6
$\begingroup$

Here is a question about the canonical momentum that I had asked some days ago, but I still have one point that I am not understand.

Considering a particle moves in a magnetic field with charge $q$ and mass $m$, its hamiltonian is $$H=\frac{\vec{P}^2}{2m}=\frac{(\vec{p}+q\vec{A})^2}{2m}$$ where $\vec{p}$ is the momentum of the particle, $\vec{A}$ is the vector potential of the magnetic field and $\vec{P}$ is the canonical momentum of the particle.

I think, because of the expression of the hamiltonian, the canonical momentum $\vec{P}$ is a conserved quantity.

But by the answer in the previous link, it seems that the canonical momentum is not conserved even in a simple example that a particle moves in a homogeneous magnetic field.

I am confused about this question. Is the canonical momentum conserved when a particle moves in magnetic field?

$\endgroup$
5
  • $\begingroup$ But why do you think that? -- not that your linked question has a very clear answer. I'd dig into L&L or similar textbooks for details rather than just re-asking the question. $\endgroup$ Feb 10 '14 at 14:04
  • $\begingroup$ @CarlWitthoft $H=\frac{P^2}{2m}$ and $[P,H]=0$, I think $P$ is conserved. Btw, I do not have this book and my textbook do not discuss this or other similar questions, could you tell me more details? $\endgroup$
    – qfzklm
    Feb 10 '14 at 14:23
  • $\begingroup$ possible duplicate of A question about canonical momentum and arbitrariness for potential in magnetism $\endgroup$
    – Kyle Kanos
    Feb 10 '14 at 15:19
  • 2
    $\begingroup$ qfzklm, you've got it wrong. The Hamiltonian function for particle in magnetic field is $H(\vec p, \vec r) = \frac{(\vec p - q\vec A(\vec r))^2}{2m}$, and lower-case $\vec p$ is called canonical momentum, while upper-case $\vec P = \vec p - q\vec A$ is kinetic momentum ($m\vec v$). $\endgroup$ Feb 10 '14 at 16:20
  • $\begingroup$ @JánLalinský Thank you very much, and I have to apologize for my fault to everyone. $\endgroup$
    – qfzklm
    Feb 14 '14 at 11:37
3
$\begingroup$

As Jan noted, the Hamiltonian should have a minus sign:

$H=\frac{(p-qA)^2}{2m}$

where $p$ is the canonical momentum, and the expression $p-qA$ is the kinetic momentum $P$.

A homogenous magnetic field is an interesting case, because the vector potential in a given gauge does not exhibit translation invariance, but the physical system clearly does. The solution to this dilemma is that you can preserve translational invariance by changing the gauge as you move the coordinates.

There is a conserved quantity associated with this symmetry, but it turns out not to be the canonical momentum (or the kinetic momentum, either). I don't know if it has a particular name, and since it is gauge-dependent there is no universal expression you can write for it. But, for example, in the gauge where $A=\frac{B}{2}(-y,x)$, it is just $(p+qA)$. If anyone has an insight into any physical interpretation of this quantity I would be interested to hear it.

There is a very nice example of all this, worked out concretely, in these notes.

$\endgroup$
0
1
$\begingroup$

Maybe an example helps. Let $B$ be a constant magnetic field. Then we can take $A=\frac12B×x$. Now $$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}(p⋅A+A⋅p)+\frac{q^2A^2}{2m},$$ and $$p⋅A+A⋅p=l⋅B$$ where $l=x×p$. Thus
$$\frac{(p+qA)^2}{2m}=\frac{p^2}{2m}+\frac{q}{2m}l⋅B+\frac{q^2A^2}{2m}.$$ Here we recognise the $l⋅B$-term as the Zeeman term.

If we now calculate the derivative of $p$ according to $$\frac{dp}{dt}=i[p,H],$$ then we obtain the Lorentz force.
This is the quantum version. In the classical case the commutator is replaced by the Poisson bracket.

$\endgroup$
2
  • $\begingroup$ You can use \frac{a}{b} to get $\frac{a}{b}$, rather than using /. I've taken the liberty of modifying your answer to use this, I hope you don't mind. $\endgroup$
    – Kyle Kanos
    Feb 11 '14 at 1:36
  • $\begingroup$ @KyleKanos. Thank you for the modification. I am not very good at LaTeX editing. $\endgroup$
    – Urgje
    Feb 11 '14 at 10:56
0
$\begingroup$

Hint: $P$ is conserved if it is not explicitly time-dependent and if its Poisson bracket with the Hamiltonian is zero.

So you just neet to check that: $$ \{P,H \}=0$$

$\endgroup$
0
$\begingroup$

Hamilton's equations state $\dot{P_i} = -\frac{\partial H}{\partial q^i}.$In this case, this is

$\dot{P_i} = -\frac{\partial H}{\partial q^i} = -\frac{\vec{P}}{m} \cdot \frac{\partial \vec{A}}{\partial q^i}.$

So the canonical momentum is not conserved.

$\endgroup$
2
  • 1
    $\begingroup$ Again, unless $ \frac {\delta A}{\delta q} $ is zero. $\endgroup$ Feb 10 '14 at 15:43
  • $\begingroup$ Carl's pointed out the main thing. Momentum in general is conserved only as long as the potential (be it vector or scalar) is uniform. We can't expect momentum to be conserved in a uniform magnetic field, any more than we can expect it to be conserved in a uniform electric field. $\endgroup$
    – Nanite
    Feb 10 '14 at 22:16
0
$\begingroup$

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. Look at a much more simple case: A particle falls down in constant gravity. The potential energy depends only on height. There is no potential momentum. The canonical momentum is just the mechanical momentum, which obviously is not constant. It increases as the particle falls.

$\endgroup$
0
$\begingroup$

This question is well answered, however, let me add one more point.

Newton's equation of motion can be written in the following form: $$ \frac{d}{dt}(\vec p + q\vec A) = -q\nabla(\phi - \vec v\cdot\vec A) $$ Here $\vec p = \gamma m \vec v$ is the kinetic momentum. $\phi$ is the electric potential. $\vec P \equiv \vec p + q \vec A$ is the canonical momentum. It is clear from this equation that the canonical momentum is only conserved if the gradient of the generalized potential $\phi - \vec v\cdot\vec A$ is zero.

As the previous answers pointed out, in the uniform magnetic field, there is a conserved quantity, owing to the translational invariance of the system. This quantity is known in the literature as the pseudo-momentum: $$ \vec K = \vec p + 2q\vec A = \vec P + q \vec A $$ You can check this by taking the time derivative of $\vec K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.