d \mathbf{A}}{d t}$


$\Delta (m\mathbf{v}) =\int q \frac{ - d \mathbf{A}}{d t} dt = - q \int d \mathbf{A} =-q \Delta \mathbf{A} $

$ \Delta ( m \mathbf{ v } +q \mathbf{A}) =0 $

$ m \mathbf{ v } +q \mathbf{A} = constant $


The whole problem starts when you try to do electromagnetism with the Lagrangian because you can't write the magnetic field in terms of a potential. However we CAN write it in terms of a vectorpotential $\vec{A}$:

$\vec{B} = \nabla\times\vec{A}$.

It seems that this is usefull and can be used to derive the appropriate Lagrangian and Hamiltonian which are given and checked here.

It seems (from the calculations given in the link) that to include the magnetic field, we need to replace our momentum with:


By replacing the momentum by this term, you are able to do Lagrangain and Hamiltonian mechanics (which work with potentials) for magnetic fields (which can't be written in terms of a potential).

For electric fields you can still include them by using the electronic potential.


Yes, it accounts for the effect of the vector potential on a moving charge. But there is also a more fundamental role that it plays: assuming position-independent vector potential, the canonical momentum is a conserved quantity, while the "normal" (or kinetic) momentum (mass times velocity) is not.

  • $\begingroup$ The canonical momentum is not conserved. Check with the Lagrangian $L= \frac{1}{2}m \vec v^2 - q\Phi(x) + q \vec v .\vec A(x)$ and Euler-Lagrange equations $\endgroup$ – Trimok Jul 25 '13 at 10:09
  • $\begingroup$ I assumed position-independent vector potential. $\endgroup$ – Frederic Brünner Jul 25 '13 at 10:35
  • $\begingroup$ But this means a zero electric and magnetic field. $\endgroup$ – Trimok Jul 25 '13 at 10:38
  • $\begingroup$ Magnetic yes, but the electric field is not necessarily zero. $\endgroup$ – Frederic Brünner Jul 25 '13 at 10:42
  • $\begingroup$ No, you have to choose $\Phi=$Constant in order to have a conserved canonical momentum, so the electric field is zero too. $\endgroup$ – Trimok Jul 25 '13 at 10:44

The canonical (total) momentum is the sum of the kinetic (mechanical) momentum and the potential momentum. Potential momentum occurs only if the potential energy depends explicitly on velocity.


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